Symmetrisation and the Feigin–Frenkel centre

Abstract

For complex simple Lie algebras of types B, C, and D, we provide new explicit formulas for the generators of the commutative subalgebra $\mathfrak z(\hat {\mathfrak g})\subset {{\mathcal {U}}}(t^{-1}\mathfrak g[t^{-1}])$ known as the Feigin–Frenkel centre. These formulas make use of the symmetrisation map as well as of some well-chosen symmetric invariants of $\mathfrak g$. There are some general results on the rôle of the symmetrisation map in the explicit description of the Feigin–Frenkel centre. Our method reduces questions about elements of $\mathfrak z(\hat {\mathfrak g})$ to questions on the structure of the symmetric invariants in a type-free way. As an illustration, we deal with type G$_2$ by hand. One of our technical tools is the map ${\sf m}\!\!: {{\mathcal {S}}}^{k}(\mathfrak g)\to \Lambda ^{2}\mathfrak g \otimes {{\mathcal {S}}}^{k-3}(\mathfrak g)$ introduced here. As the results show, a better understanding of this map will lead to a better understanding of $\mathfrak z(\hat {\mathfrak g})$.

Introduction

Let $G$ be a complex reductive group. Set $\mathfrak g=\mathrm {Lie\,} G$. As is well known, the algebra ${{\mathcal {S}}}(\mathfrak g)^{\mathfrak g}$ of symmetric $\mathfrak g$-invariants and the centre ${\mathcal {Z}}(\mathfrak g)$ of the enveloping algebra ${{\mathcal {U}}}(\mathfrak g)$ are polynomial algebras with $\mathrm {rk}\,\mathfrak g$ generators. Therefore there are several isomorphisms between them. Two of these isomorphisms can be distinguished, the one given by the symmetrisation map, which is a homomorphism of $\mathfrak g$-modules, and the Duflo isomorphism, which is a homomorphism of algebras. Both of them exist for any finite-dimensional complex Lie algebra.

The symmetrisation map is defined in the infinite-dimensional case as well. However, no analogue of the Duflo isomorphism for Lie algebras $\mathfrak q$ with $\dim \mathfrak q=\infty$ is known. Furthermore, one may need to complete ${{\mathcal {U}}}(\mathfrak q)$ in order to replace ${\mathcal {Z}}(\mathfrak q)$ with an interesting related object, see e.g. [Kac84]. In this paper, we are dealing with the most notable class of infinite-dimensional Lie algebras, namely affine Kac–Moody algebras ${{\hat {\mathfrak g}}}$, and the related centres at the critical level.

The Feigin–Frenkel centre $\mathfrak z(\hat {\mathfrak g})$ is a remarkable commutative subalgebra of the enveloping algebra ${{\mathcal {U}}}(t^{-1}\mathfrak g[t^{-1}])$. The central elements of the completed enveloping algebra $\tilde {{{\mathcal {U}}}}_{\kappa }({{\hat {\mathfrak g}}})$ at the critical level $\kappa =-{\tt h}\!^{\vee }$ can be obtained from the elements of $\mathfrak z({{\hat {\mathfrak g}}})$ by employing the vertex algebra structure [Fre07, § 4.3.2]. The structure of $\mathfrak z(\hat {\mathfrak g})$ is described by a theorem of Feigin and Frenkel [FF92], hence the name. This algebra provides a quantisation of the local Hitchin system [BD, § 2]. Elements $S\in \mathfrak z({{\hat {\mathfrak g}}})$ give rise to higher Hamiltonians of the Gaudin model, which describes a completely integrable quantum spin chain [FFR94].

The classical counterpart of $\mathfrak z({{\hat {\mathfrak g}}})$ is the Poisson-commutative subalgebra of $\mathfrak g[t]$-invariants in ${{\mathcal {S}}}(\mathfrak g[t,t^{-1}])/(\mathfrak g[t])\cong {{\mathcal {S}}}(t^{-1}\mathfrak g[t^{-1}])$, which is a polynomial ring with infinitely many generators according to a direct generalisation of a Raïs–Tauvel theorem [RT92]. Explicit formulas for the elements of $\mathfrak z({{\hat {\mathfrak g}}})$ appeared first in type A [CT06, CM09] following Talalaev's discovery [Tal06] of explicit higher Gaudin Hamiltonians. Then they were extended to all classical types in [Mol13]. The construction of [Mol13] relies on the Schur–Weyl duality involving the Brauer algebra. Type G$_2$ is covered by [MRR16]. The subject is beautifully summarised in [Mol18].

Unlike the finite-dimensional case, no natural isomorphism between the algebras ${{\mathcal {S}}}(t^{-1}\mathfrak g[t^{-1}])^{\mathfrak g[t]}$ and $\mathfrak z({{\hat {\mathfrak g}}})$ is known. Also, generally speaking, an element of $\mathfrak z(\hat {\mathfrak g})$ cannot be obtained by the symmetrisation $\varpi$ from a homogeneous $\mathfrak g[t]$-invariant in ${{\mathcal {S}}}(t^{-1}\mathfrak g[t^{-1}])$. At the same time, some of the elements do come in this way, see Example 5.2, which is dealing with the Pfaffians of $\mathfrak {so}_{2n}$. In this paper, we show that for all classical Lie algebras, $\varpi$ can produce generators of $\mathfrak z({{\hat {\mathfrak g}}})$. The symmetrisation map is not a homomorphism of algebras. However, it is a homomorphism of $\mathfrak g[t^{-1}]$-modules and it behaves well with respect to taking various limits.

According to a striking result of Rybnikov [Ryb08], $\mathfrak z({{\hat {\mathfrak g}}})$ is the centraliser in ${{\mathcal {U}}}(t^{-1}\mathfrak g[t^{-1}])$ of a single quadratic element ${\mathcal {H}}[-1]$, see § 1.1. This fact is crucial for our considerations.

Any ${\mathcal {Y}}\in {{\mathcal {U}}}(t^{-1}\mathfrak g[t^{-1}])$ can be expressed as a sum

(0.1)\begin{equation} \varpi(Y_k)+\varpi(Y_{k-1})+\cdots + Y_1 + Y_0 \quad \text{with}\ Y_j\in{{\mathcal{S}}}^{j}(t^{-1}\mathfrak g[t^{-1}]). \end{equation}

Here $Y_k=\mathrm {gr}({\mathcal {Y}})$ if $Y_k\ne 0$. Note that $\sum _{0\leqslant j\leqslant k}\varpi (Y_j)$ is a $\mathfrak g$-invariant if and only if each $Y_j$ is a $\mathfrak g$-invariant. In the following, we consider only elements with $Y_0=0$.

A polarisation of a $\mathfrak g$-invariant $F\in {{\mathcal {S}}}(\mathfrak g)$ is a $\mathfrak g$-invariant in ${{\mathcal {S}}}(t^{-1}\mathfrak g[t^{-1}])$, see § 1.5 for the definition of a polarisation. However, ${{\mathcal {S}}}(t^{-1}\mathfrak g[t^{-1}])^{\mathfrak g}$ is not generated by elements of this sort, see (3.7) for an example.

There are finite sets of elements $\{S_1,\ldots,S_{\ell }\}\subset \mathfrak z({{\hat {\mathfrak g}}})$ with $\ell =\mathrm {rk}\,\mathfrak g$, called complete sets of Segal–Sugawara vectors, see § 1.6 for the definition, that are of vital importance for the understanding of $\mathfrak z({{\hat {\mathfrak g}}})$. We prove that if $\mathfrak g$ is either a classical Lie algebra or an exceptional Lie algebra of type G$_2$, then there is a complete set $\{S_k\}$ of Segal–Sugawara vectors such that all the terms $Y_j$ occurring in presentations (0.1) for $S_k$ are polarisations of symmetric invariants of $\mathfrak g$. The map m, defined in § 1.4, plays a crucial rôle in the selection of suitable $\mathfrak g$-invariants. In particular, if $F[-1]\in {{\mathcal {S}}}^{k}(\mathfrak g t^{-1})$ is obtained from $F\in {{\mathcal {S}}}^{k}(\mathfrak g)^{\mathfrak g}$ using the canonical isomorphism $\mathfrak g t^{-1}\cong \mathfrak g$, then $\varpi (F[-1])\in \mathfrak z({{\hat {\mathfrak g}}})$ if and only if ${\sf m}(F)=0$, see Theorem 3.5 and the remark after it. More generally, if $H\in {{\mathcal {S}}}^{k}(\mathfrak g)^{\mathfrak g}$ is such that

(0.2)\begin{equation} {\sf m}^{d}(H)={\sf m}({\sf m}^{d-1}(H)) \in {{\mathcal{S}}}(\mathfrak g) \quad \text{for all} \ 1\leqslant d < k/2, \end{equation}

then there is a way to produce an element of $\mathfrak z({{\hat {\mathfrak g}}})$ corresponding to $H$, see Theorem 3.11 and (2.4).

First, for $F=\xi _1\ldots \xi _m\in {{\mathcal {S}}}^{m}(\mathfrak g)$ and $\bar a=(a_1,\ldots,a_m)\in {{\mathbb {Z}}}_{<0}^{m}$, set

(0.3)\begin{equation} \varpi(F)[\bar a]=\frac{1}{m!} \sum_{\sigma\in{\tt S}_m} \xi_{\sigma(1)} t^{a_1}\ldots \xi_{\sigma(m)} t^{a_m} \in{{\mathcal{U}}}(t^{-1}\mathfrak g[t^{-1}]), \end{equation}

then extend this notation to all elements $F\in {{\mathcal {S}}}^{m}(\mathfrak g)$ by linearity. According to Lemma 2.1, $\varpi (F)[\bar a]=\varpi (F[\bar a])$ for the $\bar a$-polarisation $F[\bar a]\in {{\mathcal {S}}}^{m}(t^{-1}\mathfrak g[t^{-1}])$ of $F$.

The expression $\varpi (\tau ^{r} F[-1]){\cdot }1$ encodes a sum of $ ({1}/{(m+r)!}) c(r,\bar a) \varpi (F)[\bar a]$, where the vectors $\bar a\in {{\mathbb {Z}}}_{<0}^{m}$ are such that $\sum _{j=1}^{m} a_j=-m-r$ and $c(r,\bar a)\in \mathbb {N}$ are certain combinatorially defined coefficients, which we do not compute explicitly. It is not clear whether any interesting combinatorial identity can be produced in this context.

For each classical Lie algebra $\mathfrak g$, there is a set of generators $\{H_1,\ldots,H_\ell \}\subset {{\mathcal {S}}}(\mathfrak g)^{\mathfrak g}$ such that ${\sf m}(H_k)\in {{\mathbb {C}}}H_j$ for some $j$ depending on $k$, see §§ 2, 4, 7 and in particular Propositions 2.3, 4.3, 7.5. In types $\mathsf{A}$ and $\mathsf{C}$, we are using the coefficients of the characteristic polynomial. In the orthogonal case, one has to work with $\det (I_n-q(F_{ij}))^{-1}$ instead. In type $\mathsf{A}_{n-1}$,

\[ {\sf m}(\tilde\Delta_k)=\frac{(n-k+2)(n-k+1)}{k(k-1)} \tilde{\Delta}_{k-2}; \]

in type C$_n$,

\[ {\sf m}(\Delta_{2k})=\frac{(2n-2k+3)(2n-2k+2)}{2k(2k-1)} \Delta_{2k-2}; \]

and finally for $\mathfrak g=\mathfrak {so}_n$, we have

\[ {\sf m}(\Phi_{2k})=\frac{(n+2k-3)(n+2k-2)}{2k(2k-1)} \Phi_{2k-2}. \]

This leads to the following complete sets of Segal–Sugawara vectors:

\begin{align*} &\biggl\{\tilde S_{k-1}=\varpi(\tilde\Delta_k[-1]) + \sum_{1\le r<(k-1)/2} \binom{n-k+2r}{2r} \varpi(\tau^{2r} \tilde\Delta_{k-2r}[-1]){\cdot}1 \mid 2\le k\le n\biggr\}\ \text{ in type } {\sf A}_{n-1};\\ &\biggl\{S_k=\varpi(\Delta_{2k}[-1]) + \sum_{1\le r<k} \binom{2n-2k+2r+1}{2r}\varpi(\tau^{2r}\Delta_{2k-2r}[-1]){\cdot}1\mid 1\le k\le n\biggr\}\ \text { in type } {\sf C}_n;\\ &\biggl\{S_{k}=\varpi(\Phi_{2k}[-1]) +\!\sum_{1\le r<k} \binom{n+2k-2}{2r} \varpi(\tau^{2r} \Phi_{2k-2r}[-1]){\cdot 1} \mid 1 \le k <\ell \biggr\} \text{ for $\mathfrak {so}_n$ with $n=2\ell-1$} \end{align*}

with the addition of $S_{\ell }=\varpi ({\rm Pf}[-1])$ for $\mathfrak {so}_{n}$ with $n=2\ell$.

The result in type $\mathsf{A}$ is not new. It follows via a careful rewriting from the formulas of [CT06, CM09]. We are not giving a new proof. Quite the contrary, we use the statement in type A in order to extend the formula to other types.

Our formulas for $\mathfrak {so}_n$ and $\mathfrak {sp}_{2n}$ describe the same elements as [Mol13], for the case of the Pfaffian-type Segal–Sugawara vector, see § 5; a more general result is recently obtained in [Mol21].

The advantage of our method is that it reduces questions about elements of $\mathfrak z({{\hat {\mathfrak g}}})$ to questions on the structure of ${{\mathcal {S}}}(\mathfrak g)^{\mathfrak g}$ in a type-free way. For example, it is possible to deal with type ${\sf G}_2$ by hand unlike [MRR16], see (6.3). It is quite probable that other exceptional types can be handled on a computer. Conjecturally, each exceptional Lie algebra possesses a set $\{H_k\}$ of generating symmetric invariants such that each $H_k$ satisfies (0.2).

One of the significant applications of the Feigin–Frenkel centre is related to Vinberg's quantisation problem. The symmetric algebra ${{\mathcal {S}}}(\mathfrak g)$ carries a Poisson structure extended from the Lie bracket on $\mathfrak g$ by the Leibniz rule. To each $\mu \in \mathfrak g^{*}\cong \mathfrak g$, one associates the Mishchenko–Fomenko subalgebra ${{\mathcal {A}}}_\mu \subset {{\mathcal {S}}}(\mathfrak g)$, which is an extremely interesting Poisson-commutative subalgebra [MF78]. In [Vin91], Vinberg proposed to find a commutative subalgebra ${\mathcal {C}}_\mu \subset {{\mathcal {U}}}(\mathfrak g)$ such that $\mathrm {gr}({\mathcal {C}}_\mu )=\langle \mathrm {gr}(Y)\mid Y\in {\mathcal {C}}_\mu \rangle _{{{\mathbb {C}}}}$ coincides with ${{\mathcal {A}}}_\mu$. Partial solutions to this problem are obtained in [NO96, Tar00]. The breakthrough came in [Ryb06], where a certain commutative subalgebra $\tilde {{\mathcal {A}}}_\mu \subset {{\mathcal {U}}}(\mathfrak g)$ is constructed as an image of $\mathfrak z({{\hat {\mathfrak g}}})$, cf. (8.1).

In [MY19, § 3.3], sets of generators $\{H_k\mid 1\leqslant k\leqslant \ell \}$ of ${{\mathcal {S}}}(\mathfrak g)^{\mathfrak g}$ such that $\tilde {{\mathcal {A}}}_\mu$ is generated by $\varpi (\partial _\mu ^{m} H_k)$, cf. (8.3), are exhibited in types B, C, and D. For the symplectic Lie algebra, $H_k=\Delta _{2k}$, in the orthogonal case $H_k=\Phi _{2k}$ with the exception of $H_\ell ={\rm Pf}$ in type ${\sf D}_\ell$. Results of this paper provide a different proof for [MY19, Theorem 3.2]. We have pushed the symmetrisation map to the level of ${{\mathcal {U}}}(t^{-1}\mathfrak g[t^{-1}])$.

In § 8.2, we briefly consider Gaudin algebras ${{\mathcal {G}}}$. If $\mathfrak g$ is a classical Lie algebra, then the two-points Gaudin subalgebra ${{\mathcal {G}}}\subset {{\mathcal {U}}}(\mathfrak g \oplus \mathfrak g)$ is generated by the symmetrisations of certain bi-homogeneous $\mathfrak g$-invariants in ${{\mathcal {S}}}(\mathfrak g \oplus \mathfrak g)$, see Theorem 8.4.

1. Preliminaries and notation

Let $\mathfrak g=\mathrm {Lie\,} G$ be a non-Abelian complex reductive Lie algebra. The Feigin–Frenkel centre $\mathfrak z({{\hat {\mathfrak g}}})$ is the centre of the universal affine vertex algebra associated with the affine Kac–Moody algebra ${{\hat {\mathfrak g}}}$ at the critical level [FF92, Fre07]. There is an injective homomorphism $\mathfrak z({{\hat {\mathfrak g}}}) \hookrightarrow {{\mathcal {U}}}(t^{-1}\mathfrak {g}[t^{-1}])$ and $\mathfrak z({{\hat {\mathfrak g}}})$ can be viewed as a commutative subalgebra of ${{\mathcal {U}}}(t^{-1}\mathfrak {g}[t^{-1}])$ [Fre07, § 3.3]. Each element of $\mathfrak {z}(\hat {\mathfrak {g}})$ is annihilated by the adjoint action of $\mathfrak g$, cf. [Mol18, § 6.2.].

1.1 The Feigin–Frenkel centre as a centraliser

We set $\mathfrak g[b]:=\mathfrak g t^{b}$ and $x[b]:=xt^{b}$ for $x\in \mathfrak g$. Furthermore, $\hat {\mathfrak g}^-:= t^{-1}\mathfrak {g}[t^{-1}]$. According to [Ryb08], $\mathfrak {z}(\hat {\mathfrak {g}})$ is the centraliser in ${{\mathcal {U}}}(\hat {\mathfrak g}^-)$ of the following quadratic element

\[ {\mathcal{H}}[-1]=\sum_{a=1}^{\dim\mathfrak g} x_a[-1]x_a[-1], \]

where $\{x_1,\ldots,x_{\dim \mathfrak g}\}$ is any basis of $\mathfrak g$ that is orthonormal with respect to a fixed $\mathfrak g$-invariant non-degenerate scalar product $(\,\,,\,)$. In this paper, a scalar product is a symmetric bilinear form.

1.2 The symmetrisation map

For any complex Lie algebra $\mathfrak q$, let $\varpi \!:{{\mathcal {S}}}^{k}(\mathfrak q)\to \mathfrak q^{\otimes k}$ be the canonical symmetrisation map. Following the usual convention, we let $\varpi$ stand also for the symmetrisation map from ${{\mathcal {S}}}(\mathfrak q)$ to ${{\mathcal {U}}}(\mathfrak q)$. Let $\mathrm {gr}(X)\in {{\mathcal {S}}}(\mathfrak q)$ be the symbol of $X\in {{\mathcal {U}}}(\mathfrak q)$. Then $\mathrm {gr}(\varpi (Y))=Y$ for $Y\in {{\mathcal {S}}}^{k}(\mathfrak q)$ by the construction.

1.3 The antipode

Let us define the anti-involution $\omega$ on ${\mathcal {U}}(\hat {\mathfrak g}^-)$ to be the $\mathbb {C}$-linear map such that $\omega (y[k])=-y[k]$ for each $y\in \mathfrak g$ and

\[ \omega(y_1[k_1]y_2[k_2]\ldots y_m[k_m])=(-y_m[k_m])\ldots (-y_2[k_2])(-y_1[k_1]). \]

Let also $\omega$ be the analogues anti-involution on ${{\mathcal {U}}}(\mathfrak q)$ for any complex Lie algebra $\mathfrak q$.

Clearly, $\omega ({\mathcal {H}}[-1])={\mathcal {H}}[-1]$. Therefore $\omega$ acts on $\mathfrak {z}(\hat {\mathfrak {g}})$. For $Y_j\in {{\mathcal {S}}}^{j}({{\hat {\mathfrak g}}}^-)$, we have $\omega (\varpi (Y_j))=(-1)^{j}\varpi (Y_j)$. A non-zero element ${\mathcal {Y}}\in {{\mathcal {U}}}({{\hat {\mathfrak g}}}^-)$ presented in the form (0.1) is an eigenvector of $\omega$ if and only if either all $Y_j$ with even $j$ or all $Y_j$ with odd $j$ are zero.

1.4 The map m

For $\mathfrak {gl}_N=\mathfrak {gl}_N({{\mathbb {C}}})={\rm End}({{\mathbb {C}}}^{N})$ and $1\leqslant r\leqslant k$, consider the linear map

\[ {\sf m}_r\!: \mathfrak{gl}_N^{\otimes k}\to \mathfrak{gl}_N^{\otimes (k-r+1)}\ \text{that sends} \ \xi_1\otimes\cdots \otimes\,\xi_k \ \text{to}\ \xi_1\xi_2\ldots \xi_r\otimes \xi_{r+1}\otimes \cdots \otimes \xi_k. \]

Note that clearly

(1.1)\begin{equation} {\sf m}_r \circ {\sf m}_s={\sf m}_{r+s-1}. \end{equation}

Via the adjoint representation of $\mathfrak g$, the map ${\sf m}_r$ leads to a map $\mathfrak g^{\otimes k}\to {\rm End}(\mathfrak g) \otimes \mathfrak g^{\otimes (k-r)}$, which we denote by the same symbol. Explicitly, the map

\[ {\sf m}_r\!: \mathfrak{g}^{\otimes k}\to {\rm End}(\mathfrak g) \otimes \mathfrak g^{\otimes (k-r)}\ \text{sends} \ y_1\otimes \cdots \otimes y_k \ \text{to}\ \mathrm{ad}(y_1)\mathrm{ad}(y_2)\ldots \mathrm{ad}(y_r)\otimes y_{r+1}\otimes \cdots \otimes y_k. \]

Observe that

\[ \mathrm{ad}(y_1)\mathrm{ad}(y_2)\ldots\mathrm{ad}(y_{2r+1})+\mathrm{ad}(y_{2r+1})\ldots\mathrm{ad}(y_2)\mathrm{ad}(y_1) \in\mathfrak{so}(\mathfrak g)\cong \Lambda^{2}\mathfrak g, \]

where $\mathfrak {so}(\mathfrak g)=\{\xi \in {\rm End}(\mathfrak g)\mid (\xi (x),y)=-(x,\xi (y)) \, \forall x,y\in \mathfrak g\}$; the isomorphism $\Lambda ^{2}\mathfrak g\cong \mathfrak {so}(\mathfrak g)$ is given by

\[ (y_1\wedge y_2)(x)=(y_2,x)y_1-(y_1,x)y_2 \]

for $y_1,y_2,x\in \mathfrak g$.

We embed ${{\mathcal {S}}}^{k}(\mathfrak g)$ in $\mathfrak g^{\otimes k}$ via $\varpi$. Set ${\sf m}={\sf m}_3$. Then ${\sf m}\!: {{\mathcal {S}}}^{k}(\mathfrak g)\to \Lambda ^{2}\mathfrak g \otimes {{\mathcal {S}}}^{k-3}(\mathfrak g)$. For example, if $Y=y_1y_2y_3\in {{\mathcal {S}}}^{3}(\mathfrak g)$, then

\begin{align*} {\sf m}(Y)&=\tfrac{1}{6}(\mathrm{ad}(y_1)\mathrm{ad}(y_2)\mathrm{ad}(y_3)+\mathrm{ad}(y_3)\mathrm{ad}(y_2)\mathrm{ad}(y_1)+ \mathrm{ad}(y_1)\mathrm{ad}(y_3)\mathrm{ad}(y_2) \\ & \quad + \mathrm{ad}(y_2)\mathrm{ad}(y_3)\mathrm{ad}(y_1) + \mathrm{ad}(y_2)\mathrm{ad}(y_1)\mathrm{ad}(y_3) + \mathrm{ad}(y_3)\mathrm{ad}(y_1)\mathrm{ad}(y_2))\in\mathfrak{so}(\mathfrak g). \end{align*}

Similarly one defines ${\sf m}_{2r+1}\!: {{\mathcal {S}}}^{k}(\mathfrak g)\to \Lambda ^{2}\mathfrak g \otimes {{\mathcal {S}}}^{(k-2r-1)}(\mathfrak g)$ for each odd $2r+1\leqslant k$. Note that each ${\sf m}_{2r+1}$ is $G$-equivariant. It is convenient to put ${\sf m}({{\mathcal {S}}}^{k}(\mathfrak g))=0$ for $k\leqslant 2$.

Suppose that $\mathfrak g$ is simple. There is a $G$-stable decomposition $\Lambda ^{2}\mathfrak g=\mathfrak g\oplus V$. This $V$ will be called the Cartan component of $\Lambda ^{2}\mathfrak g$. If $\mathfrak g$ is not of type $\mathsf{A}$, then $V$ is irreducible. For certain elements $H\in {{\mathcal {S}}}^{k}(\mathfrak g)$, we have ${\sf m}(H)\in \mathfrak g\otimes {{\mathcal {S}}}^{k-3}(\mathfrak g)$. Note that the embedding $\mathfrak g\hookrightarrow \mathfrak {so}(\mathfrak g)$ is canonical: it is given by the adjoint action of $\mathfrak g$. If ${\sf m}(H)\in {{\mathcal {S}}}^{k-2}(\mathfrak g)$, then ${\sf m}_{2r+1}(H)={\sf m_{2r-1}} \circ {\sf m}(H)$, because of (1.1). Since $(\Lambda ^{2}\mathfrak g)^{\mathfrak g}=0$, we have ${\sf m}({{\mathcal {S}}}^{3}(\mathfrak g)^{\mathfrak g})=0$.

1.5 Polarisations and fully symmetrised elements

For elements $y_1,\ldots, y_m \in \mathfrak g$ and a vector $\bar a=(a_1,\ldots,a_m)\in \mathbb {Z}_{<0}^{m}$, set $\Upsilon [\bar a]=\prod _{i=1}^{m} y_i[a_i]\in {{\mathcal {S}}}(\hat {\mathfrak g}^-)$. If we consider the product $Y=\prod _{i} y_i\in {{\mathcal {S}}}^{m}(\mathfrak g)$, then there is no uniquely defined sequence of factors $y_i$. However, the $\bar a$-polarisation $Y[\bar a]:= ({1}/{m!}) \sum _{\sigma \in {\tt S}_m} \Upsilon [\sigma (\bar a)]$ of $Y$ is well defined. We extend this notion to all elements of ${{\mathcal {S}}}^{m}(\mathfrak g)$ by linearity. Linear combinations of the elements

\[ \varpi( Y[\bar a] ) \in {{\mathcal{U}}}(\hat{\mathfrak g}^-) \]

are said to be fully symmetrised. Note that $\varpi (H)$ is fully symmetrised if $H\in {{\mathcal {S}}}^{m}(\mathfrak g t^{-1})$. If $a_i=a$ for all $i$, then $\Upsilon [\bar a]=Y[\bar a]$ and we denote it simply by $Y[a]$.

The evaluation ${\sf Ev}_1$ at $t=1$ defines an isomorphism ${\sf Ev}_1\!: {{\mathcal {S}}}(\mathfrak g[a])\to {{\mathcal {S}}}(\mathfrak g)$ of $\mathfrak g$-modules. For $F\in {{\mathcal {S}}}(\mathfrak g)$, let $F[a]$ stand for ${\sf Ev}_1^{-1}(F)\in {{\mathcal {S}}}(\mathfrak g[a])$. Then $\varpi (F)[a]:=\varpi (F[a])$ is fully symmetrised.

1.6 Segal–Sugawara vectors

Set $\tau =-\partial _t$. According to [FF92], $\mathfrak z(\hat {\mathfrak {g}})$ is a polynomial algebra in infinitely many variables with a distinguished set of ‘generators’ $\{S_1,\ldots,S_{\ell }\}$ such that $\ell =\mathrm {rk}\,\mathfrak g$ and

\[ \mathfrak z(\hat{\mathfrak{g}})= {{\mathbb{C}}}[\tau^{m} (S_k) \mid 1\leqslant k\leqslant\ell, m\geqslant 0]. \]

We have $\mathrm {gr}(S_k)=H_k[-1]$ with $H_k\in {{\mathcal {S}}}(\mathfrak g)^{\mathfrak g}$ and ${{\mathbb {C}}}[H_1,\ldots,H_\ell ]={{\mathcal {S}}}(\mathfrak g)^{\mathfrak g}$. The set $\{S_k\}$ is said to be a complete set of Segal–Sugawara vectors. The symbols of $\tau ^{m}(S_k)$ generate ${{\mathcal {S}}}({{\hat {\mathfrak g}}}^-)^{\mathfrak g[t]}$ in accordance with [RT92].

Suppose that we have $\tilde S_k\in \mathfrak z({{\hat {\mathfrak g}}})$ with $1\leqslant k\leqslant \ell$ and $\mathrm {gr}(S_k)=\tilde H_k[-1]$, where $\tilde H_k\in {{\mathcal {S}}}(\mathfrak g)^{\mathfrak g}$, for each $k$. The structural properties of $\mathfrak z({{\hat {\mathfrak g}}})$ imply that $\{\tilde S_k\}$ is a complete set of Segal–Sugawara vectors if and only if the set $\{\tilde H_k\}$ generates ${{\mathcal {S}}}(\mathfrak g)^{\mathfrak g}$.

1.7 Symmetric invariants

For a finite-dimensional Lie algebra $\mathfrak q$, we have ${{\mathcal {S}}}(\mathfrak q)\cong {{\mathbb {C}}}[\mathfrak q^{*}]$. For any reductive Lie algebra, there is an isomorphism of $\mathfrak g$-modules $\mathfrak g\cong \mathfrak g^{*}$. For $\xi \in (\mathfrak {gl}_n)^{*}$, write

(1.2)\begin{equation} \det(qI_n-\xi)=q^{n}-\Delta_1(\xi)q^{n-1} + \cdots + (-1)^{k}\Delta_k(\xi) q^{n-k} + \cdots + (-1)^{n} \Delta_n(\xi) . \end{equation}

Then ${{\mathcal {S}}}(\mathfrak {gl}_n)^{\mathfrak {gl}_n}={{\mathbb {C}}}[\Delta _1,\ldots,\Delta _n]$.

Let $\mathfrak f\subset \mathfrak g$ be a reductive subalgebra. Then there is an $\mathfrak f$-stable subspace $\mathfrak m\subset \mathfrak g$ such that $\mathfrak g=\mathfrak f\oplus \mathfrak m$, whereby also $\mathfrak g^{*} \cong \mathfrak f^{*}\oplus \mathfrak m^{*}$. Identifying $\mathfrak f$ with $\mathfrak f^{*}$, one defines the restriction $H|_{\mathfrak f}$ of $H\in {{\mathcal {S}}}(\mathfrak g)$ to $\mathfrak f$. This is the image of $H$ in ${{\mathcal {S}}}(\mathfrak g)/\mathfrak m{{\mathcal {S}}}(\mathfrak g)\cong {{\mathcal {S}}}(\mathfrak f)$.

In cases $n=2\ell$, $\mathfrak f=\mathfrak {sp}_{2\ell }$ and $n=2\ell +1$, $\mathfrak f=\mathfrak {so}_n$, the restrictions ${\Delta _{2k}}|_{\mathfrak f}$ with $1\leqslant k\leqslant \ell$ form a generating set in ${{\mathcal {S}}}(\mathfrak f)^{\mathfrak f}$. In the case $\mathfrak f=\mathfrak {so}_{n}$ with $n=2\ell$, the restriction of the determinant $\Delta _{2\ell }$ is the square of the Pfaffian and has to be replaced by the Pfaffian in the generating set.

Explicit formulas for basic symmetric invariants of the exceptional Lie algebras are less transparent.

Let $\mathfrak g$ be simple. The inclusions $\mathfrak g\subset {{\mathcal {S}}}(\mathfrak g)$ are ruled by the symmetric invariants. The key point here is that ${{\mathcal {S}}}(\mathfrak g)$ is a free module over ${{\mathcal {S}}}(\mathfrak g)^{\mathfrak g}$ [Kos63]. If $\{H_1,\ldots,H_\ell \}\subset {{\mathcal {S}}}(\mathfrak g)^{\mathfrak g}$ is a generating set consisting of homogeneous elements and $\deg H_i=d_i+1$, then to each $i$ corresponds a primitive copy of $\mathfrak g$ in ${{\mathcal {S}}}^{d_i}(\mathfrak g)$. The non-primitive copies are obtained as linear combinations of the primitive ones with coefficients from ${{\mathcal {S}}}(\mathfrak g)^{\mathfrak g}$.

1.8 Miscellaneousness

Let $\mathfrak h\subset \mathfrak g$ be a Cartan subalgebra, we let $\ell$ stand for $\dim \mathfrak h=\mathrm {rk}\,\mathfrak g$ and $W=W(\mathfrak g,\mathfrak h)$ stand for the Weyl group of $\mathfrak g$. The fundamental weights of a simple Lie algebra $\mathfrak g$ are $\pi _k$ with $1\leqslant k\leqslant \ell$ and $V(\lambda )$ stands for an irreducible finite-dimensional $\mathfrak g$-module with the highest weight $\lambda =\sum _{k=1}^{\ell } c_k \pi _k$. Please keep in mind that the Vinberg–Onishchik numbering [VO88, Tables] of simple roots (and fundamental weights) is used. If $\alpha \in \mathfrak h^{*}$ is a positive root, then $\{e_\alpha,h_\alpha,f_\alpha \}\subset \mathfrak g$ is an $\mathfrak {sl}_2$-triple associated with $\alpha$.

An automorphism $\sigma \in {\rm Aut}(\mathfrak g)$ extends to $\mathfrak g[t^{-1}]$ by setting $\sigma (t^{-1})=t^{-1}$. In this context, $\sigma$ stands also for the corresponding automorphism of ${{\mathcal {S}}}(\mathfrak g[t^{-1}])$. If we take a $\sigma$-invariant product $(\,\,,\,)$, then $\sigma ({\mathcal {H}}[-1])={\mathcal {H}}[-1]$. Therefore $\sigma$ acts on $\mathfrak z({{\hat {\mathfrak g}}})$.

If $\sigma \in {\rm Aut}(\mathfrak g)$ is of finite order $m$, then it leads to a ${{\mathbb {Z}}}/m{{\mathbb {Z}}}$-grading $\mathfrak g=\mathfrak g_0\oplus \mathfrak g_1\oplus \ldots \oplus \mathfrak g_{m-1}$. In the case $m=2$, we have $\mathfrak g_1=\{\xi \in \mathfrak g\mid \sigma (\xi )=-\xi \}$.

Throughout the paper:

• • $\{x_i\}$ is an orthonormal basis of $\mathfrak g$;

• • in the sums $\sum _{i} x_i$ or $\sum _{i,j} x_i x_j$, the ranges are from $1$ to $\dim \mathfrak g$ for $i$ and for $j$;

• • $\bar b=(b_1,b_2)\in \mathbb {Z}_{<0}^{2}$ and ${\mathcal {H}}[\bar b]$ stands for $\sum _{i} x_i[b_1]x_i[b_2]\in {{\mathcal {U}}}({{\hat {\mathfrak g}}}^-)$ and also for the symbol of this sum (in the sense of § 1.2);

• • $G_\xi$ stands for the stabiliser of $\xi$ and it is always clear from the context, which $G$-action is considered, $\mathfrak g_\xi =\mathrm {Lie\,} G_\xi$;

• • $\mathfrak q$ stands for an arbitrary unspecified complex Lie algebra;

• • if ${{\mathcal {A}}}\subset {{\mathcal {U}}}(\mathfrak q)$ is a subalgebra, then $\mathrm {gr}({{\mathcal {A}}}):=\langle \mathrm {gr}(x) \mid x\in {{\mathcal {A}}}\rangle _{{{\mathbb {C}}}}\subset {{\mathcal {S}}}(\mathfrak q)$.

2. Explicit formulas in type $\mathsf{A}$

In type $\mathsf{A}$, there are several explicit formulas for the Segal–Sugawara vectors [CT06, CM09], see also [Mol18, § 7.1]. One of them actually uses symmetrisation. One can form the matrix

\[ E[-1]+\tau = (E_{ij}[-1])+\tau I_n\]

with $E_{ij}\in \mathfrak {gl}_n$ and calculate its column determinant and symmetrised determinant. Due to the fact that this matrix is Manin, see [Mol18, Definition 3.1.1, p. 48, Lemma 7.1.2], the results are the same. The symmetrised version is more suitable for our purpose. The elements $S_j$ are coefficients of $\tau ^{n-j}$ in

\[ \det_{\rm sym}(E[-1]+\tau)=\varpi(\Delta_n[-1])+\varpi(\tau \Delta_{n-1}[-1])+\cdots +\varpi(\tau^{n-2}\Delta_2[-1])+\varpi(\tau^{n-1}\Delta_1[-1])+\tau^{n}. \]

Assume the conventions that

\[ \tau x[a] - x[a]\tau =[\tau,x[a]]=\tau(x[a])=-ax[a-1] \]

and $\tau {\cdot }1=0$. This leads for example to $\tau x[-1]{\cdot }1= x[-2]$. Note that $\varpi$ acts on the summands of $\tau ^{n-k} \Delta _{k}[-1]$ as on products of $n$ factors. It permutes $\tau$ with elements of $\mathfrak {gl}_n[-1]$.

Let $\theta$ be a Weyl involution of $\mathfrak g$, i.e. there is a $\theta$-stable Cartan subalgebra $\mathfrak h\subset \mathfrak g$ such that $\theta |_{\mathfrak h}=-{{\rm id}}_{\mathfrak h}$. As is well known, $\theta (\Delta _k)=(-1)^{k}\Delta _k$ if $\mathfrak g=\mathfrak {gl}_n$. In particular, $\theta ({\mathcal {H}}[-1])={\mathcal {H}}[-1]$ and $\theta$ acts on $\mathfrak z({{\hat {\mathfrak g}}})$. Hence one can always modify the Segal–Sugawara vectors in such a way that they become eigenvectors of $\theta$. The resulting simplified forms are

(2.1)\begin{align} S_n&= \varpi(\Delta_n[-1])+\varpi(\tau^{2}\Delta_{n-2}[-1]){\cdot}1+\cdots + \varpi(\tau^{2r}\Delta_{n-2r}[-1]){\cdot 1}+ \cdots \nonumber\\ & \quad + \varpi(\tau^{n-u}\Delta_u[-1]){\cdot}1 \quad \text{with } u=n-2\bigg[\frac{n-1}{2}\bigg], \end{align}

(2.2)\begin{align} S_k&=\varpi(\Delta_k[-1])+ \sum_{1\leqslant r< k/2} \binom{n-k+2r}{2r}\varpi(\tau^{2r}\Delta_{k-2r}[-1]){\cdot}1. \end{align}

We will see that there is a direct connection with the symmetrisation and that one could have used $\omega$ instead of $\theta$ in order to simplify the formulas. The following two lemmas are valid for all Lie algebras.

Lemma 2.1 Take $Y=y_1 \ldots y_m \in {{\mathcal {S}}}^{m}(\mathfrak g)$ and $\bar a=(a_1,\ldots,a_m)\in \mathbb {Z}_{<0}^{m}$. Then in ${{\mathcal {U}}}(\hat {\mathfrak g}^-)$, we have

\[ \mathbb{Y}[\bar a]:= \sum_{\sigma\in {\tt S}_m} y_{\sigma(1)}[a_1]\ldots y_{\sigma(m)}[a_m] =\varpi\biggl(\sum_{\sigma\in {\tt S}_m} y_{1}[a_{\sigma(1)}]\ldots y_{m}[a_{\sigma(m)}] \biggr) = m! \varpi(Y[\bar a]) \]

in the notation of § 1.5.

Proof. It suffices to show that $\mathbb {Y}[\bar a]$ is invariant under all ${\tt t}_i=(i\,i+1)\in {\tt S}_m$ with $1\leqslant i< m$. For each $\sigma \in {\tt S}_m$, both monomials

\[ y_{\sigma(1)}[a_1]{\ldots} y_{\sigma(i)}[a_{i}]y_{\sigma(i+1)}[a_{i+1}]{\ldots} y_{\sigma(m)}[a_m] \quad \text{and}\quad y_{\sigma(1)}[a_1]{\ldots} y_{\sigma(i+1)}[a_{i}]y_{\sigma(i)}[a_{i+1}]{\ldots} y_{\sigma(m)}[a_m] \]

appear in $\mathbb {Y}[\bar a]$ with the same coefficient $1$. Let $s(\sigma,i)$ stand for their sum. Then

\[ s(\sigma,i) - {\tt t}_i(s(\sigma,i)) = \cdots[y_{\sigma(i)}[a_i],y_{\sigma(i+1)}[a_{i+1}]]\cdots + \cdots[y_{\sigma(i+1)}[a_{i}],y_{\sigma(i)}[a_{i+1}]]\cdots =0, \]

because $[y_{\sigma (i)}[a_i],y_{\sigma (i+1)}[a_{i+1}]]=[y_{\sigma (i)},y_{\sigma (i+1)}][a_i+ a_{i+1}]= - [y_{\sigma (i+1)}[a_{i}],y_{\sigma (i)}[a_{i+1}]]$. Since $\mathbb {Y}[\bar a]=\frac {1}{2}\sum _{\sigma } s(\sigma,i)$ for each $i$, we are done.

Lemma 2.2 Take $F\in {{\mathcal {S}}}^{m}(\mathfrak g)$ and $r\geqslant 1$. Then $\varpi (\tau ^{r} F[-1]){\cdot }1$ is fully symmetrised and therefore is an eigenvector of $\omega$ corresponding to the eigenvalue $(-1)^{m}$.

Proof. Notice that $\varpi (\tau ^{r} (F+F')[-1]){\cdot }1=\varpi (\tau ^{r} F[-1]){\cdot }1+ \varpi (\tau ^{r} F'[-1]){\cdot }1$ for any $F'\in {{\mathcal {S}}}^{m}(\mathfrak g)$. Hence we may assume that $F=y_1\ldots y_m$ with $y_j\in \mathfrak g$. By the construction, $\varpi (\tau ^{r} F[-1]){\cdot }1$ is the sum of terms

\[ \frac{1}{(m+r)!} c(r,\bar a) \sum_{\sigma\in {\tt S}_m} y_{\sigma(1)}[a_1]\ldots y_{\sigma(m)}[a_m] \quad \text{with} \ c(r,\bar a)\in\mathbb{N}, \]

taken over all vectors $\bar a=(a_1,\ldots,a_m)\in {{\mathbb {Z}}}_{<0}^{m}$ such that $\sum a_j=-(m+r)$. The scalars $c(r,\bar a)$ depend on $(m,r,\bar a)$ in an elementary combinatorial way. Each summand here is a fully symmetrised element by Lemma 2.1. Hence the desired conclusion follows.

Let $z= ({1}/{n})I_n$ be a central element of $\mathfrak g=\mathfrak {gl}_n$ and let $\tilde \Delta _k\in {{\mathcal {S}}}^{k}(\mathfrak {sl}_n)$ denote the restriction of $\Delta _k$ to $\mathfrak {sl}_n$. Then

(2.3)\begin{equation} \Delta_k=\tilde\Delta_k+(n-k+1)z\tilde\Delta_{k-1}+\binom{n-k+2}{2}z^{2}\tilde\Delta_{k-2} +\cdots+\binom{n-2}{k-2}z^{k-2}\tilde\Delta_2+\binom{n}{k}z^{k}. \end{equation}

Fix $\mathfrak h=\langle E_{ii} \mid 1\leqslant i\leqslant n\rangle _{{{\mathbb {C}}}}$. Let $\varepsilon _i\in \mathfrak h^{*}$ be a linear function such that $\varepsilon _i(E_{jj})=\delta _{i,j}$. For $E_{ii}\in \mathfrak g$, set $\tilde E_{ii}=E_{ii}-z$.

Proposition 2.3 In type $\mathsf{A}$, we have

\[ {\sf m}_{2r+1}(\tilde\Delta_k) = \frac{(2r)! (k-2r)!}{k!} \binom{n-k+2r}{2r} \tilde\Delta_{k-2r} \]

if $k-2r > 1$ and ${\sf m}(\tilde \Delta _3)={\sf m}(\Delta _3)=0$.

Proof. Notice that the centre of $\mathfrak g$ plays a very specific rôle in m, since $\mathrm {ad}(z)=0$. In particular, ${\sf m}({{\mathcal {S}}}^{3}(\mathfrak {gl}_n))={\sf m}({{\mathcal {S}}}^{3}(\mathfrak {sl}_n))\subset \Lambda ^{2}\mathfrak {sl}_n$. Furthermore,

\[ {\sf m}(\Delta_k)\in {\sf m}(\tilde\Delta_k)+ \Lambda^{2}\mathfrak{sl}_n \otimes z{{\mathcal{S}}}^{k-4}(\mathfrak g), \]

where one can use the multiplication in either ${\rm End}(\mathfrak {gl}_{n})$ or ${\rm End}(\mathfrak {sl}_{n})$ for the definition of ${\sf m}$. Therefore we can work either with $\mathfrak {sl}_n$ or with $\mathfrak {gl}_n$, whichever is more convenient.

Suppose that $Y=E_{ij}E_{ls}E_{up}$ is a factor of a monomial of $\Delta _k$. Then

\[ i\not\in\{ l,u\}, \ j\not\in\{ s,p\}, \ l\ne u, \ \text{and } s\ne p. \]

The $\mathfrak h$-weight of $Y$ cannot be equal to either $2\varepsilon _1-\varepsilon _n-\varepsilon _{n-1}$ or $\varepsilon _1+\varepsilon _2-2\varepsilon _n$. These are the highest weights of the Cartan component of $\Lambda ^{2}\mathfrak {sl}_n$. Hence ${\sf m}(\Delta _k)\in (\mathfrak {sl}_n \otimes {{\mathcal {S}}}^{k-3}(\mathfrak g))^{\mathfrak g}$. The image in question is a polynomial function on $(\mathfrak {sl}_n \oplus \mathfrak g)^{*}\cong \mathfrak {sl}_n\oplus \mathfrak g$ of degree $1$ in $\mathfrak {sl}_n$ and degree $k-3$ in $\mathfrak g$. Note that ${\sf m}(\Delta _3)$ is a $\mathfrak {gl}_n$-invariant in $\mathfrak {sl}_n$ and is thereby zero. Suppose that $n\geqslant k>3$.

Fortunately, $G(\mathfrak {sl}_n \oplus \mathfrak h)$ is a dense subset of $\mathfrak {sl}_n \oplus \mathfrak g$. We calculate the restriction

\[ {{\boldsymbol f}}={\sf m}(\Delta_k)|_{\mathfrak{sl}_n\oplus\mathfrak h} \]

of ${\sf m}(\Delta _k)$ to $\mathfrak {sl}_n\oplus \mathfrak h$. Write ${{\boldsymbol f}}=\sum _{\nu =1}^{L} \xi _\nu \otimes {{\boldsymbol H}}_\nu$ with $\xi _\nu \in \mathfrak {sl}_n$ and pairwise different monomials ${{\boldsymbol H}}_\nu \in {{\mathcal {S}}}^{k-3}(\mathfrak h)$ in $\{E_{ii}\}$. Since ${\sf m}(\Delta _k)$ is an element of $\mathfrak h$-weight zero, $\xi _\nu \in \mathfrak h$ for each $\nu$. Thus one can say that ${{\boldsymbol f}}$ is an invariant of the Weyl group $W(\mathfrak g,\mathfrak h)\cong {\tt S}_{n}$. Without loss of generality assume that ${{\boldsymbol H}}_1= y_4\ldots y_k$ with $y_s=E_{ss}$ for all $s\geqslant 4$. In order to understand ${{\boldsymbol f}}$, it suffices to calculate $\xi _1$. Let $F$ be the polynomial obtained from $\Delta _3$ by setting $E_{ij}=0$ for all $(i,j)$ such that $i$ or $j$ belongs to $\{4,\ldots,k\}$. Then $\xi _1= ({3!(k-3)!}/{k!}) {\sf m}(F)$.

Now take $Y$ as above with $\{i,l,u\}=\{j,s,p\}=\{1,2,3\}$. Then:

• • ${\sf m}(Y)(E_{14})=0$ if $i=j$ or $l=s$ or $u=p$;

• • ${\sf m}(Y)(E_{14})=\frac {1}{6} E_{14}$ if $Y= E_{13} E_{32} E_{21}$;

• • ${\sf m}(Y)(E_{14})=\frac {1}{6} E_{14}$ if $Y= E_{12} E_{23} E_{31}$.

Besides, ${\sf m}(Y)(E_{vw})=0$ if $v,w\geqslant 4$. In the self-explanatory notation, $\eta ={\sf m}(\Delta _3^{(1,2,3)})$ is an invariant of $(\mathfrak {gl}_3\oplus \mathfrak {gl}_{n-3})$ and $\eta$ acts on $\mathfrak {gl}_3=\langle E_{vw} \mid 1\leqslant v,w\leqslant 3\rangle _{{{\mathbb {C}}}}$ as zero. Since $\Delta _3^{(1,2,3)}$ is a linear combination of $Y=E_{ij}E_{ls}E_{up}$ with $\{i,l,u\}=\{j,s,p\}=\{1,2,3\}$, the element $\eta$ acts on $\mathfrak g$ as $\frac {1}{3}(E_{11}+E_{22}+E_{33})$. This implies that $\eta =\frac {1}{3}(\tilde E_{11}+\tilde E_{22}+\tilde E_{33})$. By the construction of $F$, we now have ${\sf m}(F)=\binom {n-k+2}{2} \sum _{l\not \in \{4,\ldots,k\}} \frac {1}{3} \tilde E_{ll}$ and hence

\[ \xi_1\otimes{{\boldsymbol H}}_1=\frac{3!(k-3)!}{k!}\frac{1}{3}\binom{n-k+2}{2}\bigg(\sum_{l\not\in\{4,\ldots,k\}} \tilde E_{ll}\bigg) \otimes E_{44} \ldots E_{kk}. \]

From this one deduces that up to the scalar

\[ \frac{k-2}{3}\frac{3!(k-3)!}{k!} \binom{n-k+2}{2}, \]

the restriction of ${\sf m}(\tilde \Delta _k)$ to $\mathfrak {sl}_n\oplus \mathfrak h$ coincides with the restriction $\tilde \Delta _{k-2}|_{\mathfrak {sl}_n \oplus \mathfrak h}$, where we regard $\tilde \Delta _{k-2}$ as an element of $\mathfrak {sl}_n \otimes {{\mathcal {S}}}^{k-3}(\mathfrak g)$. In particular, ${\sf m}(\tilde \Delta _k)$ is a symmetric invariant. More explicitly,

\[ {\sf m}(\tilde\Delta_k) = \frac{(k-2)}{3}\frac{3!(k-3)!}{k!} \binom{n-k+2}{2} \tilde\Delta_{k-2} = \frac{2(k-2)!}{k!} \binom{n-k+2}{2} \tilde\Delta_{k-2}. \]

Iterating the map m, we obtain the result.

Remark Strictly speaking, ${\sf m}(\Delta _k)$ is not an element of ${{\mathcal {S}}}^{k-2}(\mathfrak {gl}_n)$. This is the reason for working with $\mathfrak {sl}_n$.

Now we can exhibit formulas for Segal–Sugawara vectors of $t$-degree $k$ that are independent of $n$, i.e. these formulas are valid for all $n\geqslant k$. First of all notice that in view of (2.3), Formula (2.2) produces an element of $\mathfrak z(\widehat {\mathfrak {sl}_n})$ if we replace each $\Delta _{k-2r}$ with $\tilde \Delta _{k-2r}$. (This statement can be deduced from (2.1) as well.) Making use of Proposition 2.3, one obtains that for $H=\tilde \Delta _k$,

(2.4)\begin{equation} \tilde S_{k-1}=\varpi(H[-1])+\sum_{1\leqslant r<(k-1)/2} \binom{k}{2r} \varpi(\tau^{2r} {\sf m}_{2r+1}(H)[-1]){\cdot}1 \end{equation}

is a Segal–Sugawara vector.

3. Commutators and Poisson brackets

In this section, we prove that Formula (2.4) is universal, i.e. that it is valid in all types, providing ${\sf m}_{2r+1}(H)$ is a symmetric invariant for each $r\geqslant 1$.

For $F\in {{\mathcal {S}}}^{m}(\mathfrak g)$, set $\mathbb {X}_{F[-1]}:=[{\mathcal {H}}[-1], \varpi (F)[-1]]$ . Note that

\[ \omega({\mathcal{H}}[-1] \varpi(F)[-1])=(-1)^{m+2}\varpi(F)[-1]{\mathcal{H}}[-1]. \]

Hence $\omega$ multiplies $\mathbb {X}_{F[-1]}$ with $(-1)^{m+1}$. This implies that the symbol of $\mathbb {X}_{F[-1]}$ has degree $m+1-2d$ with $d\geqslant 0$. Let ${\mathcal {H}}[-1]$ stand also for $\sum _{i} x_i[-1]x_i[-1]\in {{\mathcal {S}}}^{2}(\mathfrak g[-1])$.

The symmetric algebra ${{\mathcal {S}}}(\mathfrak q)$ of a Lie algebra $\mathfrak q$ is equipped with the standard Poisson bracket $\{\,\,,\,\}$ such that $\{x,y\}=[x,y]$ for $x,y\in \mathfrak q$. Using the standard filtration on ${{\mathcal {U}}}(\mathfrak q)$, one can state that

\[ \{\mathrm{gr}(X),\mathrm{gr}(Y)\}=[X,Y]+{{\mathcal{U}}}_{l+m-2}(\mathfrak q) \quad \text{if } X\in{{\mathcal{U}}}_l(\mathfrak q){\setminus}{{\mathcal{U}}}_{l-1}(\mathfrak q), Y\in{{\mathcal{U}}}_m(\mathfrak q){\setminus}{{\mathcal{U}}}_{m-1}(\mathfrak q). \]

The fact that $\{{\mathcal {H}}[-1], F[-1]\}=0$ for $F\in {{\mathcal {S}}}^{m}(\mathfrak g)^{\mathfrak g}$ follows from [FF92]. For convenience, we present a short proof here.

Lemma 3.1 Take two arbitrary $\mathfrak g$-invariants $F,F'$ in ${{\mathcal {S}}}(\mathfrak g)$. Then $\{F[-1],F'[-1]\}=0$.

Proof. The Poisson bracket of two polynomial functions can be calculated by

(3.1)\begin{equation} \{f_1,f_2\}(\gamma)=[d_\gamma f_1,d_\gamma f_2](\gamma) \quad \text{for} \ \gamma\in({{\hat{\mathfrak g}}}^-)^{*}. \end{equation}

In the case of $F[-1]$ and $F'[-1]$, the differentials $d_\gamma F[-1], d_\gamma F'[-1]$ at $\gamma$ depend only on the $(-1)$-part of $\gamma$. More explicitly, if $\gamma (x[-1])=\tilde \gamma (x)$ with $\tilde \gamma \in \mathfrak g^{*}$, then $d_\gamma F[-1]=(d_{\tilde \gamma } F)[-1]$ and the same identity hods for $F'$. We have $d_{\tilde \gamma } F,d_{\tilde \gamma }F'\in (\mathfrak g_\gamma )^{\mathfrak g_\gamma }$, since $F$ and $F'$ are $\mathfrak g$-invariants. Hence $[d_{\tilde \gamma } F,d_{\tilde \gamma }F']=0$ and also $[d_{\gamma }F[-1],d_{\gamma }F'[-1]]=0$ for any $\gamma \in ({{\hat {\mathfrak g}}}^-)^{*}$.

If $[\mathfrak g,\mathfrak g]$ is not simple, then the following assumption on the choice of the scalar product on $\mathfrak g$ is made in order to simplify the calculations.

1. (${\blacklozenge}$) There is a constant $C\in {{\mathbb {C}}}$ such that $\sum _{i=1}^{\dim \mathfrak g} \mathrm {ad}(x_i)^{2}(\xi )=C\xi$ for each $\xi \in [\mathfrak g,\mathfrak g]$.

The constant $C$ depends on the scalar product in question.

From now on, assume that $\mathfrak g$ is semisimple. As the next step we examine the difference

\[ X_{F[-1]}:= \mathbb{X}_{F[-1]}-\varpi(\{{\mathcal{H}}[-1], F[-1]\}) \]

and more general expressions, where the commutator is taken with ${\mathcal {H}}[\bar b]$. Our goal is to present $\mathbb {X}_{F[-1]}$ in the form (0.1). For any ${\mathcal {F}}_1,{\mathcal {F}}_2\in {{\mathcal {U}}}({{\hat {\mathfrak g}}}^-)$, the symbol $\mathrm {gr}([{\mathcal {F}}_1,{\mathcal {F}}_2])$ is equal to the Poisson bracket $\{\mathrm {gr}({\mathcal {F}}_1),\mathrm {gr}({\mathcal {F}}_2)\}$ if this Poisson bracket is non-zero.

3.1 Commutators, the first approximation

Fix $m\geqslant 1$. Then set $\check {j}=m-j$ for $1\leqslant j< m$.

Lemma 3.2 For $Y=\hat y_1\ldots \hat y_m\in {{\mathcal {S}}}({{\hat {\mathfrak g}}}^-)$ with $\hat y_j=y_j[a_j]$, set

\[ X_Y=X_{Y,\bar b}=[{\mathcal{H}}[b_1,b_2], \varpi(Y)]- \varpi(\{{\mathcal{H}}[b_1,b_2], Y\}). \]

Then

\begin{align*} &X_Y= \sum_{l=1}^{m}\sum_{\sigma\in \tilde{\tt S}_{m-1}}\sum_{j< p} \sum_{i,u} ({\sf m}(y_{\sigma(p)} \otimes y_{l} \otimes y_{\sigma(j)}) (x_i), x_{u}) \\ &\times (c_{2,3}(j,p) \hat y_{\sigma(1)}\ldots \hat y_{\sigma(j-1)} x_i[b_1+a_{\sigma(j)}] \hat y_{\sigma(j+1)}\ldots \hat y_{\sigma(p-1)} x_{u}[b_2+a_{l}+a_{\sigma(p)}] \hat y_{\sigma(p+1)} \ldots \hat y_{\sigma(m-1)} \\ &+c_{2,3}(j,p) \hat y_{\sigma(1)}\ldots \hat y_{\sigma(j-1)} x_i[b_2+a_{\sigma(j)}] \hat y_{\sigma(j+1)}\ldots \hat y_{\sigma(p-1)} x_{u}[b_1+a_{l}+a_{\sigma(p)}] \hat y_{\sigma(p+1)} \ldots \hat y_{\sigma(m-1)} \\ & + (-1) c_{3,2}(j,p) \hat y_{\sigma(1)}\ldots \hat y_{\sigma(j-1)} x_i[b_1+a_{\sigma(j)}+a_{l}] \hat y_{\sigma(j+1)}\ldots \hat y_{\sigma(p-1)} x_{u}[b_2+a_{\sigma(p)}] \hat y_{\sigma(p+1)} \ldots \hat y_{\sigma(m-1)} \\ & + (-1) c_{3,2}(j,p) \hat y_{\sigma(1)}\ldots \hat y_{\sigma(j-1)} x_i[b_2\!+a_{\sigma(j)}\!+a_{l}] \hat y_{\sigma(j+1)}\ldots \hat y_{\sigma(p-1)} x_{u}[b_1\!+a_{\sigma(p)}] \hat y_{\sigma(p+1)} \ldots \hat y_{\sigma(m-1)} ), \end{align*}

where $\tilde{\tt S}_{m-1}$ stands for the set of bijective maps from $\{1,\ldots,m-1\}$ to $\{1,\ldots,m\}{\setminus} \{l\}$ and we have $1\leqslant j< p\leqslant m-1$. The constants $c_{2,3}(j,p)$, $c_{3,2}(j,p)\in \mathbb {Q}$ do not depend on $Y$, they depend only on $m$. Besides,

\[ c_{2,3}(j,p)=c_{3,2}(\check{p},\check{j}), \]

$c_{2,3}(j,p)\leqslant 0$ for all $j< p$, and $c_{2,3}(j,p)< 0$ if in addition $\check {j}\geqslant p$.

Proof. Set $\hat Y=\hat y_1\ldots \hat y_m\in {{\mathcal {U}}}(t^{-1}\mathfrak g[t^{-1}])$. Let $\hat x_i^{(1)}$ stand for $x_i[b_1]$ and $\hat x_i^{(2)}$ for $x_i[b_2]$. Then

\[ [{\mathcal{H}}[b_1,b_2],\hat Y]= \sum_{j=1,i=1}^{j=m,i=\dim\mathfrak g} (\hat y_1\ldots \hat y_{j-1} \hat x_i^{(1)} [\hat x_i^{(2)},\hat y_j] \hat y_{j+1}\ldots \hat y_m+ \hat y_1\ldots \hat y_{j-1} [\hat x_i^{(1)},\hat y_j] \hat x_i^{(2)} \hat y_{j+1}\ldots\hat y_m). \]

Furthermore,

(3.2)\begin{equation} [{\mathcal{H}}[b_1,b_2],\varpi(Y)]= \frac{1}{m!} \sum_{\sigma\in{\tt S}_m} [{\mathcal{H}}[b_1,b_2],\sigma(\hat Y)]. \end{equation}

Each summand of $[{\mathcal {H}}[b_1,b_2],\sigma (\hat Y)]$ we regard as a formal non-commutative product. The symmetrisation of $P_Y=\{{\mathcal {H}}[b_1,b_2], Y\}$ resembles (3.2), but with a rather significant difference: the factor $\hat x_i^{(\upsilon )}$, which is not involved in $[\hat x_i^{(\nu )},\hat y_{\sigma (j)}]$, does not have to stay next to $[\hat x_i^{(\nu )},\hat y_{\sigma (j)}]$ (here we have $\{\nu,\upsilon \}=\{1,2\}$). The idea behind the computation of $X_Y$ is to modify each term of $\varpi (P_Y)$ in such a way that the wayward factor $\hat x_i^{(\upsilon )}$ gets back to its place as in $[{\mathcal {H}}[b_1,b_2],\varpi (Y)]$. In this process, other commutators $\pm [\hat x_i^{(\upsilon )},\hat y_l]$ will appear. It is convenient to consider the differences

\[ X_{\sigma(\hat Y)}= [{\mathcal{H}}[b_1,b_2],\sigma(\hat Y)] - \frac{1}{m+1}\sum_{j=1}^{m}\sum_{\upsilon=1}^{2} \overset{\underbrace{\phantom{1237719278}\hat x_i^{(\upsilon)}\ \text{at different places}\phantom{1733543293}}}{\hat y_{\sigma(1)}{\ldots}\hat y_{\sigma(j-1)} [\hat x_i^{(\nu)},\hat y_{\sigma(j)}] \hat y_{\sigma(j+1)}\ldots \hat y_{\sigma(m-1)}\hat y_{\sigma(m)}}, \]

where for each fixed $j$ and $\upsilon$, we add $m+1$ different formal products with $\hat x_i^{(\upsilon )}$ standing in $m+1$ different places. Then $X_Y= ({1}/{m!}) \sum _{\sigma \in {\tt S}_m} X_{\sigma (\hat Y)}$.

While modifying $\varpi (P_Y)$, one obtains products of the form

\[ X(\sigma,i,\upsilon,\nu;j,p)={\hat y_{\sigma(1)}\ldots\hat y_{\sigma(j-1)} [\hat x_i^{(\upsilon)},\hat y_{\sigma(j)}] \hat y_{\sigma(j+1)} \ldots\hat y_{\sigma(p-1)} [\hat x_i^{(\nu)},\hat y_{\sigma(p)}]\hat y_{\sigma(p+1)}\ldots \hat y_{\sigma(m)}} \]

with some coefficients; one also has to commute $\hat x_i^{(\upsilon )}$ with $[\hat x_i^{(\nu )},\hat y_j]$. The total sum of the products that correspond to $X_{\hat Y}$ and contain a double commutator $[ \hat x_i^{(\upsilon )},[\hat x_i^{(\nu )},\hat y_j]]$ as a factor is

\[ X_{{\rm dcom}}({{\rm id}},j,i):=\frac{1}{m+1}\hat y_1\ldots \hat y_{j-1} (j[\hat x_i^{(2)},[\hat x_i^{(1)},\hat y_j]] + (m-j+1) [[\hat x_i^{(2)},\hat y_j], \hat x_i^{(1)}])\hat y_{j+1}\ldots \hat y_m. \]

Observe that

\[ [\hat x_i^{(2)},[\hat x_i^{(1)},\hat y_j]]+[[\hat x_i^{(2)},\hat y_j],\hat x_i^{(1)}]=[[\hat x_i^{(2)},\hat x_i^{(1)}],\hat y_j]=0. \]

Applying $\sigma$ to all $\hat y_p$ with $1\leqslant p\leqslant m$, we obtain $X_{{\rm dcom}}(\sigma,j,i)$ from $X_{{\rm dcom}}({{\rm id}},j,i)$. Set $X_{{\rm dcom}}=\sum _{j=1}^{m}\sum _{\sigma \in {\tt S}_m}\sum _i X_{{\rm dcom}}(\sigma,j,i)$. We have $\mathrm {gr}(X_{{\rm dcom}})=0$, since

\[ \mathrm{gr}(X_{{\rm dcom}}({{\rm id}},j,i))+\mathrm{gr}(X_{{\rm dcom}}(\vartheta,m-j+1,i))=0 \]

for the transposition $\vartheta =(j(m-j+1))$.

Now consider expressions $[[\hat x_i^{(\upsilon )},[\hat x_i^{(\nu )},\hat y_j]],\hat y_l]$. In view of ($\blacklozenge$),

\[ \sum_i \mathrm{ad}(\hat x_i^{(1)})\mathrm{ad}(\hat x_i^{(2)})(y_j[a_j])=\sum_i \mathrm{ad}(\hat x_i^{(2)})\mathrm{ad}(\hat x_i^{(1)})(y_j[a_j])=Cy_j[a_j+b], \]

where $b=b_1+b_2$. This leads to

(3.3)\begin{equation} \sum_{i} [[\hat x_i^{(\upsilon)},[\hat x_i^{(\nu)},\hat y_j]],\hat y_l] = [Cy_j[a_j+b],\hat y_l]=C[y_j,y_l][a_j+a_l+b]. \end{equation}

Thus $\sum _i X_{{\rm dcom}}({{\rm id}},j,i)-\varpi (\mathrm {gr}(\sum _i X_{{\rm dcom}}({{\rm id}},j,i))) = C \sum _{l\ne j} c_{(j,l)} X_{{\rm dcom}}^{[j,l]}({{\rm id}})$, where

\[ X_{{\rm dcom}}^{[j,l]}({{\rm id}})= \begin{cases} \hat y_1\ldots\hat y_{l-1} [y_j,y_l][a_j+a_l+b]\hat y_{l+1}\ldots\hat y_{j-1}\hat y_{j+1}\ldots\hat y_m & \text{if } l< j \\ \hat y_1\ldots\hat y_{j-1}\hat y_{j+1} \ldots\hat y_{l-1} [y_j,y_l][a_j+a_l+b]\hat y_{l+1}\ldots \hat y_m & \text{if } l>j; \end{cases} \]

furthermore $c_{(j,l)}\in \mathbb {Q}$. Set $\sigma =(j\,l)$. The difference

\[ \sum_i X_{{\rm dcom}}(\sigma,j,i)-\varpi\biggl(\mathrm{gr}\biggl(\sum_i X_{{\rm dcom}}(\sigma,j,i)\biggr)\biggr) \]

has a summand

\[ Cc_{(j,l)} X_{{\rm dcom}}^{[j,l]}(\sigma)=-Cc_{(j,l)} X_{{\rm dcom}}^{[j,l]}({{\rm id}}). \]

This proves that $\sum _{\sigma \in {\tt S}_m,j,i} X_{{\rm dcom}}(\sigma,j,i)=0$, i.e. the expressions containing double commutators $[\hat x_i^{(\upsilon )},[\hat x_i^{(\nu )},\hat y_j]]$ as factors have no contribution to $X_Y$.

In the modification of $(m+1)X_{\hat Y}$, a term

\[ X({{\rm id}},i,\upsilon,\nu;j,l)=\hat y_1\ldots \hat y_{j-1} [\hat x_i^{(\upsilon)},\hat y_j] \hat y_{j+1} \ldots \hat y_{l-1} [\hat x_i^{(\nu)},\hat y_l] \hat y_{l+1} \ldots \hat y_m \]

appears $j$ times with the coefficient $1$ (these are the instances, where $\hat x_i^{(\upsilon )}$ is the wayward factor); it also appears $(m-l+1)$ times with the coefficient $(-1)$ from the instances, where $\hat x_i^{(\nu )}$ is the wayward factor. Thereby

\[ X_{\sigma(\hat Y)}= \frac{1}{m+1} \sum_{j< l} \sum_{\upsilon} \sum_{i} (j+l-m-1) X(\sigma,i,\upsilon,\nu;j,l). \]

Set $j'=m-j+1$. Then $j+j'=m+1$. Assume that $l\ne j'$ and $l>j$. For any $\sigma \in {\tt S}_m$, the symbol of $X(\sigma,i,\upsilon,\nu ;j,l)$ is the same as the symbol of $X(\tilde \sigma,i,\upsilon,\nu ;l',j')$, where

\[ \tilde\sigma(\hat Y)=\hat y_{\sigma(1)}\ldots\hat y_{\sigma(l'-1)}\hat y_{\sigma(j)}\hat y_{\sigma(l'+1)}\ldots\hat y_{\sigma(j'-1)}\hat y_{\sigma(l)}\hat y_{\sigma(j'+1)}\ldots\hat y_{\sigma(m)}; \]

furthermore, these two expressions appear in $X_{Y}$ with opposite coefficients. This shows that $X_{Y}\in {{\mathcal {U}}}_{m-1}(\hat {\mathfrak g}^-)$. In order to get a better understanding of $X_{Y}$, we modify the terms $X(\sigma,i,\upsilon,\nu ;j,l)$, which we consider as formal products.

Each $X(\sigma,i,\upsilon,\nu ;j,l)$ has factors of two sorts: elements $\hat y_p$ (depicted as points in the diagram below) and two commutators $[\hat x_i^{(\upsilon )},\hat y_{\sigma (j)}]$, $[\hat x_i^{(\upsilon )},\hat y_{\sigma (l)}]$, which are depicted as stars. We move the commutator that is closer to the middle point of the product until the expression obtains a central symmetry. In the case $l\leqslant {m}/{2}$, this looks as follows:

\[ \begin{array}{c} \cdots\cdot\star \cdots\cdots\cdot \star \cdots\cdot\cdot | \cdots\cdots\cdots\cdots\cdots\cdots\cdot \overset{\text{modification}}{\leadsto} \ \cdots\cdot \star \cdots\cdots\cdots\cdots\cdot | \cdots\cdots\cdots\cdots\cdot \star \cdots\cdot. \end{array} \]

See also Example 3.3 below. The commutator $[\hat x_i^{(\upsilon )},\hat y_{\sigma (l)}]$ is moving if and only if $l< j'$. After the modification, the products of $m$ factors annihilate each other and $X_Y$ is now a $\mathbb {Q}$-linear combination of products of $m-1$ factors, where in each term, $m-3$ factors are elements $\hat y_w$, one is a commutator $[\hat x_i^{(\upsilon )},\hat y_j]$, and another one is a commutator $[[\hat x_i^{(\nu )},\hat y_l],\hat y_p]$. A possible example in the case $\sigma ={{\rm id}}$, is

\[ \hat y_1\ldots \hat y_{j-1} [\hat x_i^{(\upsilon)},\hat y_j] \hat y_{j+1} \ldots \hat y_{l-1} \hat y_{l+1} \ldots \hat y_{p-1} [[\hat x_i^{(\nu)},\hat y_l],\hat y_p] \hat y_{p+1} \ldots \hat y_m. \]

It appears only if $l< j'$. First we deal with these expressions ‘qualitative’ and after that describe the coefficients.

Observe that for $y\in \mathfrak g$ and $a\in {{\mathbb {Z}}}_{<0}$, we have $y[a]=\sum _{i} (x_i,y) x_i[a]$. Assume for simplicity that $\{j,p,l\}=\{1,2,3\}$, disregard for the moment the other factors, and ignore the $t$-degrees of the elements. Consider the sum

(3.4)\begin{align} \sum_i [x_i,y_1] [y_3,[y_2,x_i]] &= \sum_{i,j,u} ([x_i,y_1],x_j)x_j ([y_3,[y_2,x_i]],x_u)x_u \nonumber\\ & = \sum_{i,j,u} (x_i,[ y_1,x_j] )x_j (x_i, \mathrm{ad}(y_2)\mathrm{ad}(y_3)(x_u))x_u \nonumber\\ & = \sum_{i,j,u} (( \mathrm{ad}(y_2)\mathrm{ad}(y_3)(x_u), x_i) x_i, [y_1,x_j]) x_j x_u \nonumber\\ & = \sum_{j,u} (\mathrm{ad}(y_2)\mathrm{ad}(y_3)(x_u),[y_1,x_j])x_jx_u \nonumber\\ &=\sum_{j,u}(\mathrm{ad}(y_3)\mathrm{ad}(y_2)\mathrm{ad}(y_1)(x_j), x_u)x_j x_u. \end{align}

Note that $\mathrm {ad}(y_3)\mathrm {ad}(y_2)\mathrm {ad}(y_1)(x_j)={\sf m}(y_3\otimes y_2\otimes y_1)(x_j)$. If we recall the $t$-degrees, then the product $x_j x_u$ has to be replaced with $x_j[b_\upsilon +a_1] x_u[b_\nu +a_2+a_3]$ in (3.4). The other factors $\hat y_w$ do not interfere with the transformations in (3.4).

In the process of changing the sequence of factors of $X({{\rm id}},i,\upsilon,\nu ;j,l)$ with $j< l< j'$, the term $\cdots [\hat x_i^{(\upsilon )},\hat y_j]\cdots [[\hat x_i^{(\nu )},\hat y_l],\hat y_p]\cdots$ appears with the negative coefficient $(j+l-m-1)$ as long as $l< p\leqslant j'$. This shows that indeed the constants $c_{2,3}(j,p)$ do not depend on $Y$, they depend only on $m$. Moreover, $c_{2,3}(j,p)=0$ if $p>\check {j}$ and $c_{2,3}(j,p)<0$ if $p\leqslant \check {j}$.

The symmetry $c_{2,3}(j,p)=c_{3,2}(\check {p},\check {j})$ is justified by the fact that $\omega (X_Y)=(-1)^{m-1}X_Y$. A more direct way to see this, is to notice that if a factor $[\hat x_i^{(\nu )},\hat y_l]$ moves from a place $v$ to $j'$ in some term, then $j< v< j'$ and there is a term with the apposite coefficient, where $[\hat x_i^{(\nu )},\hat y_l]$ moves from $v'$ to $j$. The first type of moves produces

\[ ({\rm coeff.}) ({\sf m}(y_{\sigma(p)}\otimes y_l\otimes y_{\sigma(j)})(x_i),x_u) \ldots x_i[b_{\upsilon}+a_{\sigma(j)}] \ldots x_u[b_\nu+a_l+a_{\sigma(p)}]\ldots \]

and the second

\[ ({\rm the\ same\ coeff.}) (x_u,{\sf m}(y_{\sigma(p')}\otimes y_l\otimes y_{\sigma(j')})(x_i)) \ldots x_u[b_\nu+a_l+a_{\sigma(p')}] \ldots x_i[b_{\upsilon}+a_{\sigma(j')}]\ldots . \]

We have $(x_i,{\sf m}(y_{\sigma (j)}\otimes y_l\otimes y_{\sigma (p)})(x_u))= - ({\sf m}(y_{\sigma (p)}\otimes y_l\otimes y_{\sigma (j)})(x_i),x_u)$ and the scalar product $(\,\,,\,)$ is symmetric. These facts confirm the symmetry of the constants and justifies the minus signs in front of $c_{3,2}(j,p)$ in the answer.

Example 3.3 Consider the case $m=6$. One obtains that

\begin{align*} X_Y&= \frac{1}{7!}\sum_{\sigma\in {\tt S}_6,i}( 4 X(\sigma,i,1,2;5,6) - 4 X(\sigma,i,1,2;1,2) + 3 X(\sigma,i,1,2;4,6) - 3 X(\sigma,i,1,2;1,3) \\ &\quad +2 X(\sigma,i,1,2;3,6) - 2 X(\sigma,i,1,2;1,4) + 2 X(\sigma,i,1,2;4,5) - 2 X(\sigma,i,1,2;2,3) \\ &\quad+ X(\sigma,i,1,2;2,6) - X(\sigma,i,1,2;1,5) + X(\sigma,i,1,2;3,5) - X(\sigma,i,1,2;2,4)) \\ &\quad+ (\text{the similar expression for } (\upsilon,\nu)=(2,1)). \end{align*}

Take $\sigma ={{\rm id}}$. Performing the modification

\[ X({{\rm id}},i,1,2;5,6)=\hat y_1\hat y_2\hat y_3\hat y_4[\hat x_i^{(1)},\hat y_5][\hat x_i^{(2)},\hat y_6] \ \overset{\text{modification}}{\leadsto} \ [\hat x_i^{(1)},\hat y_5]\hat y_1\hat y_2\hat y_3\hat y_4 [\hat x_i^{(2)},\hat y_6] , \]

we move the factor $[\hat x_i^{(1)},\hat y_5]$ to the first place producing commutators with $\hat y_4$, $\hat y_3$, $\hat y_2$, $\hat y_1$ on the way. In the same manner,

\[ X({{\rm id}},i,1,2;1,2)=[\hat x_i^{(1)},\hat y_1][\hat x_i^{(2)},\hat y_2]\hat y_3\hat y_4\hat y_5\hat y_6 \ \overset{\text{modification}}{\leadsto} \ [\hat x_i^{(1)},\hat y_1]\hat y_3\hat y_4\hat y_5\hat y_6 [\hat x_i^{(2)},\hat y_2] \]

and for $\tilde \sigma =(1\,5\,3)(2\,6\,4)$,

\[ X(\tilde\sigma,i,1,2;1,2)=[\hat x_i^{(1)},\hat y_5][\hat x_i^{(2)},\hat y_6]\hat y_1\hat y_2\hat y_3\hat y_4 \ \overset{\text{modification}}{\leadsto} \ [\hat x_i^{(1)},\hat y_5]\hat y_1\hat y_2\hat y_3\hat y_4 [\hat x_i^{(2)},\hat y_6]. \]

Performing the modification

\[ X({{\rm id}},i,1,2;4,5)=\hat y_1\hat y_2\hat y_3[\hat x_i^{(1)},\hat y_4][\hat x_i^{(2)},\hat y_5]\hat y_6 \ \overset{\text{modification}}{\leadsto} \ \hat y_1[\hat x_i^{(1)},\hat y_4]\hat y_2\hat y_3[\hat x_i^{(2)},\hat y_5]\hat y_6, \]

we move the factor $[\hat x_i^{(1)},\hat y_4]$ to the second place producing commutators with $\hat y_3$ and $\hat y_2$. The non-zero constants $c^-(j,p):=-c_{2,3}(j,p)$ are

\[ c^-(1,2)=\frac{4}{7!},\quad c^-(1,3)=\frac{7}{7!},\quad c^-(1,4)=\frac{9}{7!},\quad c^-(1,5)=\frac{10}{7!},\quad c^-(2,3)=\frac{2}{7!},\quad c^-(2,4)=\frac{3}{7!}. \]

Instead of the usual symmetrisation map, one can consider a weighted ‘symmetrisation’ or rather shuffle, where each permutation is added with a scalar coefficient assigned by a certain function $\Psi$. We will need only a very particular case of this construction. Let $\Psi \!: {\tt S}_{k+2}\to \mathbb {Q}$ be a weight function, satisfying the following assumptions:

1. (A) $\Psi (\sigma )$ depends only on $j=\sigma (k+1)$ and $p=\sigma (k+2)$, i.e. $\Psi (\sigma )=\Psi (j,p)$;

2. (B) $\Psi (j,p)=\Psi (j',p')$ if $j'=k+3-j$.

Then set

\[ \varpi_{\rm wt} (y_1\ldots y_k\otimes y_{k+1} \otimes y_{k+2}) = \sum_{\sigma\in{\tt S}_{k+2}} \Psi(\sigma)\, y_{\sigma(1)} \otimes \cdots \otimes y_{\sigma(k+2)} \]

for $y_j\in \mathfrak q$. Let also $\varpi _{\rm wt}$ stand for the corresponding map from ${{\mathcal {S}}}^{k}(\mathfrak q)\otimes \mathfrak q\otimes \mathfrak q$ to ${{\mathcal {U}}}(\mathfrak q)$. Condition (B) guaranties that $\omega (\varpi _{\rm wt}(F))=(-1)^{k} \varpi _{\rm wt}(F)$ for each $F\in {{\mathcal {S}}}^{k}(\mathfrak q)\otimes \mathfrak q\otimes \mathfrak q$. In the case $\Psi (\sigma )= {1}/{(k+2)!}$, the map $\varpi _{\rm wt}$ coincides with $\varpi$. Keep in mind that each appearing $\varpi _{\rm wt}$ may have its own weight function.

Suppose that $Y\in {{\mathcal {S}}}^{m}(\mathfrak g)$, $\bar a\in \mathbb {Z}_{<0}^{m}$, and we want to merge them in order to obtain an element of ${{\mathcal {S}}}^{m}({{\hat {\mathfrak g}}}^-)$. The only canonical way to do so is to replace $\bar a$ with the orbit ${\tt S}_m\bar a$, add over this orbit, and divide by $|{\tt S}_m\bar a|$ as we have done in § 1.5. The result is the $\bar a$-polarisation $Y[\bar a]$ of $Y$. Set

\[ X_{Y[\bar a]}=[{\mathcal{H}[b_1,b_2], \varpi(Y[\bar a])]- \varpi(\{{\mathcal{H}}[b_1,b_2]}, Y[\bar a]\}). \]

For different numbers $u,v,l\in \{1,\ldots,m\}$, let $\bar a^{u,v,l}\in \mathbb {Z}_{<0}^{m-3}$ be the vector obtained from $\bar a$ by removing $a_u$, $a_v$, and $a_l$. Let $\langle u,v,l\rangle$ be a triple such that $l< v$ and $u\ne l,v$. Write ${\sf m}(Y)= \sum _{w=1}^{L} \xi _w \otimes R_w$ with $\xi _w \in \Lambda ^{2}\mathfrak g$, $R_w\in {{\mathcal {S}}}^{m-3}(\mathfrak g)$.

Proposition 3.4 The element $X_{Y[\bar a]}$ is equal to

\[ \sum_w \sum_{\langle u,v,l\rangle }\sum_{\upsilon}\sum_{i,j} (\xi_w(x_i), x_j) \varpi_{\rm wt}(R_w[\bar a^{u,v,l}] \otimes x_i[b_\upsilon+a_u] \otimes x_j[b_\nu+a_l+a_v]), \]

where $\Psi (j,p)= 2c_{2,3}(j,p)$ if $j< p$ and $\Psi (j,p)=2c_{3,2}(p,j)$ if $j>p$ for the weight function $\Psi$.

Proof. Using the linearity, we may assume that $Y=y_1\ldots y_m$. The symmetry in $t$-degrees allows one to add the expressions appearing in the formulation of Lemma 3.2 over the triples $(y_{e}[a_{\sigma (p)}], y_{f}[a_l], e_g[a_{\sigma (j)}])$ with $\{e,f,g\}=\{\sigma (p),l,\sigma (j)\}$ while keeping $x_i[b_\upsilon +a_{\sigma (j)}]$, $x_u[b_\nu +a_{l}+a_{\sigma (p)}]$ and $x_i[b_\upsilon +a_{\sigma (j)}+a_l]$, $x_u[b_\nu +a_{\sigma (p)}]$ at their places. In this way the coefficient ${\sf m}(y_{\sigma (p)}\otimes y_l \otimes y_{\sigma (j)})$ is replaced with ${\sf m}(y_{\sigma (p)} y_l y_{\sigma (j)})$ and thereby $\xi _w$ with ${1\leqslant w\leqslant L}$ come into play. It remains to count the scalars and describe the weight function.

Suppose that $j< p$. Then

\[ \frac{2}{m!} 3! c_{2,3}(j,p) =\frac{3!} {m!} \Psi(j,p) \]

and thereby $\Psi (j,p)=2c_{2,3}(j,p)$. Analogously, $\Psi (p,j)=2c_{3,2}(j,p)$.

Theorem 3.5 For $F\in {{\mathcal {S}}}^{m}(\mathfrak g)^{\mathfrak g}$ with $m\geqslant 4$, the symmetrisation $\varpi (F)[-1]$ is an element of the Feigin–Frenkel centre if and only if ${\sf m}(F)=0$.

Proof. According to [Ryb08], $\varpi (F)[-1]\in \mathfrak z({{\hat {\mathfrak g}}})$ if and only if it commutes with ${\mathcal {H}}[-1]$. In view of Lemma 3.1, this is the case if and only if $X_{F[-1]}=0$. Lemma 3.2 describes this element. It states that $c_{2,3}(j,p), c_{3,2}(j,p)\leqslant 0$ and $c_{2,3}(j,p)<0$ if $p\leqslant \check {j}$ as well as $c_{3,2}(j,p)<0$ if $\check {p}\leqslant j$. Since $\varpi (F)[-1]$ is fully symmetrised, we can use Proposition 3.4. It immediately implies that if ${\sf m}(F)=0$, then $X_{F[-1]}=0$.

Suppose that ${\sf m}(F)\ne 0$. Write ${\sf m}(F)= \sum _{w=1}^{L} \xi _w \otimes R_w$ with $\xi _w \in \Lambda ^{2}\mathfrak g$ and linearly independent $R_w\in {{\mathcal {S}}}^{m-3}(\mathfrak g)$. If $\xi \in \Lambda ^{2}\mathfrak g$ is non-zero, then there are $i,j$ such that ${(\xi (x_i),x_j)\ne 0}$.

Set ${\boldsymbol c}=2\sum _{j< p} (c_{2,3}(j,p)+c_{3,2}(j,p))$. According to Lemma 3.2, ${\boldsymbol c}<0$. Hence

\[ \frac{m!}{(m-3)!} \, {\boldsymbol c} \sum_{w=1}^{w=L} \sum_{i,j}(\xi_w(x_i),x_j) x_i[-2]x_j[-3] R_w[-1] \]

is a non-zero element of ${{\mathcal {S}}}({{\hat {\mathfrak g}}}^-)$. In view of the same lemma, this expression is equal to $\mathrm {gr}(X_{F[-1]})$. Thus $X_{F[-1]}\ne 0$. This completes the proof.

Remark If $\mathfrak g$ is simple, then $\mathfrak g^{\mathfrak g}$ is equal to zero and ${{\mathcal {S}}}^{2}(\mathfrak g)^{\mathfrak g}$ is spanned by ${\mathcal {H}}=\sum _{i} x_i^{2}$. By our convention, ${\sf m}({{\mathcal {S}}}^{m}(\mathfrak g))=0$ if $m\leqslant 2$. Furthermore, ${\sf m}({{\mathcal {S}}}^{3}(\mathfrak g)^{\mathfrak g})\subset (\Lambda ^{2}\mathfrak g)^{\mathfrak g}=0$. Therefore Theorem 3.5 holds for $m\leqslant 3$ as well.

We will be using weighted shuffles $\varpi _{\rm wt}$ of Poisson half-brackets. If $Y=\hat y_1\ldots \hat y_m\in {{\mathcal {S}}}({{\hat {\mathfrak g}}}^-)$, then

(3.5)\begin{equation} \varpi_{\rm wt}(Y, b_1,b_2):= \sum_{j=1,i=1}^{j=m,i=\dim\mathfrak g} \varpi_{\rm wt}(Y/\hat y_j \otimes x_i[b_1] \otimes [x_i[b_2],\hat y_j] ). \end{equation}

Strictly speaking, here $\varpi _{\rm wt}$ is a linear map from ${{\mathcal {S}}}^{m}({{\hat {\mathfrak g}}}^-)$ in ${{\mathcal {U}}}({{\hat {\mathfrak g}}}^-)$ depending on $(b_1,b_2)$ and the choice of a weight function $\Psi$. The absence of ${\rm wt}$ in the lower index indicates that we are taking the usual symmetrisation.

3.2 Iterated shuffling

Another general fact about Lie algebras $\mathfrak q$ will be needed. Suppose that $Y=y_1\ldots y_m\in {{\mathcal {S}}}^{m}(\mathfrak q)$ and $x\in \mathfrak q$. Write $Y=({1}/{m}) \sum _{1\leqslant j\leqslant m} y_j \otimes Y^{(j)}$ with $Y^{(j)}=Y/y_j$. Then

(3.6)\begin{equation} \sum_{1\leqslant j\leqslant m} [x,y_j] Y^{(j)} = \{x,Y\}. \end{equation}

Proposition 3.6 Let ${\mathcal {F}}[\check {\bar a}]=\varpi (F[\bar a])\in {{\mathcal {U}}}(\hat {\mathfrak g}^-)$ be a fully symmetrised element corresponding to a polynomial $F\in {{\mathcal {S}}}^{m}(\mathfrak g)$ and a vector $\bar a=(a_1,\ldots,a_m)\in \mathbb {Z}_{<0}^{m}$. Suppose that ${\sf m}_{2r+1}(F)\in {{\mathcal {S}}}^{m-2r}(\mathfrak g)$ for all $m/2>r\geqslant 1$. Then:

1. (i) $X_{F[\bar a]}=[{\mathcal {H}}[\bar b],{\mathcal {F}}[\check {\bar a}]] - \varpi (\{ {\mathcal {H}}[\bar b], F[\bar a]\})$ is a sum of weighted symmetrisations

   \[ \varpi_{\rm wt}({\sf m}(F)[\bar a^{l,j}], b_1+a_l,b_2+a_j), \quad \varpi_{\rm wt}({\sf m}(F)[\bar a^{l,j}], b_2+a_l,b_1+a_j), \]
   where $l\ne j$ and $\bar a^{l,j}$ is obtained from $\bar a$ by removing $a_l$ and $a_j$;

2. (ii) for every weight function $\Psi$, there is a constant $c\in \mathbb {Q}$, which is independent of $F$, such that ${\mathcal {P}}_{F[\bar a]}=\varpi _{\rm wt}(F[\bar a], b_1,b_2) - c\varpi (F[\bar a],b_1,b_2)$ is a sum of $\varpi _{\rm wt}({\sf m}(F)[\bar a^{(1)}], b_\upsilon +\gamma _\upsilon, b_\nu +\gamma _\nu )$ with different weight functions, whereby $\bar a^{(1)}$ is a subvector of $\bar a$ with $m-2$ entries and $\bar \gamma \in \mathbb {Z}_{<0}^{2}$ is constructed from the complement $\bar a{\setminus} \bar a^{(1)}$ of $\bar a^{(1)}$;

3. (iii) $\mathbb {X}_{F[a]}=[{\mathcal {H}}[b_1,b_2],{\mathcal {F}}[\check {\bar a}]]$ is a sum of

   \[ C(\bar a^{(r)},\bar\gamma)\varpi({\sf m}_{2r+1}(F)[\bar a^{(r)}], b_\upsilon+\gamma_\upsilon, b_\nu+\gamma_{\nu}), \]
   where $0\leqslant r< m/2$, $\bar a^{(r)}$ is a subvector of $\bar a$ with $m-2r$ entries, $\bar \gamma \in \mathbb {Z}_{<0}^{2}$ is constructed from $\bar a{\setminus} \bar a^{(r)}$, and the coefficients $C(\bar a^{(r)},\bar \gamma )\in \mathbb {Q}$ are independent of $F$.

Proof. Since we are working with a fully symmetrised element, Proposition 3.4 applies. In the same notation, write ${\sf m}(F)=\sum _{w=1}^{L} \xi _w \otimes R_w$. By our assumptions, ${\sf m}(F)\in {{\mathcal {S}}}^{m-2}(\mathfrak g)$. In particular, $\xi _w\in \mathfrak g$ for each $w$. Observe that

\[ \sum_{i,j} (\xi_w(x_i),x_j) x_i[b_\upsilon] x_j[b_\nu] = \sum_{i,j} x_i[b_\upsilon] ([\xi_w,x_i],x_j) x_j[b_\nu] = \sum_{i} x_i[b_\nu] [\xi_w,x_i[b_\upsilon]]. \]

Thereby part (i) follows from Proposition 3.4 in view of (3.6).

(ii) Note that $\omega ({\mathcal {P}}_{F[\bar a]})=(-1)^{m+1} {\mathcal {P}}_{F[\bar a]}$, because of the assumption ($\mathsf{B}$) imposed on all weight functions. By the construction, the image of $\varpi _{\rm wt}(F[\bar a], b_1,b_2)$ in ${{\mathcal {S}}}^{m+1}({{\hat {\mathfrak g}}}^-)$ is equal to $c \sum _{i} \{x_i[b_2],F[\bar a]\}x_i[b_1]$ for some $c\in \mathbb {Q}$. This constant $c$ depends only on the weight function $\Psi$. For this $c$, we have $\deg \mathrm {gr}({\mathcal {P}}_{F[\bar a]})\leqslant m$.

The element ${\mathcal {P}}_{F[\bar a]}$ is a linear combination of products, where each product contains $m+1$ linear factors. Let us symmetrise the summands of ${\mathcal {P}}_{F[\bar a]}$ by changing the sequence of factors in them. Note that there is no need to commute factors $\hat y_j=y_j[a_{\sigma (j)}]$ and $\hat y_l=y_l[a_{\sigma (l)}]$, since ${\mathcal {P}}_{F[\bar a]}$ is symmetric in the $\hat y_{p}$. There is no sense in commuting $\hat x_i^{(1)}$ and $\hat x_i^{(2)}$ either. After this symmetrisation all products of $m+1$ factors annihilate each other and ${\mathcal {P}}_{F[\bar a]}$ becomes a linear combination of products containing $m$ factors. Now we symmetrise these new products. Because of the antipode symmetry, they disappear after the symmetrisation and now ${\mathcal {P}}_{F[\bar a]}$ is a linear combination of products containing $m-1$ factors. Furthermore, each non-zero summand must contain certain factors according to one of the types listed below:

1. (1) $[\hat x_i^{(\nu )},y_j]$ and $[[\hat x_i^{(\upsilon )},\hat y_l],\hat y_p]$;

2. (2) $[\hat y_p,[\hat y_l,[\hat y_j,\hat x_i^{(\upsilon )}]]]=\mathrm {ad}(y_p)\mathrm {ad}(y_l)\mathrm {ad}(y_j)(x_i[b_\upsilon +a_{\sigma (p)}+a_{\sigma (l)}+a_{\sigma (j)}])$ and $\hat x_i^{(\nu )}$;

3. (3) $[[\hat y_p,\hat x_i^{(\upsilon )}], [\hat y_l,\hat x_i^{(\nu )}]]=[[y_p,x_i], [y_l,x_i]][a_{\sigma (p)}+ a_{\sigma (l)}+b_1+b_2]$;

4. (4) $[\hat y_p,[\hat x_i^{(\upsilon )}, [\hat x_i^{(\nu )},\hat y_j]]]$;

5. (5) $[\hat x_i^{(\upsilon )},[ \hat y_p, [\hat y_j,\hat x_i^{(\nu )}]]]=[[\hat x_i^{(\upsilon )},\hat y_p], [\hat y_j,\hat x_i^{(\nu )}]] - [\hat y_p, [\hat x_i^{(\upsilon )}, [\hat x_i^{(\nu )},\hat y_j]]]$.

The terms of type (4) disappear if we add over all $i$ and permute $p$ and $j$, because of the properties of ${\mathcal {H}}[\bar b]$, cf. (3.3). The terms of type (3) disappear if we permute $l$ and $p$. Therefore the terms of type (5) disappear as well.

One can deal with the terms of types (1) and (2) in the same way as in Lemma 3.2 and Proposition 3.4. They lead to $\gamma =(a_j,a_l)$ and $\gamma =(a_j+a_l,0)$ as well as $\gamma = (0,a_j+a_l)$. Note that the commutators of type (2) are easier to understand, since there is no need to permute the $t$-degrees, and at the same time they give rise to half-brackets.

(iii) We are presenting $\mathbb {X}_{F[\bar a]}$ in the form (0.1) and can state at once that it has terms of degrees $m+1-2d$ only. Note that $[{\mathcal {H}}[\bar b],{\mathcal {F}}[\check {\bar a}]]$ can be viewed as a weighted symmetrisation $\varpi _{\rm wt}(F[\bar a], b_1, b_2)$ if we choose $\Psi (j,j+1)=\Psi (j+1,j)=1$ and $\Psi (j,l)=0$ in the case $|l-j|>1$. The term of degree $m+1$ is the Poisson bracket $\{{\mathcal {H}}[\bar b],F[\bar a]\}$. Here $r=0$ and $\bar \gamma =0$. In degree $m-1$, we obtain images in ${{\mathcal {S}}}^{m-1}({{\hat {\mathfrak g}}}^-)$ of the weighted symmetrisations described in part (i). Further terms, which are of degrees $m-3,m-5,m-7$, and so on, are described by the iterated application of part (ii). At all steps, we obtain combinatorially defined rational coefficients, which are independent of $F$.

Example 3.7 Suppose that $F\in {{\mathcal {S}}}^{4}(\mathfrak g)^{\mathfrak g}$ and that $\mathfrak g$ is simple. Here we have ${\sf m}(F)\in (\Lambda ^{2}\mathfrak g \otimes \mathfrak g)^{\mathfrak g}$ and $\dim (\Lambda ^{2}\mathfrak g \otimes \mathfrak g)^{\mathfrak g}=1$. This subspace is spanned by ${\mathcal {H}}=\sum x_i^{2}$. Hence ${\sf m}(F)={\mathcal {H}}$ up to a scalar. As we will proof in § 3.3, $S=\varpi (F[-1])+6\varpi (\tau ^{2}{\sf m}(F)[-1]){\cdot }1$ is a Segal–Sugawara vector. Making use of the fact that $\tau ^{2}({\mathcal {H}}[-1])\in \mathfrak z({{\hat {\mathfrak g}}})$, one can write $S$ as a sum $\varpi (F[-1])+B{\mathcal {H}}[-2]$ for some scalar $B\in \mathbb {{\mathbb {C}}}$.

The only possible vector $\bar \gamma$ that can appear in Proposition 3.6(iii) is $(-1,-1)$. Therefore the commutator $[{\mathcal {H}}[-1],\varpi (F[-1])]$ is equal to $B\varpi (\{{\mathcal {H}}[-2],{\mathcal {H}}[-1]\})=B[{\mathcal {H}}[-2],{\mathcal {H}}[-1]]$. In the orthonormal basis $\{x_i\}$, we have

(3.7)\begin{equation} \{{\mathcal{H}}[-2],{\mathcal{H}}[-1]\}=4 \sum_{i,j,s} ([x_i,x_j],x_s) x_s[-3]x_i[-2]x_j[-1]. \end{equation}

3.3 Poisson (half-)brackets

Suppose that $\hat Y=\hat y_1\ldots \hat y_m\in {{\mathcal {S}}}^{m}({{\hat {\mathfrak g}}}^-)$ and $\hat y_j=y_j[a_j]$. Then $P_{\hat Y}:= \{ {\mathcal {H}}[\bar b], \hat Y\} =P_{\hat Y}(b_1,b_2)+P_{\hat Y}(b_2,b_1)$, where

\[ P_{\hat Y}(b_\upsilon,b_\nu)= \sum_{j=1,i=1}^{j=m,i=\dim\mathfrak g} [x_i[b_\nu],\hat y_j] x_i[b_\upsilon] \hat Y/\hat y_j = \sum_{j,i,u} ([x_u,x_i],y_j) x_u[b_\nu+a_j] x_i[b_\upsilon] \hat Y/\hat y_j. \]

Note that in the case $b_\nu +a_j=b_\upsilon$, each summand $([x_u,x_i],y_j) x_u[b_\nu +a_j] x_i[b_\upsilon ] \hat Y/\hat y_j$ is annihilated by $([x_i,x_u],y_j) x_i[b_\nu +a_j] x_u[b_\upsilon ] \hat Y/\hat y_j$. Hence

(3.8)\begin{equation} P_{\hat Y}(b_1,b_2)=\sum_{j:\, a_j\ne b_1-b_2} \,\sum_{i,u} ([x_i,x_u],y_j) x_i[b_2+a_j] x_u[b_1] \hat Y/\hat y_j. \end{equation}

The product $(\,\,,\,)$ extends to a non-degenerate $\mathfrak g$-invariant scalar product on ${{\mathcal {S}}}({{\hat {\mathfrak g}}}^-)$. We will assume that $(\mathfrak g[a],\mathfrak g[d])=0$ for $a\ne d$, that $(x[a],y[a])=(x,y)$ for $x,y\in \mathfrak g$, and that

\[ (\xi_1\ldots\xi_{\boldsymbol k},\eta_1\ldots\eta_{\boldsymbol m})= \delta_{{\boldsymbol k},{\boldsymbol m}} \sum_{\sigma\in{\tt S_{\boldsymbol k}}} (\xi_1,\eta_{\sigma(1)})\ldots (\xi_{\boldsymbol k},\eta_{\sigma({\boldsymbol k})}) \]

if $\xi _j,\eta _j\in {{\mathcal {S}}}({{\hat {\mathfrak g}}}^-)$, ${\boldsymbol m}\geqslant {\boldsymbol k}$. Let ${\mathcal {B}}$ be a monomial basis of ${{\mathcal {S}}}({{\hat {\mathfrak g}}}^-)$ consisting of the elements $\hat v_1\ldots \hat v_{\boldsymbol k}$, where $\hat v_j=v_j[d_j]$ and $v_j\in \{x_i\}$. Then ${\mathcal {B}}$ is an orthogonal, but not an orthonormal basis. For instance, if $\Xi =x_1^{\gamma _1} \ldots x_{\boldsymbol k}^{\gamma _{\boldsymbol k}}$ with $\boldsymbol k\leqslant \dim \mathfrak g$, then $(\Xi,\Xi )=\gamma _1!\ldots \gamma _{\boldsymbol k}!$.

Set $M:=m+1$, ${\mathcal {B}}(M):={\mathcal {B}}\cap {{\mathcal {S}}}^{M}({{\hat {\mathfrak g}}}^-)$, and write

\[ P_{\hat Y}(b_1,b_2) = \sum_{\mathbb{V}\in{\mathcal{B}}(M)} A(\mathbb{V}) \mathbb{V} \quad \text{with}\, A(\mathbb{V})\in{{\mathbb{C}}}, \]

expressing each $\hat Y/\hat y_j$ in the basis ${\mathcal {B}}$.

Lemma 3.8 If $A(\mathbb {V})\ne 0$ and $\mathbb {V}=\hat v_1\ldots \hat v_M$, then ${\boldsymbol p}:=\{ p \mid d_p=b_1\}\ne \varnothing$. Furthermore,

(3.9)\begin{equation} A(\mathbb{V}) = \sum_{p\in{\boldsymbol p}, l\not\in{\boldsymbol p}} (\mathbb{V},\mathbb{V})^{-1} \bigg(\hat Y, \frac{\mathbb{V}}{\hat v_l\hat v_p} [v_l,v_p][d_l-b_2]\bigg). \end{equation}

Proof. The first statement is clear, cf. (3.8). It remains to calculate the coefficient of $\mathbb {V}$ in $P_{\hat Y}(b_1,b_2)$. Pick a pair $(p,l)$ with $p\in {\boldsymbol p}$ and $l\not \in {\boldsymbol p}$. If we take into account only those summands of $P_{\hat Y}(b_1,b_2)$, where $\mathbb {V}^{p,l}=\mathbb {V}/(\hat v_p\hat v_l)$ is a summand of $\hat Y/\hat y_j$ for some $j$, the factor $\hat v_l$ is $x_u[a_j+b_2]$, and $\hat v_p$ appears as $x_i[b_1]$, then the coefficient is

\[ \sum_{j=1}^{m} (\hat y_j,[v_l,v_p][d_l-b_2]) (\hat Y/\hat y_j, \mathbb{V}^{p,l}) (\mathbb{V}^{p,l},\mathbb{V}^{p,l})^{-1} = (\mathbb{V}^{p,l},\mathbb{V}^{p,l})^{-1} (\hat Y, \mathbb{V}^{p,l} [v_l,v_p][d_l-b_2]). \]

If one adds these expressions over the pairs $(p,l)$, then certain instances may be counted more than once. If $\hat v_p=\hat v_{p'}$ for some $p'\ne p$ or $\hat v_l=\hat v_{l'}$ for some $l'\ne l$, then $(p,l')$ or $(p',l)$ has to be omitted from the summation. In other words, it is necessary to divide the contribution of $(p,l)$ by the multiplicities $\gamma _p$ and $\gamma _l$ of $v_p[d_p]$, $v_l[d_l]$ in $\mathbb {V}$. Since $(\mathbb {V},\mathbb {V})=\gamma _p\gamma _l (\mathbb {V}^{p,l},\mathbb {V}^{p,l})$, the result follows.

The Poisson bracket $P_{\hat Y}$ is not multi-homogeneous with respect to ${{\hat {\mathfrak g}}}^-=\bigoplus _{d\leqslant -1} \mathfrak g[d]$. If $b_1\ne b_2$, then in general the ‘halves’ of $P_{\hat Y}$ have different multi-degrees and neither of them has to be multi-homogeneous. We need to split $P_{\hat Y}(b_1,b_2)$ into smaller pieces. For $\bar a\in {{\mathbb {Z}}}_{<0}^{m}$, set ${{\mathcal {S}}}^{\bar a}({{\hat {\mathfrak g}}}^-)=\prod _{j=1}^{m} \mathfrak g[a_j]\subset {{\mathcal {S}}}({{\hat {\mathfrak g}}}^-)$.

Let $\bar \alpha =\{\alpha _1^{r_1},\ldots,\alpha _s^{r_s}\}$ be a multi-set such that $\alpha _i\ne \alpha _j$ for $i\ne j$, $\alpha _j\in {{\mathbb {Z}}}_{<0}$ for all $1\leqslant j\leqslant s$, $\sum _{j=1}^{s} r_j=M$, and $r_j>0$ for all $j$. Set ${{\mathcal {S}}}^{\bar \alpha }({{\hat {\mathfrak g}}}^-):=\prod _{j=1}^{s} {{\mathcal {S}}}^{r_j}(\mathfrak g[\alpha _j])$, ${\mathcal {B}}(\bar \alpha ):={\mathcal {B}}\cap {{\mathcal {S}}}^{\bar \alpha }({{\hat {\mathfrak g}}}^-)$. Fix different $i,j\in \{1,\ldots,s\}$. Assume that a monomial $\mathbb {V}=\hat v_1\ldots \hat v_M\in {\mathcal {B}}(\bar \alpha )$ with $\hat v_l=v_l[d_l]$ is written in such a way that $d_l=\alpha _i$ for $1\leqslant l\leqslant r_i$ and $d_l=\alpha _j$ for $r_i< l\leqslant r_i+r_j$. Finally suppose that $F\in {{\mathcal {S}}}^{m}(\mathfrak g)$. In this notation, set

(3.10)\begin{gather} \mathbb{W}[F, \bar\alpha, (i,j)] :=\sum_{{\mathbb{V}}\in{\mathcal{B}}(\bar\alpha)} A(\mathbb{V}) \mathbb{V}\quad \text{with} \nonumber\\ A(\hat v_1\ldots \hat v_M)= (\mathbb{V},\mathbb{V})^{-1} \sum_{\substack{1\leqslant l\leqslant r_i, \nonumber\\ r_i< p \leqslant r_i+r_j}} \bigg(F,[v_l,v_p] \prod_{u\ne p,l} v_u\bigg). \end{gather}

Clearly, $\mathbb {W}[F, \bar \alpha, (j,i)] = - \mathbb {W}[F, \bar \alpha, (i,j)]$.

Proposition 3.9 Let $F\in {{\mathcal {S}}}^{m}(\mathfrak g)^{\mathfrak g}$ be fixed. Then the elements $\mathbb {W}[\bar \alpha, (i,j)]=\mathbb {W}[F,\bar \alpha, (i,j)]$ satisfy the following ‘universal’ relations:

\[ \sum_{j: j\ne i} \mathbb{W}[\bar\alpha, (i,j)] =0 \quad \text{for each} \ i\leqslant s. \]

These relations are independent of $F$.

Proof. Follow the notation of (3.10). Note that for each $1\leqslant l \leqslant r_i$,

\[ \bigg(F, \sum_{w:\, w\ne l} [v_l,v_w] \prod_{u:\, u\ne l,w} v_u\bigg) =\bigg(\{F,v_l\}, \prod_{w:\, w\ne l} v_w\bigg)=0. \]

Of course, here we are adding also over the pairs $(l,w)$ with $\hat v_w\in \mathfrak g[\alpha _i]$ if $r_i>1$. However, $[v_l,v_w]=-[v_w,v_l]$ and hence the coefficient of $\mathbb {V}$ in $\sum _{j\ne i} \mathbb {W}[\bar \alpha, (i,j)]$ is equal to

\[ (\mathbb{V},\mathbb{V})^{-1} \sum_{1\leqslant l \leqslant r_i} \bigg(F, \sum_{w:\, w\ne l} [v_l,v_w] \prod_{u:\, u\ne l,w} v_u\bigg) = \sum_{1\leqslant l \leqslant r_i} 0 =0. \]

This completes the proof.

Proposition 3.10 For $Y[\bar a]$ with $Y=y_1{\ldots } y_m\in {{\mathcal {S}}}^{m}(\mathfrak g)$ and $\bar a\in \mathbb {Z}_{<0}^{m}$ (see § 1.5 for the notation), the rescaled Poisson half-bracket

\[ {\boldsymbol P}[\bar a]:=|{\tt S}_m \bar a| P_{Y[\bar a]}(b_1,b_2) = |{\tt S}_m \bar a| \sum_{u} \{x_u[b_2],Y[\bar a]\} x_u[b_1] \]

is equal to the sum of $\mathbb {W}[Y, \bar \alpha, (i,j)]$ with $i< j$ over all multi-sets $\bar \alpha$ as above such that the multi-set $\{a_1,\ldots,a_m\}$ of entries of $\bar a$ can be obtained from $\bar \alpha$ by removing one $\alpha _j=b_1$ and replacing one $\alpha _i$ with $\alpha _i-b_2$.

Proof. For each multi-homogeneous component of ${\boldsymbol P}[\bar a]$, the multi-set of $t$-degrees is obtained from the entries of $\bar a$ by appending $b_1$ and replacing one $a_l$ with $a_l+b_2$. Moreover, here $b_1\ne a_l+b_2$, cf. (3.8). This explains the restrictions on $\bar \alpha$.

For $\bar \alpha$ and $(i,j)$ satisfying the assumptions of the proposition, we have to compare the coefficients $A(\mathbb {V})$ of $\mathbb {V}\in {\mathcal {B}}(\bar \alpha )$ given by (3.10) and (3.9). The key point here is the observation that $(Y[\bar a], \mathbb {V})=|{\tt S}_m \bar a|^{-1} (Y,v_1\ldots v_m)$ for any $\mathbb {V}=\hat v_1\ldots \hat v_m \in {\mathcal {B}}\cap {{\mathcal {S}}}^{\bar a}({{\hat {\mathfrak g}}}^-)$.

In a more relevant setup, suppose that a summand $(y_{\sigma (1)}, [v_l,v_p]) \prod _{w\ne 1,u\ne l,p} (y_{\sigma (w)},v_u)$ of the scalar product on the right hand side of (3.10) is non-zero for some $\sigma \in {\tt S}_m$ and some $l,p$. Then there is exactly one choice, prescribed by $(d_1,\ldots,d_M)$, of the $t$-degrees for a monomial of $Y[\bar a]$ such that the corresponding summand

\[ (y_{\sigma(1)}[\alpha_i-b_2],[v_l,v_p][d_l-b_2]) \prod_{w\ne 1,u\ne l,p} (y_{\sigma(w)}[d_u],\hat v_u) \]

of the scalar product on the right hand side of (3.9) is non-zero as well. By our assumptions on the scalar product, these summands are equal.

Theorem 3.11 Suppose that ${\sf m}_{2r+1}(H)$ with $H\in {{\mathcal {S}}}^{k}(\mathfrak g)^{\mathfrak g}$ is a symmetric invariant for each $r\geqslant 1$. Then (2.4) provides a Segal–Sugawara vector $S$ associated with $H$.

Proof. Since ${\sf m}_{2l+1}(H)\in {{\mathcal {S}}}(\mathfrak g)$ for any $l$, we can say that ${\sf m}_{2r+1}(H)={\sf m}^{r}(H)$, cf. (1.1). By Lemma 2.2, each $\varpi (\tau ^{2r}{\sf m}^{r}(H)[-1]){\cdot }1$ is a fully symmetrised element. It can be written as a sum of $\tilde c(r,\bar a)\varpi ({\sf m}^{r}(H)[\bar a])$, where $\bar a\in {{\mathbb {Z}}}_{<0}^{k-2r}$ and the coefficients $\tilde c(r,\bar a)\in \mathbb {Q}$ depend only on $k$, $r$, and $\bar a$. The coefficients of (2.4) depend only on $k$ and $r$. Combining this observation with Propositions 3.6(iii) and 3.10, we obtain that

\[ [{\mathcal{H}}[-1],S] = \sum C(r,\bar\alpha,i,j) \varpi(\mathbb{W}[{\sf m}_{2r+1}(H),\alpha_1^{r_1},\ldots,\alpha_d^{r_s}, (i,j)])\,, \]

where again the coefficients $C(r,\bar \alpha,i,j)\in \mathbb {Q}$ do not depend on $H$. For a given degree $k$, one obtains a bunch of $(r,\bar \alpha )$, which depends only on $k$, and each appearing coefficient depends on $k$, $r$, $\bar \alpha$, and $(i,j)$. In type $\mathsf{A}$, for each $k\geqslant 2$, we find the invariant $\tilde \Delta _k$ such that the corresponding commutator $[{\mathcal {H}}[-1],\tilde S_{k-1}]$ vanishes, cf. (2.4).

For each $F\in {{\mathcal {S}}}^{m}(\mathfrak g)^{\mathfrak g}$, the elements $\mathbb {W}[F,\bar \alpha, (i,j)]$ are linearly dependent. They satisfy the ‘universal’ relations, see Proposition 3.9. At the same time, for $m=k-2r$, the coefficients $C(r,\bar \alpha,i,j)$ provide a relation among $\mathbb {W}[\tilde \Delta _m,\bar \alpha, (i,j)]$. Our goal is to prove that this relation holds for $\mathbb {W}[{\sf m}^{r}(H),\bar \alpha, (i,j)]$ as well. To this end, it suffices to show that the terms $\mathbb {W}[\tilde \Delta _m,\bar \alpha, (i,j)]=\mathbb {W}[(i,j)]$ with fixed $m$ and fixed $\bar \alpha$ do not satisfy any other linear relation, not spanned by the ‘universal’ ones.

We consider the complete simple graph with $s$ vertices $1,\ldots,s$ and identify pairs $(i,j)$ with the corresponding (oriented) edges. Now one can say that a linear relation among the polynomials $\mathbb {W}[(i,j)]$ is given by its coefficients on the edges. Note that if $s=2$, then $\mathbb {W}[F,\bar \alpha,(1,2)]=0$ for each $F\in {{\mathcal {S}}}^{m}(\mathfrak g)^{\mathfrak g}$, cf. Example 3.12(i) below. Therefore assume that $s\geqslant 3$.

Suppose there is a relation and that the coefficient of $\mathbb {W}[(i,j)]$ is non-zero. We work in the basis

\[ \{E_{uv}[d], (E_{11}+\cdots+ E_{ww} - wE_{(w+1)(w+1)})[d] \mid u\ne v, d<0 \}. \]

Choose $\hat y_1=E_{12}[\alpha _i]$, $\hat y_2=E_{21}[\alpha _j]$ and let all other factors $\hat y_l$ with $2< l \leqslant M$ be elements of $t^{-1}\mathfrak h[t^{-1}]$. Assume that $\hat y_3=(E_{11}- E_{22})[\alpha _p]$ with $p\ne i,j$ and that $(E_{11}-E_{22},y_l)=0$ for all $l>3$. Then the monomial $\hat y_1\ldots \hat y_M$ appears with a non-zero coefficient only in $\mathbb {W}[(i,j)]$, $\mathbb {W}[(i,p)]$, and $\mathbb {W}[(p,j)]$. This means that in the triangle $(i,j,p)$ at least one of the edges $(i,p)$ and $(j,p)$ has a non-zero coefficient as well.

We erase all edges with zero coefficients on them. Now the task is to modify the relation or, equivalently, the graph, by adding scalar multiplies of the universal relations in such a way that all edges disappear.

If a vertex $l$ is connected with $j$, remove the edge $(j,l)$ using the universal relation ‘at $l$’. In this way $j$ becomes isolated. This means that there is no edge left.

Example 3.12 Keep the assumption $F\in {{\mathcal {S}}}^{m}(\mathfrak g)^{\mathfrak g}$.

1. (i) Suppose that $s=2$. Then $\mathbb {W}(F,\bar \alpha, (1,2))=0$ according to the universal relation. This provides a different proof of Lemma 3.1.

   Also in the case of $\bar a=(-3,(-1)^{m-1})$, we have $\{{\mathcal {H}}[-1],F[\bar a]\}\in \mathfrak g[-3]\mathfrak g[-2] {{\mathcal {S}}}^{m-1}(\mathfrak g[-1])$.

2. (ii) Now suppose that $s=3$. Then $\mathbb {W}(F,\bar \alpha, (1,2))=-\mathbb {W}(F,\bar \alpha, (1,3))=\mathbb {W}(F,\bar \alpha, (2,3))$.

4. Type C

There is a very suitable matrix realisation, where $\mathfrak {sp}_{2n}\subset \mathfrak {gl}_{2n}$ is the linear span of the elements $F_{ij}$ with $i,j\in \{1,\ldots,2n\}$ such that

(4.1)\begin{equation} F_{i j}=E_{i j}-\epsilon_i\,\epsilon_j\, E_{j' i'}, \end{equation}

with $i'=2n-i+1$ and $\epsilon _i=1$ for $i\leqslant n$, $\epsilon _i=-1$ for $i> n$. Of course, $F_{ij}\!=\!\pm F_{j'i'}$. Set

\[ \mathfrak h\!=\!\langle F_{jj} \mid 1 \leqslant j\leqslant n\rangle _{{{\mathbb {C}}}}. \]

The symmetric decomposition $\mathfrak {gl}_{2n}=\mathfrak {sp}_{2n}\oplus \mathfrak p$ leads to explicit formulas for symmetric invariants of $\mathfrak {sp}_{2n}$. One writes $E_{ij}=\frac {1}{2}F_{ij}+ \frac {1}{2}(E_{i j}+\epsilon _i\,\epsilon _j\, E_{j' i'})$, expands the coefficients $\Delta _k$ of (1.2) accordingly, and then sets $(E_{i j}+\epsilon _i\,\epsilon _j\, E_{j' i'})=0$. Up to the multiplication with $2^{k}$, this is equivalent to replacing each $E_{ij}$ with $F_{ij}$ in the formulas for $\Delta _{k}\in {{\mathcal {S}}}(\mathfrak {gl}_{2n})$. As is well known, the restriction of $\Delta _{2k+1}$ to $\mathfrak {sp}_{2n}$ is equal to zero for each $k$.

Until the end of this section, $\Delta _{2k}$ stands for the symmetric invariant of $\mathfrak g=\mathfrak {sp}_{2n}$ that is equal to the sum of the principal $(2k\times2k)$-minors of the matrix $(F_{ij})$.

Lemma 4.1 For each $k\geqslant 2$, we have ${\sf m}(\Delta _{2k})\in (\mathfrak g\otimes {{\mathcal {S}}}^{k-3}(\mathfrak g))^{\mathfrak g}$.

Proof. Note that $\Lambda ^{2} \mathfrak g=V(2\pi _1+\pi _2)\oplus \mathfrak g$. In the standard notation [VO88, Ref. Chapter, § 2], we have $2\pi _1+\pi _2=3\varepsilon _1+\varepsilon _2$. Assume that $y_1y_2y_3$ is a factor of a summand of $\Delta _{2k}$ of weight $3\varepsilon _1+\varepsilon _2$ and $y_s\in \{F_{i j}\}$ for each $s$. Then:

• • either $F_{1(2n)}\in \{y_1,y_2,y_3\}$ and some $y_j\ne F_{1(2n)}$ lies in the first row or the last column;

• • or all three elements $y_s$ must lie in the union of the first row and the last column.

Each of the two possibilities contradicts the definition of $\Delta _{2k}$. Thus, ${\sf m}(\Delta _k)\in (\mathfrak g \otimes {{\mathcal {S}}}^{2k-3}(\mathfrak g))^{\mathfrak g}$.

Lemma 4.2 We have:

1. (i) ${\sf m}(\Delta _{6})= ({(n-2)(2n-3)}/{15})\Delta _4$; and

2. (ii) ${\sf m}(\partial _{F_{11}}\Delta _4)= \frac {-1}{3} (2n-1)(2n-2) F_{11}$.

Proof. According to Lemma 4.1, ${\sf m}(\Delta _6)\in (\mathfrak g \otimes {{\mathcal {S}}}^{3}(\mathfrak g))^{\mathfrak g}$. Observe that ${{\mathcal {S}}}^{3}(\mathfrak g)$ contains exactly two linearly independent copies of $\mathfrak g$: one is equal to $\{\xi \Delta _2 \mid \xi \in \mathfrak {sp}_{2n}\}$, the other is primitive. Therefore

\[ (\mathfrak g\otimes{{\mathcal{S}}}^{3}(\mathfrak g))^{\mathfrak g}={{\mathcal{S}}}^{4}(\mathfrak g)^{\mathfrak g} = \langle \Delta_4,\Delta_2^{2}\rangle _{{{\mathbb{C}}}}. \]

By the construction, $\Delta _2^{2}$ contains the summand $F_{11}^{4}$. Since $F_{1 1}^{3}$ cannot be a factor of a summand of $\Delta _6$, we conclude that ${\sf m}(\Delta _6)$ is proportional to $\Delta _4$.

Let $y\otimes F_{11}^{2} F_{22}$ be a summand of ${\sf m}(\Delta _6)$. Then $y\in {{\mathbb {C}}}F_{22}$. Also $y= ({-3!3!}/{6!}){\sf m}(\partial _{F_{22}} \Delta _4^{[1]})$, where $\Delta _4^{[1]}\in {{\mathcal {S}}}^{4}(\mathfrak {sp}_{2n-2})$ stands for $\Delta _4$ of $\mathfrak {sp}_{2n-2}\subset \mathfrak g_{F_{11}}$. Next we compute $\eta ={\sf m}(\partial _{F_{22}} \Delta _4^{[1]})(F_{12})$. This will settle part (ii). So far we have shown that

(4.2)\begin{equation} \varpi(\partial_{F_{11}}\Delta_4)\in{{\mathcal{U}}}(\mathfrak g) \text{ acts as } cF_{11} \text{ on } \mathfrak g \text{ and on } {{\mathbb{C}}}^{2n} \end{equation}

and part (ii) describes this constant $c$, which is to be computed.

Recall that $F_{12}=- F_{(2n-1) 2n}$. The terms of $\Delta _4^{[1]}$ have neither $1$ nor $2n$ in the indices. Therefore a non-zero action on $F_{12}$ comes only from the following summands of $\Delta _4^{[1]}$:

(4.3)$$\begin{gather} F_{22} F_{(2n-1) j} F_{js} F_{s (2n-1)}, \quad - F_{22} F_{(2n-1) s} F_{s' s'} F_{s (2n-1)}, \end{gather}$$

(4.4)$$\begin{gather}F_{(2n-1)(2n-1)} F_{j2} F_{sj} F_{2s}, \quad - F_{(2n-1)(2n-1)} F_{s2} F_{s's'} F_{2s}. \end{gather}$$

One easily computes that

\[ {\sf m}(F_{(2n-1) j} F_{js} F_{s (2n-1)})(F_{12})= \begin{cases} \frac{1}{6} F_{12} & \text{if } j\ne s', \\[6pt] \frac{1}{3} F_{12} & \text{if } j = s', \text{ because } F_{s' s}=2E_{s' s}, \end{cases} \]

and that ${\sf m}(F_{(2n-1) s} F_{s' s'} F_{s (2n-1)})(F_{12})= -\frac {1}{6}F_{12}$. There are $2n-4$ choices for $s$ in line (4.3). If $s$ is fixed, then there are $2n-5$ possibilities for $j$, since $j\ne s$, but the choice $j=s'$ has to be counted twice. Applying the symmetry $F_{uv}=\pm F_{v'u'}$, we see that the terms in line (4.4) are the same as in (4.3). Now

\[ \eta=\frac{1}{3}( (2n-4)^{2}+(2n-4))F_{12}=\frac{(2n-4)(2n-3)}{3}F_{12}. \]

Hence $y= ({(n-2)(2n-3)}/{30})F_{22}$. Since $({(n-2)(2n-3)}/{30}) F_{11}\otimes F_{11}F_{22}^{2}$ is also a summand of ${\sf m}(\Delta _6)$, we obtain ${\sf m}(\Delta _6)= ({(n-2)(2n-3)}/{15})\Delta _4$.

Proposition 4.3 We have

\[ {\sf m}_{2r+1}(\Delta_{2k})=\frac{(2k-2r)!(2r)!}{(2k)!}\binom{2n-2k+2r+1}{2r} \Delta_{2k-2r}. \]

Proof. First we have to show that ${\sf m}(\Delta _{2k})\in {{\mathcal {S}}}^{2k-2}(\mathfrak g)$. By Lemma 4.1, ${\sf m}(\Delta _{2k})$ is a $G$-invariant polynomial function on $\mathfrak g\oplus \mathfrak g$. We use again the fact that $G(\mathfrak g\oplus \mathfrak h)$ is dense in $\mathfrak g\oplus \mathfrak g$.

Examine first the summands of $\Delta _{2k}$ that lie in ${{\mathcal {S}}}^{3}(\mathfrak g){{\mathcal {S}}}^{2k-3}(\mathfrak h)$. Such a summand has the form $y_1y_2y_3 (F_{i_1 i_1}\ldots F_{i_s i_s})^{2} F_{j_1 j_1}\ldots F_{j_u j_u}$. Here $j_a\ne j_b,j_b'$ if $a\ne b$ and the product $y_1y_2y_3$ is an element of weight zero lying in ${{\mathcal {S}}}^{3}(\mathfrak f)$, where $\mathfrak f$ is a subalgebra of $\mathfrak g$ isomorphic either to $\mathfrak {sp}_6$ or $\mathfrak {sp}_4$. Furthermore, the numbers $i_b, i_b'$ with $1\leqslant b\leqslant s$ do not appear among the indices of the elements of $\mathfrak f$ and at most three different numbers $j_b$ with $1\leqslant b\leqslant u$ can appear among the indices of the elements of $\mathfrak f$.

If $u>3$, we can change at least one $F_{j_b j_b}$ to $F_{j_b' j_b'}=-F_{j_b j_b}$ without altering the other factors and produce a different summand of $\Delta _{2k}$. These two expressions annihilate each other. Therefore $u=3$ or $u=1$. Suppose $u=3$ and that there is no way to annihilate the term via $F_{j_b j_b}\mapsto F_{j_b' j_b'}$. Then $\mathfrak f\cong \mathfrak {sp}_6$ and $y_1y_2y_3 F_{j_1 j_1} F_{j_2 j_2} F_{j_3 j_3}$ is a summand of the determinant $\Delta _6^{(\mathfrak f)}\in {{\mathcal {S}}}^{6}(\mathfrak f)$. Since ${\sf m}(\Delta _6)$ is proportional to $\Delta _4$ by Lemma 4.2 and $F_{j_1 j_1} F_{j_2 j_2} F_{j_3 j_3}$ cannot appear in $\Delta _4$, we conclude that terms $y\otimes (F_{i_1 i_1}\ldots F_{i_s i_s})^{2} F_{j_1 j_1}\ldots F_{j_u j_u}$ with $u>1$ do not appear in ${\sf m}(\Delta _{2k})$.

Fix an element ${{\boldsymbol H}}=F_{11}F_{22}^{2}\ldots F_{ll}^{2}$ with $l=k-1$. We compute the coefficient of ${{\boldsymbol H}}$ in ${\sf m}(\Delta _{2k})$. This coefficient is equal to $(-1)^{k} ({3!(2k-3)!}/{(2k)!}) {\sf m}(\partial _{F_{11}} \Delta _{4}^{(l)})$, where $\Delta _4^{(l)}$ is $\Delta _4$ of the $\mathfrak {sp}_{2n-2k+4}$-subalgebra generated by $F_{ij}$ with $i,j\not \in \{2,2',\ldots,l,l'\}$. According to Lemma 4.2, ${\sf m}(\partial _{F_{11}} \Delta _{4}^{(l)})= \frac {-1}{3} (2n-2k+3)(2n-2k+2) F_{11}$. Making use of the action of the Weyl group $W(\mathfrak g,\mathfrak h)$ on $\mathfrak h\oplus \mathfrak h$, we conclude that ${\sf m}(\Delta _{2k})$ is proportional to $\Delta _{2k-2}$, more explicitly

\[ {\sf m}(\Delta_{2k})= \frac{2(k-1)(2n-2k+3)(2n-2k+2) (2k-3)!}{(2k)! } \Delta_{2k-2} \]

and with some simplifications

\[ {\sf m}(\Delta_{2k})= \frac{(2n-2k+3)(2n-2k+2)}{ 2k(2k-1) } \Delta_{2k-2} = \binom{2k}{2}^{-1} \binom{2n-2k+3}{2} \Delta_{2k-2}. \]

Iterating the map ${\sf m}$, one obtains the result.

Theorem 4.4 For $\mathfrak g=\mathfrak {sp}_{2n}$ and $1\leqslant k\leqslant n$,

\[ S_{k} = \varpi(\Delta_{2k}[-1]) + \sum_{1\leqslant r< k} \binom{2n-2k+2r+1}{2r}\varpi(\tau^{2r}\Delta_{2k-2r}[-1]){\cdot}1 \]

is a Segal–Sugawara vector.

5. Several exceptional examples

There are instances, where our methods work very well.

Example 5.1 Suppose that $\mathfrak g=\mathfrak {so}_8$. Then ${\rm Aut}(\mathfrak g)/{\rm Inn}(\mathfrak g)={\tt S}_3$, where ${\rm Inn}(\mathfrak g)$ is the group of inner automorphisms. There are two Segal–Sugawara vectors, say $S_2$ and $S_3$, such that their symbols are $\mathfrak g$-invariants of degree $4$ in ${{\mathcal {S}}}(\mathfrak g[-1])$. Assume that $S_2$ and $S_3$ are fixed vectors of $\omega$. Then each of them is a sum $\varpi (Y_4)+\varpi (Y_2)$, cf. (0.1). Each element in ${{\mathcal {S}}}^{2}({{\hat {\mathfrak g}}}^-)^{\mathfrak g}$ is proportional to ${\mathcal {H}}[\bar b]$ for some $\bar b$. Hence it is also an invariant of ${\tt S}_3$. Without loss of generality we may assume that the symbols of $S_2$ and $S_3$ are Pfaffians ${\rm Pf}_2[-1]$, ${\rm Pf}_3[-1]$ related to different matrix realisations of $\mathfrak {so}_8$. Then for each of them there is an involution $\sigma \in {\tt S}_3$ such that $\sigma (Y_4)=-Y_4$. Replacing $S_j$ with $S_j-\sigma (S_j)$, we see that $\tilde S_2=\varpi ({\rm Pf}_2[-1])$ and $\tilde S_3=\varpi ({\rm Pf}_3[-1])$ are also Segal–Sugawara vectors. In view of Theorem 3.5, this implies that ${\sf m}({\rm Pf}_2)={\sf m}({\rm Pf}_3)=0$.

Automorphisms of $\mathfrak g$ make themselves extremely useful. We will see the full power of this devise in § 7, which deals with the orthogonal case. At the moment notice the following thing, any $\sigma \in {\rm Aut}(\mathfrak g)$ acts on ${{\mathcal {S}}}(\mathfrak g)$ in the natural way and induces a map $\sigma _{(m)}\!: {{\mathcal {S}}}^{m}(\mathfrak g)\to {{\mathcal {S}}}^{m}(\mathfrak g)$. Let $\mathfrak v_m\otimes \mathfrak g$ be the isotypic component of ${{\mathcal {S}}}^{m}(\mathfrak g)$ corresponding to $\mathfrak g$. Then $\sigma$ acts on $\mathfrak v_m$ and for this action, we have $\sigma (v)\otimes \sigma (x)=\sigma _{(m)}(v\otimes x)$, where $v\in \mathfrak v_m$, $x\in \mathfrak g$.

An interesting story is related to Pfaffians in higher ranks.

Example 5.2 (The Pfaffians)

Take $\mathfrak g=\mathfrak {so}_{2n}$. If $n<4$, these algebras appear in type A. In the case $n=4$, the Pfaffian-like Segal–Sugawara vectors are examined in Example 5.1. Suppose that $n>4$ and that $\mathfrak {so}_{2n}\subset \mathfrak {gl}_{2n}$ consists of the skew-symmetric with respect to the antidiagonal matrices. The highest weight of the Cartan component of $\Lambda ^{2}\mathfrak g$ is $\pi _1+\pi _3=2\varepsilon _1+\varepsilon _2+\varepsilon _3$. Assume that $\pi _1+\pi _3$ appears as the weight of a factor $y_1y_2y_3$ for a summand $y_1\ldots y_n$ of the Pfaffian ${\rm Pf}$. Then up to the change of indices, we must have

\[ y_1=(E_{1i}-E_{i' 1'}), \quad y_2=(E_{1j}-E_{j'1'}), \]

where $i'=2n+1-i$. If this is really the case, then the determinant $\Delta _{2n}\in {{\mathcal {S}}}^{2n}(\mathfrak {gl}_{2n})$ has a summand $E_{1i}E_{1j}\ldots$, a contradiction.

Thus ${\sf m}({\rm Pf})\in (\mathfrak g\otimes {{\mathcal {S}}}^{n-3}(\mathfrak g))^{\mathfrak g}$. If $n$ is odd, then there is no copy of $\mathfrak g$ in ${{\mathcal {S}}}^{n-3}(\mathfrak g)$ and we conclude at once that the image of the Pfaffian under m is zero.

Suppose that $n$ is even. Then we can rely on the fact that $G(\mathfrak g\oplus \mathfrak h)$ is dense in $\mathfrak g\oplus \mathfrak g$. Fix a factor ${\boldsymbol H}\in {{\mathcal {S}}}^{n-3}(\mathfrak h)$ of a summand of ${\rm Pf}$. Without loss of generality assume that ${\boldsymbol H}= \prod _{s>3} (E_{ss}-E_{s's'})$. Let ${\rm Pf}^{(3)}$ be the Pfaffian of the subalgebra spanned by

\[ E_{ij}-E_{j'i'}, \ E_{ij'}-E_{ji'}, \ E_{i'j}-E_{j'i} \quad \text{with } i,j\leqslant 3. \]

Since this subalgebra is isomorphic to $\mathfrak {so}_6\cong \mathfrak {sl}_4$ write also $\tilde \Delta _3^{(4)}$ for ${\rm Pf}^{(3)}$. By the construction, $({3!(n-3)!}/{n!}){\sf m}({\rm Pf}^{(3)})\otimes {\boldsymbol H}$ is a summand of ${\sf m}({\rm Pf})$. For a Weyl involution $\theta$ of $\mathfrak {sl}_4$, we have $\theta (\tilde \Delta _3^{(4)})=-\tilde \Delta _3^{(4)}$. Therefore $\varpi (\tilde \Delta _3^{(4)})$ acts as zero on any irreducible self-dual $\mathfrak {sl}_4$-module, in particular, on $\mathfrak {sl}_4$ and on $\Lambda ^{2}{{\mathbb {C}}}^{4}={{\mathbb {C}}}^{6}$. Now we can conclude that ${\sf m}({\rm Pf}^{(3)})=0$ and hence ${\sf m}({\rm Pf})=0$. Thus $\varpi ({\rm Pf})[-1]$ is a Segal–Sugawara vector for each $n$.

Keep the assumption that $n$ is even. Another way to see that ${\sf m}({\rm Pf})=0$ is to use an outer involution $\sigma \in {\rm Aut}(\mathfrak {so}_{2n})$ such that $\sigma ({\rm Pf})=-{\rm Pf}$. Here $\sigma (v)=-v$ for $v\in \mathfrak v_{n-1}$ such that $v\otimes \mathfrak g$ is the primitive copy of $\mathfrak g$ that gives rise to ${\rm Pf}$ and also $\sigma ({\sf m}({\rm Pf}))=-{\sf m}({\rm Pf})$. At the same time, $\sigma$ acts as ${{\rm id}}$ on $\mathfrak v_{n-3}$. Therefore $\sigma$ acts as ${{\rm id}}$ on $(\mathfrak g\otimes {{\mathcal {S}}}^{n-3}(\mathfrak g))^{\mathfrak g}$. Since ${\sf m}({\rm Pf})\in (\mathfrak g\otimes {{\mathcal {S}}}^{n-3}(\mathfrak g))^{\mathfrak g}$, it must be zero.

Explicit formulas for the Pfaffian-type Segal–Sugawara vector ${\rm Pf} F[-1]\in {{\mathcal {U}}}(\mathfrak {so}_{2n}[t^{-1}])$ are given in [Mol13, Roz14]. In the basis $\{F_{ij}^{\circ }=E_{ij}-E_{ji}\mid 1\leqslant i< j\leqslant 2n\}$ for $\mathfrak {so}_{2n}$, the vector ${\rm Pf} F[-1]$ is written as a sum of monomials with pairwise commuting factors, see [Mol13] and [Mol18, Equation (8.11)]. Hence it coincides with the symmetrisation of its symbol, in our notation, ${\rm Pf} F[-1] = \varpi ({\rm Pf})[-1]$. Example 5.2 provides a different proof for [Mol18, Proposistion 8.4].

Another easy to understand instance is provided by the invariant of degree $5$ in type ${\sf E}_6$.

Example 5.3 Suppose that $\mathfrak g$ is a simple Lie algebra of type ${\sf E}_6$. Let $H\in {{\mathcal {S}}}(\mathfrak g)^{\mathfrak g}$ be a homogeneous invariant of degree $5$. Then ${\sf m}(H)\in (\Lambda ^{2}\mathfrak g\otimes {{\mathcal {S}}}^{2}(\mathfrak g))^{\mathfrak g}$. Here $\Lambda ^{2}\mathfrak g=V(\pi _3)\oplus \mathfrak g$ and ${{\mathcal {S}}}^{2}(\mathfrak g)=V(2\pi _6)\oplus V(\pi _1+\pi _5)\oplus \mathbb {C}$. Therefore ${\sf m}(H)=0$.

Recall that we are considering only semisimple $\mathfrak g$ now and that $(\,\,,\,)$ is fixed in such a way that ${\mathcal {H}}\in {{\mathcal {U}}}(\mathfrak g)$ acts on $\mathfrak g$ as $C{{\rm id}}_{\mathfrak g}$ for some $C\in {{\mathbb {C}}}$.

Lemma 5.4 There is $c_1\in {{\mathbb {C}}}$ depending on the scalar product $(\,\,,\,)$ such that $\sum _{i} x_i[\xi,x_i] = c_1 \xi$ in ${{\mathcal {U}}}(\mathfrak g)$ for any $\xi \in \mathfrak g$. Furthermore, ${\mathcal {H}}(\xi )=\sum _{i} \mathrm {ad}(x_i)^{2} (\xi )=-2c_1 \xi$, i.e. $C=-2c_1$.

Proof. For each $\xi \in \mathfrak g$, set $\psi _1(\xi )=\sum _{i} x_i[\xi,x_i] = \sum _i x_i\xi x_i - {\mathcal {H}} \xi$. Since $[{\mathcal {H}},\xi ]=0$, we have then $\omega (\psi _1(\xi ))=-\psi _1(\xi )$. Note that

\[ \psi_1(\xi)= \sum_{i} [x_i,[\xi,x_i]] + \sum_{i} [\xi,x_i]x_i = - {\mathcal{H}}(\xi) +\omega(\psi_1(\xi))= - {\mathcal{H}}(\xi) - \psi_1(\xi). \]

Thereby $2\psi _1(\xi )=-{\mathcal {H}}(\xi )=-C\xi$.

Lemma 5.5 Let $\mathfrak g$ be a simple Lie algebra of rank at least $2$. Then ${\sf m}({\mathcal {H}}^{3})\not \in \mathfrak g\otimes {{\mathcal {S}}}^{3}(\mathfrak g)$.

Proof. Choose an orthogonal basis of $\mathfrak h$ such that at least one element in it is equal to $h_\alpha$ for a simple root $\alpha$. If the root system of $\mathfrak g$ is not simply laced, suppose that $\alpha$ is a long root. Suppose further that either $\alpha =\alpha _1$ or $\alpha =\alpha _\ell$. Changing the scalar product if necessary, we may assume that $h_\alpha \in \{x_i\}$. Consider the summand $\xi \otimes h_\alpha ^{3}$ of ${\sf m}({\mathcal {H}}^{3})$. We have

\[ \xi = \frac{3! 3!}{6!} ( 6{\sf m}(h_\alpha{\mathcal{H}}) + 8 {\sf m} (h_\alpha^{3}) ). \]

Set $\xi _0={\sf m}(h_\alpha {\mathcal {H}})$. Note that $h_\alpha {\mathcal {H}}\in {{\mathcal {U}}}(\mathfrak g)$ acts on $\mathfrak g$ as a scalar multiple of $\mathrm {ad}(h_\alpha )$. In view of Lemma 5.4, the sum $\sum _{i} x_i h_\alpha x_i$ acts on $\mathfrak g$ as another multiple of $\mathrm {ad}(h_\alpha )$. Hence $\xi _0\in \mathfrak g$.

It remains to show that $\eta =\mathrm {ad}(h_\alpha )^{3}$ is not an element of $\mathfrak g\subset \mathfrak {so}(\mathfrak g)$. Let $\alpha '$ be the unique simple root not orthogonal to $\alpha$. Observe that $\eta (e_\alpha )=8 e_\alpha$ and $\eta (e_{\alpha '})= - e_{\alpha '}$. Set $\gamma =\alpha +\alpha '$. Then $e_\gamma \ne 0$ and $\eta (e_\gamma )=e_\gamma$. Since $1\ne 8-1$, we conclude that indeed $\eta \not \in \mathfrak g$.

Proposition 5.6 Let $\mathfrak g$ be an exceptional simple Lie algebra. Suppose that $H\in {{\mathcal {S}}}^{6}(\mathfrak g)^{\mathfrak g}$. Then there is ${{\boldsymbol b}}\in \mathbb {C}$ such that ${\sf m}(H-{{\boldsymbol b}}{\mathcal {H}}^{3})\in \mathbb {C} {\mathcal {H}}^{2}\subset {{\mathcal {S}}}^{4}(\mathfrak g)$.

Proof. Let $V$ be the Cartan component of $\Lambda ^{2}\mathfrak g$. A straightforward calculation shows that $V$ appears in ${{\mathcal {S}}}^{3}(\mathfrak g)$ with multiplicity one as in the following table.

We have ${\sf m}(H)\in (V\otimes {{\mathcal {S}}}^{3}(\mathfrak g))^{\mathfrak g}\oplus (\mathfrak g\otimes {{\mathcal {S}}}^{3}(\mathfrak g))^{\mathfrak g}$. The first summand here is one-dimensional. Since ${\sf m}({\mathcal {H}}^{3})\not \in \mathfrak g\otimes {{\mathcal {S}}}^{3}(\mathfrak g)$ by Lemma 5.5, there is ${{\boldsymbol b}}\in {{\mathbb {C}}}$ such that ${\sf m}(\tilde H)\in \mathfrak g\otimes {{\mathcal {S}}}^{3}(\mathfrak g)$ for $\tilde H=H-{{\boldsymbol b}}{\mathcal {H}}^{3}$.

The degrees of basic symmetric invariants $\{H_k\mid 1\leqslant k\leqslant l\}$ indicate that ${{\mathcal {S}}}^{3}(\mathfrak g)$ contains exactly one copy of $\mathfrak g$. (This is also apparent in the table above.) Hence $(\mathfrak g\otimes {{\mathcal {S}}}^{3}(\mathfrak g))^{\mathfrak g}= {{\mathcal {S}}}^{4}(\mathfrak g)^{\mathfrak g}={{\mathbb {C}}} {\mathcal {H}}^{2}$.

Corollary 5.7 Keep the assumption that $\mathfrak g$ is exceptional. Then $\tilde H=H-{{\boldsymbol b}}{\mathcal {H}}^{3}$ of Proposition 5.6 satisfies (0.2) and there are $R(1),R(2)\in {{\mathbb {C}}}$ such that

(5.1)\begin{equation} S_2=\varpi(\tilde H)[-1] + R(1) \varpi(\tau^{2} {\mathcal{H}}^{2}[-1]){\cdot}1+ R(2) \varpi(\tau^{4}{\mathcal{H}}[-1]){\cdot}1 \end{equation}

is an element of $\mathfrak z({{\hat {\mathfrak g}}})$.

Proof. The first statement follows from Proposition 5.6 and Example 3.7. More explicitly, ${\sf m}(\tilde H)\in {{\mathbb {C}}} {\mathcal {H}}^{2}$, since there is no other symmetric invariant of degree four. Now the existence of $R(1)$ and $R(2)$ follows from Theorem 3.11.

6. Type G$_2$

Let $\mathfrak g$ be a simple Lie algebra of type G$_2$. Then $\ell =2$. The algebra ${{\mathcal {S}}}(\mathfrak g)^{\mathfrak g}$ has two generators, ${\mathcal {H}}$ and $\Delta _6\in {{\mathcal {S}}}^{6}(\mathfrak g)$. In this section, we compute the constant ${{\boldsymbol b}}$ of Proposition 5.6 for $H=\Delta _6$ and $R(1), R(2)$ of (5.1). All our computations are done by hand. A computer-aided explicit formula for a Segal–Sugawara vector of $t$-degree $6$ is obtained in [MRR16].

First we choose a matrix realisation of $\mathfrak g\subset \mathfrak {so}_7$. The embedding $\iota \!:\mathfrak {sl}_3\to \mathfrak {so}_7$ is fixed by

\[ \iota(E_{ij})=E_{ij}-E_{(7-j)(7-i)} \]

for $i\ne j$. We choose a basis of $\mathfrak h\subset \mathfrak {sl}_3$ as $\{h_1,h_2\}$ with $h_1={{\rm diag}}(1,-1,0)$, $h_2={{\rm diag}}(1,1,-2)$ and extend it to a basis of $\mathfrak {sl}_3$ by adding $e_i,f_i$ with $1\leqslant i\leqslant 3$ in the semi-standard notation, e.g. $e_3=E_{13}$, $f_1=E_{21}$, $f_3=E_{31}$. Let $\varepsilon _i\in \mathfrak h^{*}$ with $1\leqslant i\leqslant 3$ be the same as in § 2. Now it remains to describe the complement of $\mathfrak {sl}_3 = (\mathfrak {so}_6 \cap \mathfrak g)$, which is isomorphic to ${{\mathbb {C}}}^{3}\oplus ({{\mathbb {C}}}^{3})^{*}$.

Matrix (6.1) presents an element of ${{\mathbb {C}}}^{3}\oplus ({{\mathbb {C}}}^{3})^{*}\subset \mathfrak {so}_7$:

(6.1)\begin{equation} \left(\begin{array}{cccccc|c} & & & -{{\tt \beta}} & {{\tt \gamma}} & 0 & \sqrt{2}{{\tt a}} \\ & & & {{\tt \alpha}} & 0 & -{{\tt \gamma}} & \sqrt{2}{{\tt b}} \\ & & & 0 & -\alpha & {{\tt \beta}} & \sqrt{2}{{\tt c}} \\ {{\tt b}} & -{{\tt a}} & 0 & & & & \sqrt{2}{{\tt \gamma}} \\ -{{\tt c}} & 0 & {{\tt a}} & & & & \sqrt{2}{{\tt \beta}} \\ 0 & {{\tt c}} & -{{\tt b}} & & & & \sqrt{2} {{\tt \alpha}} \\ \hline \vphantom{\frac{a^l_{1}}{a^l_{1}}}-\sqrt{2}{{\tt \alpha}} & -\sqrt{2}{{\tt \beta}} & -\sqrt{2}{{\tt \gamma}} & -\sqrt{2}{{\tt c}} & -\sqrt{2}{{\tt b}} & -\sqrt{2}{{\tt a}} & 0 \end{array}\right). \end{equation}

With a certain abuse of notation, we denote the elements of the corresponding basis by the same symbols, for instance,

\[ {{\tt a}}=\sqrt{2}E_{17}-E_{42}+E_{53}-\sqrt{2} E_{76} \]

as a vector of ${{\mathbb {C}}}^{3}$.

The scalar product $(\,\,,\,)$ is such that ${\mathcal {H}}=\Delta _2$ with

\[ \Delta_2 = 2 e_1f_1 + 2 e_2f_2 + 2e_3f_3 + \tfrac{1}{2} h_1^{2} + \tfrac{1}{6} h_2^{2}-\tfrac{2}{3}({{\tt a}}{{\tt \alpha}} + {{\tt b}}{{\tt \beta}} + {{\tt c}}{{\tt \gamma}}). \]

The basic invariant of degree $6$, $\Delta _6$, is chosen as the restriction to $\mathfrak g$ of the coefficient $\Delta _6^{(7)}$ of degree 6 in (1.2) written for $\mathfrak {gl}_7$. In this case, the restriction of $\Delta _6$ to $\mathfrak {sl}_3$ is equal to $-\tilde \Delta _3^{2}$, where $\tilde \Delta _3$ is the determinant of $\mathfrak {sl}_3$. For future use, we record

\begin{gather*} [{{\tt a}}, {{\tt \alpha}}]={{\rm diag}}(-2,1,1)\in\mathfrak{sl}_3, \quad [{{\tt b}},{{\tt \beta}}]={{\rm diag}}(1,-2,1)\in\mathfrak{sl}_3, \quad [{{\tt c}},{{\tt \gamma}}]=h_2, \\ [{{\tt \alpha}},{{\tt c}}]=3f_3, \quad [{{\tt \beta}},{{\tt c}}]=3f_2, \quad [{{\tt a}},{{\tt b}}]=-2{{\tt \gamma}}, \quad [{{\tt \gamma}},{{\tt \beta}}]=2{{\tt a}}, \quad [{{\tt b}},{{\tt c}}]=-2{{\tt \alpha}}, \quad [{{\tt \beta}},{{\tt a}}]=3e_1. \end{gather*}

The decomposition $\mathfrak g=({{\mathbb {C}}}^{3})^{*}\oplus \mathfrak g\oplus {{\mathbb {C}}}^{3}$ is a ${{\mathbb {Z}}}/3{{\mathbb {Z}}}$-grading induced by an (inner) automorphism $\sigma$ of $\mathfrak g$. Note that our basis for $\mathfrak g$ consists of eigenvectors of $\sigma$.

Recall that $S_2$ is given by (5.1) and that we are computing the constants occurring there. There is an easy part of the calculation. It concerns the projection of ${\sf m}(\Delta _2^{3})$ on $(V\otimes {{\mathcal {S}}}^{3}(\mathfrak g))^{\mathfrak g}$. As we already know, the highest weight of $V$ is $3\pi _1$. Next choose a monomial of weight $3\pi _1$, for instance, $e_3^{2} f_1$.

Lemma 6.1 Let $\xi \otimes e_3^{2} f_1$ be a summand of ${\sf m}(\Delta _2^{3})$. Then $\xi (e_3)=\frac {6}{5} f_2$.

Proof. Observe that in $\Delta _2^{3}$, the factor $e_3^{2} f_1$ appears only in the summand $24 e_3^{2} f_3^{2} e_1f_1$. By the construction, we have

\[ \xi= \frac{24\times 3! \times 3!}{6!} {\sf m}(f_3^{2}e_1) = \frac{6}{5} {\sf m}(f_3^{2}e_1). \]

Note that $[e_1,e_3]=0$. Hence

\begin{align*} \tfrac{5}{6}\xi(e_3)&=\tfrac{1}{6}(2\mathrm{ad}(e_1)\mathrm{ad}(f_3)^{2}+2\mathrm{ad}(f_3)\mathrm{ad}(e_1)\mathrm{ad}(f_3))(e_3)\\ & = \tfrac{1}{6}(-2\mathrm{ad}(f_2)\mathrm{ad}(f_3)-4\mathrm{ad}(f_3)\mathrm{ad}(f_2))(e_3)=-[f_2,[f_3,e_3]]=f_2 \end{align*}

and the result follows.

In the above computation, we did not see the projection of ${\sf m}(\Delta _2^{3})$ on $(\mathfrak g\otimes {{\mathcal {S}}}^{3}(\mathfrak g))^{\mathfrak g}$, which is equally important. Set $h_3=[e_3,f_3]$. Note that $\{f_3,h_3,e_3\}$ is an $\mathfrak {sl}_2$-triple associated with the highest root of $\mathfrak g$. In the following lemma, $\mathfrak {sl}_2$ means $\langle f_3,h_3,e_3\rangle _{{{\mathbb {C}}}}$.

Lemma 6.2 Let $\eta \otimes e_3^{2}f_3$ be a summand of ${\sf m}(\Delta _2^{3})$. Then $\eta$ acts as $\frac {48}{5}\mathrm {ad}(f_3)$ on $\langle e_3,f_3,h_3\rangle _{{{\mathbb {C}}}}$ and as $\frac {42}{5}\mathrm {ad}(f_3)$ on the $11$-dimensional $\mathfrak {sl}_2$-stable complement of this subspace.

Proof. In this case, one has to pay a special attention to the summand $8e_3^{3}f_3^{3}$ of $\Delta _2^{3}$. In the product $(e_3f_3)(e_3f_3)(e_3f_3)$, there are six choices of $(e_3,e_3,f_3)$ such that $f_3$ and one of the elements $e_3$ belong to one and the same copy of $\Delta _2$; there are also three other choices. These first six choices are absorbed in $({3!3!}/{6!}) 24 {\sf m}(f_3\Delta _2)$. Note that $ ({3!\times 3! \times 24}/{6!})=\frac {6}{5}$. The contribution to $\eta$ of the three other choices is $({3!3!}/{6!}) 24 {\sf m}(e_3f_3^{2})$. Hence $\eta =\frac {6}{5}({\sf m}(f_3\Delta _2)+{\sf m}(e_3f_3^{2}))$.

The element $\varpi (\Delta _2)\in {{\mathcal {U}}}(\mathfrak g)$ acts on $\mathfrak g$ as a scalar. That scalar is $8$ in our case. According to Lemma 5.4, the sum $\sum _{i} x_i f_3 x_i\in {{\mathcal {U}}}(\mathfrak g)$ is equal to ${\mathcal {H}}f_3 -4 f_3$. Thus, ${\sf m}(f_3\Delta _2) = (8-\frac {4}{3})\mathrm {ad}(f_3)$ and $\frac {6}{5}{\sf m}(f_3\Delta _2)=8\mathrm {ad}(f_3)$.

Now consider $\eta _0=(e_3f_3^{2}+f_3^{2}e_3+f_3e_3f_3)\in {{\mathcal {U}}}(\mathfrak {sl}_2)$. Clearly, ${\eta }_0$ acts as zero on a trivial $\mathfrak {sl}_2$-module; for the defining representation on ${{\mathbb {C}}}^{2}=\langle v_1,v_2\rangle _{{{\mathbb {C}}}}$ with $e_3v_1=0$, one obtains ${\eta }_0(v_1)=v_2$ and ${\eta }_0(v_2)=0$. This suffices to state that $\frac {6}{5}{\sf m}(e_3f_3^{2})$ acts as $\frac {2}{5}\mathrm {ad}(f_3)$ on the $\mathfrak {sl}_2$-stable complement of $\langle f_3,h_3,e_3\rangle _{{{\mathbb {C}}}}$. Finally, ${\eta }_0(f_3)=0$ by the obvious reasons, ${\eta }_0(e_3)=-2h_3-2h_3=4\mathrm {ad}(f_3)(e_3)$ and ${\eta }_0(h_3)=4\mathrm {ad}(f_3)(h_3)$ as well, since ${\eta }_0$ acts on $\mathfrak g$ as an element of $\mathfrak {so}(\mathfrak g)$. All computations are done now and the proof is finished.

Let ${\sf pr}\!:\mathfrak {so}_7\to \mathfrak g$ be the orthogonal projection. In order to work with $\Delta _6$, one needs to compute the images under ${\sf pr}$ of $F_{ij}=E_{ij}-E_{(8-j)(8-i)}\in \mathfrak {so}_7$. For the elements of $\mathfrak {gl}_3\subset \mathfrak {so}_6$, this is easy, the task reduces to $F_{ii}$ with $1\leqslant i\leqslant 3$, where we have

\[ {\sf pr}(F_{11}) = \tfrac{1}{6}(3h_1+h_2), \quad {\sf pr}(F_{22}) = \tfrac{1}{6}(-3h_1+h_2),\quad {\sf pr}(F_{33}) = \tfrac{-1}{3}h_2. \]

The elements of $F_{ij}\in \mathfrak {so}_6$ with $1\leqslant i\leqslant 3$, $4\leqslant j\leqslant 6$ project with the coefficient $\frac {1}{3}$ on the corresponding letters in (6.1), e.g, ${\sf pr}(F_{14})=\frac {-1}{3}{{\tt \beta }}$, ${\sf pr}(F_{15})=\frac {1}{3}{{\tt \gamma }}$, and so on. The elements $F_{i7}$ project with the coefficient $\frac {\sqrt {2}}{3}$ on the corresponding letters, e.g, ${\sf pr}(F_{17})=\frac {\sqrt {2}}{3}{{\tt a}}$. Finally, the elements $F_{7i}$ project with the coefficient $\frac {-\sqrt {2}}{3}$ on the corresponding letters, e.g, ${\sf pr}(F_{71})=\frac {-\sqrt {2}}{3} {{\tt \alpha }}$. An explicit formula for $\Delta _6$ can be obtained by replacing first $E_{ij}$ with $F_{ij}$ in $\Delta _6^{(7)}\in {{\mathcal {S}}}^{6}(\mathfrak {gl}_7)$ and then replacing $F_{ij}$ with ${\sf pr}(F_{ij})$. We write down some of the terms of $\Delta _6$:

\begin{align*} \Delta_6 &=-\tilde\Delta_3^{2} - \tfrac{4}{27} {{\tt c}}^{3}e_3^{2}f_1 - \tfrac{4}{9} {{\tt c}}{{\tt \beta}}f_3 e_3^{2} f_1 + \tfrac{2}{3} {{\tt c}} {{\tt \alpha}}f_2 e_3^{2} f_1 + \tfrac{1}{9} {{\tt c}}{{\tt \alpha}}h_1 f_3 e_3^{2} -\tfrac{1}{27} {{\tt c}} {{\tt \alpha}}h_2 f_3 e_3^{2}\\ &\quad - \tfrac{4}{9} {{\tt b}}{{\tt \alpha}}f_2f_3e_3^{2}+\cdots . \end{align*}

With this knowledge we can attack the computation of ${\sf m}(\Delta _6)$. The first challenge is to understand the term $\tilde \xi \otimes e_3^{2} f_1$.

Lemma 6.3 For $\tilde \xi$ as above, we have $\tilde {\xi }(e_3)=\frac {5}{18}f_2$.

Proof. Once again, we rely on a direct computation. The terms of $\tilde \Delta _3$ containing $e_3$ as a factor are $e_3f_1f_2$ and $e_3f_3(\frac {1}{2}h_1-\frac {1}{6}h_2)$. Thereby the contribution of $-\tilde \Delta _3^{2}$ to $\tilde \xi$ is

\[ \tfrac{1}{20} {\sf m}\bigl(-2f_1f_2^2-2f_2f_3 \bigl(\tfrac{1}{2}h_1-\tfrac{1}{6}h_2\bigr)\bigr) \]

and this element of ${\rm End}(\mathfrak g)$ maps $e_3$ to $\frac {3}{20} f_2$.

Since $\sigma (\Delta _6)=\Delta _6$, the summands of $\Delta _6$ that contain $e_3^{2} f_1$ as a factor are of tri-degrees $(3,3,0)$ or $(1,4,1)$ with respect to the ${{\mathbb {Z}}}/3{{\mathbb {Z}}}$-grading $\mathfrak g=({{\mathbb {C}}}^{3})^{*}\oplus \mathfrak {sl}_3\oplus {{\mathbb {C}}}^{3}$. By the weight considerations, the first possibility occurs only for the monomial ${{\tt c}}^{3} e_3^{2} f_1$. Record that

\[ \mathrm{ad}({{\tt c}})^{3}(e_3)=\mathrm{ad}({{\tt c}})^{2}(-{{\tt a}})=[{{\tt c}},2{{\tt \beta}}]=-6 f_2. \]

The coefficient of ${{\tt c}}^{3} e_3^{2} f_1$ in $\Delta _6$ is equal to $\frac {-4}{27}$. The monomials of the tri-degree $(1,4,1)$ are ${{\tt c}}{{\tt \beta }}f_3 e_3^{2} f_1$ and ${{\tt c}} {{\tt \alpha }}f_2 e_3^{2} f_1$. Their coefficients are $\frac {-4}{9}$ and $\frac {2}{3}$, respectively.

Next

\[ {\sf m}({{\tt c}}{{\tt \beta}}f_3) (e_3) = \tfrac{1}{6}(\mathrm{ad}([{{\tt \beta}},{{\tt c}}])\mathrm{ad}(f_3)+2\mathrm{ad}(f_3)\mathrm{ad}([{{\tt \beta}},{{\tt c}}]))(e_3) = \tfrac{3}{2} \mathrm{ad}(f_2)\mathrm{ad}(f_3)(e_3)= -\tfrac{3}{2} f_2 \]

and

\[ {\sf m}({{\tt c}}{{\tt \alpha}}f_2)(e_3)=\tfrac{1}{2}(\mathrm{ad}(f_2)\mathrm{ad}({{\tt c}})\mathrm{ad}({{\tt \alpha}}) + \mathrm{ad}(f_2)\mathrm{ad}({{\tt \alpha}})\mathrm{ad}({{\tt c}}))(e_3)= \tfrac{1}{2}[{{\rm diag}}(1,-2,1),f_2]=\tfrac{3}{2}f_2. \]

Summing up, we have

\[ \tilde{\xi}(e_3)=\tfrac{1}{20}\bigl(3+\tfrac{8}{9}+\tfrac{2}{3}+1\bigr)f_2=\tfrac{1}{20}\bigl(5+\tfrac{5}{9}\bigr)f_2= \tfrac{1}{4}\bigl(1+\tfrac{1}{9}\bigr)f_2=\tfrac{5}{18}f_2, \]

and we are done.

Corollary 6.4 We have ${{\boldsymbol b}}= \frac {25}{108}$ and the invariant $\tilde H$ of Proposition 5.6 is equal to $\Delta _6 - \frac {25}{108} \Delta _2^{3}$.

Proof. By the definition of ${{\boldsymbol b}}$, we must have $(\tilde \xi -{{\boldsymbol b}}\xi )(e_3)=0$. From Lemmas 6.1 and 6.3 we obtain that ${{\boldsymbol b}}=\frac {5}{18}\times \frac {5}{6}=\frac {25}{108}$.

Next we deal with $\tilde \eta$ for the summand $\tilde \eta \otimes e_3^{2}f_3$ of ${\sf m}(\Delta _6)$.

Lemma 6.5 For $\tilde \eta$ as above, we have

\[ \tilde\eta({\tt a})=\tfrac{1}{20}\bigl(\tfrac{-2}{9} -\tfrac{28}{9}-\tfrac{4}{3}+\tfrac{2}{9}\bigr) \mathrm{ad}(f_3)({\tt a}) = \tfrac{-2}{9} \mathrm{ad}(f_3)({\tt a}) \]

and

\[ \tilde\eta(h_3)=\tfrac{1}{20}\bigl( \tfrac{1}{3} +\tfrac{8}{27}+\tfrac{4}{9} + \tfrac{1}{27} \bigr) \mathrm{ad}(f_3)(h_3) = \tfrac{10}{9\times 20} \mathrm{ad}(f_3)(h_3) = \tfrac{1}{18} \mathrm{ad}(f_3)(h_3). \]

Proof. We go through the relevant summands of $\Delta _6$. In $-\tilde \Delta _3^{2}$, these are $\frac {-1}{36}e_3^{2}f_3^{2}(3h_1-h_2)^{2}$ and $\frac {-1}{3} e_3^{2} f_1 f_2 f_3 (3h_1-h_2)$. The corresponding contributions to $\tilde \eta$ are

\[ \tfrac{-1}{18}{\sf m}(f_3(3h_1-h_2)^{2}) \quad\text{and}\quad \tfrac{-1}{3}{\sf m}(f_1f_2(3h_1-h_2)) \]

multiplied by $\frac {1}{20}$. We are going to keep the factor $\frac {1}{20}$ in the background. Note that ${\sf m}(f_3(3h_1-h_2)^{2})$ acts on ${{\tt a}}$ as $4\mathrm {ad}(f_3)$, and hence we add $\frac {-2}{9}$. Since $2-4+2=0$, the second of the above elements acts on ${{\tt a}}$ as zero. If we consider the action on $h_3$ instead, then the contribution of the first term is zero and ${\sf m}(f_1f_2(3h_1-h_2))$ acts as $\mathrm {ad}([f_1,f_2])=-\mathrm {ad}(f_3)$.

On account of $\sigma$, the other relevant terms have tri-degrees $(3,3,0)$, $(0,3,3)$, $(1,4,1)$, where the former two possibilities occur for $\frac {4}{27} {{\tt c}}^{2}{{\tt b}}e_3^{2}f_3$ and $\frac {4}{27} {{\tt \alpha }}^{2}{{\tt \beta }}e_3^{2}f_3$. Here

\begin{align*} {\sf m}({{\tt c}}^{2} {{\tt b}})(h_3)&=\tfrac{1}{3}(\mathrm{ad}({{\tt b}})\mathrm{ad}({{\tt c}})\mathrm{ad}({{\tt c}})+\mathrm{ad}({{\tt c}})\mathrm{ad}({{\tt b}})\mathrm{ad}({{\tt c}}))(h_3)\\ & = \tfrac{1}{3}(-2\mathrm{ad}({{\tt \alpha}})\mathrm{ad}({{\tt c}})-4\mathrm{ad}({{\tt c}})\mathrm{ad}({{\tt \alpha}}))(h_3)=(-2\mathrm{ad}(f_3)-2\mathrm{ad}({{\tt c}})\mathrm{ad}({{\tt \alpha}}))(h_3). \end{align*}

Since $[{{\tt \alpha }},h_3]={{\tt \alpha }}$ and $[{{\tt c}},{{\tt \alpha }}]=-3f_3$, the contribution in question is $\frac {4}{27}\mathrm {ad}(f_3)$. Similarly,

\begin{align*} {\sf m}({{\tt \alpha}}^{2} {{\tt \beta}})(h_3)&=\tfrac{1}{3}(\mathrm{ad}({{\tt \beta}})\mathrm{ad}({{\tt \alpha}})\mathrm{ad}({{\tt \alpha}})+\mathrm{ad}({{\tt \alpha}})\mathrm{ad}({{\tt \beta}})\mathrm{ad}({{\tt \alpha}}))(h_3)\\ & = \tfrac{1}{3}(2\mathrm{ad}({{\tt c}})\mathrm{ad}({{\tt \alpha}})+4\mathrm{ad}({{\tt \alpha}})\mathrm{ad}({{\tt c}}))(h_3)=(2\mathrm{ad}({{\tt \alpha}})\mathrm{ad}({{\tt c}})-2\mathrm{ad}(f_3))(h_3). \end{align*}

Since $[{{\tt c}},h_3]={{\tt c}}$, $[{{\tt \alpha }},{{\tt c}}]=3f_3$, we obtain again $\frac {4}{27}\mathrm {ad}(f_3)$. A slightly different story happens at $a$, namely,

\begin{align*} {\sf m}({{\tt c}}^{2} {{\tt b}})({{\tt a}})&=(-2\mathrm{ad}(f_3)-2\mathrm{ad}({{\tt c}})\mathrm{ad}({{\tt \alpha}}) +\mathrm{ad}({{\tt c}})\mathrm{ad}({{\tt c}})\mathrm{ad}({{\tt b}}))({{\tt a}}) \\ & = (-2\mathrm{ad}(f_3) -2\mathrm{ad}(f_3)+ 4\mathrm{ad}(f_3))({{\tt a}})=(-4+4)\mathrm{ad}(f_3)({{\tt a}})=0, \\ {\sf m}({{\tt \alpha}}^{2} {{\tt \beta}})({{\tt a}})&=(2\mathrm{ad}({{\tt \alpha}})\mathrm{ad}({{\tt c}}) -2\mathrm{ad}(f_3)+\mathrm{ad}({{\tt \alpha}})\mathrm{ad}({{\tt \alpha}})\mathrm{ad}({{\tt \beta}}))({{\tt a}}) \\ & =(8-2-6)\mathrm{ad}(f_3) ({{\tt a}})=0. \end{align*}

Now consider the terms of the tri-degree $(1,4,1)$. Let $\varepsilon _i-\varepsilon _j$ be the weight of the fourth element from $\mathfrak {sl}_3$. Assume first that $i\ne j$. Then $\varepsilon _3-\varepsilon _1=(\varepsilon _i-\varepsilon _j)+\varepsilon _s-\varepsilon _l$ for some $s$ and $l$. One of the possibilities is $i=3, j=1$, and $s=l$. The other two come from the decomposition $\varepsilon _1-\varepsilon _3=(\varepsilon _1-\varepsilon _2)+(\varepsilon _2-\varepsilon _3)$.

In the case $s=l$, the relevant term is $\frac {4}{9} e_3^{2}f_3^{2} {{\tt b}}{{\tt \beta }}$ and its contribution to $\tilde \eta$ is $\frac {8}{9}{\sf m}(f_3 {{\tt b}}{{\tt \beta }})$. Since both ${{\tt b}}$ and ${{\tt \beta }}$ commute with $f_3$ and $h_3$, we see that ${\sf m}(f_3 {{\tt b}}{{\tt \beta }})(h_3)=0$. Furthermore,

\[ {\sf m}(f_3 {{\tt b}}{{\tt \beta}})({{\tt a}})=\mathrm{ad}({{\tt b}})\mathrm{ad}({{\tt \beta}})({{\tt c}})-\tfrac{1}{2}{{\tt c}}= -3{{\tt c}}-\tfrac{1}{2}{{\tt c}} = \tfrac{-7}{2}\mathrm{ad}(f_3)({{\tt a}}). \]

In this way the summand $\frac {-28}{9}$ appears in the first formula of the lemma.

In the case $s\ne l$, the relevant terms are $\frac {-4}{9} {{\tt b}}{{\tt \alpha }}f_2 f_3 e_3^{2}$ and $\frac {-4}{9} {{\tt c}}{{\tt \beta }}f_1 f_3 e_3^{2}$. On $h_3$, each of the elements ${\sf m}({{\tt b}}{{\tt \alpha }}f_2)$, ${\sf m}({{\tt c}}{{\tt \beta }}f_1)$ acts as $\frac {-1}{2}\mathrm {ad}(f_3)$. Thus, $\frac {4}{9}$ appears in the second formula. Further,

\begin{align*} {\sf m}({{\tt b}}{{\tt \alpha}}f_2)({{\tt a}})&=\tfrac{1}{6}(\mathrm{ad}(f_2)\mathrm{ad}({{\tt b}})\mathrm{ad}({{\tt \alpha}})+2\mathrm{ad}({{\tt \alpha}})\mathrm{ad}(f_2)\mathrm{ad}({{\tt b}}))({{\tt a}}) \\ &= \tfrac{1}{6}(\mathrm{ad}({{\tt c}})\mathrm{ad}({{\tt \alpha}})+2\mathrm{ad}({{\tt \alpha}})\mathrm{ad}({{\tt c}}))({{\tt a}})= \tfrac{1}{2}\mathrm{ad}(f_3)({{\tt a}})+\mathrm{ad}(f_3)({{\tt a}})=\tfrac{3}{2}\mathrm{ad}(f_3)({{\tt a}}). \end{align*}

In the same fashion ${\sf m}({{\tt c}}{{\tt \beta }}f_1)({{\tt a}})=\frac {3}{2}\mathrm {ad}(f_3)({{\tt a}})$. This justifies $\frac {-4}{3}$ in the first formula.

The final term, which is $\frac {1}{27} {{\tt c}} {{\tt \alpha }} (3h_1-h_2) f_3 e_3^{2}$, fulfils the case $i=j$, $s=3$, $l=1$. Here we have ${\sf m}((3h_1-h_2){{\tt c}}{{\tt \alpha }})({{\tt a}})=6\mathrm {ad}(f_3)({{\tt a}})$, hence the last summand in the first formula is $\frac {2}{9}$. Similarly,

\[ {\sf m}((3h_1-h_2){{\tt c}}{{\tt \alpha}})(h_3)=\tfrac{1}{6}(2\mathrm{ad}(({{\tt \alpha}})\mathrm{ad}({{\tt c}})-2\mathrm{ad}({{\tt c}})\mathrm{ad}({{\tt \alpha}}))(h_3)=\mathrm{ad}(f_3)(h_3). \]

This justifies $\frac {1}{27}$ in the second formula.

Lemma 6.5 provides a different way to compute ${{\boldsymbol b}}$. Namely, $\tilde \eta -{{\boldsymbol b}}\eta$ has to act on $\mathfrak g$ as a scalar multiple of $\mathrm {ad}(f_3)$. Check

(6.2)$$\begin{gather} \bigg(\tilde\eta - \frac{25}{108}\eta\bigg)({{\tt a}})=\bigg(-\frac{2}{9}-\frac{25}{108}\times\frac{42}{5}\bigg) \mathrm{ad}(f_3)({{\tt a}}) = \frac{-13}{6}\mathrm{ad}(f_3)({{\tt a}}), \\ \bigg(\tilde\eta - \frac{25}{108}\eta\bigg)(h_3)=\bigg(\frac{1}{18}-\frac{5\times 48}{108}\bigg) \mathrm{ad}(f_3)(h_3) = \frac{-39}{18}\mathrm{ad}(f_3)(h_3) =\frac{-13}{6}\mathrm{ad}(f_3)(h_3). \nonumber \end{gather}$$

In order to compute $R(1)$ and $R(2)$, state first that according to (6.2), $\frac {-13}{6} \mathrm {ad}(f_3) \otimes e_3^{2} f_3$ is a summand of ${\sf m}(\tilde H)$. This indicates that if ${\sf m}(\tilde H)$ is written as an element of ${{\mathcal {S}}}^{4}(\mathfrak g)$, then it has a term $\frac {-13}{3} e_3^{2} f_3^{2}$, which is a summand of $\frac {-13}{12} \Delta _2^{2}$. Thus ${\sf m}(\tilde H)=\frac {-13}{12} \Delta _2^{2}$. In terms of Lemma 5.4, we have

\[ {\sf m}({\mathcal{H}}^{2}) = \frac{3!}{4!} \times 4 \bigg(-2c_1+\frac{1}{3}c_1\bigg) {\mathcal{H}} = \frac{20}{3} {\mathcal{H}}, \]

since $c_1=-4$ in our case. Making use of Theorem 3.11, we obtain the main result of this section:

(6.3)\begin{equation} S_2=\varpi\bigl(\Delta_6-\tfrac{25}{108}\Delta_2^{3}\bigr)[-1]-\tfrac{65}{4}\varpi(\tau^{2} \Delta_2^{2}[-1]){\cdot}1 - \tfrac{325}{3} \varpi(\tau^{4} \Delta_2[-1]){\cdot}1 \end{equation}

is an element of $\mathfrak z({{\hat {\mathfrak g}}})$. Furthermore, $S_1={\mathcal {H}}[-1]$ and $S_2$ form a complete set of Segal–Sugawara vectors for $\mathfrak g$.

7. The orthogonal case

Now suppose that $\mathfrak g=\mathfrak {so}_n\subset \mathfrak {gl}_n$ with $n\geqslant 7$. A suitable matrix realisation of $\mathfrak g$ uses the elements $F_{i j}=E_{i j}-E_{j' i'}$ with $i,j\in \{1,\ldots,n\}$, $i'=n-i+1$. We will be working with the coefficients $\Phi _{2k}\in {{\mathcal {S}}}^{2k}(\mathfrak g)^{\mathfrak g}$ of

\[ \det(I_n- q(F_{ij}))^{-1} = 1 + \Phi_2 q^{2}+ \Phi_4 q^{4} +\cdots + \Phi_{2k} q^{2k} +\cdots . \]

The generating invariants of this type appeared in [MY19, § 3] in connection with the symmetrisation map and they can be used in (2.4) as well. In [Mol18, MY19], the elements $\Phi _{2k}$ are called permanents, but they are not the permanents of matrices in the usual sense. Set $\mathfrak h=\langle F_{jj} \mid 1\leqslant j\leqslant \ell \rangle _{{{\mathbb {C}}}}$.

In general, $\det (I_n- qA)^{-1}=\det (I_n+qA+q^{2} A^{2}+\cdots )$ for $A\in \mathfrak {gl}_n$. In particular, ${\Phi _{2k}}|_{\mathfrak h}$ is equal to the homogeneous part of degree $2k$ of

\[ \prod_{j=1}^{\ell} (1+ F_{jj}^{2}+ F_{jj}^{4}+ F_{jj}^{6} + \cdots) . \]

By the construction, ${\sf m}(\Phi _{2k})$ is a polynomial function on $(\Lambda ^{2}\mathfrak g\oplus \mathfrak g)^{*}\cong \Lambda ^{2}\mathfrak g\oplus \mathfrak g$. Set

\[ {{\boldsymbol f}}={\sf m}(\Phi_{2k})|_{\Lambda^{2}\mathfrak g\oplus\mathfrak h} \quad \text{and write} \quad {{\boldsymbol f}}=\sum_{\nu}^{L} \xi_\nu\otimes {{\boldsymbol H}}_{\nu}, \]

where ${{\boldsymbol H}}_{\nu }\in {{\mathcal {S}}}^{2k-3}(\mathfrak h)$ are linearly independent monomials in $\{F_{jj}\}$ and $\xi _\nu \in \Lambda ^{2}\mathfrak g$. Note that each $\Phi _{2k}$ is an invariant of ${\rm Aut}(\mathfrak g)$. Since $\Phi _{2k}$ is an element of $\mathfrak h$-weight zero, each $\xi _\nu$ is also of weight zero. Hence one can say that ${{\boldsymbol f}}$ is an invariant of $W(\mathfrak g,\mathfrak h)$.

Let $\sigma \in {\rm Aut}(\mathfrak g)$ be an involution such that $\mathfrak g_0=\mathfrak g^{\sigma }\cong \mathfrak {so}_{n-1}$, $\sigma (F_{11})=-F_{11}$, i.e. $F_{11}\in \mathfrak g_1$, and $\sigma (F_{ss})=F_{ss}$ for $\ell \geqslant s>1$. Then $\mathfrak g_{0,F_{11}}:=(\mathfrak g_0)_{F_{11}}\cong \mathfrak {so}_{n-2}$. Such an involution $\sigma$ is not unique and we fix it by assuming that

(7.1)\begin{equation} \mathfrak g_1=\langle F_{1i}+F_{i'1} \mid 1< i< n\rangle _{{{\mathbb{C}}}} \oplus {{\mathbb{C}}}F_{11}. \end{equation}

The centraliser $\mathfrak g_{1,F_{11}}$ of $F_{11}$ in $\mathfrak g_1$ is equal to ${{\mathbb {C}}}F_{11}$. This property defines an involution of rank one. Set $\mathfrak h_0=\langle F_{ss} \mid \ell \geqslant s>1\rangle _{{{\mathbb {C}}}}$.

By the construction, the map ${\sf m}$ is ${\rm Aut}(\mathfrak g)$-equivariant. Here the group ${\rm Aut}(\mathfrak g)\subset {{\rm GL}}(\mathfrak g)$ acts on $\mathfrak {so}(\mathfrak g)\subset \mathfrak {gl}(\mathfrak g)$ via conjugation. In particular, $\sigma$ acts as $-{{\rm id}}$ on $\mathfrak g_0\wedge \mathfrak g_1\subset \mathfrak {so}(\mathfrak g)$ and as ${{\rm id}}$ on the subspaces $\Lambda ^{2}\mathfrak g_0$ and $\Lambda ^{2}\mathfrak g_1$. For the future use, record: ${\sf m}(F_{ii}^{3})=F_{ii}$ and if $i\ne j,j'$, then ${\sf m}(F_{ii}F_{jj}^{2})$ acts as ${{\rm id}}$ on $F_{ij}=-F_{j'i'}$, $F_{ij'}=-F_{ji'}$, as $-{{\rm id}}$ on $F_{ji}=-F_{i'j'}$, $F_{j'i}=-F_{i'j}$, and as zero on all other elements $F_{uw}$. In particular, ${\sf m}(F_{ii}F_{jj}^{2})\not \in \mathfrak g$ if $i\not \in \{j,j'\}$.

Lemma 7.1 Suppose that ${{\boldsymbol H}}_\nu =F_{11}^{\beta _1}\ldots F_{\ell \ell }^{\beta _\ell }$ and $\xi _\nu \ne 0$. Then there is exactly one odd $\beta _j$ with $1\leqslant j\leqslant \ell$. Furthermore, if $\beta _1$ is odd, then

\[ \xi_\nu\in\langle (F_{1i}-F_{i'1})\wedge (F_{1i'}+F_{i1}) \mid 1< i < n \rangle _{{{\mathbb{C}}}}\oplus \langle F_{11}\wedge F_{ss}\mid 1< s\leqslant \ell \rangle _{{{\mathbb{C}}}}. \]

Proof. Without loss of generality assume that $\beta _j$ is odd for $j\leqslant u$ and is even for $j>u$. Let $\sigma _j\in {\rm Aut}(\mathfrak g)$ with $2\leqslant j\leqslant u$ be an involution of rank one such that $\sigma _j(F_{jj})=-1$ and $\sigma _j(F_{ss})=F_{ss}$ for $s\ne j,j'$. Following the case of $\sigma _1=\sigma$, fix $\sigma _j$ by setting

\[ \sigma_j(F_{ji}+F_{i'j})=-F_{ji}-F_{i'j} \quad \text{for} \ i\not\in\{j,j'\}. \]

As we have already mentioned, $\sigma _j(\Phi _{2k})=\Phi _{2k}$ for each $j$. Thereby ${\sf m}(\Phi _{2k})$ is a $\sigma _j$-invariant as well. At the same time $\sigma _j({\boldsymbol H}_\nu )=-{\boldsymbol H}_{\nu }$ by the construction. Hence $\sigma _j(\xi _\nu )=-\xi _\nu$ for each $1\leqslant j\leqslant u$.

The above discussion has clarified, how involutions $\sigma _j$ act on $\Lambda ^{2}\mathfrak g={\sf m}({{\mathcal {S}}}^{3}(\mathfrak g))$. In particular, we must have $\xi _\nu \in \mathfrak g_0 \wedge \mathfrak g_1$. We know also that $\xi _\nu$ is an element of $\mathfrak h$-weight zero and that $F_{11}\in \mathfrak h$. Recall that $\mathfrak g_{1,F_{11}}={{\mathbb {C}}}F_{11}$. The decomposition $\mathfrak g_0=\mathfrak g_{0,F_{11}}\oplus [F_{11},\mathfrak g_1]$ indicates that

\[ \xi \in \mathfrak g_{0,F_{11}}\wedge F_{11} \oplus [F_{11},\mathfrak g_1] \wedge \mathfrak g_1. \]

Both summands here are $\mathfrak h$-stable. Furthermore, $( \mathfrak g_{0,F_{11}}\wedge F_{11})^{\mathfrak h}$ is spanned by $F_{ss}\wedge F_{11}$ with $\ell \geqslant s>1$.

The subspace $[F_{11},\mathfrak g_1]$ is spanned by $F_{1i}-F_{i'1}$, where $1< i< n$. For each $i$, the element of the opposite $\mathfrak h_0$-weight in $\mathfrak g_1$ is $F_{1i'}+F_{i1}$. Note that $(F_{1i}+F_{i'1})\wedge (F_{1i'}-F_{i1})$ is an eigenvector of $F_{11}$ if and only if $i=i'$. Thus, $([F_{11},\mathfrak g_1] \wedge \mathfrak g_1)^{\mathfrak h}$ is a linear span of

\[ \Xi(i):=(F_{1i}+F_{i'1})\wedge (F_{1i'}-F_{i1}) + (F_{1i'}+F_{i1})\wedge (F_{1i}-F_{i'1}) \]

with $1< i\leqslant i'$.

If $u>1$, then $u\geqslant 3$. The involution $\sigma _2$ acts on $F_{1i}\pm F_{i'1}$ as ${{\rm id}}$ if $2< i< n-1$. Therefore $\xi _\nu$ has to be a linear combination of $F_{22} \wedge F_{11}$ and $\Xi (2)$. At the same time, $\sigma _3$ acts as ${{\rm id}}$ on both these vectors. This contradiction proves that $u=1$.

Remark Lemma 7.1 is valid for any homogeneous $\Phi \in {{\mathcal {S}}}(\mathfrak g)^{{\rm Aut}(\mathfrak g)}$.

Now fix ${\boldsymbol H}={{\boldsymbol H}}_\nu =F_{11}^{2b_1-1} F_{22}^{2b_2} \ldots F_{\ell \ell }^{2b_\ell }$ with $b_j\in {{\mathbb {Z}}}_{\geqslant 0}$ and $b_1\geqslant 1$. The task is to compute $\xi =\xi _\nu$. Set $b_{j'}=b_j$ for $j\leqslant \ell$. In type B, set also $b_{\ell +1}=0$. Below we list the terms $Y_3$ such that $Y_3 {{\boldsymbol H}}$ is a summand of $\Phi _{2k}$:

(7.2)\begin{equation} \begin{gathered} F_{11}^{3}, \quad F_{11}F_{jj}^{2}, \quad 2(b_1+1)(b_j+1)F_{11} F_{1j} F_{j1}, \quad (b_i+1)(b_j+1)F_{11} F_{ij} F_{ji}, \\ 2b_1(b_j+1)F_{1j}F_{j1} F_{jj}, \quad 2b_1(b_i+1)(b_j+1)F_{1i} F_{ij} F_{j1}, \end{gathered} \end{equation}

where $1< i,j< n$ and $i\not \in \{j,j'\}$, and also in $F_{jj}$ we have $1< j\leqslant \ell$. When computing ${\sf m}$, one has to take into account the additional coefficients appearing from the powers of $F_{ii}$. For instance, in the case of $F_{11}^{3}$, this coefficient is $\binom {2b_1+2}{3}$, for $2b_1(b_j+1)F_{1j}F_{j1} F_{jj}$, the additional scalar factor is $2b_j+1$.

We will show that $\xi$ acts on $F_{ij}$ as $c(i,j)F_{11}$ for some constant $c(i,j)\in {{\mathbb {C}}}$, compute these constants and see that all of them are equal. Note that $[F_{11},F_{ij}]=0$ if $i,j\not \in \{1,n\}$.

Lemma 7.2 We have $\xi (F_{11})=0$, furthermore $\xi (F_{ij})=0$ if $i,j\not \in \{1,n\}$.

Proof. By a direct computation, we show that indeed $\xi (F_{11})=0$. Some expressions in (7.2) act on $F_{11}$ as zero by obvious reasons. If one takes into account that $F_{1j}F_{11}F_{j1}+F_{j1}F_{11}F_{1j}$ acts as $[F_{j1},F_{j1}]$, this covers the first line of (7.2). The same argument takes care of ${\sf m}(F_{jj}F_{1j}F_{j1})$. It remains to calculate $\eta ={\sf m}(F_{1i}F_{ij}F_{j1})(F_{11})$. Here we have $6\eta =(F_{11}-F_{ii}) + (F_{jj}-F_{11})$. If we switch $i$ and $j$, then the total sum is zero.

Since $(F_{ss}\wedge F_{11})(F_{11})=F_{ss}$ up to a non-zero scalar, Lemma 7.1 now implies that

\[ \xi \in \langle (F_{1i}-F_{i'1})\wedge (F_{1i'}+F_{i1}) \mid 1< i < n \rangle _{{{\mathbb{C}}}}. \]

Hence $\xi (F_{ij})=0$ if $i,j\not \in \{1,n\}$.

Lemma 7.3 Suppose that $n=2\ell$. Assume that $1< u< n$. Then

\[ \xi(F_{1u})=\frac{3!(2k-3)!}{(2k)!} C(1) F_{1u} \]

and $C(1)$ is equal to

\[ \binom{2b_1+2}{3} +\frac{2}{3} b_1 \sum_{j=2}^{\ell} (2b_j+1)(b_j+1) + \frac{8}{3} b_1(b_1+1) \sum_{j=2}^{\ell} (b_j+1)+ \frac{8}{3} b_1 \sum_{1< i< j\leqslant\ell} (b_i+1)(b_j+1). \]

Proof. Recall that ${\sf m}(F_{11}^{3})=F_{11}$. This leads to the summand $\binom {2b_1+2}{3}$ of $C(1)$. Consider

\[ \xi_{1}^{(j)}= {\sf m}(F_{11}F_{jj}^{2}) + {\sf m}(F_{1j} F_{j1} F_{jj}-F_{1j'} F_{j'1} F_{jj}) \]

with $1< j\leqslant \ell$. Here $\xi _1^{(j)}(F_{1u})= \frac {1}{3}F_{1u}$ for $u\not \in \{j,j'\}$ and $\xi _1^{(j)}(F_{1j})=(1-\frac {2}{3})F_{1j}$ as well as $\xi _1^{(j)}(F_{1j'})=(1-\frac {2}{3})F_{1j'}$. In $C(1)$, we have to add $\frac {1}{3}$ with the coefficients

\[ 2b_1\binom{2b_j+2}{2} = 2b_1(b_j+1)(2b_j+1). \]

The next terms are $\xi _2^{(j)}={\sf m}(F_{11} F_{1j} F_{j1})$ with $1< j< n$. Here $\xi _2^{(j)}(F_{1u})=\frac {1}{3}F_{1u}$ for $u\not \in \{j,j'\}$. Furthermore, $\xi _2^{(j)}(F_{1j})=\frac {2}{3}F_{1j}$ and $\xi _2^{(j)}(F_{1j'})=0$. Adding $\xi _2^{(j)}$ and $\xi _2^{(j')}$ with $j\leqslant \ell$ and recalling the coefficient of $\xi _2^{(j)}$, we obtain the summands $\frac {8}{3}(b_1+1)b_1(b_j+1)$.

Fix $1< i,j < n$ with $i\not \in \{j,j'\}$ and consider

\[ \xi_3^{i,j}={\sf m}(F_{1i} F_{ij} F_{j1}), \quad \xi_4^{i,j}={\sf m}(F_{11} F_{ij} F_{ji}). \]

An easy observation is that $\xi _4^{i,j}(F_{1u})=0$ if $u\not \in \{i,i',j,j'\}$. Also $\xi _3^{i,j}(F_{1u})=\frac {1}{6} F_{1u}$ in this case. Furthermore, $\xi _4^{i,j}(F_{1i})=\frac {1}{2}F_{1i}$ and $\xi _4^{i,j}(F_{1i'})=\frac {1}{2}F_{1i'}$. A more lengthy calculation brings

\begin{gather*} \xi_3^{i,j}(F_{1i})=\tfrac{1}{6}F_{1i} - \tfrac{1}{6}((\mathrm{ad}(F_{1i})\mathrm{ad}(F_{j1})+\mathrm{ad}(F_{j1})\mathrm{ad}(F_{1i}))(F_{1j})=\tfrac{1}{6}F_{1i}-\tfrac{1}{6}F_{1i}=0; \\ \xi_3^{i,j}(F_{1j})=\tfrac{1}{6}F_{1j} +\tfrac{1}{6} \mathrm{ad}(F_{ij})(F_{1i}) =0; \quad \xi_3^{i,j}(F_{1j'})= \tfrac{1}{6} \mathrm{ad}(F_{1i})\mathrm{ad}(F_{j1})(F_{1i'})=-\tfrac{1}{6}F_{1j'}; \\ \xi_3^{i,j}(F_{1i'})= \tfrac{1}{6} \mathrm{ad}(F_{ij})\mathrm{ad}(F_{1i})(F_{ji'}) =\tfrac{1}{6} \mathrm{ad}(F_{ij}) (F_{j n})=-\tfrac{1}{6} F_{1i'}. \end{gather*}

Note that $\xi _4^{i,j}=\xi _4^{j,i}=\xi _4^{i',j'}$. Now fix $1< i< j\leqslant \ell$ and consider

\[ \xi_5^{i,j}=\xi_3^{i,j}+\xi_3^{j,i}+\xi_3^{i',j}+\xi_3^{j,i'}+\xi_3^{i,j'}+\xi_3^{j',i}+\xi_3^{i',j'}+\xi_3^{j',i'}+ \xi_4^{i,j}+\xi_4^{i',j}+\xi_4^{i,j'}+\xi_4^{i',j'}. \]

Here $2b_1(b_i+1)(b_j+1)\xi _5^{i,j}$ is a summand of $ ({(2k)!}/{3!(2k-3)!}) \xi$. Moreover, $\xi _5^{i,j}(F_{1s})=\frac {4}{3}F_{1s}$ for each $1< s< n$. This justifies the last summand of $C(1)$.

Rearranging the expression for $C(1)$, one obtains

(7.3)\begin{equation} C(1)=\frac{2}{3} b_1 \biggl(\sum_{j=1}^{\ell} (b_j+1)(2b_j+1) + 4\sum_{1\leqslant i< j\leqslant\ell} (b_i+1)(b_j+1) \biggr). \end{equation}

Lemma 7.4 Suppose that $n=2\ell +1$. Assume that $1< u< n$. Then

\[ \xi(F_{1u})=\frac{3!(2k-3)!}{(2k)!} \tilde C(1) F_{1u} \]

and $\tilde C(1)$ is equal to

\[ C(1)+ \frac{4}{3}b_1(b_1+1) + \frac{4}{3}b_1\sum_{1< i\leqslant\ell}(b_i+1)=C(1)+ \frac{4}{3} b_1 \sum_{1\leqslant j\leqslant\ell} (b_j+1). \]

Proof. We have to take care of the instances, where $j=\ell +1=j'$. Here $F_{jj}=0$, thereby also $\xi _1^{(j)}=0$. By a direct calculation, $\xi _2^{(j)}(F_{1u})=\frac {1}{3}F_{1u}$ for each $u$. Recall that $\xi _2^{(j)}$ corresponds to $Y_3=2(b_1+1)(b_j+1)F_{11}F_{1j}F_{j1}$ in (7.2) and that the additional scalar factor in this case is $2b_1$. Since $b_{\ell +1}=0$, we have to add $\frac {4}{3}b_1(b_1+1)$ to $C(1)$.

The calculations for $\xi _3^{i,j}, \xi _3^{j,i}$, and $\xi _4^{i,j}$ have to be altered. The modifications are:

\[ \xi_4^{i,j}(F_{1j})=F_{1j}, \quad \xi_3^{i,j}(F_{1j})=-\tfrac{1}{3} F_{1j}, \quad \xi_3^{j,i}(F_{1j})=-\tfrac{1}{3} F_{1j}, \]

and $\xi _5^{i,j}$ with $1< i< j=l+1$ has a simpler form, here

\[ \xi_5^{i,j}=\xi_3^{i,j}+\xi_3^{j,i}+\xi_3^{i',j}+\xi_3^{j,i'} + \xi_4^{i,j}+\xi_4^{i',j}. \]

The coefficient of this $\xi _5^{i,j}$ in $({(2k)!}/{3!(2k-3)!}) \xi$ is $2b_1(b_i+1)$ and $\xi _5^{i,j}(F_{1u})=\frac {2}{3} F_{1u}$ for all $u$. This justifies the second additional summand.

Proposition 7.5 For $\mathfrak g=\mathfrak {so}_{n}$, we have ${\sf m}(\Phi _{2k})=R(k) \Phi _{2k-2}$, where

\[ R(k)=\frac{1}{k(2k-1)} \bigg(\binom{n}{2} + 2n(k-1) + (k-1)(2k-3) \bigg). \]

Proof. According to Lemmas 7.3 and 7.4, there is $c(1)\in {{\mathbb {C}}}$ such that $\xi (F_{1u})=c(1)F_{1u}$ for each $1< u< n$. Since $\xi \in \mathfrak {so}(\mathfrak g)$, we have also $\xi (F_{u1})= -c(1) F_{u1}$. Taking into account Lemma 7.2, we conclude that $\xi =c(1)F_{11}$.

Simplifying (7.3) and using Lemma 7.4, we obtain that

\begin{align*} c(1)&=\frac{2}{3}b_1\frac{3!(2k-3)!}{(2k)!} \bigg(2\bigg(\sum_{j=1}^{\ell} b_j\bigg)^{2} + (4\ell-1)\bigg(\sum_{j=1}^{\ell} b_j\bigg)+\ell+ 2\ell(\ell-1)\bigg) \\ &= \frac{b_1}{k(2k-1)(k-1)} (2(k-1)^{2}+(4\ell-1)(k-1)+\ell(2\ell-1) ) \end{align*}

in type D and that

\[ c(1)= \frac{b_1}{k(2k-1)(k-1)} (2(k-1)^{2}+(4\ell-1)(k-1)+\ell(2\ell-1) + 2(k-1)+2\ell ) \]

in type B. In both cases, the scalars $c(1)/b_1$ depend only on $k$ and $\ell$. Making use of the action of $W(\mathfrak g,\mathfrak h)$, we can conclude now that ${\sf m}(\Phi _{2k})$ is a symmetric invariant and that it is equal to $R(k) \Phi _{2k-2}$ with $R(k)\in \mathbb {Q}$. More explicitly, $R(k)$ is equal to $2(k-1) ({c(1)}/{2b_1})= ({(k-1)c(1)}/{b_1})$.

In type D, we have $2(k-1)^{2}+(4\ell -1)(k-1) = 2n(k-1)+(k-1)(2k-3)$ and $\ell (2\ell -1)=\binom {n}{2}$. Quite similarly, in type B, we have $\ell (2\ell -1)+2\ell =\ell (2\ell +1)=\binom {n}{2}$ and

\[ 2(k-1)^{2}+(4\ell-1)(k-1) + 2(k-1)=2n(k-1) + (k-1)(2k-3). \]

Therefore multiplying $c(1)$ with $(k-1)/b_1$ we obtain the desired formula for $R(k)$.

We have

\begin{align*} R(k)&=\frac{1}{k(2k-1)} \bigg(\binom{n}{2} + 2n(k-1) + (k-1)(2k-3) \bigg) \\ &=\frac{1}{k(2k-1)}\bigg((k-1)(n+2k-3)+\frac{n}{2}(2k-2+n-1)\bigg)\\ &=\frac{(n+2k-3)(n+2k-2)}{2k(2k-1)} \end{align*}

and then

\[ \binom{2k}{2r} \prod_{u=0}^{r-1} R(k-u)=\binom{2k}{2r} \prod_{u=1}^{r} \frac{(n+2k-2u)(n+2k-2u-1)}{(2k-2u+2)(2k-2u+1)}=\binom{n+2k-2}{2r}. \]

Iterating the map ${\sf m}$, sf. (2.4) and Theorem 3.11, we obtain the following result.

Theorem 7.6 For any $k\geqslant 2$,

\[ S_k=\varpi(\Phi_{2k})[-1]+\sum_{1\leqslant r< k} \binom{n+2k-2}{2r} \varpi(\tau^{2r} \Phi_{2k-2r}[-1])\cdot 1 \]

is a Segal–Sugawara vector.

8. Applications and open questions

The Feigin–Frenkel centre can be used in order to construct commutative subalgebras of the enveloping algebra in finite-dimensional cases. There are two most remarkable instances.

8.1 Quantum Mishchenko–Fomenko subalgebras

Recall the construction from [Ryb06]. For any $\mu \in \mathfrak g^{*}$ and a non-zero $u\in \mathbb {C}$, the map

(8.1)\begin{equation} \varrho_{\mu,u}\!:{{\mathcal{U}}}(t^{-1}\mathfrak g[t^{-1}])\to {{\mathcal{U}}}(\mathfrak g), \quad x t^{d} \mapsto u^{d}x +\delta_{d,-1}\mu(x),\quad x\in\mathfrak g, \end{equation}

defines a $G_\mu$-equivariant algebra homomorphism. The image of $\mathfrak z(\hat {\mathfrak g})$ under $\varrho _{\mu,u}$ is a commutative subalgebra $\tilde {{\mathcal {A}}}_{\mu }$ of ${{\mathcal {U}}}(\mathfrak g)$, which does not depend on $u$ [Ryb06, FFTL10]. Moreover, $\mathrm {gr}(\tilde {{\mathcal {A}}}_\mu )$ contains the Mishchenko–Fomenko subalgebra ${{\mathcal {A}}}_\mu \subset {{\mathcal {S}}}(\mathfrak g)$ associated with $\mu$, which is generated by all $\mu$-shifts $\partial _\mu ^{m} H$ of the $\mathfrak g$-invariants $H\in {{\mathcal {S}}}(\mathfrak g)$. The main property of ${{\mathcal {A}}}_\mu$ is that it is Poisson-commutative, i.e. $\{{{\mathcal {A}}}_\mu,{{\mathcal {A}}}_\mu \}=0$ [MF78]. If $\mu \in \mathfrak g^{*}\cong \mathfrak g$ is regular, i.e. if $\dim \mathfrak g_\mu =\mathrm {rk}\,\mathfrak g$, then ${{\mathcal {A}}}_\mu$ is a maximal with respect to inclusion Poisson-commutative subalgebra of ${{\mathcal {S}}}(\mathfrak g)$ [PY08] and hence $\mathrm {gr}(\tilde {{\mathcal {A}}}_\mu )={{\mathcal {A}}}_\mu$. Several important properties and applications of (quantum) MF-subalgebras are discussed e.g. in [Vin91, FFR10].

Mishchenko–Fomenko subalgebras were introduced in [MF78], before the appearance of the Feigin–Frenkel centre. In [Vin91], Vinberg posed a problem of finding a quantisation of ${{\mathcal {A}}}_\mu$. A natural idea is to look for a solution given by the symmetrisation map $\varpi$. For $\mathfrak g=\mathfrak {gl}_n$, the elements $\varpi (\partial _\mu ^{m}\Delta _k)\in {{\mathcal {U}}}(\mathfrak g)$ with $1\leqslant k\leqslant n$, $0\leqslant m< k$ commute and therefore produce a solution to Vinberg's quantisation problem [Tar00, FM15, MY19].

Consider ${\mathcal {F}}[\bar a]=\varpi (F)[\bar a]\in {{\mathcal {U}}}({{\hat {\mathfrak g}}}^-)$ corresponding to $F\in {{\mathcal {S}}}^{m}(\mathfrak g)^{\mathfrak g}$ in the sense of (0.3). Set $p=|\{i \mid a_i=-1\}|$. Then

(8.2)\begin{equation} \langle \varrho_{\mu,u} ({\mathcal{F}}[\bar a]) \mid u\in{{\mathbb{C}}}{\setminus}\{0\} \rangle _{{{\mathbb{C}}}} = \langle \varpi(\partial^{l}_\mu F) \mid 0\leqslant l\leqslant p\rangle _{{{\mathbb{C}}}}. \end{equation}

Combining (8.2) with (2.2), we conclude immediately that for $\mathfrak g=\mathfrak {gl}_n$, the algebra $\tilde {{\mathcal {A}}}_\mu$ is generated by $\varpi (\partial _\mu ^{m}\Delta _k)$. This observation is not new, see [MY19, § 3] and in particular § 3.2 there for a historical overview and a more elaborated proof.

In [MY19, § 3.3], sets of generators $\{H_i\mid 1\leqslant i\leqslant \ell \}$ of ${{\mathcal {S}}}(\mathfrak g)^{\mathfrak g}$ such that

(8.3)\begin{equation} \tilde{{\mathcal{A}}}_\mu={\sf alg}\langle \varpi(\partial_\mu^{m} H_i) \mid 1\leqslant i\leqslant \ell, \ 0\leqslant m<\deg H_i\rangle \end{equation}

are exhibited in types $\mathsf{B}$, $\mathsf{C}$, and $\mathsf{D}$. We rejoice to say that in type $\mathsf{C}$, $H_k=\Delta _{2k}$ in the notation of § 4. In the even orthogonal case, the set $\{H_i\}$ includes the Pfaffian. The other generators in types $\mathsf{B}$ and $\mathsf{D}$ are $\Phi _{2k}$ of § 7. Thus Theorems 4.4 and 7.6 provide a new proof of [MY19, Theorem 3.2].

In conclusion, we show that Proposition 5.6 confirms Conjecture 3.3 of [MY19] in type $\mathsf{G}_2$.

Proposition 8.1 Let $\mathfrak g$ be a simple Lie algebra of type $\mathsf{G}_2$. Let $\tilde H\in {{\mathcal {S}}}^{6}(\mathfrak g)$ be a $\mathfrak g$-invariant satisfying (0.2), cf. Corollaries 5.7 and 6.4. Then $\tilde {{\mathcal {A}}}_\mu$ is generated by $\mu, {\mathcal {H}}$, and $\varpi (\partial _\mu ^{m} \tilde H)$ with $0\leqslant m\leqslant 5$.

Proof. We work with $\mu$ as with an element of $\mathfrak g$. Clearly $\partial _\mu {\mathcal {H}}=2 \sum (x_i,\mu ) x_i=2\mu$ and

\[ \varrho_{\mu,u}({\mathcal{H}}[b_1,b_2])=u^{b_1+b_2}{\mathcal{H}}+(u^{b_2}\delta_{b_1,-1}+u^{b_1}\delta_{b_2,-1}) \mu + \delta_{b_1,-1}\delta_{b_2,-1} (\mu,\mu). \]

Let $S_2$ be the Segal–Sugawara vector provided by (5.1) and (6.3). Set also $S_1={\mathcal {H}}[-1]$. Then $\{S_1,S_2\}$ is a complete set of Segal–Sugawara vectors. A general observation is that $\tilde {{\mathcal {A}}}_\mu$ is generated by $\{\varrho _{\mu,u}(S_\nu ) \mid u\in {{\mathbb {C}}}^{{}^\times}, \nu =1,2\}$ [Mol18, Corollary 9.2.3]. We have already computed the images of $S_1$ and also of $\varpi (\tau ^{4}{\mathcal {H}}[-1]){\cdot }1$. Similarly to (8.2), the images of $\varpi (\tilde H)[-1]$ span $\langle \varpi (\partial _\mu ^{m} H)\rangle _{{{\mathbb {C}}}}$. It remains to deal with

\[ {\mathcal{Y}}=\varrho_{\mu,u}(\varpi(\tau^{2} {\mathcal{H}}^{2}[-1]){\cdot}1) = u^{-6} {\mathcal{Y}}_4+u^{-5} {\mathcal{Y}}_3 + u^{-4} {\mathcal{Y}}_2 + u^{-3} {\mathcal{Y}}_1. \]

Here ${\mathcal {Y}}_1$ is proportional to $\mu$; the term ${\mathcal {Y}}_2$ is a linear combination of ${\mathcal {H}}$ and $\mu ^{2}$. Furthermore, ${\mathcal {Y}}_3$ is a linear combination of $\mu {\mathcal {H}}$ and $\sum _{i} x_i\mu x_i$, therefore of $\mu {\mathcal {H}}$ and $\mu$, cf. Lemma 5.4. Finally, ${\mathcal {Y}}_4$ is a linear combination of ${\mathcal {H}}^{2}$ and $\sum _{i,j} x_i x_jx_j x_i=\sum _{i} x_i {\mathcal {H}} x_i={\mathcal {H}}^{2}$,

\[ \sum_{i,j} x_ix_j x_i x_j = {\mathcal{H}}^{2} + \sum_{j} c_1 x_j x_j = {\mathcal{H}}^{2}+c_1{\mathcal{H}}. \]

This completes the proof.

8.2 Gaudin algebras

Recall that elements $S\in \mathfrak z({{\hat {\mathfrak g}}})$ give rise to higher Hamiltonians of the Gaudin models, which describe completely integrable quantum spin chains [FFR94].

The underlying space of a Gaudin model is the direct sum of $n$-copies of $\mathfrak g$, and the Hamiltonians are the following sums

\[ {\mathcal{H}}_k=\sum_{j\ne k} \frac{\sum_i x_i^{(k)} x_i^{(j)}}{z_k-z_j} , \quad 1\leqslant k\leqslant n, \]

where $z_1,\ldots,z_n$ are pairwise different complex numbers. Here $\{x_i^{(k)}\mid 1\leqslant i\leqslant \dim \mathfrak g\}$ is an orthonormal basis for the $k$’th copy of $\mathfrak g$. These Gaudin Hamiltonians can be regarded as elements of ${{\mathcal {U}}}(\mathfrak g)^{\otimes n}$ or of ${{\mathcal {S}}}(\mathfrak g\oplus \cdots \oplus \mathfrak g)$. They commute (and hence Poisson-commute) with each other. Higher Gaudin Hamiltonians are elements of ${{\mathcal {U}}}(\mathfrak g)^{\otimes n}$ that commute with all ${\mathcal {H}}_k$.

The construction of [FFR94] produces a Gaudin subalgebra ${\mathcal {G}}$, which consists of Gaudin Hamiltonians and contains ${\mathcal {H}}_k$ for each $k$. Let $\Delta {{\mathcal {U}}}({{\hat {\mathfrak g}}}^-)\cong {{\mathcal {U}}}({{\hat {\mathfrak g}}}^-)$ be the diagonal of ${{\mathcal {U}}}({{\hat {\mathfrak g}}}^-)^{\otimes n}$. Then a vector $\bar z=(z_1,\ldots,z_n)\in ({{\mathbb {C}}}^{{}^\times}\!)^{n}$ defines a natural homomorphism $\rho _{\bar z}\!:\Delta {{\mathcal {U}}}({{\hat {\mathfrak g}}}^-) \to {{\mathcal {U}}}(\mathfrak g)^{\otimes n}$. In this notation, ${\mathcal {G}}={\mathcal {G}}(\bar z)$ is the image of $\mathfrak z({{\hat {\mathfrak g}}})$ under $\rho _{\bar z}$. By the construction, ${\mathcal {G}}\subset ({{\mathcal {U}}}(\mathfrak g)^{\otimes n})^{\mathfrak g}$. Since $\mathfrak z({{\hat {\mathfrak g}}})$ is homogeneous in $t$, it is clear that ${{\mathcal {G}}(\bar z)}={{\mathcal {G}}}(c\bar z)$ for any non-zero complex number $c$.

Gaudin subalgebras have attracted a great deal of attention, see e.g. [CFR10] and references therein. They are closely related to quantum Mishchenko–Fomenko subalgebras and share some of their properties. In particular, for a generic $\bar z$, the action of ${{\mathcal {G}}(\bar z)}$ on an irreducible finite-dimensional $(\mathfrak g \oplus \cdots \oplus \mathfrak g)$-module $V(\lambda _1)\otimes \cdots \otimes V(\lambda _n)$ is diagonalisable and has a simple spectrum on the subspace of highest weight vectors of the diagonal $\mathfrak g$ [Ryb20]. Applying $\rho _{\bar z}$, one obtains explicit formulas for higher Gaudin Hamiltonians from explicit formulas for the generators of $\mathfrak z({{\hat {\mathfrak g}}})$. In the following, we discuss which generators of $\mathfrak z({{\hat {\mathfrak g}}})$ one has to consider.

Let $\{S_1,\ldots,S_\ell \}$ with $\mathrm {gr}(S_k)=H_k[-1]$ be a complete set of Segal–Sugawara vectors as in § 1.6. Set $\deg H_k=:d_k$. Assume that $z_k\ne z_j$ for $k\ne j$ and that $z_k\ne 0$ for all $k$. According to [CFR10, Proposition 1], ${\mathcal {G}}(\bar z)$ has a set of algebraically independent generators $\{F_1,\ldots,F_{{{\boldsymbol B}}(n)}\}$, where ${{\boldsymbol B}}(n):= ({n-1}/{2})(\dim \mathfrak g+\ell )+\ell$, see also [Ryb06, Theorems 2&3]. Moreover, exactly $(n-1)d_k +1$ elements among the $F_j$ belong to $\langle \rho _{\bar z}(\tau ^{m}(S_k)) \mid m\geqslant 0\rangle _{{{\mathbb {C}}}}$ [CFR10, Proposition 1]. Note that $\sum _{k=1}^{\ell } d_k$ equals $(\dim \mathfrak g+\ell )/2$. Furthermore, the symbols $\mathrm {gr}(F_j)$ are algebraically independent as well and $\deg \mathrm {gr}(F_j)=d_k$ if $F_j\in \langle \rho _{\bar z}(\tau ^{m}(S_k)) \mid m\geqslant 0\rangle _{{{\mathbb {C}}}}$, see the proof of [CFR10, Proposition 1(2)] and [Ryb06, § 4].

Remark 8.2 If ${{\mathcal {A}}}\subset ({{\mathcal {S}}}(\mathfrak g)^{\otimes n})^{\mathfrak g}$ is a Poisson-commutative algebra, then ${\rm tr.deg}\,{{\mathcal {A}}}\leqslant {{\boldsymbol B}}(n)$ by [MY19, Proposition 1.1]. Combining this with [BK76, Satz 5.7], we obtain that ${\rm tr.deg}\,\tilde {{\mathcal {A}}}\leqslant {{\boldsymbol B}}(n)$ for a commutative subalgebra $\tilde {{\mathcal {A}}}\subset ({{\mathcal {U}}}(\mathfrak g)^{\otimes n})^{\mathfrak g}$. Thus ${{\mathcal {G}}}$ has the maximal possible transcendence degree. Arguing in the spirit of [PY08] and using the results of [Ryb06], one can show that ${{\mathcal {G}}}$ is also a maximal commutative subalgebra of $({{\mathcal {U}}}(\mathfrak g)^{\otimes n})^{\mathfrak g}$ with respect to inclusion.

In the case $n=2$, the application of our result looks particularly nice. Besides, this two points case has several features. Suppose that $n=2$. Set $\mathfrak l=\mathfrak g\oplus \mathfrak g$. For $H\in {{\mathcal {S}}}^{d}(\mathfrak l)$, $\xi ^{(1)}$ in the first copy of $\mathfrak g$, $\eta ^{(2)}$ in the second, and a non-zero $c\in {{\mathbb {C}}}$, write

(8.4)\begin{align} H(\xi^{(1)}+c\eta^{(2)})&=H_{d,0}(\xi^{(1)})+cH_{d-1,1}(\xi^{(1)},\eta^{(2)})+\cdots +c^{d-1}H_{1,d-1}(\xi^{(1)},\eta^{(2)})\nonumber\\ &\quad +c^{d}H_{0,d}(\eta^{(2)}). \end{align}

Here $H_{d,0}$ belongs to the symmetric algebra of the first copy of $\mathfrak g$. The symbol of $\rho _{\bar z}(\tau ^{m}(S_k))$ lies in $\langle (H_k)_{d_k-j,j} \mid 0\leqslant j\leqslant d_k\rangle _{{{\mathbb {C}}}}$. Since we must have $d_k+1$ linearly independent elements among these symbols, $\mathrm {gr}({{\mathcal {G}}})$ is freely generated by $(H_k)_{d_k-j,j}$ with $1\leqslant k\leqslant \ell$, $0\leqslant j\leqslant d_k$.

The Lie algebra $\mathfrak l$ has the following symmetric decomposition

(8.5)\begin{equation} \mathfrak l= \mathfrak l_0 \oplus \mathfrak l_1, \quad \text{where } \mathfrak l_1=\{(\xi,-\xi)\mid\xi\in\mathfrak g\} \end{equation}

and $\mathfrak l_0=\Delta \mathfrak g=\{(\xi,\xi )\mid \xi \in \mathfrak g\}$ is the diagonal. Similarly to (8.4), one polarises $H\in {{\mathcal {S}}}^{d}(\mathfrak l)$ with respect to the decomposition (8.5). Let $H_{[j,d-j]}$ with $0\leqslant j\leqslant d$ be the arising components. Then $\langle H_{j,d-j} \mid 0\leqslant j\leqslant d\rangle _{{{\mathbb {C}}}}= \langle H_{[j,d-j]} \mid 0\leqslant j\leqslant d\rangle _{{{\mathbb {C}}}}$.

On the one side, the polynomials $(H_k)_{j,d_k-j}$ generate $\mathrm {gr}({{\mathcal {G}}})$, on the other, the polynomials $(H_k)_{[j,d_k-j]}$ generate a Poisson-commutative subalgebra ${\mathcal {Z}} \subset {{\mathcal {S}}}(\mathfrak l)$ related to the symmetric pair $(\mathfrak l,\mathfrak l_0)$, which has many interesting properties [PY21]. Thus, our discussion results in the following statement.

Corollary 8.3 (cf. [PY21, Example 6.5])

The two-points Gaudin subalgebra ${{\mathcal {G}}}(z_1,z_2)$ is a quantisation of the Poisson-commutative subalgebra ${\mathcal {Z}}$ associated with the symmetric pair $(\mathfrak g \oplus \mathfrak g,\Delta \mathfrak g)$.

Let us give more information on the issue of Corollary 8.3. Observe that ${{\mathcal {G}}}(z_1,z_2)={{\mathcal {G}}}(z_1-b,z_2-b)$ if $b\in {{\mathbb {C}}}{\setminus} \{z_1,z_2\}$, see [CFR10, Proposition 1]. Hence

\[ {{\mathcal {G}}(\bar z)}={{\mathcal {G}}}\biggl(\frac {z_1-z_2}{2}, \frac {z_2-z_1}{2}\biggr)={{\mathcal {G}}}(1,-1). \]

For $\rho =\rho _{1,-1}$, we have $\rho (\xi t^{k})=\xi ^{(1)}+(-1)^{k} \xi ^{(2)}$, i.e. $\mathfrak g[-1]$, as well as any $\mathfrak g[-2k-1]$, is mapped into $\mathfrak l_1$ and each $\mathfrak g[-2k]$ is mapped into $\Delta \mathfrak g$. One can understand $\rho$ as the map from ${{\mathcal {U}}}({{\hat {\mathfrak g}}}^-)$ to ${{\mathcal {U}}}({{\hat {\mathfrak g}}}^-)/(t^{2}-1)\cong {{\mathcal {U}}}(\mathfrak l)$.

It is not difficult to see that $\mathrm {gr}(\rho (S_k))=(H_k)_{[0,d_k]}$, $\mathrm {gr}(\rho (\tau (S_k)))=(H_k)_{[1,d_k-1]}$, and in general

\[ \mathrm{gr}(\rho(\tau^{m}(S_k)))\in m!(H_k)_{[m,d_k-m]} + \langle (H_k)_{[j,d_k-j]}\mid j< m\rangle _{{{\mathbb{C}}}} \]

as long as $m\leqslant d_k$. This shows that indeed $\mathrm {gr}({{\mathcal {G}}})={\mathcal {Z}}$ and that

(8.6)\begin{equation} {{\mathcal{G}}}={\sf alg}\langle \rho(\tau^{m}(S_k))\mid 1\leqslant k\leqslant\ell, 0\leqslant m \leqslant d_k\rangle . \end{equation}

Suppose that $H_1,\ldots,H_\ell$ are homogeneous generators of ${{\mathcal {S}}}(\mathfrak g)^{\mathfrak g}$ and for each $k$ there is $k'$ such that $0\leqslant k'< k$ and ${\sf m}(H_{k})\in {{\mathbb {C}}}H_{k'}$, where $H_0=0$. If $\mathfrak g$ is simple and classical, then explicit descriptions of such sets are contained in §§ 2, 4, 7.

Theorem 8.4 If we keep the above assumption on the set $\{H_k\}$, then the two-points Gaudin subalgebra ${{\mathcal {G}}}\subset {{\mathcal {U}}}(\mathfrak l)$ is generated by $\varpi ((H_k)_{d_k-j,j})$ with $1\leqslant k\leqslant \ell$, $0\leqslant j\leqslant d_k$.

Proof. For each $k$, let $S_k$ be the Segal–Sugawara vector obtained from $H_k$ by (2.4). We will show that $V_{\rm sym}:=\langle \varpi ((H_k)_{d_k-j,j}) \mid 1\leqslant k\leqslant \ell, \, 0\leqslant j\leqslant d_k\rangle _{{{\mathbb {C}}}}$ is equal to

\[ V_{\mathcal{G}}:= \langle \rho(\tau^m(S_k))\mid 1\le k\le\ell, \, 0\le m \le d_k\rangle _{\mathbb{C}}.\]

Since $\dim V_{{{\mathcal {G}}}} =\dim V_{\rm sym}$, it suffices to prove the inclusion $V_{{{\mathcal {G}}}}\subset V_{\rm sym}$. We argue by induction on $k$. If $k=1$, then $S_1=\varpi (H_1[-1])$. Hence $\langle \rho (\tau ^{m}(S_1))\mid 0 \leqslant m \leqslant d_1\rangle _{{{\mathbb {C}}}}$ is equal to

\[ \langle \varpi ((H_1)_{d_1-j,j}) \mid 0 \leqslant j\leqslant d_1\rangle _{{{\mathbb {C}}}}.\]

If $k\geqslant 2$ and $m\leqslant d_k$, then according to the structure of (2.4) and our condition on $\{H_1,\ldots,H_\ell \}$, we have

\[ \rho(\tau^{m}(S_k))\in m!\varpi((H_k)_{[m,d_k-m]}) + V_{m,k}, \]

where

\[ V_{m,k}= \langle \varpi((H_k)_{[j,d_k-j]})\mid j< m\rangle _{{{\mathbb{C}}}} \oplus \langle \varpi((H_{k'})_{j,d_{k'}-j})\mid k'< k, \, 0\leqslant j< d_{k'}\rangle _{{{\mathbb{C}}}}. \]

Thus $\rho (\tau ^{m}(S_k))\in V_{\rm sym}$ and we are done.

8.3 Further directions

For all classical types, we find families of generators $\{H_k\}$ that behave well in terms of (0.2). The general picture is not complete yet, since the following question remains open.

Question 8.5 Does any exceptional Lie algebra $\mathfrak g$ poses a set of generators $\{H_k\}\subset {{\mathcal {S}}}(\mathfrak g)^{\mathfrak g}$ such that each $H_k$ satisfies (0.2)?

Proposition 5.6 takes care of type ${\sf G}_2$. We have seen also some partial positive answers in other types.

Question 8.6 Are there homogeneous generators $\{H_k\}$ of ${{\mathcal {S}}}(\mathfrak g)^{\mathfrak g}$ such that ${\sf m}(H_k)=0$ for each $k$?

The calculations in § 6 prove that in type G$_2$, the answer is negative. I would expect that the answer is negative in general.

As Example 3.7 shows, a set of generators $\{H_k\}$, where each $H_k$ satisfies (0.2), is not unique. For the classical Lie algebras, there is a freedom of choice in degree $4$ and there is also some freedom in degree $6$.

It is quite possible that the condition (8.3) on the set $\{H_k\}$ is less restrictive than (0.2). However, we have no convincing evidence to this point.

Remark 8.7 Probably there are some intricate combinatorial identities hidden in (2.4). In order to reveal them, one has to understand the natural numbers $c(r,\bar a)$ appearing in the proof of Lemma 2.2, the rational constants $c_{2,3}(j,p)$ of Lemma 3.2, as well as the scalars $C(\bar a^{(r)},\bar \gamma )$ of Proposition 3.4.