Let $S$ and $T$ be symmetric unbounded operators. Denote by $\overline{S+T}$ the closure of the symmetric operator $S+T$. In general, the deficiency indices of $\overline{S+T}$ are not determined by the deficiency indices of $S$ and $T$. The paper studies some sufficient conditions for the stability of the deficiency indices of a symmetric operator $S$ under self-adjoint perturbations $T$. One can associate with $S$ the largest closed $^*$-derivation $\delta_{S}$ implemented by $S$. We prove that if the unitary operators $\exp(\ri tT)$, for $t\in\mathbb{R}$, belong to the domain of $\delta_{S}$ and $\delta_{S}(\exp(\ri tT))\rightarrow0$ in the strong operator topology as $t\rightarrow0$, then the deficiency indices of $S$ and $\overline{S+T}$ coincide. In particular, this holds if $S$ and $\exp(\ri tT)$ commute or satisfy the infinitesimal Weyl relation.

We also study the case when $S$ and $T$ anticommute: $\exp(-\ri tT)S\subseteq S\exp(\ri tT)$, for $t\in\mathbb{R}$. We show that if the deficiency indices of $S$ are equal, or if the group $\{\exp(\ri tT):t\in\mathbb{R}\}$ of unitary operators has no stationary points in the deficiency space of $S$, then $S$ has a self-adjoint extension which anticommutes with $T$, the operator $S+T$ is closed and the deficiency indices of $S$ and $S+T$ coincide.

AMS 2000 Mathematics subject classification: Primary 47B25