Recall that a spanning surface for a link $L$ is by assumption a connected unoriented surface with boundary equal to $L$
 is by assumption a connected unoriented surface with boundary equal to $L$ .
.
 Theorem 1.1 and Corollary 4.9 from the paper are incorrect as stated. For example, one can construct counterexamples to Theorem 1.1 using links $L$ contained in 3-balls, so-called local links. Let $L \subset B^{3}$
 contained in 3-balls, so-called local links. Let $L \subset B^{3}$ be a link with an alternating projection on $S^{2} = \partial B^{3}.$
 be a link with an alternating projection on $S^{2} = \partial B^{3}.$ Under inclusion $B^{3} \subset \Sigma \times I$
 Under inclusion $B^{3} \subset \Sigma \times I$ , we obtain a local link $L$
, we obtain a local link $L$ in $\Sigma \times I$
 in $\Sigma \times I$ which bounds definite spanning surfaces of opposite sign. However, if the genus $g(\Sigma )>0$
 which bounds definite spanning surfaces of opposite sign. However, if the genus $g(\Sigma )>0$ , then $L$
, then $L$ does not have minimal genus.
 does not have minimal genus.
 To correct for this issue, we need to add the assumption that $L$ is not a local link in the case $g(\Sigma )>0$
 is not a local link in the case $g(\Sigma )>0$ . The corrected statement of the theorem is as follows.
. The corrected statement of the theorem is as follows.
Theorem 1.1. Let $L$ be a link in $\Sigma \times I$
 be a link in $\Sigma \times I$ , and assume that $L$
, and assume that $L$ bounds a positive definite spanning surface and a negative definite spanning surface. Then $L \subset \Sigma \times I$
 bounds a positive definite spanning surface and a negative definite spanning surface. Then $L \subset \Sigma \times I$ is a non-split alternating link which either has minimal genus or is contained in a 3-ball.
 is a non-split alternating link which either has minimal genus or is contained in a 3-ball.
 A few remarks on the proof are in order. For $g(\Sigma ) = 0$ , the proof is the same as before. For $g(\Sigma ) \geq 1$
, the proof is the same as before. For $g(\Sigma ) \geq 1$ , then arguing as before, we see that $P$
, then arguing as before, we see that $P$ and $N$
 and $N$ are not $S^{*}$
 are not $S^{*}$ -equivalent, unless the core surface $S$
-equivalent, unless the core surface $S$ of $\nu (P \cup N)$
 of $\nu (P \cup N)$ is a 2-sphere. In the latter case, $L$
 is a 2-sphere. In the latter case, $L$ is contained in a 3-ball, since $\Sigma \times I$
 is contained in a 3-ball, since $\Sigma \times I$ is irreducible, and $L$
 is irreducible, and $L$ has a connected alternating diagram on $S,$
 has a connected alternating diagram on $S,$ implying that $L$
 implying that $L$ is non-split and has an alternating diagram on $\Sigma$
 is non-split and has an alternating diagram on $\Sigma$ .
.
 Otherwise, assuming that $P$ and $N$
 and $N$ are not $S^{*}$
 are not $S^{*}$ -equivalent, then the argument goes through as before.
-equivalent, then the argument goes through as before.
Below is a corrected statement of the corollary.
Corollary 4.9. A link $L \subset \Sigma \times I$ in a thickened surface of positive genus is alternating and has minimal genus if and only if $L$
 in a thickened surface of positive genus is alternating and has minimal genus if and only if $L$ bounds definite spanning surfaces of opposite sign and is not contained in a 3-ball.
 bounds definite spanning surfaces of opposite sign and is not contained in a 3-ball.
Acknowledgements
We would like to thank the referee for bringing these issues to our attention.
 
 