Let $X$ be a smooth projective surface and $X^{(n)}:= X^n/\mathfrak{S}_n$ be the nth symmetric product of $X$.
When is the cotangent sheaf of $X^{(n)}$ reflexive?
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$\begingroup$ Welcome new contributor. For every integer $n\geq 2$, the sheaf $\Omega$ of relative differentials of $X^{(n)}$ is not reflexive. Let $p\in X$ be any point, and let $x,y\in \mathfrak{m}_p$ be generators of the ideal. In $X^n$, denote by $x_i,y_i$ the pullbacks of $x,y$ via the $i^{\text{th}}$ projection map to $X$. Consider the differential on $X^n$ that is the sum over $(i,j)$ with $1\leq i < j \leq n$ of $(y_j-y_i)d(x_j-x_i)$. This gives a local section of the reflexive hull that is not a section of $\Omega$. $\endgroup$– Jason StarrCommented Apr 27, 2022 at 16:06
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$\begingroup$ Just to say that more explicitly: for every $\mathfrak{S}_n$-invariant open subset $U$ of $X^n$, the sections of the reflexive hull of $\Omega$ on the corresponding open subset of $X^{(n)}$ equal the $\mathfrak{S}_n$-invariant differentials of $X^n$ on $U$, e.g. the one above. However, at least if $U$ is affine, the sections of $\Omega$ correspond to the linear combinations of differentials of $\mathfrak{S}_n$-invariant regular functions on $U$ that have coefficients that are $\mathfrak{S}_n$-invariant regular functions on $U$. The example above is not of this form. $\endgroup$– Jason StarrCommented Apr 27, 2022 at 20:52
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