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Let $n>1$ be a positive integer and let $A$ be an abelian variety over $\mathbb{C}$. Then the symmetric product $S^n(A)$ is a normal projective variety over $\mathbb{C}$ with Kodaira dimension zero (see for instance https://arxiv.org/pdf/math/0006107.pdf).

Let $A(n)\to S^n(A)$ be a resolution of singularities. Then, up to finite etale cover, $A(n)$ is a product of hyperkaehler varieties, an abelian variety, and simply connected strict Calabi-Yau varieties. (This should follow from the Beauville-Bogomolov decomposition theorem. Or does this require an additional hypothesis on $A(n)$.)

I am wondering how the decomposition of $A(n)$ looks like as $n$ grows. Is it always a strict Calabi-Yau variety? Could it be that $A(n)$ is an abelian variety in fact?

I am looking for examples and would appreciate any comments.

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When $\dim A = 1$, $S^nA$ is a $\mathbb{P}^{n-1}$-bundle over $A$, so its Kodaira dimension is $-\infty$.

When $\dim A = 2$, the minimal resolution of $S^nA$ is given by the Hilbert scheme $A^{[n]}$, there is a natural map $$ A^{[n]} \to A $$ (summation of points), which is smooth with fiber $K_{n-1}A$, so-called higher Kummer variety, which is hyperkahler.

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You need for the dimension of $A$ to be even in order for the canonical sheaf on the quotient to be trivial (so that the resolution has a chance to be $K$-trivial). For $\dim A =2$, the story is as Sasha described. For $\dim A $ even and at least 4, there will not exist a crepant resolution. You will encounter singularities which locally look like $\mathbb{C}^{2d}/\{\pm 1\}$ and for $d>1$ these do not admit crepant resolutions.

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