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Given a smooth map of schemes $f:X\to Y$ of relative dimension $d$, then there is a natural isomorphism $f^!\simeq f^*[d](2d)$ (in any context where the six operations are defined; see Cesinski-Deglise).

If $f$ is a smooth map of Artin stacks I imagine the same is true (there is a notion of the six operations for Artin stacks: https://arxiv.org/abs/1211.5948).

Question: Let $f:X\to Y$ be a quasismooth map of Artin stacks (say also flat to rule out things like closed embeddings). Is it still true that $f^!\simeq f^*[d](2d)$?

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In general, I think this should fail for most non-smooth local complete intersections. For a specific example, let $X = \mathbb V(xy) \subset \mathbb A^2$ and $Y = *$. Then the stalk at the dualizing complex of $X$ at the origin is $\mathbb Q_l[1] \oplus \mathbb Q_l[2]^{\oplus 2}$, contradicting the local constancy of $f^! \mathbb Q_l = \omega_X$.

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  • $\begingroup$ Thanks. Should that be $\mathbf{Q}_\ell\oplus\mathbf{Q}_\ell[1]$? $\endgroup$
    – Pulcinella
    Commented Jun 22, 2020 at 16:37
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    $\begingroup$ @Meow I just did the calculation, and I think we were both wrong. It's $H_*(X,X-0) = \mathbb Q_l[1] \oplus \mathbb Q_l [2]^{\oplus 2}$. Comes from $\mathbf D i^* \omega_X = i^! \mathbb Z_X.$ How are you getting your answer? $\endgroup$
    – Phil Tosteson
    Commented Jun 22, 2020 at 18:45
  • $\begingroup$ Sorry, that was a guess not a computation on my part. Thanks! $\endgroup$
    – Pulcinella
    Commented Jun 22, 2020 at 19:27

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