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Let $X$ be a smooth algebraic variety over $\mathbb{C}$ and let $M^\bullet\in\mathsf{D}^b_\text{h}(\mathcal{D}_X)$ be a complex of D-modules with holonomic cohomologies. We define the support of $M^\bullet$, as usual, as $$\bigcup_{i\in \mathbb{Z}}\operatorname{Supp}(\mathscr{H}^i(M^\bullet)),$$ where $\operatorname{Supp}(\mathscr{H}^i(M^\bullet))$ has its usual interpretation. (The set of points in which the stalk is non-zero.)

Given a morphism of algebraic varieties $f:X\to Y$, we can form functors $f_+:\mathsf{D}^b_\text{h}(\mathcal{D}_X)\to \mathsf{D}^b_\text{h}(\mathcal{D}_Y)$ and $f^!:\mathsf{D}^b_\text{h}(\mathcal{D}_Y)\to \mathsf{D}^b_\text{h}(\mathcal{D}_X)$. The former is the usual direct image of left D-modules and the latter is the functor which, up to a shift, coincides with the derived inverse image of O-modules.

I wonder if it's true that $\operatorname{Supp}(f_+ M^\bullet)=f(\operatorname{Supp}(M^\bullet))$ and $\operatorname{Supp}(f^! P^\bullet)= f^{-1}(\operatorname{Supp}(P^\bullet))$.

By a remark in section VI.4 in Borel's et al book on algebraic D-modules, I know that $\operatorname{Supp}(f^! P^\bullet)\subset f^{-1}(\operatorname{Supp}(P^\bullet))$ always holds. Is there a counter-example for the equality? And what about the direct image? Do we always have $\operatorname{Supp}(f_+ M^\bullet)=f(\operatorname{Supp}(M^\bullet))$? If not, is one side always contained in the other?

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Regarding $f_+$: neither inclusion holds in general, as the following two examples show.

  • Let $j$ be the inclusion of $U:=\mathbb A^1 - \{0\}$ into $\mathbb A^1$. Then $j_+(\mathcal O_U)$ has support all of $\mathbb A^1$, which contains $U=j(Supp(\mathcal O_U))$ as a proper subset.

  • On the other hand, consider the projection $p: U \to pt$, and let $M$ be the $D$-module $\mathcal O_U z^{\lambda}$ corresponding to a rank 1 local
    system with non-trivial monodromy. Then $p_+(M) =0$, so its support is empty (which is a proper subset of $p(Supp(M))=pt$).


Here is a counter example for the equality in the $f^!$ case.

  • Let $i: \{0\} \hookrightarrow \mathbb A^1$, and let $N=j_+(\mathcal O_U)$, using the notation above. Then $i^!(N)=0$ so has empty support (whereas $i^{-1}(Supp(N))=\{0\}$).
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    $\begingroup$ But do we have $Supp(f_+M)\subset \overline{f(Supp(M))}$ in general? $\endgroup$
    – Doug Liu
    Jan 5, 2023 at 17:44

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