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We consider D-modules over the affine line $\mathbb{A}_{\mathbb{C}}^{1}$, i.e. modules of Weyl algebra $A_1(\mathbb{C})=\mathbb{C}[x,\partial]$. There is a set of microfunctions (ref. A Primer of Algebraic D-modules, S.C.Coutinho), denoted by $\mathcal{M}$, which is an $A_1(\mathbb{C})$-module. Let $\delta\in \mathcal{M}$ be the Delta microfunction, and the D-module generated by it can be identified to $\mathbb{C}[\partial]$.

Consider $\delta^{(m)}=\partial^{m} \delta$, and let $$ A_1(\mathbb{C}) \cdot \delta^{(m)} $$ be the D-modules generated by $\delta^{(m)}\in \mathcal{M}$. It's not hard to varify that $$ A_1(\mathbb{C})\cdot \delta^{(m)}\simeq A_1(\mathbb{C})/(x^{m},x\partial +m). $$

I have two questions about those D-modules.

  1. I find that $x\cdot \delta^{(m)}=x\partial^{m}\delta=(\partial^{m}x-m\partial^{m-1})\delta=-m \delta^{(m-1)}$ as $x\cdot \delta=0$. Therefore, I think there are isomorphisms $$ A_1(\mathbb{C}) \cdot \delta=A_1(\mathbb{C}) \cdot \delta^{(m)}. $$ But this seems unreasonable when considering the quotient modules.
  2. I want to know which perverse sheaf corresponds with D-module generated by $\delta$. Consider the resolution $$ A_1(\mathbb{C}) \xrightarrow{\cdot x} A_1(\mathbb{C}). $$ After derived tensoring the canonical sheaf, this complex becomes (after taking cohomology sheaf) $$ 0 \rightarrow \mathbb{C}. $$ Do I get the correct result? (Edit: This is wrong. It should be skyscraper sheaf as Sawin said in comment.)
Martin Tang
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    The $A_1(\mathbb C)$-module generated by $\delta$ is simple. So it is generated by any of its non-zero elements, in particular, $\delta^{(m)}$. – Friedrich Knop May 18 '23 at 03:47
  • @FriedrichKnop That's right. This module has multiplicty 1. – Martin Tang May 18 '23 at 04:56
  • I think you're supposed to get the complex $\mathcal O_{\mathbb A^1} \to_{\cdot x} \mathcal O_{\mathbb A^1}$, which is quasi-isomorphic to the skyscraper sheaf $\mathbb C$ at $0$. (Is that what your $\mathbb C$ was meant to denote)? – Will Sawin Jun 03 '23 at 14:10
  • @WillSawin Actually, I made a mistake here. Now I know that it is quasi-isomorphic to skyscraper sheaf, because the cokernel is the same as the cokernel of $\mathcal{O}(-1\cdot 0)\rightarrow \mathcal{O}$. – Martin Tang Jun 04 '23 at 12:16
  • @WillSwain, so your quasi-isomorphism lies at the basis to the answer to my question https://mathoverflow.net/questions/97361/explaining-mukai-fourier-transforms-physically, right? – Tom Copeland Jun 04 '23 at 18:44

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