Let $X$ be an algebraic variety, $M \in Mod(\mathcal{D}\_X)$. I am studying the characteristic variety associated to this module, and I am trying to understand why all the different definitions coincide. Fix $F$ a good filtration of $M$, we define:


$Ch(M) = supp \left( \widetilde{gr^{F}(M)} := \mathcal{O}\_{T^{\*}X} \otimes\_{\pi^{-1}(\pi\_{\*}(\mathcal{O}\_{T^{\*}X}))} \pi^{-1}(gr^{F}M) \right)$


Being the $\mathcal{O}\_{T^{\*}X}$-module in the right-hand side coherent, we have: $Ch(M) = V(\mathcal{Ann}(\widetilde{gr^{F}(M)})$


I don't understand why for an affine open subset $U \subset X$ we have:


$Ch(M) \cap T^{\*}U = V \left( \; Ann\_{gr^FD\_U(U)}(gr^{F}M(U)) \; \right)$


The problem that I face is that: $\Gamma(U, \mathcal{Ann}( \widetilde{gr^{F}(M)}) = Ann\_{\mathcal{O}\_{T^{\*}U}(T^{\*}U)}(\Gamma(U, \widetilde{gr^{F}(M)})$, which, even though $\mathcal{O}\_{T^{\*}U}(T^{\*}U) = gr^{F}D\_{U}(U)$, I don't see why should equal $Ann\_{gr^FD\_U(U)}(gr^{F}M(U))$. If the last equality were true than the conclusion would follow:


$Ch(M) \cap T^{\*}U = V(\mathcal{Ann}(\widetilde{gr^{F}(M)} \rvert\_{T^{\*}U}) = V \left( \; Ann\_{gr^FD\_U(U)}(gr^{F}M(U)) \; \right)$


**EDIT:** To clarify, what I don't understand is why


$Ann\_{\mathcal{O}\_{T^{\*}U}(T^{\*}U)}(\Gamma(U, \widetilde{gr^{F}(M)}) = Ann\_{gr^FD\_U(U)}(gr^{F}M(U))$


It is easy to verify that the RHS is included in the LHS, but the reverse inclusion doesn't seem obvious to me because the sections of the tensor product are not the tensor product of the sections of the two sheaves.



Characteristic variety of a D-module
Although your cautionary statement about tensor products is true sometimes, in this case it’s not. This is because $\pi$ is *affine*, so over your affine open $U$, everything commutes with taking global sections. 


