2
$\begingroup$

I am learning D-modules recently, and my question might be technical. It arises from Lemma 2.6.13 in Hotta-Takeuchi-Tanisaki's book, which states that there exists a canonical isomorphism $$ R\mathcal {H}om_{D_X}(M^\cdot, D_X) \otimes^L_{D_X} N^\cdot \xrightarrow{\cong} R\mathcal {H}om_{D_X}(M^\cdot, N^\cdot) $$ where $M^\cdot$ and $N^\cdot$ are coherent complexes of $D_X$-modules. And I get stuck with its proof: how to find such a canonical morphism?

I tried to find this in Kashiwara-Schapira's book "Sheaves on Manifolds". What I found was the following: (page 112, eq.(2.6.11)) let $\mathcal S\to \mathcal R$ be a morphism of sheaves of rings on $X$ whose image is contained in the center of $\mathcal R$ and $\mathcal S$ is commutative, then we have $$ R\mathcal {H}om_{\mathcal R}(F^\cdot, G^\cdot) \otimes^L_{\mathcal S} H^\cdot \xrightarrow{\cong} R\mathcal {H}om_{\mathcal R}(F^\cdot, G^\cdot\otimes^L_{\mathcal S} H^\cdot) $$ which is very similar to the above but cannot really imply it, since $D_X$ is not commutative. So, how to fix this gap? Moreover, I will appreciate it if you have any idea in understanding the above canonical isomorphism.

Improve this question
$\endgroup$
3
  • 1
    $\begingroup$ Have you tried to make such a morphism in the non-derived setting? Hint: if $M, N$ are complexes of left $D$-modules, do you see how to make a map from $\text{Hom}(M, D)\otimes N\to \text{Hom}(M, N)$? $\endgroup$
    – Daniel Litt
    Commented Apr 1, 2018 at 0:50
  • $\begingroup$ You may have a canonical morphism $\mathcal {H}om_{\mathcal R}(F,G)\otimes_{\mathcal S}H \to \mathcal H om_{\mathcal R}(F, G\otimes_{\mathcal S} H)$ (see Corollary 2.2.10 in the second book I mentioned above) $\endgroup$
    – Hang
    Commented Apr 1, 2018 at 1:02
  • 1
    $\begingroup$ It might be helpful to re-write $N^\bullet$ as $Hom_{D_X}(D_X,N^\bullet)$, with the module structure comming from $D_X$ being a bi-module. Then I guess the pairing is more obvious to see. $\endgroup$
    – S. carmeli
    Commented Apr 1, 2018 at 6:31

1 Answer 1

Reset to default
3
$\begingroup$

A map of sheaves is an isomorphism if and only if it's an isomorphism on stalks. Thus, it's enough to check this for the stalk of these sheaves, and so on can just check that for a non-commutative ring $R$ and two complexes of modules, you have $RHom(M,R)\otimes^{L}_RN\cong RHom(M,N)$. Since modules have free resolutions, you can assume that $M\cong R^m$ and $N\cong R^n$ are free modules. Now the statement just says that $n\times m$ matrices are the tensor product of row vectors and column vectors over $R$.

Improve this answer
$\endgroup$
2
  • $\begingroup$ A remark: you need to assume that $M$ is a perfect complex to make $R\operatorname{Hom}(M,R)\otimes_R^{\mathbb L}N\simeq R\operatorname{Hom}(M,N)$ valid (see Lurie, Higher Algebra, Proposition 7.2.4.4, which says that this equivalent is true if and only if $M$ is a perfect complex). $\endgroup$
    – user20948
    Commented Jun 7, 2020 at 9:21
  • $\begingroup$ Fair point. Luckily, D-modules on a smooth variety are always perfect. $\endgroup$
    – Ben Webster ♦
    Commented Jun 8, 2020 at 14:06

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .