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The Riemann-Hilbert correspondence, as proved by Kashiwara and Mebkhout, says that for X a smooth algebraic variety over $\mathbb{C}$ there is an equivalence of triangulated categories

$D^b_c(X,\mathbb{C})\cong D^b_\mathrm{rh}(\mathcal{D}_X)$

between the bounded derived category of complexes of $\mathbb{C}$-modules on $X$ with constructible cohomology sheaves, and the bounded derived category of complexes of coherent $\mathcal{D}_X$-modules with regular holonomic cohomology sheaves.

Moreover, this equivalence respects the 6 operations $f^* , \mathbf{R}f_* , f^!, \mathbf{R}f_!, \boxtimes, \mathbb{D} $ of usual and extraordinary direct and inverse images, exterior tensor product and duality.

$\mathbf{Question:}$ Does the Riemann-Hilbert correspondence also preserve the interior tensor product?

On the constructible side, the interior tensor product is $\Delta_X^*(-\boxtimes-) $ where $\Delta_X$ is the diagonal immersion, but on the holonomic side the interior tensor product is $\Delta^!_X(-\boxtimes-)[d_X]$. So its seems to a novice like me that we're getting different operations. Or is there some comparison between $f^!$ and $f^*$ for a closed immersion $f$ that I've missed?

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up vote 6 down vote accepted

This is correct. Verdier duality does not preserve tensor products in general. Another point of view is that each of these categories has two versions of tensor product, $\otimes ^\ast$ and $\otimes ^!$, which are interchanged by Verdier duality. It just happens that for $D$-modules the shriek version (or a shift of it) is most natural to define, whereas for constructible sheaves the $\ast$-version is more natural.

In Bernstein's notes on $D$-modules on page 28, he talks about the ``!-tensor product'', and remarks that it is just a shift of the naive tensor product of $D$-modules (over $\mathcal O$).

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By the way, there continues to be an online link to Bernstein's notes: math.columbia.edu/~khovanov/resources/Bernstein-dmod.pdf –  Jim Humphreys Dec 1 '12 at 14:32

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