Tensor product of $\mathcal{D}$-modules and constructible sheaves

Question

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?

Answer 1

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 $\mathcal O$).