Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Considering $f:X\to \mathbb{C}$, $g:X\to \mathbb{C}$ and $f\oplus g:(x,y)\mapsto f(x)+g(y)$. The Sebastiani-Thom isomorphism is an isomorphism $\Phi_{f\oplus g}(M\boxtimes N) = \Phi_{f}(M) \otimes \Phi_{g}(N)$ compatible with monodromies.

The original theorem was for constant coefficient $M = \mathbb{C}_X$, $N = \mathbb{C}_Y$. David Massey gave a proof for general constructible coefficients. Is there an algebraic proof for D-modules?

All proofs use topological arguments that don't seem to translate. In his article "On microlocal b-funtions" Saito mentions a result to be published but I couldn't find it.

share|cite|improve this question
BTW, it would be interesting to see Deligne's original proof, which was apparently never published. (Reference for Delignes's proof is Demazure, Sem. Bourbaki no. 443, p.7; a similar proof technique: Scholl "Vanishing Cycles ...", Inventiones 124) – Thomas Riepe Dec 13 '09 at 13:05
I don't know of a reference beyond the work of Saito that you know of. In one of their recent preprints "Cohomological Hall algebra, exponential Hodge structures and motivic Donaldson-Thomas invariants" Kontsevich and Soibelman discuss the vanishing cycle functor, and note in Section 7.4 (page 97) that there is a Thom-Sebastiani theorem in the mixed Hodge module case referring to a paper of Saito "Thom-Sebastiani theorem for Hodge modules" which is listed as a 2010 Kyoto University preprint. – Kevin McGerty Mar 1 '11 at 2:02
Couldn't find the preprint on the web (I tried the Department of Math server and the RIMS preprint server). – AFK Mar 1 '11 at 18:12
I believe this fails for the case $X = \mathbb C$, $f=g=id$, when $M$ and $N$ have non-regular singularities, e.g. already at the kernel of the differential operator $\left(\frac{d}{x} - \frac{1}{x^2}\right)$. I know it fails for perverse sheaves in characteristic $p$, and these usually behave the same as $D$-modules. – Will Sawin Oct 1 '13 at 20:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.