Let $X$ be a nonsingular algebraic variety over a field $k$ of characteristic zero. (We may assume $k$ algebraically closed if need be, but I want to avoid specifically demanding $k = \mathbb{C}$.) Let $\mathcal{D}_X$ be the sheaf of $k$-linear differential operators on $X$, and let $\mathcal{M}$ be a left $\mathcal{D}_X$-module. Since giving a left $\mathcal{D}_X$-module structure on $\mathcal{M}$ is equivalent to giving an integrable connection $\nabla$ on $\mathcal{M}$, we can use $\nabla$ to construct the de Rham complex $\mathcal{M} \otimes \Omega^{\bullet}_{X/k}$ of $X$ with coefficients in $\mathcal{M}$. The corresponding de Rham cohomology is the hypercohomology $\mathbb{H}^{\bullet}(X, \mathcal{M} \otimes \Omega^{\bullet}_{X/k})$ of this complex. This is the definition given in Hartshorne's 1975 paper "On the de Rham cohomology of algebraic varieties", section III.4.

My main question is the following:

If $\mathcal{M}$ is assumed to be a holonomic $\mathcal{D}_X$-module, are all the de Rham cohomology spaces (as defined above) finite-dimensional $k$-spaces?

Certainly such finiteness statements are well-known in many cases. For instance, taking $k = \mathbb{C}$ and $X$ to be affine, the finiteness of de Rham cohomology is proved in Bjork's 1979 book "Rings of Differential Operators". My interest is in whether I can get away without assuming $k = \mathbb{C}$ and with taking $X$ to be nonsingular but nothing more (not necessarily affine or projective).

In the analytic case, where $X$ is a complex manifold, this finiteness statement is a corollary of a much stronger claim, Kashiwara's constructibility theorem. In the book "$\mathcal{D}$-Modules, Perverse Sheaves, and Representation Theory" by Hotta, Takeuchi, and Tanisaki, the proof of a special case of the constructibility theorem in the algebraic setting (Theorem 4.7.7) uses analytifications everywhere, and so seems to depend in a crucial way on the ground field being $\mathbb{C}$. They remark (Remark 4.7.3) that for smooth algebraic varieties it "is not a good idea" to consider the de Rham complex as defined above, but rather one should analytify everything first. For my own reasons (coming from the study of local cohomology) I am specifically interested in the de Rham cohomology whose purely algebraic definition is given above, rather than passing to analytifications.

Is it necessary to pass to analytifications in the definition of de Rham cohomology (and in particular, to insist on the ground field being $\mathbb{C}$) in order to obtain the finiteness results for holonomic $\mathcal{D}$-modules?


1 Answer 1


Your question is a special case of the theorem that states that $D$-module pushforward preserves holonomicity. Note that $\pi_\ast \mathcal M$ is computed by the de Rham cohomology, (where $\pi:X \to pt$) - this fact is purely algebraic.

Preservation of holonomicity is proved for example in Bernstein's notes (he doesn't assume $k=\mathbb C$).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.