In Hotta, Takeuchi, Tanisaki's book on "D-modules, Perverse Sheaves, and Representation theory", for a morphism of smooth algebraic varieties $f:X \to Y$, they use the notation $$ \int_f:D^b(D_X^{op}) \to D^b(D_Y^{op}) $$ for the derived pushforward of right $D$-modules. What motivates this notation with the integral? I have hear this has something to do with "integrating along the fibers", but wikipedia gives an article about partially integrating differential forms.

  • 3
    $\begingroup$ At least for de Rham cohomology (with compact support in the vertical), it's integration along the fibers. For ordinary cohomology theories it's the inverse of the Thom isomorphism. For a sheaf recall that the stalks of $R^i f_{!} F$ are the cohomology with compact support, so when you have a locally constant sheaf the stalks are exactly ordinary cohomology with values in this local system $H_c^i (X_y, L)$ and by applying integration along the fiber (or composing with Poincaré duality and it's inverse), you get $H^{i - ({d_X - d_Y})} (Y, L) $. $\endgroup$
    – user40276
    Oct 7, 2016 at 0:35
  • 1
    $\begingroup$ So, at least for locally constant sheaves, this map is integration along the fibers. Now for more general sheaves or complexes of sheaves we just generalize this procedure, which a priori has nothing to do with integration. $\endgroup$
    – user40276
    Oct 7, 2016 at 0:41
  • $\begingroup$ @user40276 But what does this mean in terms of a differential equation? $\endgroup$
    – 54321user
    Oct 8, 2016 at 20:27
  • $\begingroup$ For example, if I consider the projective $f:\mathbb{A}^2 \to \mathbb{A}^1$, and have the distribution giving a differential equation $\int (\partial^2_x + \partial^2_y)(-)dx\wedge dy = 0$, what would its pushforward mean? $\endgroup$
    – 54321user
    Oct 8, 2016 at 20:31
  • $\begingroup$ My knowledge about D-modules is very poor. I just tried to clarify what you said that was written on Wikipedia. For the case you mention you should try computing the cohomology with compact support of the fibers of your local system given by the connection corresponding to your diff. equation. Then the pushfoward will be integration of the derived functors of these solutions (the covariant global sections). I'm not sure if the de Rham cohomology of the flat bundle is the same as the singular cohomology of the local system, but if this is true, this map will be integration of these diff. forms. $\endgroup$
    – user40276
    Oct 12, 2016 at 5:15

2 Answers 2


I think, for a locally constant $D$-module it basically cashes out to

  1. Defining a formal solution to the differential equation.
  2. Defining a formal integral along the fibers.
  3. Writing down all the differential relations satisfied by the formal integral that follow by integration by parts from the differential equation applied to the formal solution.
  4. Converting this to a differential equation satisfied by the formal solution.

More precisely we are taking the de Rham cohomology, so there may be more than one integral over more than one cycle, all of which satisfy the same differential equation. To calculatee this you take a formal ensemble of integrals of an arbitrary differential operator applied to $f$ (which is just the D-module as a module) and find all the formal differential relations between them.

Unfortunately in your example you have a single PDE in two variables, so the solutions are not a holonomic D-module. I don't know if this still works in that case. If it does, you can observe that $\int \partial_x^2f dx =0$ (by integration by parts) forces $\partial_y^2\int f df dx = \int \partial_y^2 f dx =0$ (by the differential equation).

Let's do a holonomic example instead. If we have the equations $\partial f/ \partial x=2xy f$, $\partial f/\partial y=x^2 f$, which is satisfied by $e^{x^2 y}$, whose integral is proportional to $\sqrt{y}$, we can find the differntial equation satisfied by $\sqrt{y}$ by observing

$$ y \partial_y \int f dx = \int y \partial_y f dy = \int y x^2 f dx= \int x \partial_x f dx/2 = \int \partial_x ( xf)dx/2 - \int fdx/2 dx= 0 - \int f dx/2$$

which is the correct differential equation, and I think you can check that this algebra is exactly what the definition of D-module cohomology by de Rham cohomology is doing.


(All functors and categories are derived, all $\mathcal{D}$-modules are holonomic and $f: X \to Y$ is proper for simplicity).

We may use the adjunction $\int_f \dashv f^!$ to get some interpretation

Let $\mathcal{M}$ be a $\mathcal{D}_X$-module and $\mathcal{S}$ a $\mathcal{D}_Y$-module. Then the solution complex of $\int_f \mathcal{M}$ in $\mathcal{S}$ satisfies:

$$\mathcal{Hom}_{\mathcal{D}_X}(\int_f\mathcal{M},\mathcal{S}) \cong f_*\mathcal{Hom}_{\mathcal{D}_Y}(\mathcal{M,f^!\mathcal{S}})$$

So morally at least the solutions of $\int_f\mathcal{M}$ in some function space $\mathcal{S}$ on $Y$ correspond to the integrals (in the sense of sheaf theory) of solutions to $\mathcal{M}$ in functions pulled back from $\mathcal{S}$ along $f$.

In terms of hom-sets in the derived category one perhaps obtains an even clearer picture:

$$Hom_{\mathcal{D}_X}(\int_f\mathcal{M},\mathcal{S}) \cong Hom_{\mathcal{D}_Y}(\mathcal{M},f^!\mathcal{S})$$

(Very-)simple illustrating example: Suppose $Y = pt := Spec \mathbb{C}$, $\mathcal{M} = \mathcal{O}_X$, $\mathcal{S}=\mathcal{O}_Y=\mathbb{C}$. In that case, since $X \to pt$ is flat we have $f^!\mathcal{S} = \mathcal{O}_X$. Moreover, in general $\mathcal{End}_{\mathcal{D}_X}(\mathcal{O}_X) \cong \mathbb{C}_X$ and therefore:

$$\mathcal{Hom}_{\mathcal{D}_X}(\int_f\mathcal{O}_X,\mathbb{C}) \cong f_*\mathcal{Hom}_{\mathcal{D}_Y}(\mathcal{O}_X,f^!\mathbb{C}) \cong f_*\mathcal{End}_{\mathcal{D}_X}(\mathcal{O}_X) \cong f_*\mathbb{C}_X \cong H^{\bullet}_{dR}(X)$$

In other words, the solution complex of the de-Rham $\mathcal{D}$-module associated with the structure sheaf $\mathcal{O}_X$ is the de-Rham cohomology of $X$.

In general when $Y$ is not a point and $\mathcal{M}$ arbitrary you will still have a Spencer-like projective resolution of $\mathcal{M}$ of the form $(\mathcal{M}\otimes \bigwedge^{\bullet} \Theta_X, \partial_M)$ and then $\mathcal{Hom}_{\mathcal{D}_X} (\mathcal{M},f^! \mathcal{S})$ will resemble a de-Rham complex. Then applying $f_*$ is a sheaf theoretic operation which corresponds to simply taking global sections of the resulting complex.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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