Musings on D-modules, part 2

So we have moved along in our D-modules seminar. It’s been surprisingly successful. I would have never guessed we would get 8 people to come to a seminar in the middle of summer. I want to share with you some of what we have learned.

When we left off last time (see my last post), I explained that a D-module on a smooth complex variety X was a sheaf of modules M over the sheaf of differential operators D_X. Then I gave some motivation. Now, I want to get into what you actually do with these D-modules.

One first point of interest is to explain how to pull-back and push-forward D-modules. Let’s review first how to pull-back and push-forward coherent sheaves (ie O-modules). Suppose that f : Y \rightarrow X is a morphism of varieties and that F is a coherent sheaf on Y. Then we can do a sheaf-theoretic push-forward, f_*F . We need to functions on X to act here, but that is easy since we have a map from functions on X to functions on Y (that is part of the data of f) and functions on Y already act on F. So we can take this f_*F as our push-forward.

On the other hand, suppose we have a sheaf F on X. Then we can do a sheaf pullback f^* F which is a sheaf on Y. We need to get functions on Y to act on this sheaf. To do so, we will have to modify this pull-back to f^*F \otimes_{f^* \mathcal{O}_X} \mathcal{O}_Y and then functions on Y will act.

Now let us think about D-modules. For D-modules, we need to be able to make both functions and vector fields act when we push-forward and pull-back. For functions, we do as above. Now, what about vector fields? Well, notice that when f: Y \rightarrow X , then we get a map from vector fields on Y to vector fields on X (ie it goes in the opposite direction as functions). So let F be a sheaf on X. Then D_Y automatically acts on the pull-back f^*F \otimes_{f^* \mathcal{O}_X} \mathcal{O}_Y as defined above. On the other hand, if F is a sheaf on Y, then to get vector fields to act on the push-forward, we will have to modify the definition of the push-forward to f_* F \otimes_{\mathcal{O}_Y} \mathcal{D}_Y. At this moment, some confusions between left and right D-modules enter, which I don’t want to go into…

What is remarkable about these definitions of push-forward and pull-back is that you can relate them to systems of differentially equations. Let’s look at an example. Consider Y=\mathcal{C} =X and f : Y \rightarrow X by f(y) = y^2 . Then, we can consider the D-module M = \mathcal{D}/\mathcal{D} \partial which corresponds to the differential equation f' = f . We can pull-back M to Y to get the D-module \mathcal{D}/\mathcal{D} (\partial - 2y) which corresponds to the differential equation f' = 2yf . Note that when you pull-back the function e^x by f you get the function e^{y^2} which is a solution to this differential equation.

As a challenge, try pushing forward M along the same map.

Correction: the push-forward for D-modules as defined above is a wrong. It should be more like: f_* (F \otimes_{\mathcal{D}_Y} \mathcal{O}_Y \otimes_{f^* \mathcal{O}_X} f^* \mathcal{D}_X ) . What a mouthfull!