# 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!