## D-Module Basics II

In my first post, we showed how important information about a PDE on $\mathbb{C}^n$ could be turned into a certain module over the ring $D_{\mathbb{C}^n}$ of regular differential operators on $\mathbb{C}^n$.  Our focus today will be about extending the idea of differential operators to other kinds of objects.

The first step will be to define the ring of regular differential operators on an arbitrary complex affine variety.  The construction is a bit mysterious, but for smooth varieties it is the ring generated by multiplying by regular functions and differentiating along regular vector fields.  Then, we will show how there is a natural restriction map from the differential operators on an affine variety to the differential operators on a Zariski open set in that variety.  With this important tool, we can compare differential operators in two different affine subsets of a non-affine variety, and see if we can patch them together into a differential operator over the union.  Effectively, this will define the sheaf of differential operators on any variety.  We will close by computing the global differential operators on $\mathbb{P}^1$, and showing that they are in agreement with the universal enveloping algebra of the Lie algebra of regular vector fields on $\mathbb{P}^1$.

I should apologize a bit, since my choice to consider regular differential operators (the theory of algebraic D-modules) over holomorphic differential operators (the theory of analytic D-modules) means that it isn’t sufficient to understand differential operators only on subsets of $\mathbb{C}^n$.  This is because complex analytic manifolds are locally open subsets of $\mathbb{C}^n$, while algebraic varieties are not locally open subsets of the Zariski version of $\mathbb{C}^n$.  As such, we have to skip right to the abstract definition of differential operators.

Differential Operators for an Affine Variety

Let $X$ be an complex affine variety whose coordinate ring is $R$.  I will construct the ring of differential operators on $X$ in pieces as follows.   Let $D^0(X)=R$, and define $D^i$ to be the $\mathbb{C}$-linear maps from $R$ to itself such that

$\gamma\in D^i\;\; \Leftrightarrow \;\; \forall r\in R, \;[\gamma, r]\in D^{i-1}$

where $r$ is thought of as a multiplication operator acting on $R$.  Define $D(X)$, the ring of differential operators on latex $X$, to be the union over all $D^i$ (it is alternatively called $D(R)$).  While each $D^i$ is just a vector space, $D^iD^j\subset D^{i+j}$, so $D(X)$  is a subring of $End_\mathbb{C}(R)$.  Notice that it is a filtered ring by construction, an element in $D^i$ but not in $D^{i-1}$ is said to have order $i$.

What on earth does this ring have to do with actual differential operators?  Well, notice that every derivation on $R$ will be in $D^1$.  Since a derivation on $R$ determines a regular vector field on $X$, this shows that differentiation with respect to any regular vector field is in $D(X)$

The problem is really that by talking about arbitrary affine varieties, singularities can enter the picture, which can behave oddly with respect to differential operators.  For instance, consider the ring of polynomials in $z$ whose derivative at 0 is equal to zero (this is the coordinate ring of the cusp $y^2=x^3$).  $\frac{d}{dz}$ doesn’t exist, but $z\frac{d}{dz}$ does, as well as the weirder fellow $z\frac{d^2}{dz^2}-\frac{d}{dz}$.  No notion of ‘vector field’ will generate an operator of the latter form.

However, in my examples this will be irrelevant, because for a smooth affine variety, the ring of differential operators is generated by the multiplication operators and the derivations (vector fields).  I am trying to sneak this theorem in a little under the radar, because the only proof I know is a bit round-about and would take too long (showing that the derivations generate the associated graded algebra to the differential operators).  Unfortunately, this fact is extremely important, since it lets us reduce the study of rings of differential operators to studying derivations (particularly, their universal enveloping algebras).

Differential operators and localization

The next goal is to try to patch together differential operators on affine pieces of a non-affine variety to get some global definition of a differential operator.  To do this, we need an open affine subset $V$ of an affine $U$ to determine a map from $D(U)$ to $D(V)$  that tells us how to differentiate functions defined on open subsets.  Let $R$ be the coordinate ring of $U$ and let $S$ be the coordinate ring of $V$.

On coordinate rings, the inclusion $U\hookrightarrow V$ is given by a localization map $R\hookrightarrow S$, where $S$ is gotten from $R$ by inverting some collection of elements.  Here’s where it would be awesome if $D:R\rightarrow D(R)$ was a functor, since the localization map $R\rightarrow S$ would induce a map $D(R)\rightarrow D(S)$.  However, (and this is important) $D$ is not a functor!

This should not stop us from our original task, which we can now regard as proving that $D$ is a ‘functor on localization maps’.  The magic trick is the simple calculus fact that if I know how to differentiate $f$, I automatically know how to differentiate $f^{-1}$.  Specifically, if $\partial$ is a derivation, than $\partial(f^{-1})=-f^{-2}\partial(f)$ (this also follows from the definition of a derivation).  Thus, there is a canonical map $Der(R)\rightarrow Der(S)$ for derivations.  If $U$ is smooth, then this map, together with the inclusion $R\rightarrow S$ generates the desired map $D(R)\rightarrow D(S)$, which is what we wanted.

As far as the non-smooth case goes, none of the literature I have found addresses this problem.  Bernstein goes so far as to declare that the above definition gives the wrong category of D-modules of a singular variety, and uses something called Kashiwara’s Theorem to construct a functional alternative.

An Example: Differential Operators on $\mathbb{P}^1$

Now we should be able to define global differential operators as a choice of a differential operator in each open affine set that agrees with every other choice on intersections.  Let’s try to find the ring of global differential operators on $\mathbb{P}^1$.

First, we need a convenient open cover to work with.  We can construct $\mathbb{P}^1$ out of two copies of the affine line $\mathbb{A}^1$ and gluing them along the punctured affine line $\mathbb{A}^\times$.  On coordinate rings, this corresponds to the rings $\mathbb{C}[x]$ and $\mathbb{C}[x^{-1}]$, glued along $\mathbb{C}[x,x^{-1}]$ with the obvious inclusions.

A global differential operator is then given by differential operators on $\mathbb{C}[x]$ and $\mathbb{C}[x^{-1}]$ which map to the same differential operator on $\mathbb{C}[x,x^{-1}]$.  Since these are all smooth varieties, we know that the rings of differential operators are

$\mathbb{C}[x,\partial_x]$,

$\mathbb{C}[x^{-1},\partial_{x^{-1}}]$

and

$\mathbb{C}[x,x^{-1},\partial_x]=\mathbb{C}[x,x^{-1},\partial_{x^{-1}}]$,

where $\partial_x$ denotes differentiation with respect to $x$.  Also, I have used hard brackets illegimately, since these are not commutative rings.  The hard brackets are meant to denote that these variables commutate as the symbols dictate, ie, $[\partial_x,x]=1$.

So, a global differential operator of order 1 is a pair $(f_1(x)\partial_x +f_0(x),\; g_1(x^{-1})\partial_{x^{-1}} +g_0(x^{-1}))$ which agrees on overlaps.  However, both of these terms act on $\mathbb{C}[x,x^{-1}]$ exactly as their appear.  The only important fact is the relationship:

$\partial_x=-x^{-2}\partial_{x_{-1}}$

This implies that in the above terms,

$-x^{-2} f_1(x)=g_1(x^{-1})$

$f_0(x)=g_0(x^{-1})$

The second equation says that $f_0$ and $g_0$ are both constant, which shouldn’t be a surprise.  However, the additional factor of $-x^{-2}$ in the first equation gives us some room for fun.  It implies that $f_1(x)$ is of degree at most 2, and that $g(x^{-1})$ is totally determined by such an $f_1(x)$.  Explicitly, a global differential operator of order 1 must look like $([a+bx+cx^2]\partial_x+d, [c+bx^{-1}+ax^{-2}]\partial_{x^{-1}}+d)$ for some constants $a,b,c,d$.

Because $\mathbb{P}^1$ is smooth, these differential operators generate the whole ring of global differential operators.  Thus we see that this ring is generated by the three elements $\partial_x$, $x\partial_x$, and $x^2\partial_x$.  Each of these three elements defines a regular vector field on $\mathbb{P}^1$, and these three span the entire space of regular vector fields on $\mathbb{P}^1$.  Thus, the ring of global differential operators is the universal enveloping algebra of the Lie algebra of vector fields.  In this case, this Lie algebra is isomorphic to $\mathfrak{sl}_2$, so the differential operators corresponds to a very well understood ring, and its category of D-modules is the same as the category of representations of $\mathfrak{sl}_2$.

The last paragraph might have been a bit quick, but that was in part because it wasn’t very D-module theoretic.  In essence, we used the unproven theorem about smooth varieties to reduce it to a question of derivations/vector fields, and the previous computation was a verification that our vector fields glued together as vector fields should.

I haven’t decided what to talk about next time.  Most likely, I will define differential operators on bundles, and how they relate to covariant derivatives and connections.  I also might try to move toward some more noteworthy facts like the Riemann Hilbert correspondence.  The latter is more exciting, but I am shakier on some of the constituent elements, perverse sheaves in particular.