ag.algebraic geometry - What is a twisted D-Module intuitively?

NOTE: I ended up writing out a long version of this and correcting a mistake, so this is substantially changed from the old answer.

If you're taking twisted differential operators in a complex power of a line bundle, $L^c$, then you should think of them as vector bundles/sheaves on the total space $T$ of $L$ minus its zero section, endowed with a flat connection that behaves specially along the fibers of the bundle projection map.

Special how? The action of $mathbb C^*$ by fiber rotation has a differential, which is a vector field on the total space that looks like $tfrac{d}{dt}$ for any trivialization, where $t$ is the coordinate on the fiber. One should take a connection where differentiating along this vector field integrates to an equivariant structure for $mathbb C^*$ (that is, it has integral eigenvalues).

To see this, note that $mathbb C^*$ -invariant functions on $T$ are the same as functions on $X$. However, there are more $mathbb C^*$-invariant vector fields; there's a Lie algebra map from the sheaf $Y$ of $mathbb C^*$-invariant vector fields on $T$ to vector fields on $X$, but the kernel is given by functions times the vector field $t frac{d}{dt}$ (the action vector field for $mathbb C^*$). This element is central, since we're looking at $mathbb C^*$-invariant vector fields (which exactly means they commute with the action vector field). This is a central extension of Lie algebras, and Chern-Weil theory tells us that the first Chern class of the line bundle is the obstruction to splitting this extension. If you want to get very fancy, this gives a Lie algebroid over $X$, of a special type called a Picard Lie algebroid.

Now, we can think of sections of powers of this line bundle as functions on $T$ with a fixed weight under $mathbb C^*$: the sections of $L^c$ have weight $c$. So, on the sections of $L^c$, I have the relation $(tfrac{ d}{dt} -c) s = 0$, so the differential operators twisted by $L^c$ are given by taking $mathcal O_X$, the vector fields $Y$, letting them commute past each other in the usual way, and then imposing $tfrac{ d}{dt} -c=0$.

Now, if $c$ isn't an integer, then there aren't going to be any functions that satisfy this equation, but the description I gave of the TDO is fine. There's just nothing interesting to act on.

Well, except there might be. I could take some other D-module on $T$ instead, and I would still get an action of the TDO on the sheaf of solutions to $(tfrac{ d}{dt} -c) s = 0$ in that D-module; this gives a functor from D-modules on T to twisted D-modules on X. This functor is an equivalence when restricted to the subcategory of D-modules where $tfrac{ d}{dt} -c$ integrates to a $mathbb C^*$-action (since you can always pull back a twisted D-module on X and get a D-module of this form on Y). Note that whether this vector field integrates only depends on the class of $c$ modulo $mathbb Z$; if you track through these functors, the resulting equivalence between modules of TDOs is tensoring with the correct power of the line bundle.

Now, assume I have a simple holonomic twisted D-module; the corresponding $mathbb C^*$-equivariant D-module on $T$ is also simple, so this is an intermediate extension of a $mathbb C^*$-equivariant local system on a locally closed subvariety $T'$ which is the preimage of some $X'subset X$. The monodromy of a flat section of this local system around the fiber must be $e^{2 pi i c}$, so if $c$ is irrational, the fundamental group of the fiber must inject into that of $T'$, whereas if $c=a/b$ in lowest form, then the kernel of this map of fundamental groups can only contain loops at go around the fiber a multiple of $b$ times (even then, you could have trouble depending on the structure of the fundamental group).

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