MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When I think about $\mathcal{D}$-Modules, I find it very often useful to envison them as vectorbundles endowed with a rule that decides whether a given section is flat. Or alternatively a notion of parallel transport.

Now my question is, what are good ways to think about modules over sheavers of twisted differential operators?

share|cite|improve this question
up vote 13 down vote accepted

One way to think of twisted $D$-modules that I like is to view them as monodromic $D$-modules (see Beilinson, Bernstein A Proof of Jantzen Conjectures section 2.5, available as number 49 on Bernstein's web page ). Let $T$ be a torus, and let $\pi: \tilde{X} \to X$ be a $T$-torsor. The sheaf of algebras $\tilde{D} = (\pi_* D_{\tilde{X}})^T$ has center $U(\mathfrak{t}) = S(\mathfrak{t})$, and its category of modules is the category of weakly $T$-equivaraint $D$-modules on $\tilde{X}$. For any $\chi \in \mathfrak{t}^\vee$, there is a maximal ideal $m_\chi \subset S(\mathfrak{t})$, and the algebra of $\chi$-twisted differential operators is $\tilde{D}/m_\chi \tilde{D}$.

If you want to twist by a fractional power $c$ of a line bundle $L$, then you can let $T$ be the usual one dimensional split torus, $\tilde{X}$ be the total space of $L$ with the zero section removed, and $\chi = c$. For ordinary differential operators, set $\chi = 0$. Intuitively, I think of a ($\mathcal{O}$-coherent) twisted $D$-module as a vector bundle on the total space of the torsor such that flat sections obey a fixed monodromy when parallel transported in the torus direction.

share|cite|improve this answer
Thanks Scott. It was hard to decide which answer to accept. I like the reduction perspective as well as the curvature picture. I decided for accept your answer, because you added a reference :) – Jan Weidner Sep 16 '10 at 12:20

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 $t\frac{d}{dt}$ 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 $(t\frac{ d}{dt} -c) s = 0$, 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 $t\frac{ d}{dt} -c=0$.

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 $(t\frac{ d}{dt} -c) s = 0$ 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 $t\frac{ d}{dt} -c$ 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).

share|cite|improve this answer
Thanks for your answer. This looks like TDOs are hamiltonian reductions of the differentialoperators on L-0, right? – Jan Weidner Sep 8 '10 at 6:39
@Jan Weidner: Generally speaking, differential operators on X/G are obtained by Hamiltonian reduction from differential operators on X (perhaps one should say quantum Hamiltonian reduction, because the ring of diff. operators is non-commutative). In order to get TDOs, we take the Hamiltonian reduction for non-zero value of the moment map. So the answer is `yes'. – t3suji Sep 8 '10 at 20:21
@Jan Weidner- That would be a concise way of summing up both my answer and Scott's. – Ben Webster Sep 8 '10 at 21:50
Thanks, t3suji and Ben! – Jan Weidner Sep 16 '10 at 12:12
I have some kind of mental block about TDOs, so I was excited to read Ben's lucid explanation. But can I ask him to clarify further ? Should the connection be flat ? So is it locally the usual trivial connection $\nabla$ plus $c\frac{dt}t$? Globally we might not have a function $t$, so how do we know such a connection exists ? Sorry if I've got the wrong end of a stick. – Richard Thomas Feb 17 at 21:15

Here's another perspective to complement the homogeneous approach given in Ben's and Scott's answers. One can look at twisted $D$-modules as connections with fixed scalar curvature. This is particularly powerful if you think complex-analytically: you can describe all possible twistings as follows:

Suppose $M$ is a complex manifold. In general, twistings of the sheaf of differential operators are parametrized by the hypercohomology of the truncated de Rham complex $$\Omega^1_{hol}\to\Omega^2_{hol}\to\dots.$$ (I use the lower index `hol' to distinguish the sheaf of holomorphic sections from the sheaf of $C^\infty$-sections.) If we use Dolbeault's complex to compute the hypercohomology, you see that twistings are represented by a closed 2-form $\omega$ whose $(0,2)$-part vanishes. $\omega$ matters only up to a differential of a $(1,0)$-form.

Let $\omega$ be such a closed 2-form. We can then consider vector bundles $F$ (or, more generally, quasicoherent sheaves) on $M$ equipped with a connection $$\nabla:F\to F\otimes\Omega$$ whose curvature is $\omega$. More precisely, suppose $F$ is a holomorphic vector bundle. The sheaf of $C^\infty$-sections of $F$ carries an action of $$\overline\partial:F\to F\otimes\Omega^{0,1}.$$ An action of the TDO corresponding to $\omega$ on $F$ is the same as extension of $\overline\partial$ to $\nabla$ with prescribed curvature.

Remark. If one works algebraically (for instance, over fields other than ${\mathbb C}$), only some TDO's can be viewed in similar way; namely those whose class belongs to the image of the space $H^0(\Omega^2)$.

share|cite|improve this answer
Thanks, I like your perspective! – Jan Weidner Sep 16 '10 at 12:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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