# What is a twisted D-Module intuitively?

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?

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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.

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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).

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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 $H^0(\Omega^2)$.

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Thanks, I like your perspective! – Jan Weidner Sep 16 '10 at 12:15