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 2010 at 12:20

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 of L minus its zero section, endowed with a 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 multiplies by c.

This follows immediately from the fact that the ring of twisted differential operators in $L^c$ are exactly $\mathbb{C}^*$-invariant differential operators on the total space minus its zero section modulo $t\frac{d}{dt}-c$.

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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 2010 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 2010 at 20:21
@Jan Weidner- That would be a concise way of summing up both my answer and Scott's. – Ben Webster Sep 8 2010 at 21:50
Thanks, t3suji and Ben! – Jan Weidner Sep 16 2010 at 12:12

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 2010 at 12:15