What is the relation between the Lie bracket on $TX$ as commutator and that coming from the Atiyah class?

Question

Let X be a complex manifold and $TX$ its tangent bundle. The Atiyah class $\alpha(E)\in \text{Ext}^1(E\otimes TX, E)$ for a vector bundle $E$ is defined to be the obstruction to the global existence of holomorphic connections on $E$. In particular if we take $E=TX$ to be the tangent bundle of $X$ itself, we can prove that the Atiyah class $\alpha(TX)\in \text{Ext}^1(TX\otimes TX, TX)$ and this gives a Lie algebra structure $$ TX[-1]\otimes TX[-1]\rightarrow TX[-1] $$ in the derived category $D(X)$.

For the details see M. Kapranov's paper "Rozansky–Witten invariants via Atiyah classes" (arXiv) and N. Markarian's paper "The Atiyah class, Hochschild cohomology and the Riemann-Roch theorem" (arXiv).

And it is well-known that we have the Lie bracket as commutator on the tangent vector fields $$ \Gamma(X,TX)\otimes \Gamma(X,TX)\rightarrow\Gamma(X,TX). $$

Of course these two kinds of Lie bracket are very different. $\textbf{My question}$ is: is there any relation between them? More precisely, could we regard the Lie bracket of Atiyah class as a kind of "higher lift" of the naive Lie bracket as commutators?

Answer 1

It would seem foolish to say there's no relation between them, since everything is related, but that's kind of what I'd like to say. As Pavel explained, one (the usual Lie bracket on vector fields) has to do with the Lie algebra of symmetries of your space, while the other (Atiyah) has to do with the Lie algebra of symmetries of POINTS of your space, i.e., the Lie algebra of the loop space. For example take $X=BG$ --- then the Atiyah algebra is just the Lie algebra of $G$ itself, while the usual Lie bracket of vector fields vanishes.

A more fancy way to say this is to do a Koszul dual formulation --- replace Lie algebras by the deformation problems they represent, or the formal moduli spaces you build out of them (by taking solutions of the Maurer-Cartan equations). This makes the distinction between the two very clear: the usual Lie algebra of vector fields on $X$ represents the deformation theory of the space $X$, via Kodaira-Spencer theory (take the derived global sections of the tangent sheaf you get a derived Lie algebra representing deformations of $X$). On the other hand the Atiyah bracket represents the space $X$ itself --- i.e., the deformation theory of a point in $X$ (or if you want to vary the point, of the diagonal in $X\times X$). From this perspective it's also clear why one is linear (local in $X$) and one isn't.

In the example of $BG$, we find the formal moduli space associated to the Lie algebra $g$ is the classifying space of the formal group of $G$ (by definition -- there's nothing in degree one to consider solving Maurer-Cartan with, but we can formally exponentiate $g$ to give automorphisms of this trivial solution).

Another fun example is the case of a singular point of a variety - eg a simple node. We're used to the fact that smooth affine varieties have no deformations, so smooth schemes have no local deformations, so to see something out of Kodaira-Spencer theory you need to pass to something global. But if you take something affine but singular like a node then the tangent bundle (taken correctly, ie the tangent complex) has something in degree one even locally, so you CAN see interesting solutions to the Maurer-Cartan equation even locally, which exactly correspond to the existence of local deformations -- in this case, smoothing out the node.

Answer 2

This is more of a comment than an answer.

Given a Lie algebra $T_X[-1]$, you have the universal enveloping algebra $U(T_X[-1])$, which is also a cocommutative coalgebra. Taking $Spec$ of the dual gives you a family of groups (or $E_1$ spaces) over $X$, which is the free loop space $\mathcal{L}X\rightarrow X$.

The commutator of vector fields is Koszul dual to the de Rham differential, which is represented by the $S^1$ action on the loop space $\mathcal{L}X$. So your question can reformulated as follows: what is the compatibility between the $S^1$ action on the loop space and the group structure on $\mathcal{L} X\rightarrow X$?

(The failure of the commutator to be $\mathcal{O}_X$-linear is related to the fact that the $S^1$ action does not preserve the projection $\mathcal{L} X\rightarrow X$.)

Answer 3

Starting from $X$ one can construct a sequence $(X_0,X_1,\dots)$ of groupoid objects in spaces, all having $X$ as space of objects (I am writing "space" in order not to specify the geometric context I am working with, but I actually mean "derived scheme"):

1. $X_0$ is the pair groupoid $X\times X$ of $X$. The Lie algebroid of this groupoid is $TX$.

2. $X_1=X\times^h_{X_0}X$ is the derived self-intersection of the diagonal in $X_0$ (i.e. the derived loop space of $X$). The Lie algebroid of this groupoid is $TX[-1]$ with the Atiyah Lie bracket. It happens to be a Lie algebra (because the derived loop space is actually a group).

3. In general $X_{n+1}=X\times^h_{X_n}X$ and its Lie algebroid $\mathfrak g_{n+1}$ is $TX[-n-1]$, and is an abelian Lie algebra as soon as $n\geq1$.

In other words, we have that $\mathfrak g_n=\Omega^n_0(TX)=TX[-n]$. Hence, in particular $TX[-1]$ is a group in Lie algebroids, and thus must be a Lie algebra. You get that the Atiyah-Kapranov Lie bracket on $TX[-1]$ is completely dertmined by the Lie algebroid structure on $TX$.

One also gets that the Lie algebroid $\mathfrak g_2$ of $X_2$ is going to be $\Omega_0(TX[-1])$, and thus is a group in Lie algebras... hence it must be an abelian Lie algebra.

The hierarchy Lie algebroid -> Lie algebra -> abelian Lie algebra -> ... one gets is analogous to the following one we get in algebraic topology: set -> group -> abelian group -> ...

Answer 4

I'd say, if you take the vector field $v \in \Gamma(TM)$, then $At_v \in Ext^1(E, E)$. First extensions correspond to the first order deformations, and this deformation is just given by the infinitesimal pullback along the field $v$.