## Tannakian construction of the fundamental group and Kapranov’s fundamental Lie algebra March 17, 2008

Posted by Joel Kamnitzer in D-modules, differential geometry.

This post is a report on a talk that Mikhail Kapranov gave in Berkeley a few weeks ago on the “fundamental Lie algebra” — it is a Lie algebra associated to a manifold and a point on the manifold much like the fundamental group. Before telling about what Kapranov said, I’d like to start with some background.

Let $M$ be a manifold and $x \in M$. Suppose we consider pairs $\mathcal{E} = (E, \nabla)$ where $E$ is a vector bundle and $\nabla$ is a flat connection on $E$. Two such pairs $\mathcal{E}, \mathcal{E'}$ may be tensored together to produce a new vector bundle with flat connection $\mathcal{E}\otimes \mathcal{E'}$.

Define a group $G$ by the following procedure. An element $g \in G$ is a collection $(g_\mathcal{E})$ of linear isomorphisms $g_{\mathcal{E}} : E_x \rightarrow E_x$ for each $\mathcal{E}$ which are natural with respect to maps $\mathcal{E} \rightarrow \mathcal{E'}$ and which obey the rule $g_{\mathcal{E} \otimes \mathcal{E'}} = g_{\mathcal{E}} \otimes g_{\mathcal{E'}}$.

In other words, an element of $G$ is a linear automorphism of the fibre at $x$ of any vector bundle with flat connection. So, what is $G$?

As you might have guessed, $G$ is the fundamental group of $M$ with respect to the basepoint $x$. To see why this is reasonable, note that any loop in $M$ based at $x$ gives rise to a monodromy map $E_x \rightarrow E_x$ by integrating the connection. However, homotopic loops give rise to the same map since the connection is flat.

Now, what happens when we change our setup to include all connections (not necessarily flat). Then we end up with a much bigger group which is roughly the group of loops in $M$ based at $x$ modulo reparmetrization and going back and forth. This is a kind of infinite dimensional Lie group. Since this group is a bit hard to study, we will consider its Lie algebra.

To define this Lie algebra, consider the vector space $P(M,x)$ whose elements are collections $X_\mathcal{E} :E_x \rightarrow E_x$ of linear maps where now $\mathcal{E}$ ranges over all vector bundles with connection. We will impose the naturality condition and also that $X_{\mathcal{E} \otimes \mathcal{E'}} = X_{\mathcal{E}} \otimes 1_{E'_x} + 1_{E_x} \otimes X_{\mathcal{E'}}$. This vector space is naturally a Lie algebra. Kapranov calls this Lie algebra the fundamental Lie algebra.

To see some elements of $P(M,x)$, recall that if $\mathcal{E}$ is a vector bundle with connection, then we have a curvature $F$ which is an $ad_E$ valued 2-form on $M$. Hence if we have any element of $\Lambda^2 T_xM$ then we can pair it against the value of $F$ at $x$ to get a endomorphism of $E_x$. This construction gives rise to a map $\Lambda^2 T_xM \rightarrow P(M,x)$.

Kapranov has proven the following theorem which describes $P(M,x)$. Before stating it, we need to recall the notion of the free Lie algebra associated to a vector space $V$, $FLie(V)$. As you might expect it is just the free Lie algebra generated by elements of $V$. It is graded with graded pieces $FLie_d(V), d \ge 1$. For example $FLie_1(V) = V$ and $FLie_2(V) = \Lambda^2 V$.

Theorem
$P(M,x)$ is isomorphic (not canonically) to $FLie_{\ge 2}{T_x M}$.

In particular it is a free Lie algebra, since every Lie subalgebra of a free Lie algebra is free.

I should mention in closing is that what we have been doing is Tannakian constructions of groups (and Lie algebras). Namely we started with a tensor category with fibre functor and then we considered automorphisms of this fibre functor. Those who know me well will understand that I like this procedure since it is the key idea in the proof of the geometric Satake correspondence.

1. Urs Schreiber - March 18, 2008

Thanks for this interesting summary.

I am wondering to which extent the terminology “fundamental Lie algebra” is suggestive. By that theorem you mentioned, the only information it contains is the dimension of the manifold. Or did I get this wrong?

The following remark is probably clear to everybody, but maybe it deserves to be said anyway in this context here.

There is (also) a natural and obvious Lie version of the fundamental group:

Namely the fundamental group itself is slightly evil, and naturally rather to be thought of as coming from the fundamental groupoid. That, in turn, is Lie (if the space is smooth) and hence has a Lie algebroid linearizing it: that Lie algebroid is just what is usually called the “tangent Lie algebroid” of X (the anchor is the identity map), but which I wouldn’t hesitate to address as the “fundamental Lie algebroid” of X.

Being a Lie algebroid, this is is also a Lie-Rinehart pair, given by an ordinary Lie algebra and an associative algebra, which are modules of each other in a compatible way.

For the tangent Lie algebroid that Lie algebra of its Lie-Rinehart incarnation is just the Lie algebra of vector fields on X (acting on the associative algebra of functions on X).

So then, it wouldn’t be all too radical to propose to address the Lie algebra of vector fields as the “fundamental Lie algebra” of X. Because it is still immediately the Lie version of the fundamental Lie groupoid, in the general sense in which Lie groupoids have Lie versions.

2. Urs Schreiber - March 18, 2008

Sorry, I realize that my terminology is a bit a ambiguous: by “Lie version” of a Lie groupoid I mean its linearization to a Lie algebroid.

3. Joel Kamnitzer - March 18, 2008

There is a Lie algebroid version of Kapranov’s work as well which he explained during his talk as well. In the Tannakian description, instead of endomorphisms of the fibre of the vector bundle you should look at the endomorphisms of the whole vector bundle. This will give a sheaf of Lie algebras on M. Note that it will contain vector fields, which act through the connection. However, their Lie bracket in this Lie algebra will not be their ordinary Lie bracket — you get another term corresponding to the connection. Kapranov calls this the Lie algebra of non-commutative vector fields on M. It is a Lie algebroid and the fundamental Lie algebra is its fibre.

4. James - March 18, 2008

Is it true that in the original example with flat connections you get the fundamental group on the nose? I would have expected that you get the pro-algebraic completion. I think that’s the way it works in algebraic geometry.

5. Scott Carnahan - March 18, 2008

I guess it shouldn’t be a surprise that the isomorphism type doesn’t depend on anything but the dimension of M, since the Lie algebra of the group in question is given by equivalence classes of infinitesimally small loops based at x.

I imagine the Lie algebroid in question can be given as some free object on the universal formal groupoid of M (morphisms given by the completion of the diagonal in M x M).

6. Scott Carnahan - March 18, 2008

James,

I think the existence of connections with infinite order monodromy (e.g. the differential equation zd/dz – sqrt(2) on the punctured line) destroys the universality of the proalgebraic completion. This is not really my field, though.

7. David Ben-Zvi - March 18, 2008

Scott – I don’t think infinite monodromy is a problem for
the proalgebraic completion, just for the profinite completion:
in the proalgebraic completion we’re taking a limit
over representations into GL_n for all n, which can
certainly have infinite monodromy.

James – I think the question is what you mean by a flat vector bundle.
If you allow infinite rank vector bundles I think you see the full pi_1 —
after all every group has a faithful representation on a vector space
(namely the regular representation), however unpleasant that
representation may be. I may be missing something more subtle though.

8. Urs Schreiber - March 18, 2008

I guess it shouldn’t be a surprise that the isomorphism type doesn’t depend on anything but the dimension of M

It’s not a surprise given the definition. I was wondering about the motivation for addressing the Lie algebra given by that definition as the “fundamental” Lie algebra, given than there is another choice which might rather deserve this name and which does capture more information.

But it’s just a matter of taste, not really important.

9. James - March 18, 2008

David – That must be it. But you need to add that the representation is not just faithful but has closed image or something (which maybe meant). You can have faithful finite-dimensional representations of infinite fundamental groups, as Scott points out. Similarly, you can have faithful actions of infinite discrete groups on pro-finite sets. So, if you allow infinite-rank vector bundles then the corresponding version of pro-algebraic completion probably agrees with the identity functor, and all is OK.

10. Bruce Bartlett - March 19, 2008

> I should mention in closing is that what we have been doing is Tannakian constructions of groups (and Lie algebras). Namely we started with a tensor category with fibre functor and then we considered automorphisms of this fibre functor.

That’s indeed a nice way of thinking about it. Another small point to bear in mind is that Urs and Konrad Waldorf (arXiv:0705.0452) have made sense of the notion that a vector bundle with connection on a manifold X can be -defined- as a “smooth functor” from the path groupoid P_1(X) of X into Vect. This is the “look mom, no total space and no infinitesimals” viewpoint of a vector bundle with connection.

Then as you say, for every x we have a monoidal functor

fib_x: [P_1(X), Vect] –> Vect

and we are looking at the automorphisms of this fiber functor. I’m not really sure how to sneak infinite dimensional vector bundles into this game though.

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