7
$\begingroup$

If $A \to M$ is a Lie algebroid over a smooth manifold $M$ then a representation of $A$ is a vector bundle $E \to M$ with a flat $A$-connection $$ \nabla : \Gamma(E) \to \Gamma(E\otimes A^*). $$ If $G$ is a Lie groupoid over $M$ then a representation is a vector bundle $E\to M$ and a Lie groupoid homomorphism $G \to GL(E)$, where $GL(E)$ is the groupoid over $M$ whose arrows are linear isomorphisms of the fibers of $E$.

Now suppose $G$ is a Lie groupoid that integrates $A$.

Under what sort of conditions can every representation of $A$ be integrated to a representation of $G$?

Two extreme cases are well-known:

  • If $M$ is a point then we're just talking about Lie groups and algebras and every representation of the algebra integrates to a representation of the group if the group is simply connected.

  • If $A = TM$ is the tangent Lie algebroid and $G$ is the fundamental groupoid then the Riemann-Hilbert correspondence says that every representation of $TM$ (i.e. flat vector bundle over $M$) corresponds to a representation of $G$. On the other hand, the pair groupoid also integrates $A$ but it doesn't seem like it has the same representations (I think a representation of the pair groupoid is given by a vector bundle $E\to M$ and a section of $p_1^* E^* \otimes p_2^* E \to M\times M$, where $p_j : M\times M \to M$ is the projection on the $j$th factor).

I would be interested in seeing any references that talk about this.

$\endgroup$

1 Answer 1

6
$\begingroup$

The usual condition that one uses to ensure that all representations of $A$ integrate to representations of $G$ is that $G$ be "source 1-connected", meaning that all of the fibres of the source map $s : G \to M$ are connected and simply connected.

That the source 1-connected condition is sufficient is a consequence of a more general theorem which asserts that if $G$ and $H$ are Lie groupoids with $G$ source 1-connected, and if $\phi : Lie(G) \to Lie(H)$ is a morphism between their Lie algebroids, then $\phi$ integrates to a unique Lie groupoid morphism $G \to H$. You can find this result, for example, in the papers

  • "Integrations of Lie bialgebroids" by Mackenzie and Xu (Topology 39 (2000), no. 3, 445–467)

  • "On integrability of infinitesimal actions" by Moerdijk and Mrčun (Amer. J. Math. 124 (2002), no. 3, 567–593)

Your question deals with the specific case when $H = GL(E)$ is the gauge groupoid of a vector bundle $E \to M$, since a flat $A$-connection on $E$ is the same thing as a morphism $E \to \mathfrak{gl}(E)$ to the Atiyah algebroid.

Regarding the specific case $A = TM$: it's a good exercise to a) convince yourself that the fundamental groupoid $\Pi_1(M)$ is source 1-connected, and b) convince yourself that a flat connection will integrate to a representation of the pair groupoid if and only if it has trivial holonomy along every loop. One way to think about b) is that the natural map $\Pi_1(M) \to Pair(M)$ expresses $Pair(M)$ as a quotient of $\Pi_1(M)$ and you're looking for representations that descend to the quotient.

Finally, I should point out that, while source 1-connectedness is sufficient to conclude that every representation of $A$ integrates to a representation of $G$, it is, in general, not necessary. The example I know of is when the manifold $M$ is the Riemann sphere $\mathbb{C}P^1$ and the Lie algebroid $A$ is generated by the real and imaginary parts of the holomorphic vector fields that vanish at $0 \in \mathbb{C}P^1$. A flat connection for this algebroid is basically a flat connection in the usual sense, but it is allowed to have a singularity of logarithmic type at $0$. In this case the source 1-connected groupoid integrating $A$ is not Hausdorff, but it has a quotient that is Hausdorff, and every representation of $A$ integrates to a representation of this Hausdorff quotient.

$\endgroup$
1
  • $\begingroup$ You're welcome, @Eric; I hope it helps! $\endgroup$
    – Brent Pym
    Jun 28, 2014 at 19:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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