Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

First, if this post must be broken up in separate questions, please tell me so. I thought it would be better if I simply posed my questions in one thread, as they are directly related to each other.

I am currently trying to understand the construction of the Hitchin moduli space, constructed in this paper (pdf). I'll jump right in and point to the passages in the text where my questions arise, using the same notation as in the text. Below, all principal and vector bundles will be over the same compact riemann surface $M$.

Let's look at the beginning of §2. Here, we consider an $SO(3)$ principal bundle $P$ and make a case distinction based on whether the 2nd Stiefel-Whitney class vanishes or not. There are a few things I do not understand:

  1. How do we obtain a $SU(2)$ or $U(2)$ principal bundle covering $P$?
  2. What is $V$, and why does a connection on $P$ induce a connection on $V$ (or am I missing that $V$ is associated not only to a cover of $P$ as above, but to $P$ itself)?

Next is a question that is most likely the result of my ignorance concerning the nature of $V$.

I was under the impression that the result that one wants to show is that the moduli space of solutions to the self-duality equations on $P$ modulo gauge equivalence is a smooth manifold, for some restriction on $M$.

Theorem (5.7) proves this for solutions on a rank 2 vector bundles of odd degree, as long as the genus of $M$ is greater than 1. This is somehow equivalent to looking at $P$, at least if I interpret the introduction correctly. But how are these viewpoints equivalent? Is there a bijection between the moduli spaces of solutions of the self-duality equations on P, respective V?

I am trying to construct this bijection from the results of the paper, but I fear I'm a tad lost at the moment. I would be grateful for answers to any of my questions, or references to literature that might help me progress.

Thank you very much!

share|cite|improve this question

1 Answer 1

up vote 2 down vote accepted

Hitchin is using some standard arguments from the general theory of principal fiber bundles and connections (see e.g. Kobayashi-Nomizu; there is also a good appendix in Lawson-Michelsohn). I will sketch the arguments he is using.

A homomorphism of Lie groups $\rho:\hat G\to G$ induces a map between $\hat G$-principal fiber bundles and $G$-principal fiber bundles. Also, if $P$ is the image of $\hat P$ under this map, then with any $\hat G$-connection $\hat A$ on $\hat P$ there is associated canonically a $G$-connection $A$ on $P$. If $d\rho: \hat{\mathfrak{g}}\to \mathfrak g $ (the derivative of $\rho$ at $e\in \hat G$) is injective then the map $\hat A \mapsto A$ is injective.

The inverse map to $\hat P\mapsto P$ is called "lifting" a $G$-bundle to $\hat G$ and in general does not always exist or unique.

If $\rho:\hat G \to G$ is surjective with kernel $A\subset \hat G$ which is abelian, then there is an "exact sequence" relating the equivalence classes of principal bundles of the three groups over some base manifold $M$:

$$ H^1(M; A)\to H^1(M;\hat G)\to H^1(M; G)\to H^2(M;A)$$

So the obstruction for lifting a $G$-bundle to $\hat G$ lies in $H^2(M;A)$.

For $SO(3)=SU(2)/\mathbb Z_2$ this is the second Stiefel-Whitney class $w_2(P)\in H^2(M; \mathbb Z_2)$, while for $SO(3)=U(2)/\mathbb C^*$ there is no obstruction, and the lift is unique, since $H^k(M;\mathbb C^*)=0$ for $k>0$ (this is because the sheaf of $\mathbb C^*$-valued smooth functions is a so-called fine sheaf; this is explained in any textbook on Čech cohomology).

Once a principal $SO(3)$-bundle $P$ is lifted to a principal $\hat G=SU(2)$ or $\hat G=U(2)$-bundle $\hat P$, you form $V$ by taking the vector bundle associated with the standard representaion of $\hat G$ on $\mathbb C^2$, $V=\hat P\times \mathbb C^2/\hat G$.

If $\hat P$ is a lift of $P$ to $U(2)$ and is equipped with some $U(2)$-connection, say $\hat A$, then this connection induces an hermitian connection on $V$, also denoted by Hitchin by $\hat A$; it contains exactly the same information as the connection on $\hat P$, since the standard representation of $U(2)$ is faithful. $\hat A$ also induces connections $A$ on $P$ and $A_0$ on $\Lambda^2 V$. If you think of $\hat A$ as an antihermitian 2 by 2 matrix of 1-forms on $\hat P$, then $A,A_0$ are just the tracelss part and trace of this matrix.

share|cite|improve this answer
Thank you for your help. I just noticed I messed up the description - in the paper, there isn't a $U(2)$ covering, but a $U(2)$ bundle associated to $P$. Do you see how we induce a connection then? –  David Hornshaw Mar 28 '14 at 14:18
You have to follow carefully what he says in the first 3 lines of p. 67 (he has a small error in line 3; you lift from $P$ to $\hat P$, not the other way around). In general: a $\hat G$-connection on a principal $\hat G$-bundle $\hat P$ induces tautologically connections on everything you wish to associate with $\hat P$. He Uses this for $\hat G=U(2)$, with 3 associated bundles: $V$, $\Lambda^2(V)$, $P$. Then, there is a bijection between 3 things: (1) connections on $\hat P$, (2) connections on $V$, (3) pairs $(A,A_0)$ of a connection $A$ on $P$ and a connection $A_0$ on $\Lambda^2(V)$. –  Gil Bor Mar 28 '14 at 15:34
I see. It makes more sense to me now. Yet I thought that in general, there isn't a bijection between connection forms on a principal bundle and the connections on an associated vector bundle (I was under the impression that's why frame bundles are so nice, because here you do have bijections if you take the right kind of connections). Is this covered in Kobayashi or Michelsohn too? Thanks again. –  David Hornshaw Mar 28 '14 at 18:09
Probably in K-N (don't have it with me now). Eventually, it's just elementary representation theory (linear algebra). –  Gil Bor Mar 28 '14 at 20:57
Expanded a bit my answer. –  Gil Bor Mar 28 '14 at 22:27

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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