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I have questions about the definition of representation variety. In François Labourie's book "Lectures on representations of surface groups", Section 3.5, the author gives four models of the representation variety. I am confused about the model using the language of vector bundles.

Definition 1 (representation variety):

A representation variety of $S$ is gauge equivalences of pairs ($G$-vector bundles $L$ over the surface $S$, flat $G$-connection on $L$).

Definition 2 (Gauge equivalence):

Two connections on the same vector bundle are said to be gauge equivalent if the can be connected using the pullback of some lift of the identity map.

What confused me is definition 2. In order to use gauge equivalence in the definition of representation variety, why do we need to restrict the definition to the same vector bundle?

In other words, can we say that every flat $\bf R$-vector bundle over the surface is trivial (where $\bf R$ is the real number field)?

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  • $\begingroup$ Certainly not! Flat $\mathbb{R}$-vector bundles of rank $r$ correspond to homomorphisms from $\pi _1(S)$ to $\operatorname{GL}(r,\mathbb{R}) $, up to conjugacy. Unless $S$ is simply connected, there are plenty! $\endgroup$ – abx Aug 15 '18 at 13:58
  • $\begingroup$ Sorry to say that the question about trivial bundle is a little bit ambiguity. Here when I say flat R-vector bundle, I mean it is a real vector bundle with R to be its structure group, that is the G in definition 1, R, with multiplication, is the lie group. $\endgroup$ – BiM Aug 15 '18 at 14:09
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    $\begingroup$ You're right, you should consider isomorphism classes of flat $G$-vector bundles, and not just "gauge" equivalence classes as referred to here. $\endgroup$ – Andy Sanders Aug 15 '18 at 14:14
  • $\begingroup$ $\mathbb{R}$ with multiplication is not a group. $\endgroup$ – Ben McKay Aug 15 '18 at 15:25
  • $\begingroup$ @BenMcKay Yes I see, I just want to clarify the ambiguity of my question, actually we need use the positive part of real number field. Thanks $\endgroup$ – BiM Aug 15 '18 at 15:29
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Any flat $\mathbb{R}^{+}$-bundle is precisely given, as in Tsemo Aristide's answer, by a representation of the fundamental group of the surface into $\mathbb{R}^+$. As $\mathbb{R}^+$ is abelian, this descends to an element of $H_1(S,\mathbb{R}^+)$, whose logarithm is a uniquely determined element of $H_1(S,\mathbb{R})=\mathbb{R}^{2g}$ where $g$ is the genus (assuming $S$ is orientable). So there are many such bundles which are not trivial.

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  • $\begingroup$ This homology group actually corresponding to the change of flat connections, but the underlying vector bundle is the same trivial bundle. $\endgroup$ – BiM Aug 15 '18 at 15:47
  • $\begingroup$ Your question was whether the flat $\mathbb{R}^+$-bundles are trivial. They are not. But as you say the associated vector bundle is smoothly trivializable, but not by parallel sections, for any representation of $\mathbb{R}^+$. $\endgroup$ – Ben McKay Aug 15 '18 at 15:55
  • $\begingroup$ Hah, Thanks, anther point of my question need to be clarify, anyway, so the answer is those vector bundle always can be trivializable. I do not understand your words ´´not by parallel sections´´, then generally, how to trivialize such kind of vector bundle without using parallel sections? $\endgroup$ – BiM Aug 15 '18 at 16:12
  • $\begingroup$ Calculate for yourself as a simple example the parallel transport for the trivial rank 1 bundle on the circle, with nonzero constant connection coefficient. You can see that the parallel transport is not the identity (indeed it is exponential growth), so there is no global parallel nonzero section. $\endgroup$ – Ben McKay Aug 15 '18 at 18:49
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A flat connection defined on a vector bundle $p:E\rightarrow S$ whose typical fibre is $V$ is defined by a representation $\rho:\pi_1(S)\rightarrow V$ which is the holonomy of the flat connection. Let $\hat S$ be the universal cover of $S$, $E$ is the quotient of $\hat S\times V$ by the diagonal action of $\pi_1(S)$ where $\pi_1(S)$ acts on $\hat S$ by the Deck representations.

The bundles, $p:E\rightarrow S$ is isomorphic to $p':E\rightarrow S$ if there exists an isomorphism of flat $f:E\rightarrow E'$ of vectors bundles. This morphism lifts to a morphism $\hat f:\hat S\times V\rightarrow \hat S\times V'$ which commutes with the action of $\pi_1(S)$, we deduce that there exists an isomorphism $g:V\rightarrow V'$ such that $g\circ \rho(\gamma)=h(\rho(\gamma))\circ g$.

We can always assume that $V=V'$ by taking any isomorphism $l:V'\rightarrow V$ between and replace $\rho'$ by $l\circ \rho'$, you obtain equivalent bundles in the sense above. And if you fix $V=V'$, the relation above means that $\rho$ and $\rho'$ are conjugated.

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