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This question is somehow related to this one.

Let $M$ be a smooth (compact, if you wish) connected manifold.

Then, it is well known that there is an equivalence between the isomorphism classes of pairs $(E,\nabla)$, where $E\to M$ is a complex vector bundle of rank $r$ and $\nabla$ a flat connection on $E$, and the isomorphism classes of complex linear representations of dimension $r$ of the fundamental group $\pi_1(M)$ of $M$.

Here, $(E,\nabla)\simeq (F,\nabla')$ iff there exists an isomorphism of differentiable vector bundles $f\colon E\to F$ over $M$ such that $f^*\nabla'=\nabla$.

Question. Are there necessary and/or sufficient conditions such that given two representations $\rho,\rho'\colon\pi_1(M)\to\operatorname{Gl}(r,\mathbb C)$, the associated complex vector bundles $E_\rho,E_{\rho'}\to M$ are isomorphic?

Of course, in the question we are forgetting the flat connections naturally given by these two representations, and we are asking the two complex vector bundles to be isomorphic just as complex vector bundles.

More specifically, can we for instance give an answer in terms of connected components of the space of the representations - or, if you want, in terms of (some well-chosen notion of) homotopies between the representations?

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My impression is that a necessary and sufficient condition is that $\rho$ and $\rho'$ are conjugate, but I don't have a reference or proof, so I'll leave this as a comment. –  Peter Samuelson Mar 26 '12 at 15:01
Dear Peter, in fact this is the necessary and sufficient condition for the vector bundles $E_\rho$ and $E_{\rho'}$ to be isomorphic together with their natural flat connections defined by the representations. I am looking for a weaker condition since i would like just the two vector bundle to be isomorphic without caring about the flat connections. –  diverietti Mar 26 '12 at 16:12
Ah, I see, I misread the question. –  Peter Samuelson Mar 26 '12 at 16:23
It is a trivial remark, but one should note that representations $\rho, \rho'$ which are in the same component of $Hom(\pi_1(M), G)$ (in your case, $G=GL(r, {\mathbb C})$ but it works in general as well) are isomorphic as vector bundles. This gives a sufficient, but, of course, far from necessary, condition for an isomorphism of vector bundles. In particular, you get finiteness of the number of isomorphism clases of vector bundles. E.g., in the case of $r=1$, it shows that all line bundles are trivial since $Hom$ is connected. –  Misha Mar 27 '12 at 3:16
A correction: $Hom(\pi, U(1))$ is connected if $H_1(M)$ is torsion-free, otherwise it is disconnected. –  Misha Mar 27 '12 at 3:30
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3 Answers

up vote 7 down vote accepted

I have two comments:

(1). It is a trivial remark, but one should note that representations $\rho, \rho'$ which are in the same component of $Hom(\pi,G)$ (in your case, $G=GL(r, {\mathbb C}))$ yield flat bundles $E_{\rho}, E_{\rho'}$ which are isomorphic as vector bundles. Here and below $\pi=\pi_1(M)$.

Proof. It suffices to work with principal $G$-bundles. Let $\rho_t$ be a path of representations between $\rho, \rho'$. Then $\rho_t$ determines a (flat) principal $G$-bundle $P$ over $M\times [0,1]$, whose restrictions $P_0, P_1$ to $M\times 0, M\times 1$ are isomorphic to the principal $G$-bundles associated with $\rho, \rho'$. I claim that this bundle is just the product $P_0\times I$. Indeed, construct a section of the bundle $Hom(P, P_0\times I)$ starting with the identity on $M\times 0$ and then extend it to the rest of $M\times I$ (path-lifting property for Serre fibrations). QED.

This gives a sufficient, but, of course, far from necessary, condition for an isomorphism of vector bundles. In particular, you get finiteness of the number of isomorphism classes of vector bundles. As a simple example consider the case $r=1$ and $H_1(M)$ torsion-free. Then $Hom(\pi, {\mathbb C}^\times)= ({\mathbb C}^\times)^n$ is connected (here $n=rank(H_1(M))$). In particular, all flat line bundles in this case are trivial.

(2). A far less trivial is the result of Deligne and Sullivan "Fibres vectoriels complexes a groupe structural discret", C. R. Acad. Sci. Paris 281 (1975), 1081-1083. They prove that for every finite cell complex $M$ and every $\rho: \pi=\pi_1(M)\to GL(r, {\mathbb C})$ there exists a finite cover $\tilde{M}\to M$ so that the pull-back of the associated flat bundle $E_\rho$ to $\tilde{M}$ is trivial (as a bundle). Their proof is constructive and (from what I can tell by reading Math Review of their paper since I lost my copy) gives a sufficient condition for triviality of $E_\rho$: Let $A$ denote the subring of ${\mathbb C}$ generated by the matrix entries of $\rho(\pi)$. Suppose that there are two maximal ideals $m_1, m_2$ in $A$ so that:

a. $A/m_i$ have distinct characteristics and

b. $\rho(\pi)$ maps trivially to $GL(r, A/m_i), i=1,2$.

Then $E_\rho$ is trivial as a vector bundle.

From this one can extract a sufficient condition for an isomorphism of bundles $E_\rho, E_{\rho'}$ (by considering the flat bundle $E_\rho^*\otimes E_{\rho'}$).

If I remember correctly, a proof of this theorem by Deligne and Sullivan was redone by Eric
Friedlander, in "Étale homotopy of simplicial schemes". Annals of Mathematics Studies, 104. Princeton University Press, 1982. I cannot tell if it is easier to read than the original.

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Here's an answer for line bundles. A flat connection on the trivial bundle is the same as a closed one-form $\eta$. The corresponding local system is the sheaf of logarithmic antiderivatives of $\eta$. What is the representation attached to a closed one-form? A loop $\gamma$ acts on a vector $v$ via multiplication by $\exp(\int_\gamma \eta)$. So a $\pi_1$ character gives rise to the trivial line bundle (with possibly non-trivial flat connection) iff there is a closed one-form $\eta$ on $X$ such that the character is given by integrating as above, and two arbitrary $\pi_1$ characters give rise to isomorphic line bundles (with possibly different flat connection) iff their quotient gives rise to the trivial line bundle.

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This reduces the problem to deciding when $E_\rho$ is trivial. The group of characters is $ G:=Hom(\, H_1(M,\mathbb{Z}, \mathbb{C}^*)$ and we have a Chern class group morphism $$ c_1: G\to H^2(M\mathbb{Z})_{tors},\;\;\rho\masto c_1(E_\rho).$$ The question is to decide what is the kernel of this map. This is not entirely obvious to me. –  Liviu Nicolaescu Mar 26 '12 at 16:38
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Clearly one must have

$$c_k(E_\rho)=c_k(E_{\rho'} ), $$

for any $k\leq r$. When $r=1$ this is a necessary and sufficient condition. This raises an even more concrete question, namely how does one compute the Chern classes of $E_\rho$ which are obviously torsion classes.

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The equality of all Chern classes for higher rank vector bundle won't never give a sufficient condition, anyhow. That's why I was asking about something which can be read directly on the representation... –  diverietti Mar 26 '12 at 14:15
There is a theoretical method involving the classifying space of $\pi_1(M)$. The representation gives a map $B\pi_1(M)\to BU(r)$, while the universal covering gives a map $M\to B\pi_1(M)$ and one has to understand the homotopy type of the composition $M\to BU(r)$. Things simplify a bit when $M$ is a $K(\pi,1)$ so that $M=B\pi_1(M)$ the induced map $M\to B\pi_1(M)$ is the identity. Ultimately you will have to deal with the cohomologies of the group $\pi_1(M)$. –  Liviu Nicolaescu Mar 26 '12 at 14:30
Unfortunately I really don't know this language... Even notations are obscure for me! Sorry... –  diverietti Mar 26 '12 at 14:39
All questions concerning isomorphisms of vector bundles ultimately reduce to questions concerning classifying spaces. For example $BU(r)$ is homotopic to the grassmannian of $r$-dimensional spaces in an infinite-dimensional vector space. –  Liviu Nicolaescu Mar 26 '12 at 16:07
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