# pairs of matrices up to similiarity and vector bundles over punctured torus

I would like to construct 2D vector bundles over the punctured torus, but I don't know a lot of K-theory. Over the square, there can only be the trivial bundle, but now since $\pi_1(\mathbb{T}^2\backslash \{ pt\})= \langle x,y \rangle$ (the free group on two elements) I simply need two matrices $x,y \in GL(V)$ where $V = \mathbb{R}^2$ is the fiber over the bundle.

It looks like I have shown that vector bundles over a surface are parameterized by representations of the fundamental group $\pi_1(\mathbb{T}^2 \backslash \{ pt\})$ over $V$. I wonder that many of these representations are the same... that you can find some element $U \in GL(V)$ so that if $x,y$ determine a vector bundle the, $UxU^{-1}, UyU^{-1}$ define an equivalent bundle.

In turn, this looks like I am asking about representations of a quiver $\circ \to^2 \circ$ of pairs of matrices up to simultaneous similarity.

Is this the right way to classify spaces vector bundles over Riemann surfaces? What is the relationship between vector bundles over Riemann surfaces, the representation variety and quiver representations?

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A punctured torus up to homotopy-equivalence is a wedge of two circles. So your bundle is determined by its classifying map $S^1 \vee S^1 \to B(O_2)$. $O_2$ fits into an extension $SO_2 \to O_2 \to \mathbb Z_2$, and orientable $2$-dimensional bundles over circles (and wedges of circles) are trivial. So the only data that goes into your bundle is whether or not it is orientable over each circle wedge summand. So you don't have to worry about issues like simultaneous diagonalizability. – Ryan Budney Jul 24 '11 at 17:39
2-dimensional bundles over a punctured torus (by the above) are completely determined by their 1st Stiefel-Whitney class, $\omega_1$. Moreover, every element of $H^1(S^1 \vee S^1;\mathbb Z_2)$ is realizable as a $\omega_1$ for some 2-dimensional bundle over $S^1 \vee S^1$. – Ryan Budney Jul 24 '11 at 17:46
post this as an answer... but I don't know what 1st Steifel-Whitney class is. Maybe I'd rather take this naive approach... – john mangual Jul 24 '11 at 18:24
Well... it sounds like since $\mathbb{T}^2\backslash \{ pt\}$ deforms to $S_1 \wedge S^1$, bundles over once-punctured torus deform to bundles over wedge of two circles. Over each circle you are checking either the bundle is a Mobius band or strip (or 3D version thereof). – john mangual Jul 24 '11 at 18:34

A punctured torus up to a homotopy-equivalence is a wedge of two circles. So your bundle is determined by its classifying map $S^1\vee S^1 \to B(O_2)$. $O_2$ fits into an extension $SO_2\to O_2\to \mathbb Z_2$, and orientable $2$-dimensional bundles over circles (and wedges of circles) are trivial. So the only data that goes into your bundle is whether or not it is orientable over each circle wedge summand. So you don't have to worry about issues like simultaneous diagonalizability. i.e. there are precisely four $2$-dimensional vector bundles over your punctured torus up to an isomorphism of bundles.
Another way to phrase the above result is that $2$-dimensional bundles over a punctured torus (by the above) are completely determined by their 1st Stiefel-Whitney class, $\omega_1 \in H^1(\mathbb T^2 \setminus \{*\};\mathbb Z_2)$. Moreover, every element of $H^1(\mathbb T^2 \setminus \{*\};\mathbb Z_2)$ is realizable as an $\omega_1$ for some $2$-dimensional bundle over $\mathbb T^2 \setminus \{*\}$. The Stiefel-Whitney class is a cohomology class which (in this case) measures whether your bundle is orientable over any given $1$-cycle in your space. In general the story is a little more complicated than that, but if you're interested the Wikipedia page gives a decent description of them.