TITLE: Is there a way to establish a correspondence between the vector bundles over a torus and some kind of homotopy groups, just as we do to spheres
QUESTION [7 upvotes]: By introducing the "clutching function" one can relate the complex (real) vector bundles on a sphere with homotopy groups of $GL_n(\Bbb C)$ ($GL_n^+(\Bbb R)$ for oriented ones). Can we do a similar thing to torus? The motivation is that I'm curious about how many vector bundles of rank 1,2,3,... are there over our familiar spaces, at least for some specified low rank. (BTW, It seems in the case of 2, plane bundles correspond to $H^2(X,\Bbb Z)$?)

REPLY [6 votes]: To actually answer the question, according to Grigory M. here: How to compute homotopy classes of maps on the 2-torus?, the homotopy classes of maps from $T^2$ into $X$ are in 1-1 correspondence with $\{(a,b,s)\in \pi_1(X)^2\times \pi_2(X)| ab=ba\}$.
What is the appropriate $X$ here?
Well, to every rank $k$ vector bundle $E$, after choosing a Riemannian structure on it, there is an principal $O(k)$ bundle for which $E$ is the associated bundle under the usual action of $O(k)$ on $\mathbb{R}^k$.
Hence, rank $k$ vector bundles over $T^2$ are in bijection with principal $O(k)$ bundles on $M$, and these are in bijection with $[T^2, BO(k)]$.  By the above, these are in bijection with $\{(a,b,s)\in \pi_1(BO(k))^2\times \pi_2(BO(k))| ab=ba\}$.
So, what are $\pi_1(BO(k))$ and $\pi_2(BO(k))$?  Well, we have a fibration $O(k)\rightarrow EO(k)\rightarrow BO(k)$ with $E$ contractible.  The associated long exact sequence of homotopy groups shows that $\pi_k(BO(k)) = \pi_{k-1}(O(k))$.
So, such bundles are classified by an element of $\pi_0(O(k))\times\pi_0(O(k))\times\pi_1(O(k))$. (the condition $ab=ba$ is automatic).  So, when $k=1$, they're classified by an element of $(\mathbb{Z}/2\mathbb{Z})^2$, rank 2 bundles are classified by $(\mathbb{Z}/2\mathbb{Z})^2\times\mathbb{Z}$, and rank $\geq$ 3 bundles are classified by $(\mathbb{Z}/2\mathbb{Z})^3$.
In the oriented case, one changes all of the $O(k)$s to $SO(k)$s, so rank $k$ oriented bundles over $T^2$ are in bijection with $\pi_0(SO(k))\times\pi_0(SO(k))\times\pi_1(SO(k)) = \pi_1(SO(k))$.  Thus we see that there is a unique oriented rank 1 bundle, the rank 2 bundles are in bijection with $\mathbb{Z}$, and the rank $\geq 3$ bundles are in bijection with $\mathbb{Z}/2\mathbb{Z}$.