How to classify continuous/differentiable maps from $T^2$ to $U(N)$?

Question

I read a physics paper, of which the main idea is based on the topological classification of maps from 3-torus to the space of $N\times N$ unitary matrices. To quote their equation (4), which gives a winding number: $$W(U)=\frac{1}{8\pi^2}\int_{T^3} dtdk_xdk_y U^{-1}\partial_t U\cdot[U^{-1}\partial_{k_x} U,U^{-1}\partial_{k_y} U],$$ where $(t,k_x,k_y)\in S^1\times S^1\times S^1\cong T^3$, and $U:=U(t,k_x,k_y)$ taking values of N-dimensional unitary matrices. The formula comes from this math paper, which is not really comprehensible to me given my limited math background.

Now I'm interested in the classification (up to homotopy) of maps from 2-torus to the space of $N\times N$ unitary matrices, is this already well developed? Any explanation or reference will be appreciated.

PS. I'm familiar with the language of fundamental groups (but not many sophisticated theorems) and a little bit homology. Also some basics on differentiable geometry. If possible, please explain in a way accessible to me, but if this means too much compromise, just go ahead and shower me with the proper mathematics.

Answer 1

You are asking about the structure of the set $M=[T^2,U(N)]$ of homotopy classes of maps from $T^2$ to $U(N)$. The functor $H_1$ gives a map $M\to\text{Hom}(H_1(T^2),H_1(U(n)))$, and standard calculations show that $H_1(T^2)=\mathbb{Z}^2$ and $H_1(U(N))=\mathbb{Z}$, so we end up with a map $\phi\colon M\to\mathbb{Z}^2$. We can use the group structure on $U(N)$ to make $M$ into a group, and it is standard that $\phi$ is a homomorphism with respect to this structure.

If we let $p_i$ denote the $i$'th projection $T^2\to S^1=U(1)\subset U(N)$, then $\phi(p_1)=(1,0)$ and $\phi(p_2)=(0,1)$, so $\phi$ is surjective.

In fact, $\phi$ is injective as well. One way to see this is to check that the multiplication map $U(1)\times SU(N)\to U(N)$ is a homeomorphism, so $[T^2,U(n)]=[T^2,U(1)]\times[T^2,SU(N)]$. The first factor is $\mathbb{Z}^2$ again. The second factor is trivial, because $T^2$ is $2$-dimensional, and $SU(N)$ is $2$-connected (as one can prove using $SU(2)=S^3$ and the standard fibrations $SU(N)\to SU(N+1)\to S^{2N+1}$).

For another point of view, one can argue using the above fibrations that $M$ is independent of $N$, so we can take $N=\infty$, in which case $M$ becomes the complex $K$-theory group $KU^1(T^2)$.

UPDATE: We can write an integral formula as follows. For $k=1,2$ let $i_k\:S^1\to T^2$ be the $k$'th axis inclusion. Then the $k$'th component of $\phi(f)\in\mathbb{Z}^2$ is just the ordinary winding number of the composite $$ g_k = (S^1 \xrightarrow{i_k} T^2 \xrightarrow{f} U(N) \xrightarrow{\det} U(1)=S^1). $$ If we identify $S^1$ with the unit circle in $\mathbb{C}$, this is just $$ \frac{1}{2\pi i} \oint_{S^1} \frac{g'_k(z)}{g_k(z)}\, dz. $$