I read a [physics paper](https://arxiv.org/abs/1212.3324), 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](https://projecteuclid.org/journals/communications-in-mathematical-physics/volume-62/issue-3/Some-remarks-on-the-paper-of-Callias--Axial-anomalies/cmp/1103904396.full), 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.



How to classify continuous/differentiable maps from $T^2$ to $U(N)$?
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
$$ \frac{1}{2\pi i} \oint\_{S^1} \frac{g'\_k(z)}{g\_k(z)}\, dz. $$


