Consistency condition for Yang-Mills on a Torus So I was recently studying 't Hooft's paper on self-dual solutions to Yang-Mills on $\mathbb{T}^4$. So the basic idea is that you consider a box with periodic boundary conditions and then you impose the condition that gauge invariant quantities are periodic. This means that the gauge field is periodic modulo a gauge transformation. So you associate four gauge functions with each wall of the box given by $\Omega_{\mu}(x_{\mu}=a_{\mu})$. So, if you cross a wall at some point $x_{\mu}=a_{\mu}$, then the gauge field comes back on the other side after being acted upon by the corresponding gauge transformation. At least, that is what I understand at the moment. Now, my confusion is regarding the consistency condition he imposes in order for this system to be consistent at the corners of the box. It's the idea that if you start at one corner, the gauge field should change by the same amount whichever wall you move along to reach the opposite corner. This, he claims, gives rise to the condition: $$ \Omega_2(x_1=a_1; x_2=0)\Omega_1(x_1=0; x_2=0)
 =\Omega_1(x_1=0; x_2=a_2)\Omega_2(x_1=0; x_2=0)$$ 
But now consider starting at the origin. Consider a square for simplicity with the origin being at the bottom left corner. If you move along $x_1$ to reach $(x_1=a, x_2=0)$ ($a$ is the side of the square), the gauge field at this point must be the same as at the point $(x_1=0, x_2=a)$. This just gives us $$\Omega_1(x_1=0; x_2=0)=\Omega_2(x_1=0; x_2=0)$$ Correct me if I'm misunderstanding what's going on here but I don't see why this can't be the case. Any help regarding this is highly appreciated. Also, let me know if any clarifications are required. I know my explanation is probably not very clear. So, I am attaching 't Hooft's paper here for reference.
https://projecteuclid.org/euclid.cmp/1103920245
 A: I am guessing a bit here without reading the paper. Apologies if I am on the wrong track. 
$$Ω_2(x_1=a_1;x_2=0)Ω_1(x_1=0;x_2=0)=Ω_1(x_1=0;x_2=a_2)Ω_2(x_1=0;x_2=0)$$
implies he is applying two gauge transformations. 
Start near the origin just inside the box. Move along the $x_1$ axis to just outside the box. This crosses the $x_2$ axis boundary, so you apply $Ω_2(x_1=a_1;x_2=0)$. 
Now move along the $x_2$ axis until you cross the $x_1$ boundary, putting you just outside the opposite corner. You now apply $Ω_1(x_1=a_1;x_2=a_2)$. Is there are requirement that 
$$Ω_1(x_1=a_1;x_2=a_2) = Ω_1(x_1=0;x_2=0)?$$
If so, the total transformation is 
$$Ω_2(x_1=a_1;x_2=0)Ω_1(x_1=0;x_2=0)$$
You do the same trip along opposite edges to get the right half of the equation.

Edit - Start from just outside the box near the origin, where $x_1$ and $x_2$ are both slightly negative. 
Move along the $x_1$ axis. There is an immediate gauge transformation as you cross the $x_1$ boundary, $Ω_1(x_1=0;x_2=0)$. Continue to just short of $x_1 = a_1$
Move along the $x_2$ axis. There is an immediate gauge transformation as you cross the $x_2$ boundary, $Ω_2(x_1=a_1;x_2=0)$. Continue to just short of $x_2 = a_2$
The total gauge transformation is 
$$Ω_2(x_1=a_1;x_2=0)Ω_1(x_1=0;x_2=0)$$
Do the same trip along opposite edges to get the right half of the equation.
