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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.

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.

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 $Ω_1(x_1=a_1;x_2=a_2)$. This is a torus, so periodicity implies that $$Ω_1(x_1=a_1;x_2=a_2) = Ω_1(x_1=0;x_2=0)$$

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.

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.

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 $Ω_1(x_1=a_1;x_2=a_2)$. Does periodicity imply 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.

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 $Ω_1(x_1=a_1;x_2=a_2)$. This is a torus, so periodicity implies that $$Ω_1(x_1=a_1;x_2=a_2) = Ω_1(x_1=0;x_2=0)$$

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.