Let A and B r × r matrices. Consider the matrix valued one form the positive reals: $$ \omega := (A/x) dx + (B/y) dy$$ with $x,y$ in the positive reals Which condition on A and B guarantees that $$\nabla = d + \omega$$ defines a flat connection?
So i have a couple of problems by figureing this out. First I know that for a connection to be flat we want $F(\nabla)=d\omega+\omega\wedge\omega=0$. But I am not sure on how to compute both of these parts.
This are my ideas so far:
For $d\omega$ I was thinking $d((A/x) dx)= (-A/x^2)dx\wedge dx$ idem for B but I am confused since i thought $dx\wedge dx=0$ since the wegde seems anti-symmetric to me.
For $\omega\wedge\omega$ I was thinking the following: $$\omega\wedge\omega=?=(A/x\wedge B/y)dx\wedge dy +(B/y\wedge A/x)dy\wedge dx $$ here I again thought the AA and BB part are zero since I have terms of the form $a\wedge a$. Now I could use anti symmetry and reduce the above to $$((A/x\wedge B/y)-(B/y\wedge A/x))dx\wedge dy$$ but now I have no clue anymore on how to continue
Any help would be great.
Simplifying the expression and using commutator notation gives $\frac{1}{xy} [A,B] dx \wedge dy$. If you want this to be $0$, you see that $[A,B]=0$. Everything else in that expression is nonvanishing on the domain. That is to say $A$ and $B$ should commute.