Physics Asked on May 11, 2021
I know that the Yang-Mills equation is $delta F=0$ where $delta$ is the adjoint operator of the covariant derivative and $F$ is the curvature of a principal connection. In a trivial $U(1)$-bundle (for example $mathbb{R}^4times U(1)$ as the one used in electrodynamics), $F=dA$ for some 1-form $A$, so the Yang-Mills equation is $delta dA=0$, but considering the inner product $langle-,-rangle$ we have for all 1-form $B$ $$0=langle B,delta Frangle=langle B,delta dArangle=langle dB,dArangle$$ in particular for $B=A$, so $$0=langle dA,dArangle Longrightarrow 0=dA=F,$$ i.e., if a connection satisfies the Yang-Mills equation then its curvature is zero. But if this is correct, in $mathbb{R}^4times U(1)$ we have that the electromagnetic field is always zero (F=0) which is clearly wrong. What is going on?? I can’t see it!! Is this apparent contradiction because in $mathbb{R}^4$, $langle-,-rangle$ is not defined for all 1-forms?? or because it is not an inner product (taking the Minkowski metric)??
How does $int_M F wedge star F = 0$ imply that $F = 0$? $int_M F wedge star F $ corresponds to the Lagrangian of the field, which is the integral of $E^2 - B^2$. That can be $0$ even if $F$ isn't $0$. This is actually satisfied for a plane wave of light.
In Lorentzian signature, the inner product $int_M (cdot) wedge star (cdot)$ is not positive definite, so the fact that its zero doesn't mean the input is $0$.
Answered by user1379857 on May 11, 2021
The answer is in fact contained in the last line of the question: the reason is that it is not an inner product (taking the Minkowski metric). If we were studying Yang-Mills theory in a Euclidean space (with appropriate boundary conditions), then indeed F=0 would be the only solution (waves exist only for indefinite metric)
Answered by Mikhail Skopenkov on May 11, 2021
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