# A doubt about Yang-Mills equation in trivial $U(1)$-bundles

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 $$\mathbb{R}^4\times U(1)$$ 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 $$B$$ $$0=\langle B,\delta F\rangle=\langle B,\delta dA\rangle=\langle dB,dA\rangle$$ in particular for $$B=A$$, so $$0=\langle dA,dA\rangle \Longrightarrow 0=dA=F,$$ i.e., if a connection satisfies the Yang-Mills equation then its curvature is zero. But if this is correct, in $$\mathbb{R}^4\times U(1)$$ 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)??

• The E&M equation of motion is $d \star F = J$, so in vacuum $d \star F = 0$. If you got a different result then there must be a mistake somewhere. I'm not seeing the Hodge star $\star$ in what you've written here. Commented Apr 23, 2021 at 16:19
• @user1379857 the Hodge star is implicit in the co-differential: $\delta\sim\star d\star$ Commented Apr 23, 2021 at 16:22
• @user1379857 I know, but I can't see it. My problem is not the equation of motion, my problem is that apparently, just F=0 satisfies the equation of motion, which is wrong but I don't know why.
– GaSa
Commented Apr 23, 2021 at 16:22

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 $$0$$.

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)

The bundle $$\mathbb{R}^4\times U(1)$$ is trivial, and thus admits a flat connection $$A$$, namely the Maurer-Cartan form $$g^{-1}dg$$. The fact that this is flat, just means that $$dA=0$$, and this is certainly a solution to the Yang-Mills equations in a vacuum. However, it is a trivial solution, and we are, in general, interested in non trivial solutions.

What you are saying is like looking at the Maxwell field equations in a vacuum, and then concluding that $$E=B=0$$, which is certainly a solution to that set of PDE's, but it is not the only solution, and is indeed a trivial solution.

Edit:

Furthermore, your inner product is really:

\begin{align*} \langle \omega,\eta\rangle_{L^2}=\int_{\mathbb{R}^4}\langle \omega,\eta\rangle\text{dvol}_g \end{align*} where: \begin{align*} \langle \omega,\eta\rangle=\frac{1}{k!}\omega_{i_1\cdots i_k}\eta^{i_1\cdots i_k} \end{align*} and $$\text{dvol}_g$$ is your volume form. For general pseudo-Riemannian metrics, this inner product does not satisfy: \begin{align*} \langle \omega,\omega\rangle_{L^2}\geq 0 \end{align*} It is however non-degenerate, meaning that if: \begin{align*} \langle \omega,\eta\rangle_{L^2}=0 \end{align*} for all $$\eta$$, then $$\omega=0$$. In your case, we have that $$\delta dA=0$$, so your reasoning that: \begin{align*} \langle A,\delta dA \rangle_{L^2}=0\Rightarrow \langle dA, dA \rangle_{L^2}=0 \end{align*} is correct. However, you have not shown that $$\langle dA,\omega\rangle_{L^2}=0$$ for all $$\omega\in \Omega^2(M)$$. You have shown that for all exact $$2$$ forms, i.e. those that can be written as $$dB$$, that the $$L^2$$ inner product is zero, but this is simply not the case in general.