# The existence and uniqueness of the curvature of a Yang-Mills connection.

I am reading Jost's Riemannian Geometry and Geometric Analysis (7-Ed) and having a question about the Yang-Mills functional (page 183).

Let $$M$$ be a compact manifold and $$E$$ be a metric bundle with dimension $$n = 2$$ whose structure group is $$SO(2)$$. For a metric connection $$D$$, its curvature $$F_D = D\circ D$$ can be viewed as a $$Ad(E)$$-valued 2-form where $$Ad(E)$$ means that each fiber is skew-symmetric. Since the Lie algebra $$o(2)$$ is trivial and the structure group is $$SO(2)$$, $$Ad(E)$$ is a trivial line bundle: $$Ad(E) \simeq M \times \mathbb{R}$$, meaning that $$F_D$$ is just an ordinary $$2$$-form (am I right?).

Jost showed that $$F_D$$ is harmonic if and only if $$D$$ is a Yang-Mills connection. Then he concluded that the existence and uniqueness of the curvature of a Yang–Mills connection are consequences of Hodge theory.

I don't understand this conclusion. Based on my understanding, the Hodge theory says that given a cohomology class $$\mu \in H^p_{dR}(M,\mathbb{R})$$, there exists a unique harmonic $$p$$-form $$f$$ such that $$\mu = [f]$$. Therefore, I would say that if $$D_1$$ and $$D_2$$ are two Yang-Mills connections such that the closed forms $$F_{D_1}$$ and $$F_{D_2}$$ are cohomologous, then $$F_{D_1} = F_{D_2}$$. Can we say or is there a theorem stating that the curvature is independent of the choice of Yang-Mills connections?

As you have noticed, $$\mathrm{Ad}(E)\cong M \times \mathbb{R}$$. We also have $$d_A=d$$ in this case. Given any connection $$1$$-form $$A$$, you have $$F_A=d_AA(=dA+[A,A])=dA$$ and the Yang-Mills equations become $$d_A^*F_A=d^*F_A=0$$ and the Bianchi-Identity (which is always true) $$d_AF_A=dF_A=0.$$ This is a partial answer to your last question: Regardless of the choice of connection $$A$$, you have $$d_A=d$$, but not necessarily $$d^*F_A=0$$. Now, for a the curvature of a connection to solve the Yang-Mills equation is equivalent to the curvature, as a "normal" 2-form, being harmonic.
But now Hodge theory tells you that each cohomology class contains exactly(!) 1 harmonic form. This is existence and uniqueness. If the curvature satisfies the YM-equations, it has to be the harmonic representative of this cohomology class. On thing which you are maybe(?) not aware of is that $$\Delta F_A=(d^*+d)^2F_A=0 \iff d^*F_A=0 \text{ and }dF_A=0.$$ You can prove this by identity by manipulating the integral $$\int_M \langle F_A, \Delta F_A \rangle dvol_g$$.
• Thank you for the answer. So we can say that given a harmonic $2$-form $f$, there exists a unique connection $1$-form $A$ such that $F_A = f$. Commented Jun 8, 2023 at 15:24
• You should re-rephrase that in terms of cohomology. You need $dA=f$ to be solveable (for $A$) and for that, you need $f$ to be exact. Commented Jun 8, 2023 at 16:20
• But the expression $D=d+A$ is local, right? And any closed form is locally exact by Poincare lemma. Could you elaborate what you mean by “this is the existence and uniqueness” a little bit more? Commented Jun 8, 2023 at 17:11
• What I’m thinking is that if we starts at a YM connection, then its curvature (which always exists) is harmonic. On the other hand, if we start at a harmonic $2$-form $f$, can we find a connection such that its curvature is $f$? This is what I have in mind when saying existence and uniqueness. Commented Jun 8, 2023 at 17:24
• No, we only showed here that $F_A$ is harmonic $\iff$ $A$ is a Yang-Mills connection. You are just sort of resolving the difference between $d_A$ and $d$. The existence(!) of a Yang-Mills connection(!) is some entirely difficult, much more different question. We only showed that, in this case, if it exists, its curvature is the unique harmonic representative of its cohomology class. Commented Jun 8, 2023 at 22:23