12-08 Notes

We consider the Yang-Mills equations in four dimensions.

Let $(X,g)$ be a Riemannian $4$-manifold. Let $\mathcal{A}$ be the space of connections for some principal bundle over $X$. For $A\in \mathcal{A}$, we have two quantities: the Yang-Mills energy $$\mathcal{E}(A) = -\int_X \mathrm{tr}\left(F_A \wedge *F_A\right) = \|F_A\|^2_{L^2}$$ and a characteristic number independent of $A$ (if $X$ is closed) $$c(A) = -\int_X \mathrm{tr}\left(F_A \wedge F_A\right).$$

Here, $*$ is the Hodge star operator $*:\Omega^2(X)\rightarrow \Omega^2(X)$ induced by the metric $g$. At a point $x \in X$, we have $*:\Lambda^2 T^*_x X \rightarrow \Lambda^2 T_x X$. We have $*^2 = 1$, so we get a splitting into $\pm 1$-eigenspaces of $*$: $$\Lambda^2 = \Lambda^2_+ \oplus \Lambda^2_-.$$ We call the former self-dual (SD) and the latter anti-self-dual (ASD). If $e^1$, $e^2$, $e^3$, $e^4$ is an orthonormal frame for $T^* X$, then \begin{eqnarray*} e^1\wedge e^2 \pm e^3 \wedge e^4 \\ e^1\wedge e^3 \mp e^2 \wedge e^4 \\ e^1\wedge e^4 \pm e^2 \wedge e^3 \end{eqnarray*} form a basis for $\Lambda^2_{\pm}$ (i.e. are self-dual and anti-self-dual respectively).

Applying this decomposition to the curvature, we have $F_A = F_A^+ + F_A^-$. Because $\Lambda^2_+$ and $\Lambda^2_-$ are orthogonal, and thus $F_A^+$ and $F_A^-$ are pointwise orthogonal, we have $$\mathcal{E}(A) = \|F_A^+\|^2_{L^2} + \|F_A^-\|^2_{L^2}$$ and in addition we get $$c(A) = \|F_A^+\|^2_{L^2} - \|F_A^-\|^2_{L^2}$$ by examination. Thus we have

\begin{lem} $\mathcal{E}(A) \geq |c(A)|$ with equality if and only if $F_A$ is purely self-dual (if $c(A) \geq 0$) or purely anti-self-dual (if $c(A) \leq 0$). \eop \end{lem}

From this we see that purely SD or ASD solutions are Yang-Mills (YM) solutions:

\begin{lem} If $F_A$ is purely self-dual or purely anti-self-dual, then it is Yang-Mills. \end{lem}

\begin{proof} Method 1: If $F_A = \mp *F_A$ then $$d_A^* F_A = -*d_A*F_A = \mp *d_AF_A = 0$$ (by Bianchi), so $F_A$ is a Yang-Mills solution. Method 2: If $F_A$ is purely self-dual or purely anti-self-dual, then by the above Lemma, $A$ is a minimal point for $\mathcal{E}(A)$, and thus is a critical point of $\mathcal{E}(A)$, i.e. a Yang-Mills solution. \end{proof}