# 09-24 Notes

\section{Injectivity} Suppose that $A\in \mathcal{A}^p_k$ and $(k+1)p>n$. For $\forall \epsilon>0$, define $$S_{A,\epsilon}=\{A+a|\ d_A^*a=0,\ \norm{a}_{L^p_k}<\epsilon\}$$ and the map $$m:S_{A,\epsilon}\times \mathcal{G}^p_{k+1}\To \mathcal{A}_k^p,\ m(A+a,u)=A+u^{-1}d_Au+u^{-1}au.$$ This induces a map $$\bar{m}: S_{A,\epsilon}\times\mathcal{G}^p_{k+1}/Stab_A\To \mathcal{A}^p_k.$$ Thus we have the following property: \begin{prop} There exists an $\epsilon_0>0$, such that $\forall \epsilon<\epsilon_0$, the map $\bar{m}$ is a diffeomorphism onto a neighborhood of $A$. \end{prop} \begin{proof} Due to the discussion in the former lectures, it remains to show that $\bar{m}$ is injective. Suppose $A+a,\ A+b\in S_{A,\epsilon}$ with the same image under the map $m$. Thus \begin{equation*} u^{-1}au+u^{-1}d_Au=b\Rightarrow d_Au=ub-au. \end{equation*} Composing both sides with $d_A^*$, we have \begin{eqnarray*} d_A^*d_Au&=&-*d_A*(ub-au)\\ &=&-*d_A(u*b-*au)\\ &=&-*(d_Au\wedge*b+u\wedge d_A*b-d_A*a\wedge u+*a\wedge d_Au)\\ &=&-*(d_Au\wedge*b)-*(*a\wedge d_Au). \end{eqnarray*} Take the inner product with $u^0$, the component of $u$ which is $L^2$-orthogonal to $Ker d_A$. Thus H\"{o}lder inequality, $$\norm{d_Au^0}_{L^2}^2\leq \norm{d_Au^0}_{L^2} (\norm{a}_{L^n}+\norm{b}_{L^n}) \norm{u^0}_{L^{\frac{2n}{n-2}}}.$$ Denote the norm of the Sobolev embedding $L^2_1\hookrightarrow L^{\frac{2n}{n-2}}$ by $C_{sob}$, thus we have $$\norm{d_Au^0}_{L^2}\leq C_{sob} (\norm{a}_{L^n}+\norm{b}_{L^n})\norm{u^0}_{L^2_1}.$$ Since $d_A: L^2_1(X, End(E))\To L^2(X,End(E))$ has closed range with finite dimenstional kernel, thus $$\norm{u^0}_{L^2_{1,A}}\leq C(\lambda_1)\norm{d_Au^0}_{L^2},$$ where the constant depends only on the fist positive eigenvalue $\lambda_1$ of $d_A$, that is \begin{equation*} \lambda_1=inf\frac{\norm{d_Au^0}_{L^2}} {\norm{u^0}_L^2}. \end{equation*} Recall the Kato inequality, that is, if $A$ is the metric connection on the bundle $E\To X$ and $s$ is a section of $E$, then $|d|s||\leq|\nabla_As|,\ a.e.$. Hence $\norm{u^0}_{L^2_1}\leq\norm{u^0}_{L^2_{1,A}}$ and then $$\norm{d_Au^0}_{L^2}\leq C_{sob}\cdot C(\lambda_1)\norm{d_Au^0}_{L^2}(\norm{a}_{L^n} +\norm{b}_{L^n}).$$ Thus if we choose $\epsilon>0$ sufficiently small, then we could get contradiction from above unless $u\in Stab_A$. Therefore the proposition holds for sufficient small positive $\epsilon$. \end{proof} \begin{rem} Observe that the map $\bar{m}$ is also injective on a small $L^n$-ball in the slice not just the $L^p_k$-ball, which follows from the Multipication theorem. And the $L^n$ norm of $1$-form is scaling invariant. Indeed, let $\delta_\lambda: \Real^n\To\Real^n$ be $\delta_\lambda(x)=\lambda x$. For $a\in\Omega^1_c(\Real^n)$, \begin{eqnarray*} \int_{\Real^n} |\delta_\lambda^*a|^ndx^1\wedge\cdots\wedge dx^n&=&\int_{\Real^n}\lambda^n|a(\lambda x)|^ndx^1\wedge\cdots\wedge dx^n\\ &=&\int_{\Real^n}|a|^ndx^1\wedge\cdots\wedge dx^n. \end{eqnarray*} Thus for $\forall\epsilon>0$, define $$\tilde{S}_{A,\epsilon}=\{A+a|\ d_A^*a=0,\ a\in L^p_k,\ \norm{a}_{L^n}\leq\epsilon\}.$$ Then we can define $m:\tilde{S}_{A,\epsilon}\times\mathcal{G}^p_{k+1} \To\mathcal{A}^p_k$ and $\bar{m}:\tilde{S}_{A,\epsilon}\times \mathcal{G}^p_{k+1}/Stab_A \To\mathcal{A}^p_k$ identically with the former maps, and have the following property, \begin{prop} $\bar{m}$ is a diffeomorphism onto an $L^n$ ball about $A$ in $\mathcal{A}^p_k$. \end{prop} \end{rem} \section{Yang-Mills Functional} We now consider the equation involving curvature. First define the Yang-Mills functional. Given $A\in \mathcal{A}^p_k$ a metric connection on the bundle $E\To X$, $(k+1)p>n$, and the curvature $F_A=dA+A\wedge A$. \begin{defn} Yang-Mills functional $\mathcal{E}(A)=\int_X|F_A|^2dvol$. \end{defn} Especially, in dimension $2$ and $4$, the Yang-Mills functional has nice properties: in dimension $4$, it is scaling invariant, since the $L^2$ norm of $2$-form is scaling invariant and is invarant under the conformal change of metric.\\ Now we would like to derive the Euler-Lagrangian equation for Yang-Mills functional. By definition, for $t\in \Real$ and $a\in L^p_k$, \begin{eqnarray*} \frac{d}{dt}\mathcal{E}(A+ta)|_{t=0} &=&\frac{d}{dt}\int_X-tr(F_{A+ta}\wedge *F_{A+ta})dvol|_{t=0}\\ &=&-2\int_Xtr(F_A\wedge*d_Aa)dvol\\ &=&<F_A, d_Aa>_{L^2}.\\ \end{eqnarray*} Therefore the Euler-Lagrangian equation of $\mathcal{E}$ is $d_A^*F_A=0$, which is also called Yang-Mills equation.\\
Observe that the Bianchi identity gives that $d_AF_A=0$, and thus the solution $F_A$ of the Yang-Mills equation is harmonic $2$-form. Locally the Yang-Mills equation is $$d^*_{a^\tau}(da^\tau+a^\tau\wedge a^\tau)=0,$$ where $A=d+a^\tau$. Thus $$d^*da^\tau+Q(a^\tau,\nabla a^\tau)+C(a^\tau,a^\tau,a^\tau)=0,$$ where $Q$ is some quardratic term and $C$ is some cubic term. Note that in dimension greater or equal to $2$, the linearized equation of the Yang-Mills equation is not elliptic.\\
In dimension $2$, the solution $F_A$ of Yang-Mills equation satisfies that $*F_A\in\Omega^0(X, End(E))$ is parallel. Suppose that $\xi$ is a paralle section of skew Hermitian bundle $U(E)\To X$. Decompose $E$ into direct sum of the eigenspace $E_j$, which corresponds to the eigenvalue $i\lambda_j$ and $\lambda_j\in\Real$. Thus $\xi|_{E_j}=i\lambda_j$. If $e_\lambda$ is a local eigensection of $E$, $\xi e_\lambda=i\lambda e_\lambda$, then \begin{equation*} \nabla_X^A(\xi e_\lambda)=\nabla^A_X\xi\cdot e_\lambda+\xi\nabla^A_Xe_\lambda =i\lambda(\nabla^A_Xe_\lambda). \end{equation*} Hence $\nabla^A$ preserves the eigenspace. It follows that $*F_A$ is a scaling endomorphism. And the eigenvalues of a parallel endomorphisms are constants and the eigenspaces are vector bundle which is projections of flat bundles.