# 11-19 Notes

section{Uhlenbeck's lemma:$L^n$ estimates: Continued} Recall the notation $\mathcal A_1^p$, $n/2< p< n$, $\pt X$ maybe nonempty, $S_A(\eps)=\set{A+a|\rd_A^* a=0,*a|_{\pt X}=0,\|a\|_{L^n}\leq\eps}$. The map $m$ defined on $\mathcal U(\eps):=\mathcal G_2^p\times S_A(\eps)\bigg/\operatorname{stab}(A)$ is diffeomorphic to its image. For $\delta$ small enough, $B_A(\delta)\subset \mathcal U(\eps)\subset\mathcal A_1^p$, where $B_A(\delta):=\set{A|A+a,\|a\|_{L^n}< \delta}$.

Set $W(\delta)=\mathcal U(\eps)\cap B_A(\delta)$, if we can show $W(\delta)$ is a component of $B_A(\delta)$, then $W(\delta)=B_A(\delta)$.

Since $W(\delta)$ is an intersection of open sets, so it is open. To show it is also closed in $B_A(\delta)$, we have the following Lemma: \begin{lem} If $u(A+b)=A+a\in S_A(\eps)$, then $$\label{eq:2} \|\rd_Au\|_{L^n}+\|a\|_{L^n}\leq\kappa\|b\|_{L^n},$$ provided that $\|a\|_{L^n}\ll1$.

If furthermore, $\|b\|_{L^n}\ll1$, then we have $$\label{eq:3} \|\rd_Au\|_{L_1^p}+\|a\|_{L^p_1}\leq\kappa\|b\|_{L_1^p}, \quad\forall n/2< p< n.$$ \end{lem} Now, if $\set{A_i=A+b_i}\subset W(\delta)$ and $b_i\to b$ in $L_1^p$, then $A+b\in W(\delta)$. In fact, there exists $u_i$, $a_i$, such that $u_i\cdot(A+a_i)=A+b_i$ and $\|a_i\|_{L^n}< \eps$, then \eqref{eq:2} implies $\|\rd_Au_i\|_{L^n}+\|a_i\|_{L^n}\ll1$ , and \eqref{eq:3} implies that $\|\rd_Au_i\|_{L^p_1}+\|aa_i\|_{L_1^p}\ll1$ for any $n/2< p< n$. Passing to weakly convergence subsequence, we have \begin{itemize} \item $a_i\weakto a$ in $L^p_1\hookrightarrow L^n$ compactly, thus $a_i\to a$ strongly in $L^n$; \item $u_i\weakto u$ in $L^p_2\hookrightarrow L^n_1$ compactly, thus $u_i\to u$ strongly in $L_1^n$; \item $u_i\weakto u$ in $L^p_2\hookrightarrow C^0$ compactly, thus $u_i\to u$ strongly in $C^0$; \end{itemize} Now $\rd_A^*a_i=0$, thus $\rd_A^*a=0$, $\|a\|_{L^n}\leq\eps$, using \eqref{eq:2} again to obtain $\|a\|_{L^n}\leq\kappa\delta< \eps$.

Need to check that $A+a$ is gauge equivalent to $A+b$. In fact, $u_i\cdot(A=a_i)=b_i$, take limits in $L^n$, we have $\rd_Au=\lim_{i\to\infty}\rd_A u_i=\lim_{i\to\infty}u_ib_i-a_iu_i=$ub-au.

If $A$ is a $C^\infty$-connection, then there exists $\delta >0$ such that for any $A+b\in\tilde B_A(\delta)\subset\mathcal A_1^n$ , then there exists $u\in\mathcal G_2^n$ such that $u\cdot(A+b)\in\tilde S_A(\eps):=\set{A+a|\rd_A^*a=0,\|a\|_{L^n< \eps}}$. \begin{proof} We can use smooth $b_i$ to approximate $A+b$ by $A+b_i$, with $\|b_i\|_{L^n}< 2\delta$. Now $A+b_i$ are all gauge-equivalent into $S_A(\eps)$, by the previous result, There exists $u_i$, $a_i$ such that $u\cdot(A+a_i)=A+b_i$, and $\|\rd_Au_i\|_{L^n}+\|a_i\|_{L^n}\leq\kappa\|b_i\|_{L^n}$, thus $a_i\weakto a$ in $L^n$ with $\|a_i\|_{L^n< \eps}$, and $u_i\weakto u$ in $L_1^n$ which implies $u_i\to u$ strongly in $L^q$ for any $q< +\infty$. This together with $b_i\to b$ in $L^n$ strongly implies that $u_ib_i\to ub$ strongly in $L^{q-\eps}$, similarly $a_iu_i\to au$ strongly in $L^{n-\eps}$. Now, since $\rd_Au_i=u_ib_i-a_iu_i$, take weak limits in $L^n$, we have $\rd_Au_i\weakto \rd_Au$ and $u_ib_i\weakto\tilde \psi$, $a_iu_i\weakto \psi$. \end{proof} \begin{thm} There exists $\eta >0$ such that for any $A\in\mathcal A_1^2(B^4)$ with $\int_{B^4}|F_A|^2< \eta$, then there exists $u\in\mathcal G_2^2$, $\Gamma+a\in\mathcal A_1^2(B^4)$ with $\rd_A^*a=0$, $*a|_{\pt B^4}=0$ and such that $u(\Gamma+a)=A$ and $\int|\nabla_\Gamma a|^2+|a|^2\leq\kappa\int|F_A|^2$. \end{thm} \begin{proof} Set $V(\eta)=\set{A\in\mathcal A^4(B^4)|\int_{B^4}|F_A|^2}< \eta$.

\emph{Claim.} $V(\eta)$ is contractible.

In fact, if we set $A_\lambda=i_\lambda^*(A)|_{B^4(1)}$, for $\lambda >0$, then $A_\lambda\to \Gamma$ on $B^4$, since $\|a_\lambda\|_{L^4(B_1)}=\|a\|_{L^4(B_\lambda)}$, $\int_{B^4}|F_{A_\lambda}|^2\leq\int_{B^4}|F_A|^2< \eta$, thus $A\to A_\lambda$ is a diffeomorphic retration of $V(\eta)$ to a point.

Set $\tilde U(\eps):=(\mathcal G_1^4\times\tilde S_4(\eps))$.

\emph{Claim.} For $\eps$ small enough, $\tilde U(\eps)$ is an open set in $\mathcal A^4(B^4)$ We shall defer the proof of the claim.

To show $\tilde U(\eps)\cap V(\eta)$ is closed in $V(\eta)$, let $\Gamma+b_i\in \tilde U(\eps)\cap V(\eta)$ and $\Gamma+b_i\to \Gamma+b\in V(\eta)$ strongly in $L^4$, then there exists $u_i\in L_1^4$ and $a_i\in L^q$, such that $u_i\cdot(\Gamma+a_i)=\Gamma+b_i$, $\|a\|_{L^4}< \eps$, $\rd_A^*a_i=0$, $*a_i|_{\pt B}=0$. Thus $\int_{B^4}|F_{\Gamma+a_i}|^2< \eta$ implies $\int_{B^4}|\nabla_{\Gamma a_i}|^2+|a_i|^2\leq\kappa\int_{B^4}|F_{\Gamma+a_i}|^2< \kappa\eta$. We can pass to a weakly $L_1^2$ convergent sequence $a_i\weakto a$ in $L_1^2$, $u_i\weakto u$ in $L_2^2$, $u\cdot(\Gamma+a)=\Gamma+b$, $\rd_\Gamma^*a=0$, $*a|_{\pt B}=0$, and if $\eta$ is small enough, $\kappa \eta< \eps$, $\|a\|_{L^4}< \eps$, then $\tilde U(\eps)\cap V(\eta)$ is closed in $V(\eta).$ \begin{align*} C\|a\|_{L^2_1(B^4)}^2&\leq\int_{B^4}|\nabla_\Gamma a|^2\\ &=\int_{B^4}|(\rd+\rd^*)a|^2=\int_{B^4}|\rd a|^2\\ &=\int_{B^4}|F_{\Gamma+a}-a\wedge a|^2\leq\int_{B^4}|F_{A}|^2+\|a\|_{L^4}^2\|a\|_{L_1^2}^2. \end{align*} We already have Palais-Smale sequence $A_i\in A_{L_1^2}(\Sigma)$, since $\int_\Sigma|F_{A_i}|^2< \eta$, and $\rd_{A_i}^*F_{A_i}\to0$ in $L^2$, we have on sufficiently small $B_\alpha$, there exists $u_{i,\alpha}$, $a_{i,\alpha}$, such that $u_{i\alpha}\cdot(\Gamma+a_{i,\alpha})=A_i|_{\pt B_\alpha}$, $a_{i,\alpha}\to a_\alpha$ in $L^2_2(B_\alpha)$.

Set $g_{i,\beta\alpha}=(u_{i,\beta}^{-1}u_{i,\alpha})^{\pm q}$ , we want to show that $g_{i,\beta\alpha}$ convergent. $g_{i,\beta\alpha}(\Gamma+a_{i,\alpha})=\Gamma+a_{i,\beta}$. Then $g^{-1}_{i,\beta\alpha}\rd g_{i,\beta\alpha}+g^{-1}_{i,\beta\alpha}a_{i,\alpha}g_{i,\beta\alpha}=a_{i,\beta}$, thus $\rd g_{i,\beta\alpha}=g_{i,\beta\alpha}a_{i,\beta}-a_{i,\alpha}g_{i,\beta\alpha}$ on $B_\alpha\cap B_\beta$. As $a_{i,\alpha}\to a_\alpha$ in $L_2^2$ and $a_{i,\beta}\to a_\beta$ in $L_2^2$ , we conclude that $\rd g_{i,\beta\alpha}$ is bounded in $L^p$ for any $p$, futhermore $g_{i,\beta\alpha}\weakto g_{\beta\alpha}$ in $L_1^p$ thus $g_{i,\beta\alpha}\to g_{\beta\alpha}$ in $L^q$ for any $q$, which implies $\rd g_{i,\beta\alpha}\to\rd g_{\beta\alpha}$ in $L^p$ for any $p$, thus $g_{i,\beta\alpha}\to g_{\beta\alpha}$ in $L_1^p$ , that is $g_{i,\beta\alpha}a_{i,\beta}-a_{i,\alpha}g_{i,\beta\alpha}$ converges strongly in $L_1^p$, thus $\rd g_{i,\beta\alpha}\to g_{\beta\alpha}$ strongly in $L^p_2$. \end{proof}