11-12 Notes
\section{Uhlenbeck's Lemma: Continued} \begin{lem}[Uhlembeck's Lemma] Suppose $n\leq4$, then there exists $\eps$, $C >0$, such that if $A\in L_1^2(B^n,\wedge^1\otimes g)$ and $\int_{B^n}|F_A|^2*1\leq \eps$, then $u\in\mathcal{G}_{L_2^2}$, and satisfies $u\cdot A=\Gamma+a$, $\rd^*a=0$, $*a|_{\pt B}=0$ and $\int_{B^n}|\nabla_\Gamma a|^2+|a|^2\leq C\int_{B^n}|F_A|^2$. \end{lem} \begin{proof} Step1. Given $A$ as a $L_k^2$-connection on $M^n$, there exists $\eps,\delta >0$, such that if $A'=A+a$, $\|a\|_{L^n}\leq \delta$, then there exists a $u\in\mathcal{G}_{L_1^n}$ such that $u\cdot A'=A+a'$, and $\rd^*_A a'=0$, $*a'|_{\pt B^n}=0$, and $\|a'\|_{L^n}\leq\eps$.
There is a concern about the proof, so we will defer it. \end{proof} \begin{rem} Note that $\int_{B^n}|F_A|^2< \eps$ implies that \end{rem} In fact, \begin{align*} \int_{B^n}|F_A|^2*1&=\int_{B(1)}|F_{s^*_\rho(A)}|^2s^*_\rho(*1) =\int_{B^n}\tr(s^*_\rho(F_A)\wedge s^*_\rho(*)s^*_\rho(F_A))\\ &=\int_{B^n}\tr(s^*_\rho(F_A)\wedge*s_\rho^*(F_A)\rho^{n-4}, \end{align*} since $*:\wedge^k\to\wedge^{n-k}$ on $\R^n$, $*(\rd x^{i_1}\wedge\rd x^{i_2}\cdots\wedge\rd x^{i_k})=\pm\rd x^{j_1}\wedge\rd x^{j_2}\cdots\wedge\rd x^{j_{n-k}}$, and $s_\rho^*(*)\rho^k\rd x^{i_1}\wedge\rd x^{i_2}\cdots\wedge\rd x^{i_k})=\eps \rho^{n-k}\rd x^{j_1}\wedge\rd x^{j_2}\cdots\wedge\rd x^{j_{n-k}},$ we have, $$ s_\rho^*(*)=\rho^{n-2}**. $$ On $B(\rho)$, we need $$ \int_{B(\rho)}\tr(F_A\wedge*F_A)\leq\eps\rho^{n-4}. $$ If $n< 4$, then constant increases as $\rho\to0$. \begin{cor} If $n< 4$ and we have a sequence of connections with $\int_M|F_A|^2< \kappa$, then there exists a cover of $M$ by balls where $\int_B|F_A|^2\leq \eps\rho^{n-4}$, so on each ball, $\int_B|\rd_\Gamma a_i|^2+|a_i|^2\leq C\kappa$. \end{cor} If $n=4$ and $\int_M|F_{A_i}|^2\leq\kappa$, try to cover $M$ by balls of radius $\rho$, $\set{B_{x_i}(\rho)|i\in\set{1,2,\ldots,n}}$. Choose the cover so that each point at most belongs to $N$ balls. For each ball, where $\int_{B_{n_i}(\rho)}|F_{A}|^2\geq\eps$, call this bad. $$ \kappa\geq\int_M|F_{A_1}|^2\geq\int_{\text{Bad balls}}|F_A|^2\geq\frac{1}{N}\sum_{\text{bad balls}}\int_{B}|F_A|^2\geq\frac{\eps}{N}{}^\#\set{\text{bad balls}}, $$ thus $$ {}^\#\set{\text{bad balls}}\leq\frac{\kappa N}{\eps}, $$ which is independent of $\rho$!
Now cover the union of bad balls by balls of $1/2$ radius, again at most $\frac{\kappa N}{\eps}$ are bad for new cover. Iterate, the sequence of centers of balls converge to some limit points $\set{x_1,\ldots,x_l}$. In complement of $\set{x_1,\ldots,x_l}$, each point is contained in a ball where $\int_B|F_{A_i}|^2\leq\eps$.
Image $A_i$ on $\Sigma$, such that $\int|F_{A_i}|^2\leq\kappa$ and $\|D_{A_i}^*F_{A_i}\|_{L^2}\to0$. Choose a ball $B(\rho)\subset\Sigma$, such that Uhlenbeck's Lemma holds for $M$, after a gauge transformation, $a_i$ satisfies: $\rd^*a_i=0$, $*a_i|_{\pt B}=0$ and $\int_B|\nabla_ia_i|^2+|a_i|^2\leq C\int|F_{A_i}|^2$. Passing to a subsequence to obtain $a_i\weakto a$ in $L_1^2(B)$. Since $u_i\cdot A_0=\Gamma+a_i$, then $F_{u_i\cdot A_i}=\rd a_i+a_i\wedge a_i$, and \begin{align*} D_{u_i\cdot A_i}^*F_{u_i\cdot A_i}&=(\rd +a_i)^*(\rd a_i+a_i\wedge a_i)\\ &=\rd^*\rd a_i-*d*(a_i\wedge a_i)-*a_i\wedge(*\rd a_i)-*a_i\wedge (*(a_i\wedge a_i))\\ \end{align*} which is tends to $0$ in $L^2$. We have $\Delta a_i-(*d*(a_i\wedge a_i)+*a_i\wedge(*\rd a_i)+*(a_i\wedge*(a_i\wedge a_i)))\to 0$, in $L^2$. We want to show that $\Delta a_i\in L^2$, thus $a_i\in L_2^2$. In fact, $$
- d*(a_i\wedge a_i)+*a_i\wedge(*\rd a_i)\to\nabla a_i\cdot a_i\in L^2\otimes L_1^2,
$$ and note that $L_1^2\hookrightarrow L^p$ (but not $L^\infty$), then $$
- (a_i\wedge*(a_i\wedge a_i))\to a_i^3\in L^2.
$$ Now $L^{2-s}\hookrightarrow L^2_{-\frac{1}{2}}:=(L^2_{-\frac{1}{2}})^*$ is compact for $s$ is small enough. In $L^2_{-\frac{1}{2}}$, we can pass to a subsequence where $\nabla a_i\cdot a_i$ and $a_i^3$ converge strongly. Thus $a_i$ converge strongly in $L^2_{\frac{3}{2}}$. $\nabla a_i$ is strongly converge in $L^2_{\frac{1}{2}}$, as is $\nabla a_i\cdot a_i$.
Now, $\Delta a_i+b_i\to0$ in $L^2$ implies that $\delta a_i+b_i\to 0$ in $L^2_{-\frac{1}{2}}$, and $b_i\to b$ in $L^2_{-\frac{1}{2}}$, thus $a_i\to a$ in $L^2_{-\frac{3}{2}}$. Finally, we conculde that $a_i\to a$ in $L^2_2$, since $b_i\to b$ in $L^2_{\frac{1}{2}}$ and $\Delta a_i+b_i\to 0$ in $L^2$.
