# 11-17 Notes

\section{Uhlenbeck's lemma:$L^n$ estimates} Let $E$ be a vector bundle over $M^n$ with fiber $\C^d$. Fix a $\C^\infty$ connection $A$ in $\mathcal A_1^{n/2^*}$, set $S_A(\eps)=\set{A+a|\rd_A^*a=0,\|a\|_{L^n}< \eps}\subset \mathcal A_1^{n/2^*}$. We know that the map: $$m\mathpunct{:}(S_A(\eps)\times\mathcal G_2^{n/2^*})\bigg/\operatorname{stab}(A)\to\mathcal A_\eps^{n/2^*}$$ is injective with injective differential (i.e., diffeomorphism onto its image).

We want to show that image contains a $\mathcal G_1^n$ acts $L^n$-ball about $A$. Note that $\dim M=4$, $L_1^4\hookrightarrow L^4$. Set $W_\delta=m(S_A(\eps)\times\mathcal G_2^{n/2^*})\cap U_\delta$, where $U_\delta$ is a $L^n$-ball of radius $\delta$ in $\mathcal A_1^{n/2}$. We want to show that $W_\delta=U_\delta$ when $\delta$ is small enough.

Clearly $W_\delta\subset U_\delta$, $U_\delta$ is connected and open. We'll show that $W_\delta$ is open and closed in $U_\delta$.

Since $W_\delta$ is the intersection of open sets in the $L_1^{n/2^+}$-topology, it is open. $W_\delta$ is closed in the $L_1^{n/2^+}$ topology, i.e., $A_i=A+b_i$ is a sequence of connections in $W_\delta$, then there exists $u_i\in\mathcal G_2^{n/2^+}$, and $a_i\in L_1^{n/2^+}$, such that $u_i\cdot (A+a_i)=b_i$ and $A+a_i\in S_A(\eps)$. Since $b_i\weakto b$ in $L_1^{n/2^+}$, we can choose $u_i$, $a_i$ to be converge in $(L_2^{n/2^+}, L_1^{n/2^+})$ topology to $u\cdot a$, $\|a\|_{L^n}$ is small.

Note that $u\cdot(A+a_i)=A+b_i$, $\rd_A^*a_i=0$, $\rd_Au_i=-a_iu_i+u_ib_i$, we have a closed decomposition \begin{align*} L_k^p(X,\wedge^1\otimes\ad P)&=\rd_A(L_{k+1}^p(X,\ad P))\\ &\qquad+\ker(\rd_A^*\mathpunct{:}L_k^p(X,\wedge^1\otimes\ad P)\to L_{k-1}^p(X,\ad P))). \end{align*} Given $\theta\in L_k^p(X,\wedge^1\otimes\ad P)$, we can write $\theta=\theta^e+\theta^c$, with $\theta^e\in\operatorname{Im}\rd_A$ and $\theta^c\in\ker\rd_A^*$. \begin{lem} With the above notation, we have $$\|(ua)^c\|_{L^n}\leq C\|\rd u\|_{L^n}\|a\|_{L^n}, \quad\|(au)^c\|_{L^n}\leq C\|\rd u\|_{L^n}\|a\|_{L^n}.$$ \end{lem} \begin{lem} With the above notation, we have $$\|(ua)^e\|_{L_1^{n/2^+}}\leq C\|\rd u\|_{L_1^{n/2^+}}\|a\|_{L^n}, \quad\|(au)^e\|_{L_1^{n/2^+}}\leq C\|\rd u\|_{L_1^{n/2^+}}\|a\|_{L^n}.$$ \end{lem} \begin{proof} Since $(ua)^e=\rd_A(\rd_A^*\rd_A)^{-1}\rd_A^*(ua)$, and $\rd_A(\rd_A^*\rd_A)^{-1}:L_k^p\to L_{k+1}^p$. For \ref{lem:1}, we need to estimate the term $\|\rd_A^*(ua)\|_{L^n_{-1}}$.

Since $\rd^*_A(ua)=-*\rd_A*(ua)=-*\rd_A(u*a)=-*\rd_Au*a-*u(\rd_A*a)$, and $\rd_A*a\to0$. Thanks to $L^n\times L^n\hookrightarrow L^n_{-1}$, $L^{n/n-1}_1\hookrightarrow L^{n/n-2}$ and $L^n\times L^n\times L^{n/n-1}_1\hookrightarrow L_1^1$, we have $$\|\rd u\cdot a\|_{L_{-1}^n}\leq\|\rd u\|_{L^n}\|a\|_{L^n}.$$ \end{proof}

\begin{lem} With the above notation, we have $$\|(ua)^e\|_{L_1^{n/2^+}}\leq C\|\rd_A^*(ua)\|_{L^{n/2+}} \leq C\|\rd u\cdot a\|_{L^{n/2^+}}\leq C\|\rd u\|_{L^{n/2^+}_1}\|a\|_{L^n}.$$ \end{lem} Here we need $+$ small, such that $L^{n/2^+}_1\times L^n\hookrightarrow L^{n/2^+}$. \begin{lem} There exists $\kappa,\delta_1 >0$ such that if $A+a\in S_A(\delta_1)$, $u\in\mathcal G_2^{n/2^+}$, $u\cdot (A+a)=A+b\in\mathcal A_1^{n/2^+}$, and $$\|\rd_Au\|_{L^n}+\|a\|_{L^n}\leq\kappa\|b\|_{L^n}.$$ \end{lem} \begin{proof} Since $\|u\|=1$ a.e., \begin{align*} \|\rd_Au\|_{L^n}+\|a\|_{L^n}&=\|\rd_Au\|+\|ua\|_{L^n}\\ &\leq\|\rd_Au\|_{L^n}+\|(ua)^c\|_{L^n}+\|(ua)^e\|_{L^n}\\ &=C\|\rd_Au+(ua)^c\|_{L^n}+\|(ua)^e\|_{L^n}\\ &\leq C\|\rd_Au+ua\|_{L^n}+(C+1)\|(ua)^e\|_{L^n}\\ &\leq C\|bu\|_{L^n}+(C+1)\|(ua)^a\|_{L^n}\\ &\leq C\|b\|_{L^n}+(C+1)C'\|\rd_Au\|_{L^n}\|a\|_{L^n}. \end{align*} If $(C+1)C'\|a\|_{L^n}\leq 1/2$, then $$\frac{1}{2}\|\rd_Au\|_{L^n}+\|a\|_{L^n}\leq C\|b\|_{L^n}.$$ Now $L_1^{n/2^n}$. \begin{align*} &\|\rd_Au\|_{L_1^{n/2^+}}+\|a\|_{L_1^{n/2^+}}=\|\rd_Au\|_{L^{n/2^+}}+\|\nabla a\|_{L^{n/2^+}}+\|a\|_{L^{n/2^+}}\\ &\qquad=\|\rd_Au\|_{L^{n/2^+}}+\|u\nabla a\|_{L^{n/2^+}}+\|ua\|_{L^{n/2^+}}\\ &\qquad=\|\rd_A u\|_{L^{n/2^+}}+\|ua\|_{L^{n/2^+}}+\|\nabla u\cdot a\|_{L^{n/2^+}}\\ &\qquad\leq\|\rd_A u\|_{L^{n/2^+}}+\|(ua)^c\|_{L^{n/2^+}}+\|(ua)^e\|_{L^{n/2^+}}+\|\rd u\|_{L^{n/2^+}}\|a\|_{L^n}\\ &\qquad\leq C\left(\|\rd_Au+(ua)^c\|_{L^{n/2^+}}\right)+\|(ua)^e\|_{L^{n/2^+}}+\|\rd u\|_{L^{n/2^+}}\|\\ &\qquad\leq C\|\rd Au+ua\|_{L^{n/2^+}}+(C+1)\|(ua)^*\|_{L^{n/2^+}}+\|\rd u\|_{L^{n/2^+}}\|a\|_{L^n}\\ &\qquad\leq\left(\|b\|_{L^{n/2^+}}+\|\rd u\|_{L^{n/2^+}}\|b\|_{L^n}\right)\\ &\qquad\qquad+(C+1)C'\|\rd u\|_{L^{n/2^+}}\|a\|_{L^n}+\|\rd u\|_{L^{n/2^+}}\|a\|_{L^n}, \end{align*} thus we obtain the Lemma. \end{proof} Now , for $A+b_i\to A+b$ in $L_1^{n/2^+}$, $\|b_i\|_{L^n}=\delta_1$, there exists $u_i\in\mathcal G_1^{n/2+}$, $u_i\in S_A(C)$, such that $u_i(A+a_i)=A+b_i$, taking $\delta_i$ small enough , so that $\kappa\delta_1< \eps$, then $c_i$