11-05 Notes

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\section{Uhlenbeck Gauge Fixing Lemma} Let $A(t)$ be a solution to $\frac{\partial A}{\partial t}=D^*_AF_A$ and thus $\mathcal{E}(A(t_1))-\mathcal{E}(A(t_0))= \int_{t_1}^{t_0}\norm{D^*_AF_A}^2_{L^2}dt$. \begin{thm} (Uhlenbeck Foundamental Lemma) Let $P=B^n\times G\To B$ and $n\leq 4$. There exist $C,\ \epsilon>0$ so that: if $A\in\mathcal{A}_{L^2_1}$ and $\int_B|F_A|^2<\epsilon$, then there exists $u\in\mathcal{G}_{L^2_1}$ such that if $u\cdot A=\Gamma+a$, where $\Gamma$ is the trivial connection, then (a) $\int_B|\nabla_{\Gamma}a|^2+|a|^2\leq C\int_B|F_A|^2$; (b) $d^*_{\Gamma}a=0 and *a|_{\partial B^n}=0$. \end{thm} \begin{rem} (b) is the "slice" condition. And (b) is equivalent to $\int<d_{\Gamma}\xi,a>=0$, $\forall \xi\in\Omega^0(B,adP)$. \end{rem} Since $d-tr(\xi*a)=-tr(d_A\xi\wedge *a)=tr(\xi\wedge d_A*a)$, thus $\int_{\partial B}-tr(\xi*a)=-\int_Btr(d_A\xi\wedge *a)=\int_Btr(\xi\wedge *(*d_A*)a)$. If $A\To\Gamma+a$, $d_A^*a=0$, $*a|_{\partial B}=0$ and $\norm{a}_{L^2_1}$ is small, then the estimate (a) follows. Note that $\norm{F_A}_{L^2(B)}=\norm{F_{\Gamma+a}}_{L^2(B)}$, $F_{\Gamma+a}=da+a\wedge a$, thus $\int_B|da+a\wedge a|^2\geq\int_B|da|^2-C\int_B|a|^4$. Since $n\leq 4$, thus $L^2_1\hookrightarrow L^4$ and $\int_B|a|^4\leq C_1\norm{a}_{L^2_1}^4$.

If $\int_B|da|^2\geq C_2\norm{a}_{L^2_1}$, denoted by (*), then \begin{eqnarray*} \int_B|F_A|^2&\geq& C_2\norm{a}_{L^2_1}^2-CC_1\norm{a}^4_{L^2_1}\\ &\geq&(C_2-CC_1\norm{a}^2_{L^2_1})\norm{a}_{L^2_1}^2. \end{eqnarray*} Now the question is that when does (*) hold. \begin{thm} Suppose $M$ is a manifold with boundary and $H^1(M)$ is represented by harmonic form, i.e. $\{\theta|(d+d^*)\theta=0 and *\theta|_{\partial M}=0\}$. Find $\theta$ by minimizing $\norm{\theta+df}_{L^2}$ with $d\theta=0$. If $H^1(M)=0$, then the map $d+d^*:\ \{\alpha|*\alpha|_{\partial M}=0\}\To\Omega^0_{L^2}+\Omega^2_{L^2}$ is injective. \end{thm} Since $(d+d^*)^2=\nabla^*\nabla+$Riemannian curvature, which is zero in flat ball, thus $\int|(d+d^*)a|^2=\int|\nabla a|^2$, which holds on the $B^n$ with $*a|_{\partial B}=0$.

There are two problems:(1)There might be a small energy connection without gauge representative near trivial connection; (2)Assume the connection was gauge-equivalent into the slice.

We showed that if we fix $\Gamma\in\mathcal{A}_{L^2_k}$, then there exists an $\epsilon$ such that \begin{equation*} \tilde{S}_{\Gamma,\epsilon}=\{\Gamma+a|d_A^*a=0, *a|_{\partial M}=0, \norm{a}_{L^n}<\epsilon\} \end{equation*} is injective into the quotient space $\mathcal{A/G}$, that is $\tilde{S}_{\Gamma,\epsilon}\times \mathcal{G}_{stab(\Gamma)}\To\mathcal{A}$ is an injection.

Hope: if $A$ is an $L^n-$connection, then there exists $u\int\mathcal{G}_{L^n_1}$ so that $u\cdot A=\Gamma+a$ and $a\in\tilde{S}_{\Gamma,\epsilon}$. \begin{lem} There exists a $\mathcal{G}-$invariant neighborhood $U$ of $\Gamma$ in the $L^n$ topology so that every connection $A\in U$ is gauge-equivalent into the slice $\tilde{S}_{\Gamma,\epsilon}$. \end{lem} Note that rescaling decreases the Yang-Mills energy. We want to prove for $L^n$ ball of radius $\delta$, but we cannot use implicit function theorem since we are on the border line. Instead, we use open/closed argument. We want to sharpen result to show that \begin{equation*} m: (\tilde{S}_{\Gamma,\epsilon}\times \mathcal{G}_{L^2_{k+1}})/stab(\Gamma)\To \mathcal{A}_{L^2_k} \end{equation*} is a diffeomorphism onto its image. We need to show that the differential is an isomorphism. This map is equivariant, so it suffices to check this at $(\Gamma+a,1)$. note that $\mathcal{D}_{(\Gamma+a,1)}m(\alpha,\xi) =d_{\Gamma}\xi+[a,\xi]+\alpha$.

For simplicity, given that $\ker(d_{\Gamma})=0$, prove the injectivity. Suppose $\mathcal{D}_{(\Gamma+a,1)}m(\alpha,\xi)=0$. Take inner product with $d_A\xi$,thus

\begin{eqnarray*} \int|d_A\xi|^2+\int<[a,\xi],d_A\xi> +\int<\alpha,d_A\xi>&=&0.\\ \end{eqnarray*} Since $\alpha$ is in slice and $\frac{1}{n}+\frac{n-2}{2n}+\frac{1}{2}=1$, thus \begin{eqnarray*} \int|d_{\Gamma}\xi|^2&\leq&\norm{a}_{L^n} \norm{\xi}_{L^{\frac{2n}{n-2}}} \norm{d_{\Gamma}\xi}_{L^2}. \end{eqnarray*} Since $\ker(d_\Gamma)=0$ and $\norm{\xi}_{L^2_1}\leq C\norm{d_\Gamma\xi}_{L^2}$, denoted by (**), thus \begin{eqnarray*} \int|d_\Gamma\xi|^2&\leq& C\norm{a}_{L^n}\norm{d_\Gamma\xi}^2_{L^2}. \end{eqnarray*} Hence $\norm{d_\Gamma}_{L^2}=0$ if $\norm{a}_{L^n}$ is small in the case $n\neq 0$.

If $\ker(d_\Gamma)\neq 0$, then we can assume that $(\alpha,\xi)$ is a representative of the tangent space of the quotient $(\tilde{S}_{\Gamma,\epsilon}\times\mathcal{G}) /stab(\Gamma)$ at $(\Gamma+a,1)$, that is, $(\alpha,\xi)$ is orthogonal to $([\alpha,\xi],\xi_0)$, $\xi_0\in\ker(d_\Gamma)$, and thus $\xi\perp\ker(d_\Gamma)$. For such $\xi$, the estimate (**) holds and the argument above works. Therefore $\xi=0$ implies that $\alpha=0$ and $\mathcal{D}_{(\Gamma+a,1)}m$ is injective. $\mathcal{D}_{(\Gamma+a,1)}m$ is a continuously varying family of Fredholm opertors. Since $\mathcal{D}_{(\Gamma,1)}m$ is invertible, thus $index=0$, and the injectivity implies the isomorphism.

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