11-12 Notes

Uhlenbeck’s Lemma: Continued

Lemma 1 (Uhlembeck’s Lemma). Suppose \(n\leq4\), then there exists \(\epsilon\), \(C >0\), such that if \(A\in L_1^2(B^n,\wedge^1\otimes g)\) and \(\int_{B^n}|F_A|^2*1\leq \epsilon\), then \(u\in\mathcal{G}_{L_2^2}\), and satisfies \(u\cdot A=\Gamma+a\), \(\mathrm{d}^*a=0\), \(*a|_{\partial B}=0\) and \(\int_{B^n}|\nabla_\Gamma a|^2+|a|^2\leq C\int_{B^n}|F_A|^2\).

Proof. Step1. Given \(A\) as a \(L_k^2\)-connection on \(M^n\), there exists \(\epsilon,\delta >0\), 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 \(\mathrm{d}^*_A a'=0\), \(*a'|_{\partial B^n}=0\), and \(\|a'\|_{L^n}\leq\epsilon\).

There is a concern about the proof, so we will defer it. ◻

Remark 1. Note that \(\int_{B^n}|F_A|^2< \epsilon\) implies that

In fact, \[\begin{aligned} \int_{B^n}|F_A|^2*1&=\int_{B(1)}|F_{s^*_\rho(A)}|^2s^*_\rho(*1) =\int_{B^n}\operatorname{tr}(s^*_\rho(F_A)\wedge s^*_\rho(*)s^*_\rho(F_A))\\ &=\int_{B^n}\operatorname{tr}(s^*_\rho(F_A)\wedge*s_\rho^*(F_A)\rho^{n-4}, \end{aligned}\] since \(*:\wedge^k\to\wedge^{n-k}\) on \(\mathbb{R}^n\), \(*(\mathrm{d}x^{i_1}\wedge\mathrm{d}x^{i_2}\cdots\wedge\mathrm{d}x^{i_k})=\pm\mathrm{d}x^{j_1}\wedge\mathrm{d}x^{j_2}\cdots\wedge\mathrm{d}x^{j_{n-k}}\), and \(s_\rho^*(*)\rho^k\mathrm{d}x^{i_1}\wedge\mathrm{d}x^{i_2}\cdots\wedge\mathrm{d}x^{i_k})=\epsilon\rho^{n-k}\mathrm{d}x^{j_1}\wedge\mathrm{d}x^{j_2}\cdots\wedge\mathrm{d}x^{j_{n-k}},\) we have, \[s_\rho^*(*)=\rho^{n-2}**.\] On \(B(\rho)\), we need \[\int_{B(\rho)}\operatorname{tr}(F_A\wedge*F_A)\leq\epsilon\rho^{n-4}.\] If \(n< 4\), then constant increases as \(\rho\to0\).

Corollary 1. 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 \(\int_B|F_A|^2\leq \epsilon\rho^{n-4}\), so on each ball, \(\int_B|\mathrm{d}_\Gamma a_i|^2+|a_i|^2\leq C\kappa\).

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 \(\int_{B_{n_i}(\rho)}|F_{A}|^2\geq\epsilon\), call this bad. \[\kappa\geq\int_M|F_{A_1}|^2\geq\int_{\text{Bad balls}}|F_A|^2\geq\frac{1}{N}\sum_{\text{bad balls}}\int_{B}|F_A|^2\geq\frac{\epsilon}{N}{}^\#\set{\text{bad balls}},\] thus \[{}^\#\set{\text{bad balls}}\leq\frac{\kappa N}{\epsilon},\] which is independent of \(\rho\)!

Now cover the union of bad balls by balls of \(1/2\) radius, again at most \(\frac{\kappa N}{\epsilon}\) 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 \(\int_B|F_{A_i}|^2\leq\epsilon\).

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: \(\mathrm{d}^*a_i=0\), \(*a_i|_{\partial B}=0\) and \(\int_B|\nabla_ia_i|^2+|a_i|^2\leq C\int|F_{A_i}|^2\). Passing to a subsequence to obtain \(a_i\rightharpoonup a\) in \(L_1^2(B)\). Since \(u_i\cdot A_0=\Gamma+a_i\), then \(F_{u_i\cdot A_i}=\mathrm{d}a_i+a_i\wedge a_i\), and \[\begin{aligned} D_{u_i\cdot A_i}^*F_{u_i\cdot A_i}&=(\mathrm{d}+a_i)^*(\mathrm{d}a_i+a_i\wedge a_i)\\ &=\mathrm{d}^*\mathrm{d}a_i-*d*(a_i\wedge a_i)-*a_i\wedge(*\mathrm{d}a_i)-*a_i\wedge (*(a_i\wedge a_i))\\ \end{aligned}\] which is tends to \(0\) in \(L^2\). We have \(\Delta a_i-(*d*(a_i\wedge a_i)+*a_i\wedge(*\mathrm{d}a_i)+*(a_i\wedge*(a_i\wedge a_i)))\to 0\), in \(L^2\). We want to show that \(\Delta a_i\in L^2\), thus \(a_i\in L_2^2\). In fact, \[*d*(a_i\wedge a_i)+*a_i\wedge(*\mathrm{d}a_i)\to\nabla a_i\cdot a_i\in L^2\otimes L_1^2,\] 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 \(L^2\).