11-19 Notes

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

Set \(W(\delta)=\mathcal U(\epsilon)\cap B_A(\delta)\), 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:

Lemma 1. If \(u(A+b)=A+a\in S_A(\epsilon)\), then \[\label{eq:2} \|\mathrm{d}_Au\|_{L^n}+\|a\|_{L^n}\leq\kappa\|b\|_{L^n},\] provided that \(\|a\|_{L^n}\ll1\).

If furthermore, \(\|b\|_{L^n}\ll1\), then we have \[\label{eq:3} \|\mathrm{d}_Au\|_{L_1^p}+\|a\|_{L^p_1}\leq\kappa\|b\|_{L_1^p}, \quad\forall n/2< p< n.\]

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 \(\|a_i\|_{L^n}< \epsilon\), then [eq:2] implies \(\|\mathrm{d}_Au_i\|_{L^n}+\|a_i\|_{L^n}\ll1\) , and [eq:3] implies that \(\|\mathrm{d}_Au_i\|_{L^p_1}+\|aa_i\|_{L_1^p}\ll1\) for any \(n/2< p< n\). Passing to weakly convergence subsequence, we have

• \(a_i\rightharpoonup a\) in \(L^p_1\hookrightarrow L^n\) compactly, thus \(a_i\to a\) strongly in \(L^n\);

• \(u_i\rightharpoonup u\) in \(L^p_2\hookrightarrow L^n_1\) compactly, thus \(u_i\to u\) strongly in \(L_1^n\);

• \(u_i\rightharpoonup u\) in \(L^p_2\hookrightarrow C^0\) compactly, thus \(u_i\to u\) strongly in \(C^0\);

Now \(\mathrm{d}_A^*a_i=0\), thus \(\mathrm{d}_A^*a=0\), \(\|a\|_{L^n}\leq\epsilon\), using [eq:2] again to obtain \(\|a\|_{L^n}\leq\kappa\delta< \epsilon\).

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 \(\mathrm{d}_Au=\lim_{i\to\infty}\mathrm{d}_A u_i=\lim_{i\to\infty}u_ib_i-a_iu_i=\)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 \(u\cdot(A+b)\in\tilde S_A(\epsilon):=\set{A+a|\mathrm{d}_A^*a=0,\|a\|_{L^n< \epsilon}}\).

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 \(S_A(\epsilon)\), by the previous result, There exists \(u_i\), \(a_i\) such that \(u\cdot(A+a_i)=A+b_i\), and \(\|\mathrm{d}_Au_i\|_{L^n}+\|a_i\|_{L^n}\leq\kappa\|b_i\|_{L^n}\), thus \(a_i\rightharpoonup a\) in \(L^n\) with \(\|a_i\|_{L^n< \epsilon}\), and \(u_i\rightharpoonup u\) 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 \(L^{q-\epsilon}\), similarly \(a_iu_i\to au\) strongly in \(L^{n-\epsilon}\). Now, since \(\mathrm{d}_Au_i=u_ib_i-a_iu_i\), take weak limits in \(L^n\), we have \(\mathrm{d}_Au_i\rightharpoonup\mathrm{d}_Au\) and \(u_ib_i\rightharpoonup\tilde \psi\), \(a_iu_i\rightharpoonup\psi\). ◻

Theorem 1. 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 \(\mathrm{d}_A^*a=0\), \(*a|_{\partial B^4}=0\) and such that \(u(\Gamma+a)=A\) and \(\int|\nabla_\Gamma a|^2+|a|^2\leq\kappa\int|F_A|^2\).

Proof. Set \(V(\eta)=\set{A\in\mathcal A^4(B^4)|\int_{B^4}|F_A|^2}< \eta\).

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 \(\tilde U(\epsilon):=(\mathcal G_1^4\times\tilde S_4(\epsilon))\).

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

To show \(\tilde U(\epsilon)\cap V(\eta)\) is closed in \(V(\eta)\), let \(\Gamma+b_i\in \tilde U(\epsilon)\cap V(\eta)\) 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 \(u_i\cdot(\Gamma+a_i)=\Gamma+b_i\), \(\|a\|_{L^4}< \epsilon\), \(\mathrm{d}_A^*a_i=0\), \(*a_i|_{\partial B}=0\). 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 \(a_i\rightharpoonup a\) in \(L_1^2\), \(u_i\rightharpoonup u\) in \(L_2^2\), \(u\cdot(\Gamma+a)=\Gamma+b\), \(\mathrm{d}_\Gamma^*a=0\), \(*a|_{\partial B}=0\), and if \(\eta\) is small enough, \(\kappa \eta< \epsilon\), \(\|a\|_{L^4}< \epsilon\), then \(\tilde U(\epsilon)\cap V(\eta)\) is closed in \(V(\eta).\) \[\begin{aligned} C\|a\|_{L^2_1(B^4)}^2&\leq\int_{B^4}|\nabla_\Gamma a|^2\\ &=\int_{B^4}|(\mathrm{d}+\mathrm{d}^*)a|^2=\int_{B^4}|\mathrm{d}a|^2\\ &=\int_{B^4}|F_{\Gamma+a}-a\wedge a|^2\leq\int_{B^4}|F_{A}|^2+\|a\|_{L^4}^2\|a\|_{L_1^2}^2. \end{aligned}\] We already have Palais-Smale sequence \(A_i\in A_{L_1^2}(\Sigma)\), since \(\int_\Sigma|F_{A_i}|^2< \eta\), and \(\mathrm{d}_{A_i}^*F_{A_i}\to0\) in \(L^2\), we have on sufficiently small \(B_\alpha\), there exists \(u_{i,\alpha}\), \(a_{i,\alpha}\), such that \(u_{i\alpha}\cdot(\Gamma+a_{i,\alpha})=A_i|_{\partial B_\alpha}\), \(a_{i,\alpha}\to a_\alpha\) 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 \(g^{-1}_{i,\beta\alpha}\mathrm{d}g_{i,\beta\alpha}+g^{-1}_{i,\beta\alpha}a_{i,\alpha}g_{i,\beta\alpha}=a_{i,\beta}\), thus \(\mathrm{d}g_{i,\beta\alpha}=g_{i,\beta\alpha}a_{i,\beta}-a_{i,\alpha}g_{i,\beta\alpha}\) 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 \(\mathrm{d}g_{i,\beta\alpha}\) is bounded in \(L^p\) for any \(p\), futhermore \(g_{i,\beta\alpha}\rightharpoonup g_{\beta\alpha}\) in \(L_1^p\) thus \(g_{i,\beta\alpha}\to g_{\beta\alpha}\) in \(L^q\) for any \(q\), which implies \(\mathrm{d}g_{i,\beta\alpha}\to\mathrm{d}g_{\beta\alpha}\) 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 \(\mathrm{d}g_{i,\beta\alpha}\to g_{\beta\alpha}\) strongly in \(L^p_2\). ◻