# 11-24 Notes

In this lecture we continue the problem from last time of proving that given a Palais-Smale sequence of connections in a bundle, if we cover the base manifold by open balls and trivialize the bundle over each element of the cover then the transition functions which come from applying Uhlenbeck's lemma to these connections converge in an appropriate norm.

\section{The two-dimensional case}

Let $\Sigma$ be a Riemann surface with $\mathbb{C}^{n}$-bundle $E\to\Sigma$ and a set of connections $A_{i}$ in $E$ satisfying the conditions \begin{eqnarray*} \int_{\Sigma}|F_{A_{i}}|^{2}\leq M, & & d_{A_{i}}^{*}F_{A_{i}}\stackrel{L^{2}}{\to}0. \end{eqnarray*} We recall that there is a cover $\{U_{\alpha}\}$ of $\Sigma$ and a sequence of trivializations $\tau_{\alpha,i}:E|_{U_{\alpha}}\to U_{\alpha}\times\mathbb{C}^{n}$ such that $(\tau_{\alpha,i})_{*}(A_{i})=\Gamma+a_{\alpha,i},$ where the $a_{\alpha,i}$ satisfy $d^{*}a_{\alpha,i}=0$ and $*a_{\alpha,i}|_{\partial U_{\alpha}}=0$ and there are $a_{\alpha}$ such that $a_{\alpha,i}\stackrel{L_{2}^{2}}{\to}a_{\alpha}$. From this we get transition functions $g_{\alpha\beta,i}$ satisfying $(x,g_{\alpha\beta,i}(x)v)=(\tau_{\alpha,i}\circ\tau_{\beta,i}^{-1})(x,v)$, and there are functions $g_{\alpha\beta}$ such that $g_{\alpha\beta,i}\stackrel{L_{3}^{2}}{\to}g_{\alpha\beta}$. (Note that $L_{3}^{2}$ embeds into $C^{1,\alpha}$ for $0<\alpha<1$.) We would like to find a sequence $u_{i}$ of gauge transformations so that $u_{i}^{*}(A_{i})\stackrel{L_{2}^{2}}{\to}A$ for some $A$.

If the gauge group $G$ were abelian, then given an exponential map $\exp:\mathfrak{g}\to G$ we could write $g_{\alpha\beta,i}=g_{\alpha\beta}\exp(h_{\alpha\beta,i})$. Let $\{\varphi_{i}\}$ be a partition of unity supported by $\{U_{\alpha}\}$ and define $\lambda_{\alpha,i}=\exp(\sum_{\beta}\varphi_{\beta}h_{\alpha\beta,i})$. Then we can construct new trivializations $\tilde{\tau}_{\alpha,i}=-\lambda_{\alpha,i}\tau_{\alpha,i}$ for which the transition functions satisfy \begin{eqnarray*} \tilde{g}_{\alpha\beta,i} & = & \lambda_{\alpha,i}^{-1}g_{\alpha\beta,i}\lambda_{\beta,i}\\

& = & \exp(-\sum_{\gamma}\varphi_{\gamma}h_{\alpha\gamma,i})\cdot g_{\alpha\beta}\exp(h_{\alpha\beta,i})\cdot\exp(\sum_{\delta}\varphi_{\delta}h_{\beta\delta,i})\\
& = & g_{\alpha\beta}\exp(\sum_{\gamma}\varphi_{\gamma}h_{\gamma\alpha,i}+\sum_{\delta}\varphi_{\delta}h_{\beta\delta,i}+h_{\alpha\beta,i}).


\end{eqnarray*} Now the functions $h$ satisfy the cocycle condition $h_{\alpha\beta}+h_{\beta\gamma}+h_{\gamma\alpha}=0$, so the argument to the exponential function vanishes and we are left with $\tilde{g}_{\alpha\beta,i}=g_{\alpha\beta}$.

Note that we cheated slightly in this argument: the cocycle condition is really $g_{\alpha\beta}g_{\beta\gamma}g_{\gamma\alpha}=1$, and we'd like to be able to take logarithms to get $h_{\alpha\beta}+h_{\beta\gamma}+h_{\gamma\alpha}=0$ as above. This may not actually be possible for the given cover, but if we refine it so that we can control the $C^{0}$ norm of $g_{\alpha\beta}$ then we will be able to control $\log(g_{\alpha\beta,i})$ enough to get the cocycle condition for $h_{\alpha\beta}$. Our goal is to understand the analogous argument in four dimensions.

\section{The four-dimensional case}

Suppose now that we have a bundle on a 4-manifold $X$ together with a sequence of connections satisfying \begin{eqnarray*} \int_{\Sigma}|F_{A_{i}}|^{2}\leq M, & & d_{A_{i}}^{*}F_{A_{i}}\stackrel{L^{2}}{\to}0. \end{eqnarray*} We know that after passing to a subsequence, there exists a finite set $\{x_{1},\dots,x_{n}\}\subset X$ and a countable cover $\{U_{\alpha}\}$ of $X\backslash\{x_{1},\dots,x_{n}\}$ so that $\int_{U_{\alpha}}|F_{A_{i}}|^{2}<\epsilon_{0}$ for each $\alpha$. Thus there are gauge transformations $u_{\alpha,i}$ such that $u_{\alpha,i}(A_{i})=\Gamma+a_{\alpha,i}$ is in Coulomb gauge and $||a_{\alpha,i}||_{L_{1}^{2}}\leq\int_{U_{\alpha}}|F_{A_{i}}|^{2}$. We wish to analyze the equation \begin{eqnarray*} d_{u_{\alpha,i}\cdot A_{i}}^{*}F_{u_{\alpha,i}\cdot A_{i}} & = & (d+a_{\alpha,i})^{*}(da_{\alpha,i}+a_{\alpha,i}\wedge a_{\alpha,i})\\

& = & d^{*}da_{\alpha,i}-*a_{\alpha,i}*\wedge da_{\alpha,i}-*d*(a_{\alpha,i}\wedge a_{\alpha,i})-*(a_{\alpha,i}\wedge*(a_{\alpha,i}\wedge a_{\alpha,i})),


\end{eqnarray*} using the invertibility of $\Delta:L_{k}^{p}\to L_{k-2}^{p}$ and a bootstrapping argument to get the appropriate convergence.

Assuming that $a_{\alpha,i}$ has small $L_{1}^{2}$ norm, we can bound the nonlinear terms on the right as $||\nabla a_{\alpha,i}\#a_{\alpha,i}||_{L_{-1}^{2}}\leq c||a_{\alpha,i}||_{L_{1}^{2}}^{2},$ since we have a continuous multiplication map $L^{2}\times L_{1}^{2}\times L_{1}^{2}\to L^{1}$ and hence $L^{2}\times L_{1}^{2}\to L_{-1}^{2}$; and $||a_{\alpha,i}\#a_{\alpha,i}\#a_{\alpha,i}||_{L_{-1}^{2}}\leq c||a_{\alpha,i}||_{L_{1}^{2}}^{3},$ since multiplication $L_{1}^{2}\times L_{1}^{2}\times L_{1}^{2}\times L_{1}^{2}\to L^{1}$ is continous and hence so is $L_{1}^{2}\times L_{1}^{2}\times L_{1}^{2}\to L_{-1}^{2}$. We wish to show using this that the sequences $a_{\alpha,i}$ are Cauchy sequences in $L_{1}^{2}$. In fact, for any $i,j$ the condition $d_{A_{i}}^{*}F_{A_{i}}\stackrel{L^{2}}{\to}0$ gives us $b_{i},b_{j}$ such that $b_{i}\stackrel{L^{2}}{\to}0$ and $\Delta(a_{\alpha,i}-a_{\alpha,j})=(\nabla a_{\alpha,i}\#a_{\alpha,i}-\nabla a_{\alpha,j}\#a_{\alpha,j})+(a_{\alpha,i}\#a_{\alpha,i}\#a_{\alpha,i}-a_{\alpha,j}\#a_{\alpha,j}\#a_{\alpha,j})+(b_{i}-b_{j}).$ Taking $L_{-1}^{2}$ norms, and using the invertibility of $\Delta$ and the relation $x^{3}-y^{3}=(x-y)(x^{2}+xy+y^{2})$, we get $||a_{\alpha,i}-a_{\alpha,j}||_{L_{1}^{2}}\leq c||a_{\alpha,i}-a_{\alpha,j}||_{L_{1}^{2}}\cdot((||a_{\alpha,i}||_{L_{1}^{2}}+||a_{\alpha,j}||_{L_{1}^{2}})+(||a_{\alpha,i}||_{L_{1}^{2}}+||a_{\alpha,j}||_{L_{1}^{2}})^{2})+||b_{i}-b_{j}||_{L_{-1}^{2}}.$ Since $||b_{i}-b_{j}||_{L_{-1}^{2}}\to0$ and the norms $||a_{\alpha,i}||_{L_{1}^{2}}$ are small, we see that $||a_{\alpha,i}-a_{\alpha,j}||_{L_{1}^{2}}\to0$ as $i,j\to\infty$ and thus we have a Cauchy sequence as desired.

Unfortunately, we need a stronger convergence result for the $a_{\alpha,i}$. The reason for this is that $a_{\alpha,i}\stackrel{L_{1}^{2}}{\to}a_{\alpha}$ on the intersection $U_{\alpha}\cap U_{\beta}$ gives $L_{2}^{2}$ convergence for the transition functions $g_{\alpha\beta,i}$, but we do not have an inclusion $L_{2}^{2}\subset C^{0}$ and so there is no guarantee that the transition functions defining our bundle are continuous. We would like to show instead that $a_{\alpha,i}\stackrel{L_{2}^{2}}{\to}a_{\alpha}$, since this will suffice to make the $g_{\alpha\beta}$ continuous. This is not immediate, since if we proceed as before then we will need an inequality of the form $||a_{\alpha,i}\#a_{\alpha,i}\#a_{\alpha,i}||_{L^{2}}\leq c||a_{\alpha,i}||_{L_{2}^{2}}||a_{\alpha,i}||_{L_{1}^{2}}^{2},$ and in general this would require a multiplication map $L_{2}^{2}\times L_{1}^{2}\times L_{1}^{2}\to L^{2}$. In this case, however, the $a_{\alpha,i}$ aren't arbitrary elements of $L_{2}^{2}$ or $L_{1}^{2}$: we'll see later that the condition $d^{*}a_{\alpha,i}=0$ is enough to establish this bound.