Uhlenbeck's Theorem

Setting

For fixed $n$, on the vector bundle over $B^4$¹, our connection $A$ is $U\left(n\right)$ connections. Our connection matrix over $B^4$ is supposed to be smooth on the open ball, and over ${\bar{B}}^4$ is smooth up to the boundary².

Denote $A_r$ the radial component $\sum \left(x_r/r\right)A_i$ of the connection matrix, defined on the punctured ball.

$F_A$ is a curvature 2-form.

Main Theorem

There are constants ${ϵ}_1,M$ such that any connection $A$ on the trivial bundle over ${\bar{B}}^4$ with ${|F_A\Vert }_{L^2}<{ϵ}_1$ is gauge equivalent to a connection $\tilde{A}$ over ${\bar{B}}^4$ with

1. $d^{\ast }\tilde{A}=0$³
2. ${\lim }_{|x|\to 1}{\tilde{A}}_r=0$
3. ${\Vert \tilde{A}\Vert }_{L_1^2}\le M{\Vert F_{\tilde{A}}\Vert }_{L^2}.$⁴

Moreover, for suitable constants ${ϵ}_1,M$, the connection is unique up to Gauge transformation $u_0\in U\left(n\right)$ such that $\tilde{A}\to u_0\tilde{A}{u}_0^{-1}.$

With Anti Self Dual Connection

There is a constant ${ϵ}_2>0$ such that if $\tilde{A}$ is any ASD connection on the trivial bundle over $B^4$ which satisfies the Coulomb gauge relative to the product connection, i.e. $d^{\ast }\tilde{A}=0$ and ${\Vert \tilde{A}\Vert }_{L^4}\le {ϵ}_2$,

then for any interior domain $D⋐B^4$ and any $l\ge 1$ we have ${\Vert \tilde{A}\Vert }_{L_l^2}\left(D\right)\le M_{l,D}{\Vert F_A\Vert }_{L^2\left(B^4\right)},$ for a constant $M_{l,D}$ depending only on $l$ and $D$.

The Key Point

The small curvature regime the ASD equations in Coulomb gauge, we can use the Elliptic theory. Also, all of derivatives of the connection is controlled by the $L^2$ norm of the curvature norm.

Hence, for the small curvature regime, any ASD connection over the unit ball, we can change the connection in a smooth Coulomb gauge over interior domain $D$.

What is the point

The most interesting thing is compactness.

For any sequence of ASD connections $A_{\alpha }$ over ${\bar{B}}^4$ with ${\Vert F\left(A_{\alpha }\right)\Vert }_{L^2}\le ϵ$, there is a subsequence ${\alpha }^{\prime }$ and gauge equivalent connections ${\tilde{A}}_{\alpha }$, which converge in $C^{\infty }$ on the ball.

Proof

Let ${\left\{A_{\alpha }\right\}}_{\alpha \in I}$ be a ASD connection sequence in $B^4$ with ${\Vert F\Vert }_{L^2}\le ϵ=\text{min}\left({ϵ}_1,{ϵ}_2/CM\right)$, where $C$ is the Sobolev constant. Let the domain $D_n=B_4\left(0,1-\frac{1}{n}\right)$.

Uniformly Bounded in $D_n$

By the main theorem, we can choose the $\left\{{\tilde{A}}_{\alpha }\right\}$ such that $d^{\ast }{\tilde{A}}_{\alpha }=0$ and ${\Vert {\tilde{A}}_{\alpha }\Vert }_{L^2\left(D_n\right)}\le {\Vert {\tilde{A}}_{\alpha }\Vert }_{L^4\left(B^4\right)}\le C{\Vert {\tilde{A}}_{\alpha }\Vert }_{L_1^2\left(B^4\right)}\le CM{\Vert F_{{\tilde{A}}_{\alpha }}\Vert }_{L^2\left(B^4\right)}=CM{\Vert F_{A_{\alpha }}\Vert }_{L^2}\le ϵ.$

Equi-continuity on $C^0$

For $n=1$, clear.

First do for $n=2,r=0$.

By the mean value theorem, $\left|{\tilde{A}}_{\alpha }\left(x\right)-{\tilde{A}}_{\alpha }\left(y\right)\right|={\Vert \nabla A_{\alpha }\left(\xi \right)\Vert }_{L^{\infty }}\left|x-y\right|$.

Since our connections are $ASD$ connection, we can find the upper bound, ${\Vert \widetilde{A_{\alpha }}\Vert }_{L_l^2\left(D_2\right)}\le M_{l,D_n}{\Vert F_{\tilde{A}}\Vert }_{L^2\left(B^4\right)}.$ So, $\left|{\tilde{A}}_{\alpha }\left(x\right)-{\tilde{A}}_{\alpha }\left(y\right)\right|={\Vert \nabla A_{\alpha }\left(\xi \right)\Vert }_{L^{\infty }}\left|x-y\right|\le M_{l,D_2}\left|x-y\right|ϵ$.

By the Arzela-Ascoli theorem, there is a $C^0$-convergent subsequence $\left\{{\tilde{A}}_{{\alpha }^{\prime }}\right\}$ in $D_2$.

Do the diagonal argument in $n$, and after do the same diagonal argument in $r$. Then we have a $C^{\infty }$-convergent subsequence.

Where do we use?

TBA

Footnotes

1. Actually, since $B^4$ is contractible, it is trivial bundle. ↩

2. Distribution Sense. ↩

3. Coulomb Gauge fixing relative to product connection. ↩

4. Sobolev space notation is $W^{k,p}=L_k^p$. ↩