Elliptic Bootstrapping for Gauge Transformations I'm reading John Morgan's book "Seiberg-Witten equations and applications to the topology of smooth manifolds". I'm stuck in the proof of Lemma 4.5.3, which says the following. Suppose $(a_n,\psi_n)$ and $(b_n,\mu_n)$ are sequences of $W^{2,2}$ (connection,spinor) on a closed $4$-manifold which converge in $W^{2,2}$ to $(a,\psi)$ and $(b,\mu)$ respectively. If $\sigma_n$ is a sequence of $W^{3,2}$ gauge transformations such that $(a_n,\psi_n)\cdot \sigma_n = (b_n,\mu_n)$, then there exists a subsequence of the $\sigma_n$ converging in $W^{3,2}$ to an element $\sigma$ of the gauge group.
The proof starts as follows. Define $\tau_n = \det(\sigma_n)$, and note that we have $d\tau_n = \tau_n(b_n-a_n)$. We have (obviously) a uniform bound on $\|\tau_n\|_{L^6}$, which using the multiplication $L^6\otimes W^{2,2}\to L^5$ gives a uniform bound on $\|\tau_n\|_{W^{1,5}}$. Following this strategy once again, we can show (using Sobolev embeddings) that there is a uniform bound on $\|\tau_n\|_{W^{2,4}}$. After this, it is claimed we can iterate this argument once again to get a uniform bound on $\|\tau_n\|_{W^{3,3}}$.
It is at this step that I'm having trouble. My understanding is that, we need to use a bounded multiplication $W^{2,4}\otimes W^{2,2}\to W^{2,3}$ to execute this step. But I suspect this is not true since $W^{2,2}$ does not embed in $W^{2,3}$ (since the latter embeds in $L^\infty$). Am I missing something?
Can someone say where my mistake lies, or how to get the $W^{3,3}$ bound on $\tau_n$? Once this step is done, the rest of the proof looks okay to me.
 A: I agree that you can't get a bound in $W^{3,3}$ for the reason you give: that multiplication doesn't exist (certainly $1 \in W^{2,4}$, so $1 \otimes W^{2,2} \to W^{2,3}$ would be the identity, but as you say $W^{2,2}$ does not sit inside $W^{2,3}$). I don't have a copy of Morgan's book available, but if you did have a $W^{3,3}$ bound I guess you would want to conclude using the compactness of the inclusion $W^{3,3} \hookrightarrow W^{3,2}$. I tried for a little while but do not see how to fix this argument.
Freed-Uhlenbeck prove the same theorem for connections on a principal $G$-bundle and connections in $W^{\ell, 2}$, $\ell \geq 2$ as their A.5 in "Instantons and Four-Manifolds". They only use the conclusion that $\|\tau_n\|_{W^{3,2}}$ is uniformly bounded, which you outline the proof of in your answer (there is, after all, a product $W^{2,4} \otimes W^{2,2} \to W^{2,2}$).
They conclude by noting that the compactness of inclusions implies that a subsequence $\tau_n'$ converges in eg $W^{2,3}$; then using the Sobolev multiplication $W^{2,3} \otimes W^{2,2} \to W^{2,2}$ you see using your equation that $d\tau_n'$ converges in $W^{2,2}$. This is enough to conclude that $\tau_n'$ converges in $W^{3,2}$, as desired.
