2
$\begingroup$

Let $X$ be a compact, oriented, four dimensional Riemannian manifold and $Q\longrightarrow X$ be a principal $G$-bundle over $X$ for a smooth, compact Lie group $G$. Let $M$ be a smooth, Riemannian manifold admitting a left action of $G$. Denote, for simplicity, the associated bundle by $E(M):= Q\times_{G}M$. Then $\Gamma(X, E(M))\cong C^{\infty}(Q,M)^{G}$, the latter being the space of smooth equivariant maps from $Q \longrightarrow M$. By Nash embedding theorem, there exists an $N\in \mathbb{N}$, such that the embedding $\iota:E(M)\hookrightarrow \mathbb{R}^N$ is an isometric embedding. We define the $L^{p}_{k}$-norm of $u\in \Gamma(X, E(M))$ as $||u||_{L^{p}_{k}}:= ||\iota\circ u||_{L^{p}_{k}}$. We now define $L^{p}_{k}(X,E(M))$ to be the completion of $\Gamma(X, E(M))$ in the $L^{p}_{k}$-norm and for any $\hat{u} \in C^{\infty}(Q,M)^{G}$, $||\hat{u}||_{L^{p}_{k}} := ||u||_{L^{p}_{k}}$, where $u$ is the section associated to the map $u$.

Let $u\in L^{p}_{k}(Q,M)^{G}$ and $A\in\Omega^{1}(Q,\mathfrak{g})^{G}_{L^{q}_{l}}$ be an $L^{q}_{l}$-connection on $Q$. Define $A \cdot u := K^{M}_{A}|_{u}$, where $K^{M}_{A}$ is the fundamental vector field on $M$ due to $A$ and along $u$, given by $K^{M}_{A}|_{u}(v) = K^{M}_{A(v)}|_{u}$.

My question is, in what Sobolev space, does $K^{M}_{A}|_{u}$ belong? Does the Sobolev Multiplication Theorem $L^{p}_{k} \times L^{q}_{l} \longrightarrow L^{r}_{m},$ where, $\frac{1}{r}-\frac{m}{4} > \frac{1}{p}-\frac{k}{4} + \frac{1}{q}- \frac{l}{4}$ (below borderline case) and $r = min(k,l)$, hold true in this setting?

$\endgroup$
  • $\begingroup$ If $M$ is just a manifold admitting an action, how are you defining a metric on the total space $E(M)$ for your embedding? Also, you seem to be treating $\Gamma(X, E(M))$ as if it is a vector space. $\endgroup$ – Paul Reynolds Sep 15 '15 at 16:31
  • $\begingroup$ Thanks Paul for pointing it out! $M$ is a smooth, Riemannian manifold. I don't understand your second remark though. $\endgroup$ – Varun Sep 16 '15 at 4:41
2
$\begingroup$

I have a solution in mind. But for that the embedding $E(M) \to \mathbb R^N$ is unnatural. So, I proceed in a slightly different way. Firstly, in any case it is required that $kp>4$, because the problem will involve left composing a Sobolev function with a smooth function. These operations are well-behaved only above the Sobolev borderline. Making that assumption, $X$ has a cover $X=\cup_\alpha U_\alpha$, such that the bundle $Q|_{U_\alpha}$ is trivial; and after choosing a trivialization, $u(U_\alpha)$ is contained in a chart of $M$. Then, locally we may assume $u$ is a map $u:U_\alpha \to \mathbb R^m$ (where $m$ is the dimension of $M$).

The infinitesimal action of $\mathfrak g$ on $M$ induces a smooth section $\Theta:M \to \operatorname{End}(\mathfrak g,TM)$. Locally, this is a smooth section $\Theta:V \to \operatorname{End}(\mathfrak g,\mathbb R^m)$, where $V$ is a chart of $M$. By composition of functions, we get $\Theta \circ (u|_{U_\alpha}):U_\alpha \to \operatorname{End}(\mathfrak g,\mathbb R^m)$ is in $L^p_k$. Note that for this step, you either need $kp>4$ or that $u \in L^p_k \cap C^0$. Now, $A.u$ is the multiplication of sections of vector bundles, namely it is $(\Theta \circ u)\cdot A$, for which relevant results in Sobolev multiplication can be used.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks a ton! That helps. For the case I study, I have that apriori, $u \in L^{1}_{2}\cap L^{\infty}$. From your answer, I gather that it won't suffice to obtain an $L^2$-bound on $\Theta\circ (u|_{U_{\alpha}})$. $\endgroup$ – Varun Sep 16 '15 at 6:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.