Problem about multiplication of functions in Sobolev spaces and regularity related to gauge theory

Question

Let $X$ be a closed Riemannian 4-manifold.

Let $f\in L^2_k(X)$ (Sobolev space of maps $X\to \mathbb{C}$ of regularity $k$), with $k>2$. Suppose also that $f|_{\mathcal O} \not\equiv 0$ for any open set $\mathcal{O}\subseteq X$.

Let $G_r := \{u: X\to \mathbb{C} \ | u\in L^2_r(X), \ |u(x)|= 1 \ \forall x\in X\}$ where $r\geq 3$, so that $G_r$ consists of continuous functions and is also a Lie group.

Now, $G_{k+1}$ acts (smoothly) on $L^2_k(X)$ by multiplication.

Suppose that $u\in G_3$ and $uf \in L^2_k$ (so we gained regularity if $k>3$), does this imply that $u \in G_{k+1}$?

To understand the motivation behind this question, think of $G_{k+1}$ as the "gauge group" acting on $L^2_k(X)$, then $f$ is gauge equivalent to $\tilde {u} f$ if $\tilde u \in G_{k+1}$. Then the question asks if the orbits of the $G_{k+1}$ action are sensitive to the regularity of the gauge transformations $G_r$.

Answer 1

If I understand your question right - the answer is no. That's basically because the product $uf$ only carries partial information about the behavior of $u$.

If

• $u$ is smooth except some small ball $B$,
• $f$ is smooth everywhere,
• $f \equiv 0$ on a neighborhood of $B$,

then the product $uf$ is also smooth, in particular it's in any space $L^2_k$. However, $u$ can have arbitrary regularity on $B$, so one can have $u \in L^2_3$ and not more.