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Let $E\to X$ be a vector bundle with an inner product and fix a reference connection $A_0$ on $E$. Then for $1\leq p < \infty$ and $k\geq 0$ we can define the Sobolev space $W^{k,p}(E)$ as the completition of $C^\infty(X, E)$ with the norm

$$ \lVert f \rVert_{W^{k,p}} = \int_X |f|^p + |\nabla_{A_0}f|^p + \cdots + |\nabla^k_{A_0} f|^p ) d\mathrm{vol}. $$ My question is about Sobolev multiplication:

If we have two elements $f\in W^{1,2}(E)$ and $g\in W^{1,2}(F)$, under what conditions is it true that $f\otimes g\in W^{1,2}(E\otimes F)$?*.

I am interested mainly at the case when $E$ is a rank-2 vector bundles and $n=\dim(X)=4$ considering that additional conditions may be needed.

Note: I found these notes about the borderline case $kp=n$ for traditional Sobolev spaces. They suggest an $L^\infty$-bound condition. But I am not very familiar with these results and I do not know if they can be applied here.

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    $\begingroup$ No. This is not even true for two copies of the trivial $\mathbb{R}$ bundle with the usual $|\cdot|$. $\endgroup$ Apr 27, 2015 at 8:48
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    $\begingroup$ Working locally by trivialisation you see that at the very least you need the exponents $k,p$ and $n = \dim X$ to be above typical threshold for Sobolev spaces to form an algebra, that is, $n < kp$. $\endgroup$ Apr 27, 2015 at 8:51
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    $\begingroup$ General multiplications results can be found in R.Palais' book Foundations of Global Non-Linear Analysis, especially Thm. 9.5. vmm.math.uci.edu/PalaisPapers/… $\endgroup$ Apr 27, 2015 at 13:56

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The rank of $E, F$ essentially doesn't enter the discussion, since given a basis of $e_i$ and a basis $f_j$ of $E$ and $F$, you have a basis $e_i\otimes f_j$ of $E\otimes F$ and representing your connection against this basis you are down to dealing with scalar functions.

Suppose first that you can choose bases which are locally parallel. Then you see that exactly the usual theory applies: in dimension $\dim X = 4$ you have that $W^{1,2} \hookrightarrow L^4$ by Sobolev so you have that

$$ W^{1,2}\otimes W^{1,2} \hookrightarrow L^2 $$

So you need to control

$$ \nabla (f\otimes g) = \nabla f \otimes g + f \otimes \nabla g $$

and you know that

$$ (\nabla f) \in L^2, (\nabla g)\in L^2 $$

so you see that if you assume $f,g \in L^\infty$ also then you are all set.

When the bases are not locally parallel, you are in the general situation of having chosen a local coordinate system relative to which the connection $\nabla_A f = \partial f + A f$ where $A$ is the connection coefficients. For the second part of the argument nothing changes at all. For the first part where we used the Sobolev inequality we have to observe that $$ \|\partial f + Af\|_{L^2} \geq \|\partial f\|_{L^2} - \|A f\|_{L^2}$$ so as long as $A$ has a uniform $L^\infty$ bound (across a set of different coordinate patches that cover $X$) you have that $$ \|\partial f + Af\|_{L^2} + \|f\|_{L^2} \gtrsim \|\partial f\|_{L^2} + \|f\|_{L^2} $$ with a constant depending on the size of $\|A\|_{L^\infty}$. So we can still apply Sobolev.

In short, a sufficient set of conditions is that

$f,g\in W^{1,2} \cap L^\infty$ and that the reference connection $A$ has a uniform $L^\infty$ bound over $X$ in local coordinates. This condition on the connection would be satisfied if the connection is smooth and $X$ is compact, for example.

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