Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $M$ be a compact manifold and let $E_i \xrightarrow{\Large \pi_i} M$, $i = 1, 2$, be two (real or complex) vector bundles of the same rank $k$ over $M$. Assume we have metrics $g_1, g_2$ on $E_1$, $E_2$ respectively. We can then form Banach spaces

$$\mathcal{B}_i = \Gamma^0(E_i) = \{ s : M \xrightarrow{\large C^0} E_i~\vert~ \pi_i \circ s = \mathrm{Id}_M \}$$ of continuous sections, equipped with the sup norm: $$\|s\|^i_0 = \sup_{x \in M} g_i\big(s(x), s(x)\big).$$

Question: under which conditions are $\mathcal{B}_1$ and $\mathcal{B}_2$ isomorphic as topological vector spaces?

My thoughts: I started naively trying to prove that any two such spaces are isomorphic, to see if I could find any obstruction. This is equivalent to proving that any such $\Gamma^0(E)$ is isomorphic to $C^0(M, \mathbb{R}^k)$, that is, sections of the trivial bundle.

I can construct maps from $\Gamma^0(E)$ to $C^0(M, \mathbb{R}^k)$ and vice-versa by choosing a trivializing cover $\big(U_i\big)_{i = 1}^m$ for $E$ and a partition of unity $\big(\rho_i\big)_{i = 1}^m$ associated to it and then decomposing/gluing a section in the obvious way. But these maps don't seem to be inverses of each other in general, although under some special conditions they are. I can give more details later, if required.

Thanks in advance!

share|improve this question
    
Well, obviously, $\Gamma^0 (E_1)$ is isomorphic to $\Gamma^0 (E_2)$ if $E_1$ and $E_2$ are isomorphic. Less obviously, if you know $\Gamma^0 (E_1)$ and you know the action of $C^0(M, \mathbb{R})$ on $\Gamma^0 (E_1)$, then you can reconstruct $E_1$ up to isomorphism. –  Zhen Lin Feb 3 '13 at 18:28
    
@ZhenLin You're right, but the trivial case $E_1 \simeq E_2$ is not very interesting. I'm sure that the $\Gamma^0(E_i)$'s could be isomorphic even though the bundles are not. By what you said, the isomorphism would necessarily fail to preserve the action of $C^0(M, \mathbb{R})$. I don't see where to go from here. –  student Feb 3 '13 at 18:46

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.