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Let's define a smooth family of vector spaces to be the following data $(E,M,p,a,s,z)$, where

(1) $p:E\to M$ is a smooth submersion of smooth manifolds

(2) There is some fixed integer $n\ge 0$ such that the fibre $E_m = p^{-1}(m)$ has the structure of an $n$-dimensional vector space over $\mathbb R$ for each point $m\in M$. [The smooth structure of $E_m$ as a vector space coincides with its smooth structure as a submanifold of $E$.]

(3) For this fibrewise vector space structure, the addition map $a:E\times_ME\to M$ is smooth, the scaling map $s:\mathbb R\times E\to E$ is smooth and the zero section $z:M\to E$ is a smooth embedding.

Then, is $p:E\to M$ a locally trivial vector bundle (with $a,s,z$ being its addition, scaling and zero section maps)?

Just by virtue of $p:E\to M$ being a smooth submersion, we have a locally trivial vector bundle $T_{E/M}$ on $E$ (the "vertical tangent bundle" of $p$), and we can pull it back to $M$ to get a locally trivial vector bundle $z^*T_{E/M}$.

Now, define a map $\phi:E\to z^*T_{E/M}$ given by $e\mapsto\frac{\partial}{\partial t}|_{t=0}s(t,e)$. This is a smooth map (since $s,z$ are smooth) that commutes with the projections to $M$ and is a (fibrewise linear) bijection.

Finally, if we look at the map $\phi$ along the image $z(M)$ of the zero section, it quite clearly has full rank. Thus, it has full rank on a neighborhood of $z(M)\subset E$. But now, since $\phi$ commutes with scaling, it follows that $\phi$ has full rank on all of $E$ and thus, it is a diffeomorphism by the inverse function theorem. Thus, $E$ being isomorphic (as a smooth family of vector spaces) to $z^*T_{E/M}$ means it is locally trivial.

My question is: Is there any (possibly subtle) mistake in this proof (and if so, is local triviality true nevertheless)?

Curiously, this argument does not seem to need even the smoothness of $a$ (it follows as a consequence of the isomorphism $\phi$, which itself depends only on the smoothness of $s$ and $z$). Also, the same argument seems to work if we replace all the objects involved by complex manifolds, holomorphic maps and complex vector spaces. The bracketed assumption in (2) seems to be used when we show that $\phi$ is a fibrewise linear bijection and that it has full rank along the zero section. I think it can be removed and that the proof should still go through - is this true?

If you think the answer to the first question is NO, then I'd still like to hear about it in the comments. My actual motivation is to apply this in the setting of infinite dimensional Banach manifolds, and an answer to the second question covering this case would also be appreciated.

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Here is a short answer (sorry I am a bit pressed for time, and it seems to be too long for a comment). If I see it correctly the answer should be that indeed the construction gives you a vector bundle as stated. The reason why that should be right is now interesting and explains also why one should not be surprised that the additive structure is irrelevant in the construction. I think this is akin to a recent and (apparently yet not very well known) theorem by Grabowski and Rotkiewicz:

In 1 they investigate (among many other things) how smooth actions of the multiplicative monoid of positive reals on a manifold uniquely determine vector bundle structures. I think your setting gives you exactly what is needed to apply Theorem 2.1 from their paper 1 (the unique vector bundle structure they get should be exactly the one you constructed). So in a nutshell I think your proof is aspecial case of their result.

Unfortunately I do not recall if their methods use some decidedly finite dimensional structure or if it goes through on Banach manifolds. Short addendum: The cited paper works with the reals, it is not clear to me if you can just replace it with the complex numbers... (but reading it carefully one can certainly find out)

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  • $\begingroup$ My argument, if correct, seems insensitive to whether we are working in the real or complex category. Thanks for the reference to Grabowski and Rotkiewicz, I'll take a look at that. $\endgroup$ Jun 7 '19 at 20:24
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You are right. Also, it can be added that a submersion with every fibre merely diffeomorphic to R^n is always a locally trivial fibration (my paper "Submersions, fibrations and bundles", Trans. Amer. Math. Soc. 354 (2002), 3771-3787).

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