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edited Nov 16 2009 at 8:44

Edit: I finally remembered the classic example of this: $L^2$-functions on a Lie group. For simplicity, let's take $S^1$. Then $S^1$ acts in the obvious way on $L^2(S^1,\mathbb{C})$ (hereinafter $L^2$). The action $S^1 \times L^2 \to L^2$ is jointly continuous but the associated map $S^1 \to GL(H)$ is most assuredly not. Indeed, if $\lambda \ne \mu \in S^1$ then $\|R_\lambda - R_\mu\| = 2$ so the image is discrete. So if we have an $S^1$-principal bundle $P \to B$ then we can form a new space by taking the quotient $P \times_{S^1} L^2$. This will be locally trivial, and the transition maps will be fibrewise linear as they are of the form $x \mapsto R_{\lambda(x)} : L^2 \to L^2$ where $x \mapsto \lambda(x)$ are the transition functions of the $S^1$-bundle. But the associated map $x \to GL(H)$ is not continuous and so it won't be a genuine vector bundle in the sense Lang defines.

(What's particularly embarrassing about how long it took me to remember this example is that in a recent paper I go into great detail about the different "levels" one can require for continuity of the action of subgroups of $Diff(S^1)$ on various loop spaces. The paper in question is this one in case anyone's interested.)

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answered Nov 11 2009 at 8:09

Andrew Stacey
16k●1●40●110

I suspect that the issue here is actually to do with continuity, not differentiability. Without more details on the exact definition of vector bundle as given, I can't be sure (and I don't have a copy of Lang's book to hand to check). If the definition is a "top down" one, then continuity is certainly an issue. By "top down" then I mean that a vector bundle consists of two smooth manifolds, $E$ and $B$, and a smooth map $p : E \to B$ satisfying the above conditions. (The other approach is to build it up from the transition functions.)

I'll assume that this is so.

Then from local triviality and the fibrewise description, we obtain the transition functions $\psi : U \times F \to U \times F$. Now, from the properties we can find a function $\theta : U \to GL(F)$ with the property that $\psi(x,v) = (x,\theta(x)v)$. There is no difficulty in simply defining this function.

The problem is that continuity of $\psi$ (even higher differentiability) is not sufficient to guarantee the continuity of $\theta$ in infinite dimensions. The map $\theta$ is continuous if $GL(F)$ is given the weak topology where $(A_n) \to A$ if $(A_nv) \to Av$ and $(A_n^{-1}v) \to A^{-1}v$ for all $v$. But normally we ask for the strong (norm) topology on $GL(F)$.

In finite dimensions, this topology agrees with the standard topology but in infinite dimensions they are very different. For example, if we take $\ell^2$ and let $P_n$ be the projection onto the first $n$-coordinates then $(P_n) \to I$ in the weak topology but not in the strong topology (this is an example in $L(H)$ but can be easily tweaked to give an example in $GL(H)$).