"Vector bundle" with non-smoothly varying transition functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T14:47:57Z http://mathoverflow.net/feeds/question/4943 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4943/vector-bundle-with-non-smoothly-varying-transition-functions "Vector bundle" with non-smoothly varying transition functions Zack 2009-11-10T22:44:24Z 2009-11-16T08:44:42Z <p>I'm working my way through Lang's <em>Fundamentals of Differential Geometry</em>, and when he introduces vector bundles, he states that for finite dimensional bundles, the third axiom is redundant. I'm hoping someone can give a counterexample in infinite dimensions.</p> <p>His axioms (for a $C^p$ bundle) are (1) local triviality, (2) transition maps are Banach space isomorphisms (linear homeomorphisms), and (3) the maps $x\mapsto A_x$ are $C^p$ (where $A_x$ is a transition map).</p> <p>Essentially, I'm looking for a $C^p$ automorphism of $B\times V$ of the form $(x,v)\mapsto(x,A_xv)$ such that $x\mapsto A_x$ is not $C^p$. Here, $B$ is open (say the unit ball) in some Banach space, $V$ is another Banach space, and the linear maps are given the operator norm.</p> <p>All derivatives are Fréchet derivatives.</p> http://mathoverflow.net/questions/4943/vector-bundle-with-non-smoothly-varying-transition-functions/4997#4997 Answer by Andrew Stacey for "Vector bundle" with non-smoothly varying transition functions Andrew Stacey 2009-11-11T08:09:29Z 2009-11-16T08:44:42Z <p>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.)</p> <p>I'll assume that this is so.</p> <p>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 <em>defining</em> this function.</p> <p>The problem is that continuity of $\psi$ (even higher differentiability) is <em>not</em> sufficient to guarantee the continuity of $\theta$ <strong>in infinite dimensions</strong>. The map $\theta$ is continuous if $GL(F)$ is given the <strong>weak</strong> 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 $GL(F)$.</p> <p>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 $GL(H)$).</p> <p>Further reading:</p> <ul> <li>Topological Vector Spaces, Schaefer. Contains lots about the different topologies.</li> <li>A Convenient Setting of Global Analysis, Kriegl and Michor. Contains lots about the intricacies of infinite dimensional differential topology.</li> </ul> <p><strong>Edit:</strong> 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.</p> <p>(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 <a href="http://www.math.ntnu.no/~stacey/Research/Papers/smooth.html" rel="nofollow">this one</a> in case anyone's interested.)</p>