User slow student - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T13:47:41Z http://mathoverflow.net/feeds/user/12774 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54550/the-third-axiom-in-the-definition-of-infinite-dimensional-vector-bundles-why The third axiom in the definition of (infinite-dimensional) vector bundles: why? slow student 2011-02-06T18:56:43Z 2012-07-21T12:39:21Z <p>Serge Lang's <em>Differential and Riemannian Manifolds</em> is a no doubt the best available reference for the theory of not-necessarily-finite-dimensional differential manifolds, but unfortunately it suffers the defect of containing no exercises and few examples. This makes it difficult to learn the subject from this book, especially if one is say a graduate student who is also still in the process of learning functional analysis.</p> <p>One place where an example would really have been helpful is in the context of the definition of vector bundle (pp. 40-41), which involves three axioms that Lang labels VB1 - VB3. The third one, VB3, states that, in coordinate overlaps, the mapping of points of the base space into the automorphisms of the fibers induced by coordinate changes should be a morphism. As Lang notes, this axiom is redundant in the finite-dimensional case because of the following result (p.42): </p> <p><strong>Proposition 1.1.</strong> Let $\mathbf{E}$, $\mathbf{F}$ be finite-dimensional vector spaces. Let $U$ be open in some Banach space. Let $f: U \times \mathbf{E} \to \mathbf{F}$ be a morphism such that for each $x \in U$, the map $f_x : \mathbf{E} \to \mathbf{F}$ given by $f_x(v) = f(x,v)$ is a linear map. Then the map of $U$ into $L(\mathbf{E},\mathbf{F})$ given by $x \mapsto f_x$ is a morphism.</p> <p>However, this result is apparently false in the infinite-dimensional case. The problem is that Lang does not provide an example showing this; and nor does he discuss why smoothness of the map from $U$ into $L(\mathbf{E},\mathbf{F})$ (or, in the specific case of interest, from $U_i \cap U_j$ into $Laut(\mathbf(E))$ is necessary or convenient for whatever purposes such infinite-dimensional bundles are used for.</p> <p>So if I may ask: what would be a counterexample to the above proposition in the infinite-dimensional case? Even more to the point, where does one <em>go looking</em> for such a counterexample? Can I take $U = \mathbf{F} = \mathbb{R}$, and $\mathbf{E} = \ell_2$? Can we make even <em>continuity</em> fail, i.e. is VB3 necessary even for $C^0$ - manifolds? I know we can't make $f$ <em>bilinear</em>, since $L^2(\mathbf{E}, \mathbf{F}; \mathbf{G}) \cong L(\mathbf{E}, L(\mathbf{F},\mathbf{G}))$ -- but this is what makes the question mysterious to me, because my understanding is that "continuity failures" in infinite dimensions arise from non-convergence of sequences (so that you can't just "write everything in a matrix and see that the entries are continuous/smooth"), in which case you ought to be able to exhibit the phenomenon in the simplest case of (bi)linear maps; but the aforementioned isomorphism blocks this. So why does a fundamental difference between finite- and infinite- dimensional spaces suddenly appear when we switch from linear to nonlinear maps? Why doesn't the fact that $f$ is a two-argument morphism provide bounds that would force $x \mapsto f_x$ to be a morphism as well, just like in the bilinear case?</p> <p>Also, why can't we just "do without" Lang's VB3 in the case of infinite-dimensional manifolds?</p> http://mathoverflow.net/questions/54550/the-third-axiom-in-the-definition-of-infinite-dimensional-vector-bundles-why/54706#54706 Comment by slow student slow student 2011-02-08T19:40:42Z 2011-02-08T19:40:42Z I'm afraid I don't see where you used VB3. For the dual bundle to make sense, you need $u \mapsto f(p)^*(u) = u\circ f(p)$ to be continuous, certainly -- but this follows from the continuity of $f(p)$, right? Where do you need continuity/smoothness of $p \mapsto f(p)$?