Serge Lang's Differential and Riemannian Manifolds 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.

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):

Proposition 1.1. 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.

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.

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 go looking for such a counterexample? Can I take $U = \mathbf{F} = \mathbb{R}$, and $\mathbf{E} = \ell_2$? Can we make even continuity fail, i.e. is VB3 necessary even for $C^0$ - manifolds? I know we can't make $f$ bilinear, 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?

Also, why can't we just "do without" Lang's VB3 in the case of infinite-dimensional manifolds?


I would refer you to the remark B in supplement 3.4A on Manifolds, Tensor Analysis and Applications of Abraham Marsden Ratiu. It hope it could be useful, and so I quote:

"The following counterexample is due to A.J. Tromba. Let $h: [0, 1]\times L^2[0, 1]\rightarrow L^2[0, 1]$ be given by $h(x,\phi)=(h'(x))(\phi) =\int_0^1{\sin(\frac{2\pi}{x})\phi(t)} dt$, if $x\neq 0$, and $h(0,\phi)=0$. Continuity at each $x\neq 0$ is obvious and at $x = 0$ it follows by the Riemann-Lebesgue lemma (the Fourier coefficients of a uniformly bounded sequence in $L^2$ relative to an orthonormal set converge to zero). Thus $h$ is $C^0$. However, since $h(x, \sin(\frac{2\pi t}{x}))=\frac{1}{2}-\frac{x}{4\pi}\sin(\frac{4\pi}{x})$, we have $h(\frac{1}{n},\sin(2\pi nt))=\frac{1}{2}$ and therefore its $L^2$-norm is $\frac{1}{\sqrt 2}$; this says that $\|h'(\frac{1}{n})\|\geq \frac{1}{\sqrt 2}$ and thus $h'$ is not continuous."

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  • 1
    $\begingroup$ Shouldn't $h$ goes into $R$ instead of $L^2[0,1]$ ? $\endgroup$ – Auguste Hoang Duc Feb 6 '11 at 21:03

The main reason for including condition VB3 obviously is to make it possible to obtain various associated tensor bundles from a given vector bundle $\mathcal E\to M$ . The simplest case probably is the formation of the dual bundle $\mathcal E'\to M$ . Suppose that we are given a chart overlap $U=U_1\cap U_2$ on the base manifold, and the corresponding vector bundle charts $\theta_i:W\to U\times E_i$ . Then we have $\theta_1\circ\theta_2^{-1}:U\times E_2\to U\times E_1$ given by $(p,y)\mapsto(p,f(p)(y))$ , where $f:U\to\mathcal L_b(E_2,E_1)$ is a morphism by VB3. Here $\mathcal L_b(E_2,E_1)$ denotes the Banach(able) space of continuous linear maps equipped with the topology of uniform convergence on bounded sets. Letting $E_i'=\mathcal L_b(E_i,\mathbb R)$ , then the corresponding chart change map for the dual bundle is $U\times E_1'\to U\times E_2'$ given by $(p,u)\mapsto(p,u\circ(f(p)))$ . This should be a morphism in order for the dual bundle construction to make sense. The scheme for proving this is to decompose the map as

$U\times E_1'\to U\times(\mathcal L_b(E_2,E_1)\times E_1')\to U\times E_2'$ by

$(p,u)\mapsto(p,(f(p),u))=(p,(v,u))\mapsto(p,u\circ v)=(p,{\rm{comp}}(v,u))$ .

Here ${\rm{comp}}:\mathcal L_b(E_2,E_1)\times E_1'\to E_2'$ is a continuous bilinear map, hence smooth, and so a morphism. Since also all other occurring maps are morphisms, so is the composite, and the construction of the dual bundle succeeds at least for the part of getting the chart change a morphism in the appropriate category.

To obtain condition VB3 also for the dual bundle atlas, one needs to know that the map $\mathcal L_b(E_2,E_1)\to\mathcal L_b(E_1',E_2')$ given by $u\mapsto\langle z\mapsto z\circ u\rangle$ is smooth. Since this is a continuous linear map, it is smooth.

I think I have somewhere in my files an unpublished example where for each nonzero natural number $k$ there is a $C^k$−map $\mathbb R\times c_0\to\mathbb R\times c_0$ of the form $(t,x)\mapsto(t,f(t)(x))$ with $f(t):c_0\to c_0$ a continuous linear map and $f:\mathbb R\to\mathcal L_b(c_0,c_0)$ not a $C^k$−map, but I have not had time to check this.

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  • $\begingroup$ 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 $p \mapsto f(p)$? $\endgroup$ – slow student Feb 8 '11 at 19:40
  • $\begingroup$ For the dual bundle to make sense, one needs $(p,u)\mapsto u\circ(f(p))$ to be a morphism, i.e. some $C^k$−map. It is **not** sufficient just to have $u\mapsto u\circ(f(p))$ continuous for fixed $p$ . Having $f$ , i.e. $p\mapsto f(p)$ , a morphism is needed in getting the first factor $(p,u)\mapsto(p,(f(p),u))$ in the decomposition a morphism. The image $(p,(f(p),u))$ has three coordinate functions of which all are morphism, and hence so is the whole image. In fact, $f$ here is the only one whose being a morphism is problematic without explicitly assuming it, and VB3 gives this. $\endgroup$ – TaQ Feb 9 '11 at 0:49

In fact, the underlying question is the one of the STRUCTURE (algebraic, differential) that you allow on the group of isomorphisms of the fibers. As you certainly know, for a vector bundle E of rank n, there is a (canonical) principal bundle GL(E) with fibers the linear group GL_n. This Lie group is analytic, and with very few possible topologies (up to equivalences), where as infinite dimensional Lie groups have many pathologies (this is an actual research subject). The bundles come AFTER Lie groups in the construction so...

So that, the book of Lang is perhaps very good (if you say so...) but not sufficiently up-to-date and I should say very formal. Try more recent references, e.g. Kriegl, Michor, "the convenient setting for global analysis" (1997) to be convinced of that. I do not say that it is THE reference, but this is a serious one to have a rigorous viewpoint of ONE actual approach of the subject.

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