Let $X$ be a Frechet or Banach manifold. We can define tangent vectors by equivalence classes of smooth curves. But, we could also define them as derivations of germs of smooth functions. Do these two notions agree in this infinite dimensional setting?
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3 Answers
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To expand on Gjergji's answer a little:
According to Kriegl and Michor, the key for the identification is two properties on the model space:
- It be reflexive
- It have the bornological approximation property
Reflexivity is needed because, essentially, of what André mentions: linear maps in to a space are not always the same as linear maps out of linear maps out of the space! The derivational definition of tangent vectors is as operators on cotangent vectors (even if dressed up a derivations on germs) so it's a "dual of a dual" and you need reflexivity to get back where you started.
Approximation properties (of which there are many) are a way of saying that linear maps from one space to another aren't too messy, by which we mean that they behave a little like matrices. Matrices themselves correspond to things in $E \otimes E'$, but as we're in infinite dimensions then that's only finite rank operators. Still, as we're Grown Ups, we can handle approximations and so it's enough for us that any linear transformation can be approximated by a matrix. With the strong (or operator) topology, this only gets us compact operators, but with a weak topology we have a chance of getting the whole thing. The different types of approximation property correspond to different topologies on linear maps.
There are spaces (even Banach spaces) that don't have the requisite approximation property, but they're "not nice" spaces and also not ones that are usually encountered. All the "most common" spaces have the approximation property.
In particular, nuclear Frechet spaces are reflexive and have the approximation property, so - as Gjerji says - that's sufficient. However, Hilbert spaces are also reflexive and have the approximation property.
The last thing I have to say on this is that if your real question is "What's the right notion of tangent vector in infinite dimensions?" then you should look at the chart in Kriegl and Michor's book in section 33.21 (p351 in the current version online). From that chart, it's clear that the "right" definition is as equivalence classes of curves.
answered Oct 23, 2010 at 7:41
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$\begingroup$ Mind you, the way that this is expressed in K&M, it is as a "sufficient" theorem, not a "necessary" one. However, I would be surprised to learn that a space without the BAP identified derivations with tangent vectors. $\endgroup$ Commented Oct 23, 2010 at 7:44
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I believe the condition for these two definition to agree is that the Frechet manifold be nuclear. See section 28 (proposition 28.7) of "The convenient setting of global analysis" by A. Kriegl and P.W. Michor.
answered Oct 22, 2010 at 16:58
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$\begingroup$ Thanks! This section is enlightening. My question: How do I see that for Frechet spaces, being nuclear is the same as the "bornological approximation property"? $\endgroup$ Commented Oct 22, 2010 at 20:05
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$\begingroup$ I'm not sure if they are the same. I know that the bornological approximation property is implied by nuclear. $\endgroup$ Commented Oct 22, 2010 at 20:28
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$\begingroup$ Sufficient but not necessary. Hilbert manifolds work too. $\endgroup$ Commented Oct 23, 2010 at 7:41
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Let $X$ be a Banach or Fréchet vector space.
Then the following two notions already don't agree:
1) linear map $\mathbb R\to X$
2) linear functional on the space of linear functions.
answered Oct 22, 2010 at 16:48
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$\begingroup$ True, but that is not relevant. Instead of your 2) what is relevant for the question asked is the space of point derivations at $0$ of the ring $\Gamma$ of germs of smooth functions at $0$, i.e., the linear maps of $\ell: \Gamma \to \mathbb R$ satisfying the derivation identity $\ell(fg) = \ell(f) g(0) + f(0) \ell(g)$. $\endgroup$ Commented Oct 22, 2010 at 18:02
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$\begingroup$ Actually, it is relevant. Kriegl and Michor show that with the bornological approximation property, derivations are the same as the double dual. Thus reflexivity is a key property. $\endgroup$ Commented Oct 23, 2010 at 7:43
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$\begingroup$ Right, this seems to be a key place where things go wrong... $\endgroup$ Commented Oct 23, 2010 at 11:04