Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given an infinite dimensional manifold modeled on a Banach space, what does it mean for it to have a Riemannian metric? Does it necessarily mean that it is actually a Hilbert manifold?

My understanding is that Hilbert Spaces have inner products, whereas Banach Spaces just have norms. If a Banach manifold has a Riemannian metric, it means that at each tangent space, which is a Banach space, it has an inner product... so wouldn't that make any Banach manifold a Hilbert Manifold?

share|improve this question
    
I think one of the things I was confusing was terminology -- when someone says "$E$ is a Banach space that is not Hilbert space," I originally thought it meant that $E$ does not admit a complete inner product. It probably actually means that there is no inner product on $E$ that corresponds to the norm on $E$. –  Braindead Oct 31 '10 at 17:40

2 Answers 2

up vote 5 down vote accepted

There are two notions of what it means for a(n infinite dimensional) manifold to have a Riemannian structure. A strong Riemannian structure means a (smooth) choice of inner product on each tangent space which induces an isomorphism of each tangent space with its corresponding cotangent space. A weak Riemannian structure simply means a (smooth) choice of inner product on each tangent space.

Strong Riemannian structures only exist if the manifold is modelled on a Hilbert space, and even then they have to be chosen correctly (so the usual $L^2$-metric on the space of $L^{1,2}$-loops is not a strong structure, even though the manifold is Hilbertian). Weak Riemannian structures exist much more widely. For a weak Riemannian structure you only need to know that the manifold admits smooth partitions of unity and that the model spaces admit continuous inner products. So, for example, continuous loops in a smooth manifold admit a weak Riemannian structure but not a strong one.

Although strong Riemannian structures are very good for generalising much of ordinary differential geometry to infinite dimensions, there are occasions where the requirement of having a Hilbert manifold is too strong, and one can get away with merely having a weak Riemannian structure. I've written an article where having a weak Riemannian structure on the space of smooth loops was an essential step and where the construction would not have worked on a Hilbertian manifold (though actually it was a co-Riemannian structure that I needed).

(Declaration of interests: I've actually proposed a refinement of the "weak/strong" classification as I found it too harsh. See my article here for this, and the above-mentioned result, and a load of examples of spaces with different types of Riemannian structure.)

share|improve this answer
    
Just thought of another example where a weak structure is not just needed but it is essential that it is weak: Wiener integration on path spaces. –  Loop Space Oct 30 '10 at 20:34
    
This is great! I've been "working" on an infinite dimensional manifold for years without knowing or understanding all this. Thank you so much for pointing these things out, and I'll also take a look at the paper. –  Braindead Oct 31 '10 at 17:35

Lang has written several (very similar) books that discuss these topics. Basically, you're right: Riemannian metrics only make sense when the manifold is modeled on a Hilbert space. There is, however, a notion of psuedo-Riemannian manifold, which makes sense when the manifold is modeled on a self-dual Banach space. Take a look at Chapter 7 of Lang's book, visible through Google Books:

http://books.google.com/books?id=VfxGB5nYv1MC&printsec=frontcover#v=onepage&q&f=false

share|improve this answer
    
I got my hands on the book, thank you for the reference. –  Braindead Oct 31 '10 at 17:36

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.