Fréchet manifolds vs ILH manifolds

Question

What is the precise relation between ILH manifolds and Fréchet manifolds? Specifically:

1. Does any ILH manifold has a canonical structure of a Fréchet manifold?

2. If so, is it true that any ILH submanifold is a Fréchet sumanifold?

Background. The notion of an ILH (Inverse Limit Hilbert) manifold was developed by Omori in his studies of diffeomorphism groups. Very roughly it is a manifold modelled on a topological vector space that is the inverse limit of a countable family of Hilbert spaces. This object is easier to deal with than the usual Fréchet manifolds due to the avaiability of the inverse function theorem.

For all I know the inverse limit of Hilbert spaces need not be Fréchet. More precisely, it seems that the inverse limit inherits a countable family of seminorms from the Hilbert spaces, but I see no reason for the inverse limit to be complete (as a uniform space), and I am not sure if the inverse limit is always Hausdorff.

The ILH manifold I am trying to understand is the diffeomorphism group of a compact manifold, so in particular, it is what Omori called a strong ILH manifold, which perhaps makes a difference in answering 1-2. The diffeomorphism group is also a Fréchet manifold (indeed, it is the so called Fréchet-Lie group), but I am not sure how ILH and Fréchet manifold structures interact.

Answer 1

It's been a while since I looked at this so take this with the proverbial pinch of salt ...

The key thing (as far as I remember) with ILH manifolds is not the model space but the transition functions.

Firstly, though, let us consider the model space. It must be Hausdorff since if there are two distinct points in the limit then there must be a Hilbert space in the sequence which distinguishes between them, whence we can find the required two disjoint open sets. In the definition of an ILH space, the inverse limit is over a countable family, and so is metrisable. It is complete as well, a quick-and-dirty proof being that as we have $X = \varprojlim H_n$ then we have maps $X \to H_n$ for each $n$. As $H_n$ is complete, these extend uniquely to maps $\bar{X} \to H_n$ from the completion. Thus there is a map $\bar{X} \to \varprojlim H_n$ which is the identity on $X$; hence $X$ must be complete. Thus the model space is Fréchet.

But an ILH manifold is not just a manifold modelled on ILH-spaces. The transition functions must also respect the ILH structures. In a particular chart, I can imagine that I can see all the surrounding Hilbert spaces: if $M$ is my manifold and $\phi \colon X \supseteq U \to V \subseteq M$ a suitable chart map then for most $n$, there is an open subset $U_n \subseteq H_n$ such that $U = U_n \cap X$ (I may need to shrink $U$ to get this to work). So at a point $x \in U$ I can imagine all the $U_n$s around me with more points than I thought I had.

On an arbitrary Fréchet manifold there is not reason whatsoever for this ambient structure to be well-defined independent of the charts. But on an ILH manifold, it does because the transition functions are restricted to preserve it. I don't remember the exact conditions, but it is something like that the transition function $\phi_{ab} \colon U_a \to U_b$ must also extend to all the $H_n$s, and there's also some compatibility between the $H_n$s (this is where I'm really hazy).

This means that an ILH-manifold has considerably more structure than just a mere Fréchet manifold, but it is still - underneath - a Fréchet manifold. Thus (I believe) there is a forgetful functor from ILH-manifolds to Fréchet manifolds and, in particular, a sub-ILH-manifold will be a sub-Fréchet manifold.

(But as I said at the start, it has been a while since I thought about these things.)

Answer 2

Dear Igor

In order to answer your questions we first need to ensure what we mean by a Fréchet manifold. Obviously a Fréchet manifold should be modeled on Fréchet spaces, so the transition maps should be smooth maps between Fréchet spaces. Thats where the problem lies. How is differentiability of maps between Fréchet spaces defined ? There is no natural (canonical) way to extend the classical notion of differentiability from Banach spaces to Fréchet spaces. (because there is no norm in a Fréchet space) There are many inequivalent ways to define differentiability in Fréchet spaces, and the choice may depend o the applications in mind. For some of them there are even Nash-Moser type Inverse Function Theorems.

That was the main reason we considered the ILH structure when studying diffeomorphism groups, or the groups of pseudo differential operators or Fourier integral operators.

For $Diff^{\infty}(M)$ the Fréchet spaces are given as spaces of smooth sections of certain vector bundles and the Fréchet topology of $Diff^{\infty}(M)$ is known as the smooth compact-open topology. In other words, we don't start with a sequence of Hilbert spaces to get as inverse limit our Fréchet space, but we have the Fréchet space to begin with and consider it as inverse limit of Hilbert spaces by changing the topologies. This way we have to our proposal the whole powerful theory of Hilbert spaces.

Rudolf Schmid