# Fréchet manifolds vs ILH manifolds

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.

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This is more of a question along these line. How does ILH differ from Hofer-Wysocki-Zehnder's scale calculus? –  user30532 Jan 10 '13 at 3:38

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.)

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Thank you, Andrew! –  Igor Belegradek Aug 30 '11 at 16:26
@Igor: No problem! I'm sorry that it's so vague on the details, but hopefully it's enough for you to find the right details in Omori's books (or in other work, such as Kriegl and Michor's book). –  Andrew Stacey Aug 30 '11 at 18:02
Kriegl-Michor's book doesn't discuss ILH structures; instead they refer to Omori's book, which I think doesn't deal with the trival matters I am concerned with. I like the presentation in Omori's paper "On the group of diffeomorphisms of a compact manifold" in Global Analysis 1970, where he mentions in passing that a strong ILH manifold is a Frechet manifold whose transition functions are linear isomorphisms of the model space that extend to linear isomorphisms of the Hilbert spaces "converging" to the model space, and moreover this structure is integrable. –  Igor Belegradek Aug 30 '11 at 18:54
Which book of Omori's are you looking at? I thought that there was a nice exposition in one of them, but I may be misremembering. That last sentence that you say sounds familiar, certainly. –  Andrew Stacey Aug 30 '11 at 20:55
I know two Omori's books: "Infinite dimensional Lie transformation groups" (LNM 427) and "Infinite dimensional Lie groups" (translations of Mathematical monographs, 158. In the former one on page 3 he defines an ILH group $G$ as the inverse limit of topological groups $G^s$ that satisfy seven technical conditions. –  Igor Belegradek Aug 30 '11 at 21:28
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.