# Orthogonal complements in Hilbert bundles

It's a standard fact that for a finite-dimensional vector bundle with an inner product, the othogonal complement of any subbundle is itself a locally trivial vector bundle.

What is known about the corresponding question for infinite-dimensional bundles?

Specifically, say $X$ is a smooth Riemannian manifold (modeled on a separable Hilbert space) and we are given a smoothly varying inner product (i.e. a metric) on the tangent bundle $TX$, which induces the correct topology on each tangent space (i.e. the metric is strong). If $Y\subset X$ is a locally closed, smooth submanifold, then $TY$ is a subbundle of $TX|_Y$. Is it known whether this guarantees that $TY^\perp$ is locally trivial? What if $Y$ has finite codimension?

Note that if the orthogonal projection operators associated to $T_y Y \subset T_y X$ vary smoothly, in an appropriate sense, we obtain a smooth surjection of vector bundles $TX|_Y \to TY$ and hence its kernel, $TY^\perp$, is locally trivial.

Of course one could ask similar questions over any base space.

Here is some background:

David Ebin has a paper from 1970 (The manifold of Riemannian metrics, MR0267604) in which he considers an orbit $O$ of the diffeomorphism group on the manifold $M$ of Riemannian metrics (on some underlying finite-dimensional smooth manifold). (To be precise, Ebin works with $L^2_s$ versions of these spaces, rather than the usual $C^\infty$ versions.) He considers a weak metric on $TM$ (one that does not induce the correct topology on $T_m M$), and goes to some effort to show that the (weak) orthogonal complement to $TO$ is a smooth subbundle of $TM$. It seems to me that he's doing this because when working with the weak metric, it's easier to see that the orthogonal projections vary smoothly. There is a strong metric in the picture here, and if the orthogonal complement with respect to the strong metric were automatically a smooth subbundle, then I don't know why Ebin would be using the weak metric (but I may be missing something). In this situation, the orbits have infinite codimension, so I'm hoping the assumption of finite codimension is helpful.

Kondracki and Rogulski go through a similar process to Ebin's in their article MR0866577.

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When the vector bundles are not finite dimensional (say they are modeled on Banach Spaces), there are splitting conditions on the morphisms cf. Lang Fundamentals of Differential Geometry. That is, an injective (resp. surjective) morphism of vector bundles not only has to be injective (resp. surjective) and continuous, but also has to induce a split exact sequence on each vector space. This is satisfied trivially in the finite dimensional case (with the discrete topology). –  Harry Gindi Mar 8 '10 at 23:51
I don't have access to either of the two articles, so I can't check the details, but I would be very cautious about claiming the existence of a strong metric in this case. Diffeomorphisms certainly aren't a Hilbert manifold and I strongly doubt that Riemannian metrics form a Hilbert manifold either. A quick check in the Canonical Source (Kriegl and Michor's book) backs this up. If I'm right, the notion of a strong metric doesn't make sense. –  Andrew Stacey Mar 9 '10 at 8:39
Andrew, Ebin is working with a Sobolev ($L^2_s$ for some $s$) version of the space of Riemannian metrics (and a corresponding version of the diffeomorphism group). So that's where the strong metric comes from. He does eventually prove results about the space of smooth metrics and smooth diffeomorphisms, using the Sobolev completions as tools in his proof. I'll edit the question to make this clear. –  Dan Ramras Mar 9 '10 at 17:47
fqpc: You're right about these splitting conditions. In the case of submanifolds, though, the splittings are built in to the definition, essentially: a map $f: X \to Y$ is an immersion if and only if $Tf$ is injective and splits at each point. This is Proposition 2.2 (and/or Proposition 2.3) in Chapter II of Lang's Differential and Riemannian Manifolds. (For whatever reason, I seem to like this book better than his other differential geometry texts, although they are all largely cut-and-pasted from one another...) –  Dan Ramras Mar 9 '10 at 17:52

I shall prefix this with my standard "rubber stamps":

1. This is not really an answer, but is a bit longer than a comment allows.
2. You should read "A Convenient Setting of Global Analysis" by Kriegl and Michor. In particular, section 45 (Manifolds of Riemannian Metrics) and section 27.11ff (Submanifolds, in particular there's a good discussion of the necessity of the splitting condition).

I should also say that, as I mentioned in the comments above, I don't have access to the two articles cited so I can only speculate on what they are trying to achieve.

Firstly, many times in infinite dimensional analysis one wants to work with a space $X$ but it's tricky, so we work instead with a space $Y$. Only there's not one particular choice for $Y$, there's lots. And sometimes if we can say something for every such $Y$ in a compatible way then we can deduce that it also holds for our original $X$. The key here is the "in a compatible way". A common example is studying some infinite dimensional Frechet manifold (such as here) by expressing it as an inverse limit of Hilbert manifolds. Now each of those Hilbert manifolds has its own Hilbertian structure, and so (for example) is diffeomorphic to an open subset of some Hilbert space. But just because these exist, doesn't mean that they exist nicely with respect to each other. So to make some general statement about all of them, sometimes you have to sacrifice the really strong structure you have and work with something in the middle.

In the case of loop spaces, to take an example I do know something a little about, then one can always consider the various Sobolev spaces of loops, say $L^s M$. Each of those is a Hilbert manifold and its tangent bundle is thus a Hilbert bundle and can be given a strong metric. However, the "natural" structure group of $T L^s M$ is $LO_n$, loops on the orthogonal group. This only acts orthogonally on the "usual" Hilbert completion of $L\mathbb{R}^n$. So any other orthogonal structure requires changing the structure group. Admittedly, the space of all the choices is contractible, but that's homotopy theory and by passing to Hilbertian manifolds one is essentially declaring that one wants to do analysis so one would have to keep track of all the homotopies involved and keep taking them into account. (To make the point a little clearer, it's the difference between knowing that a solution exists and actually going out and finding it. If you really need to know the solution, knowing that it exists gives you a little hope but doesn't really help you actually write it down.)

I would also like to say that the usual classification of orthogonal structures into just "weak" and "strong" is a little simplistic. There is almost always some addition structure in the background (usually related to the structure group) and taking it into account can give a much finer picture. I have such a finer classification in my paper How to Construct a Dirac Operator in Infinite Dimensions together with examples of the different types.

So to return to the actual question. Let me see if I can simplify it a little. We can work locally, and in the actual question you are only concerned with what happens over the submanifold. So we have some open subset of a model space, $U$, and a Hilbert space, $H$, two trivial bundles over $U$ modelled on $H$, say $E_1$ and $E_2$, and an inclusion of bundles $E_1 \to E_2$ such that the image of each fibre is closed. Then we want to know if $E_2^\top$ is locally trivial. We can, if we choose, impose some codimensionality conditions on the inclusions.

Adjointing, we have a smooth map $\theta : U \to \operatorname{Incl}(H,H)$ where $\operatorname{Incl}(H,H)$ is the space of closed linear embeddings of $H$ in itself.

Now we see where the crux of the matter lies. What topology do we have on $\operatorname{Incl}(H,H)$? We have lots of choices. The two most popular are the strong and weak topologies. Without making further assumptions, we can only assume that we have the weak topology! Where this distinction comes into play is that the weak topology is very badly behaved. With the weak topology, $\operatorname{Incl}(H,H)$ is not an ANR so there's no nice extension results. With the strong topology, it's a CW-complex and so lots of nice things follow - in particular, orthogonal complementation will be continuous and the complement will be a locally trivial bundle.

Imposing finite codimensionality doesn't help either. That's a bit like saying that you know that you end up in Fredholm operators. Again, in the strong case then that's okay since the index is well-defined and so the complement will have constant dimension and be locally trivial. With the weak topology, the index is not continuous as, with the weak topology, the space of Fredholm operators is contractible.

So there's where to find your counterexample: find such an inclusion $E_1 \to E_2$ such that the adjoint map is only continuous into the weak topology, not the strong one. A good source of such examples is with the obvious representation of a Lie group $G$ on $L^2(G)$. I expect that with a bit of bundle-crunching, this could be turned into an example.

It is just possible that the bits of your question that I threw out would save you (namely that the bundles were tangent bundles) but I doubt it since I expect that you could take an example as I outlined above and consider that as the inclusion of manifolds, then the tangent bundles of that inclusion would have the same properties of the inclusion itself. That is, if you can find a counterexample, $E_1 \to E_2$, to my version of your question then viewing $Y = E_1$ and $X = E_2$ then I think you get a counterexample to your original setting.

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Thanks, Andrew. It sounds like I need to just buckle down and prove what I need in the specific case I'm looking at, rather than hope for a general theorem. I'll take a look at Kriegl and Michor as well. –  Dan Ramras Mar 10 '10 at 23:45