I shall prefix this with my standard "rubber stamps":
- This is not really an answer, but is a bit longer than a comment allows.
- 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.
For sake of completeness, I am writing a full answer based on the suggestion of
Pietro Majer.
The following are equivalent:
1) $A$ is an isometry w.r.to some norm.
2) $A$ is diagonalizable (over $\mathbb{C}$) , with all eigenvalues of modulus 1.
3) All orbits of $A$ are bounded ( $\sup_{k\in\mathbb{Z}}\|A^k x\|<+\infty$ for any $x\in \mathbb{R}^n$).
4) $A$ is an isometry w.r.to some inner product.
$(1)\iff (4)$ leads to an interesting point: The union of all isometries of all norms equals the union of all isometries of all inner products.
Proof:
$(1) \Rightarrow (2):$
Assume $||$ is a norm preserved by $A$. Then the operator norm of $A$ w.r.t to $||$ equals 1. Also $||A^n||_{op}=1$. By the spectral radius formula: $\rho(A)=\lim_{n\rightarrow\infty}||A^n||^{1/n}=1 $. The same argument for $A^{-1}$ implies $\rho(A^{-1})=1$. This implies all the eigenvalues (including the complex ones) are of absolute value one.
Also, it is easy to see that If $A$ is an isometry of the norm $| |_1$ , $P∈GL(\mathbb{R^n})$, $P^{−1}AP$ is an isometry of the norm $||_2$ where $||x||_2=||Px||_1$. Thus, the property that a given matrix admit such a norm is invariant under similarity.
So it is enough to focus upon matrices of Jordan form. (which is available to us since we work over $\mathbb{C}$).
Now assume $A$ is not diagonalizable. By looking at Jordan form of a non-diagonalizable matrix, we can see there is a vector $v∈\mathbb{C}^n$ such that $||A^kv||_{Euclidean}$ diverges. (Look at the example of $\begin{pmatrix} 1 & 1 \\\ 0 & 1 \end{pmatrix}$ given in the question).
Since all the norms are equivalent This implies that $||A^kv||$ diverges. But since $v=x+iy$ with $x$ and $y$ in $\mathbb{R}^n$, and $A^kv=A^kx+iA^ky$, either $||A^kx||$ or $||A^ky||$ diverge. This is impossible if $A$ is an isometry of $||$.
$(2) \Rightarrow (4):
$ This is proved here (The basic idea is to look at each Jordan block separately).
$(4) \Rightarrow (1):$ Obvious.
It remains to prove $(1) \iff (3)$:
$(3) \Rightarrow (1):$ If all orbits of $A$ are bounded, then we can renorm $\mathbb{R}^n$ by $\|x\|_A:=\sup_{k\in \mathbb{Z}} \|A^k x\|$, which is an $A$-invariant equivalent norm.
$(1) \Rightarrow (3):$ This follows immediately by the fact all norms on a finite dimensional vector space are equivalent. The orbits of $A$ are all of constant norm ($\|x\|$) w.r.t to the norm $A$ preserves, hence are bounded. (w.r.t any other norm).
Best Answer
Yes, there are global obstructions. Consider $M= S^1 \times \mathbb R$ and $\phi$ acting by an irrational rotation on $S^1$ and by multiplication by $2$ on $\mathbb R$. There are no finite orbits, but also no invariant metrics: The function that takes a point $P$ on $S^1$ to the length of a vector pointing along $\mathbb R$ at $(P,0)$ is a function on $S^1$ that, when rotated irrationally, is multiplied by $2$. No nonzero such functions exist - consider what happens to the max and min when you rotate and multiply by $2$.
The general obstruction is that if you have an orbit whose closure has positive dimension, and take a sequence of powers $n$ such that $\phi^n(x)$ converges to $x$, then the absolute values of the eigenvalues of $d\phi^n(x)$ must converge to $1$. This doesn't depend on your choice of basis for the tangent space near $x$ because the eigenvalues depend continuously on the matrix and as $\phi^n(x)$ converges to $0$ you get a good result.