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.
This is called natural bundle. Apparently, all known information is in Kolár, Slovák, Michor: Natural operations in differential geometry (recommended by Stefan Waldmann).
From the description, it is also the best categorical textbook on differential geometry and topology.
Third in the beginning of this book we try to give an introduction to the fundamentals of differential geometry (manifolds, flows, Lie groups, differential forms, bundles and connections) which stresses naturality and functoriality from the beginning and is as coordinate free as possible. Here we present the Frölicher–Nijenhuis bracket (a natural extension of the Lie bracket from vector fields to vector valued differential forms) as one of the basic structures of differential geometry, and we base nearly all treatment of curvature and Bianchi identities on it. This allows us to present the concept of a connection first on general fiber bundles (without structure group), with curvature, parallel transport and Bianchi identity, and only then add G-equivariance as a further property for principal fiber bundles. We think, that in this way the underlying geometric ideas are more easily understood by the novice than in the traditional approach, where too much structure at the same time is rather confusing.
Best Answer
As Piotr remarks, these kind of questions quickly lead to studying reflexive sheaves. I would add that one also better get acquainted with Serre's condition $S_2$. For more on this see this and this MO answers.
Perhaps the first remark is that besides the singularities of the surface $X$ you also have to take into account the singularities of the sheaf you are considering.
You are asking about normal surfaces. Normal implies $S_2$ and pretty much everything that follows is OK for $S_2$ in arbitrary dimensions.
A coherent sheaf on an open set can always be extended as a coherent sheaf on the ambient space. Furthermore if $X$ is $S_2$ and $j:U\hookrightarrow X$ is an open set such that $\mathrm{codim}_X(X\setminus U)\geq 2$, then a coherent sheaf $\mathscr F$ on $X$ such that $\mathscr F|_U$ is locally free is reflexive if and only if $$\mathscr F \simeq j_*(\mathscr F|_U).$$ Notice that this means that if $X$ is $S_2$, then a locally free sheaf $\mathscr E$ on $U$ can be extended as a locally free sheaf to $X$ if and only if $j_*\mathscr E$ is locally free. (The point is, that since $X$ is $S_2$, $j_* \mathscr E$ is reflexive and if there is a locally free sheaf extending $\mathscr E$, then that is also a reflexive sheaf which agrees with $j_*\mathscr E$ on an open set with at least codimension $2$ complement, so they have to agree).
This also tells you how to produce locally free sheaves that cannot be extended as a locally free sheaf: Take any reflexive sheaf that is not locally free, then take the open set where it is locally free. By the above, this sheaf on that open set is a locally free sheaf that does not have an extension as a locally free sheaf on the ambient space. Piotr's example is probably the simplest such sheaf.
So now the question is: When is a reflexive sheaf locally free?
In some contexts one defines the singularity set of a coherent sheaf $\mathscr F$ as the locus where it is not locally free. Then there are various results that say that the singularity set of certain sheaves have to be at least such and such codimension. Here is a short list of those:
Sample statement: If $\mathscr F$ is bluh, then the singularity set of $\mathscr F$ has codimension at least boo.
Actual statement If $X$ is smooth, we can substitute the following into the above sentence:
You may recognize theorems that you know as simple consequences of these:
Remark You may also note that this does not require $X$ smooth, but that's essentially because the conclusion is invariant under first restricting to the open set where $X$ is smooth.
Remark We know that this fails if the variety is either not smooth or has dimension at least $2$.
Remark This is the reason why the statement you mention at the start is true. If $X$ is a smooth surface, then a sheaf that's locally free on the complement of finitely many points pushes forward to a reflexive sheaf which by this theorem has to be locally free. Piotr's example or any Weil divisor that's not Cartier shows that this fails if the surface is not factorial. (Meaning that every local ring is a UFD. This is a weaker condition than being smooth.) It is also true that a reflexive sheaf of rank $1$ on a smooth variety is always locally free, so you need to go to higher ranks to get a counter-example.
To complete the picture here is an example of a reflexive sheaf on a smooth $3$-fold that is not locally free.
Example Consider the Euler sequence of $\mathbb P^3$: $$ 0\to \mathscr O_{\mathbb P^3}\to \bigoplus_{i=0}^3 \mathscr O_{\mathbb P^3} (1) \to T_{\mathbb P^3} \to 0 $$ and consider (one of) the induced maps $$ \alpha: \mathscr O_{\mathbb P^3} (1) \to T_{\mathbb P^3}. $$ This is clearly injective and let the cokernel of $\alpha$ be $\mathscr F$. The original (Euler) sequence shows that $\alpha$ cannot be a vector bundle embedding and hence $\mathscr F$ cannot be locally free. I leave it for you to prove that it is reflexive. (This is not absolutely trivial, but it is a good exercise).
So, to answer your question, the statement you cite is not true for singular varieties, at least not in the way it is stated. You can cook up some versions that work by adding additional assumptions. In any case, these are likely the notions that will help you do that.
Finally, here are some references: