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I have spelled out the proofs that over a paracompact Hausdorff space every vector sub-bundle is a direct summand, and that over a compact Hausdorff space every topological vector bundle is a direct summand of a trivial bundle, here
I have spelled out further elementary detail at topological vector bundle.
In (what is now) the section Transition functions I have added a detailed argument that the thing which is glued from the transition functions of a vector bundle is indeed isomorphic to that vector bundle.
Then in (what is now) the section Basic properties I have spelled out a detailed proof that a homomorphism of topological vector bundles is an isomorphism as soon as it is a fiberwise linear isomorphism.
(I was trying to be really explicit, maybe in contrast to what Hatcher offers. The only thing I should still add for completeness is at general linear group the statement that the inclusion GL(n,k)↪Maps(kn,kn) into the mapping space with its compact-open topology is continuous.)
The only thing I should still add for completeness is at general linear group the statement that the inclusion GL(n,k)↪Maps(kn,kn) GL(n,k) \hookrightarrow Maps(k^n, k^n) into the mapping space with its compact-open topology is continuous.)
I wonder if one can see this using the fact GL(n,k) is an open subspace of End(kn), End(kn)≃k⊗k*, and the resulting linear map k→k⊗Maps(kn,kn). Here we’d need Maps(kn,kn) as a topological vector space. But, hmm, what sort of fields k are you allowing? Just ℝ and ℂ? (and perhaps ℍ…)
Ah, I see you did this on the other thread!
I have spelled out in some detail the proof that topological vector bundles are classified by the relevant Cech cohomology: here.
I have spelled out more statements and proofs in the section Over closed subspaces
Sorry for not reacting earlier. I’d rather we fix an explanation than just removing it. So I have made it this:
(like every finite dimensional vector space, by the Heine-Borel theorem)
you had added this sentence:
Constructions in Vect(X) can be acheived by means of smooth functors, which represent the constructions on vector spaces that can be applied fiberwise to vector bundles.
I am not sure that “smooth functor” is a good term here in the page on topological vector bundles.
Probably you want to refer to “natural operations” (to be created) as in Kolar-Michor-Slovak?
I have moved the sentence to a Remark with that title (now here).
Added:
This result is apparently due to Steenrod, see Theorem 11.4 in \cite{Steenrod}.
Added:
The original reference for many results about bundles, including the theorem that concordance implies isomorphism, is
12: In traditional literature on (topological) vector bundles over paracompact Hausdorff spaces there is a discussion of “continuous functors” (on several covariant and several contravariant vector space variables) on vector spaces (definition in terms of graph of the functor) like exterior product etc. which automatically induce functors on products of categories of vector bundles on the space. This is not the same (I think) as a (newer definition of a) natural operation in the sense of Kolar-Michor-Slovak as the tangent bundle does not make sense in that generality (no manifolds in the game!). I guess some considered also smooth functor in the same sense. Most examples are the same, but I believe “continuous” allows some more possibilities.
12, 14 Milnor, Stasheff, Characteristic classes, page 32, the definition (which is not fully expanded there) and theorem 3.6.
BTW, you can easily grab the formatted bibitem from places like here.
But thanks for insisting. So I have removed the previous remark and instead added one now titled Fiberwise Operations.
Currently it reads as follows:
The category FinVect of finite dimensional vector spaces over a topological ground field is canonically a Top-enriched category, and so are hence its product categories FinVectn, for n∈ℕ. Any Top-enriched functor
F:FinVectn⟶FinVectinduces a functorial construction of new topological vector bundles ˆF(𝒱1,,⋯,𝒱n) from any n-tuple (𝒱1,𝒱2,⋯,𝒱n) of vector bundles over the same base space B, by taking the new fiber over a point b∈B to be (e.g. Milnor & Stasheff 1974, p. 32):
F(𝒱1,⋯𝒱n)b≔F((𝒱1)b,⋯,(𝒱n)b).For example:
if F≔(−)*:FinVect⟶FinVect is the operation of forming dual vector spaces, then ˆF constructs the fiberwise dual vector bundle;
if F≔det:FinVect⟶FinVect is the operation of forming determinants, then ˆF is the construction of fiberwise determinant line bundles;
if F≔⊕:FinVect2⟶FinVect is the direct sum of vector space, then ˆF constructs the fiberwise direct sum of vector bundles (“Whitney sum”);
if F≔⊗:FinVect2⟶FinVect is the tensor product of vector spaces, then ˆF constructs the fiberwise tensor product of vector bundles.
There is currently a problem with the Slice category construction. I will list the problems in order of importance:
“for some n”. Supposedly this reads as “there exists an open cover U≔⊔i∈IUi (equipped with disjoint union of inclusion maps because we are in Top/X) such that there exists a non-negative integer n such that there exists a linear isomorphism”. However, this is equivalent to saying that the rank of the vector bundle is constant across the module, which isn’t necessarily true.
Since we are not working in Top/X, but in the category Vectk internal to Top/X, which I will call VTX, we should say that the above isomorphism is in (VTX)/U with U→X being a zero-dimensional vector space. Right now, the concept that we have a “linear isomorphism” is ambiguous because we have to go through the same ordeal of defining the canonical vector space structure on objects in Top/U as we already did for Top/X and then having to somehow prove or visualise that this is the same structure, when we could just carry everything over from VTX by composing an object in VTX/U→X with U→X itself to form an object of VTX. This is using that we have an equivalence of categories (Top/X)/(U→X)=Top/U.
Here is how I would construct the category of topological vector bundles over X. I’m still learning and, apart from potential beginner mistakes, I want to resolve before editing the page: defining an open cover categorically.
Let VTX be the category of k-vector spaces (Vectk) internal to the ambient category of Top/X for some topological space X∈Top. Vect(X) is defined to be a full subcategory of VTX, where an object E∈VTX is also in Vect(X) if there exists an open cover of X as the family {Ui⊆X}i∈I regarded in Top/X as the family of inclusion maps {Uiinci↪X}i∈I and also a family of non-negative integers indexed by the same set {ni}i∈I such that there exists an isomorphism (of vector spaces) in VTX: Ui×Topkni≃Ui×XE. The LHS has vector space structure from Kni and has the map to X given by projection to Ui composed with incli; the RHS has vector space structure transfered from E and is the pullback of functions incli and E.
I believe this to be a full construction of Vect(X). One of the advantages of this construction is that some properties of Vector bundles are quite easily derived from properties of Vect, for example, once you prove that Vect(X) is a cartesian category with the cartesian product given by the cartesian product of Top/X, then, since Vect is a cartesian closed category, we have the Vect(X) is also cartesian closed! We should also be able to generalise pretty easily to the global category of F.D topological vector bundles over all topological spaces by using the arrow category of Top rather than a slice of Top.
Please implement some or all of these suggestions! Thanks.
Thanks for looking into this.
Regarding the “problem”:
That U denotes the disjoint union of patches in an open cover is stated in the item just before, so I am not sure on this point what issue you see. (?) But I admit I have only glanced over the entry and your message for now. (It’s late here, need to call it a day.)
Regarding the definition of “n”: True, this was a little mix-up. I have now edited to declare n:I→ℕ and made the product with ℝn a fiber product over I.
Apart from this I don’t see a substantial issue?
I guess your motivation is to give a more streamlined account, for which there is probably room.
You could go ahead and add your version below the existing one, introduced by a line like like:
Another way to say this is the following, which has the advantage that…
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