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It seems to me that vector bundles are useful because they allow us to bring to bear all of the linear algebra we know to aid in the study of topological spaces. Now, I've read somewhere that it is an important and difficult problem to classify all of the vector bundles of a space. I'm willing to accept that the problem is difficult, but why is it important? What are some applications of such a classification?

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  • $\begingroup$ Some algebraic geometers spend a lot of time on constructing moduli of vector bundles over a variety. I am told it has applications to Gromov-Witten theory, etc.. $\endgroup$
    – Regenbogen
    Feb 24, 2010 at 20:26
  • $\begingroup$ This question might be relevant: mathoverflow.net/questions/7836/… $\endgroup$ Feb 25, 2010 at 2:05

7 Answers 7

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If your space is a manifold, knowing the vector bundles over that space amounts to knowing all of its tubular neighbourhoods when you embed the space in another manifold. This frequently allows you to find many relationships between the two manifolds.

One classical application would be the proof that all smooth embeddings $S^n \to S^{n+2}$ (co-dimension two embedding of a sphere in a sphere) has a Seifert surface -- meaning there is an embedded, orientable $n+1$-manifold $M \to S^{n+2}$ whose boundary is the $n$-sphere. One of the main steps is showing that the $n$-sphere in $S^{n+2}$ has a trivial tubular neighbourhood.

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    $\begingroup$ Could you point out a reference (preferably with lots of details) where the classical application you mention is explained a bit more, please? Thanks in advance! $\endgroup$
    – D1811994
    Apr 19, 2017 at 7:34
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    $\begingroup$ @D1811994: Cameron Gordon "Some aspects of classical knot theory" SLN 685, 1977. $\endgroup$ Apr 20, 2017 at 2:36
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I would say that the importance of the classification of vector bundles comes first from the fact that it leads naturally to the "characteristic classes" and their complete description. Characteristic classes are computable and powerful invariants of vector bundles. Look at the book Milnor-Stasheff "Characteristic classes" which is the wonderful classic in the domain. There you find first applications of the classification of vector bundles, the fundamental classification theorem being that isomorphism classes of rank n vector bundle E-->X over a space X are in 1-1 correspondance with homotopy classes of maps g:X-->BO(n) where BO(n) is an explicit topological space, namely it is the grasmannian of n-vector spaces inside Rinifinity. The usage of this theorem is then that, using the cohomology of BO(n), you can build and classify very interesting and computable invariants of vector bundles (the more famous being the Euler class, the Stiefel-Whitney classes, the Pontrjagin classes and (for complex vector bundles) the Chern classes.) This is whar are called characteristic classes of vector bundles.

Among applications of this given in Milnor-Stasheff are the following:
- you can prove that the real projective spaces Pn(R) (which is an n-dimensional manifold) does not embed in Rn+k for low values of k (depending on n). For example the 8-dimensional projective space does not embed in R14.
-you can prove that it is impossible to define a bilinear multiplication on Rn without zero divisor (i.e. such that x.y≠0 when x≠0 and y≠0) unless n is a power of 2 (but when n=2 you have the complex multiplication, when n is 4 you have the quartenions, for n=8 you have octonions). This was latter improved by Adams who showed that n=1,2,4,8 are actyually the only possible values.
- you can classify compact smooth manifolds "up to cobordism" (Thom theorem). This was a first step to the classification of smooth manifolds up to diffeomorphism (surgery theory). The classification up to cobordism is very directly related to the cohomology of BO(n).
- another striking application which is not completely detailled in Milnor-Stasheff is the construction of exotic spheres, i.e. smooth manifolds homeomorphic but not diffeomorphic to a sphere (for example in dimension 7). A key ingredient in the proof is Hirzebruch signature formula which implies that some vector bundles cannot be the tangent bundle of a compact smooth manifold.
-extending this you can also build examples of topological manifolds which cannot have a smooth structure
- and this was only the beginning in the late 1950's :-). Then came K-theory, the complete classification of exotic spoheres, the classification of smooth manioflds, etcetera...

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    $\begingroup$ So your answer is that it is important to classify vector bundles because it leads to invariants of vector bundles, which are, in turn, useful invariants of the manifold? If so, why not just take a shortcut and calculate characteristic classes without bothering about the classification of vector bundles? (playing devil's advocate) $\endgroup$ Feb 24, 2010 at 10:15
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This is a variation on Ryan's answer.
In PL topology, from which all of geometric topology sprang, every object X comes equipped with an embedding into Rn for n sufficiently large. With the rise of homological algebra, and the shift of focus towards the smooth and topological categories, it became useful to separate between the chain complex of X over some ring (which you might use to calculate homology), which tells you something like how X is glued together out of simplexes, and between the information about how X might be locally embedded, which is the bundle theory over X. These two pieces of information about X are disjoint (although I'm now reading about quadratic structures on chain complexes, a formalism which combines them in some sense). In particular, the algebraic theory of chain complexes is a long way from the geometric topology of manifolds, even in the presence of Poincare duality structure.
All of this is part of the effort to view a space "intrinsically". Vector bundles are an abstraction of the tubular neighbourhood of X in some big Euclidean space, in the sense that chain complexes are an abstraction of stacking simplexes to build X. In this way, they become the natural tool to study "intrinsic external" structure in such settings as surgery problems, passage between categories (smooth, PL, and topological), differential topology over X, and so on.

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Topological K-theory is quite useful: for example, it played a key role in understanding the index theorem.

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    $\begingroup$ The question was why CLASSIFICATION of bundles is useful, not so much why bundles and their invariants are useful. $\endgroup$ Feb 24, 2010 at 14:47
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    $\begingroup$ Topological K-theory depends (e.g.) on the classification of vector bundles modulo stable equivalence. And the elements of K-groups themselves classify vector bundles. $\endgroup$ Feb 24, 2010 at 15:23
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While not specific to vector bundles as such, classifying any kind of structure usually gives some sort of deeper understanding of the structure, and more importantly, tends to yield good and often compact descriptors.

For instance, the classification of modules over (nice enough) algebras gives us all sorts of interesting invariants and decompositions: the classification of modules over PIDs gives us the Jordan normal form, with all the power it brings to linear algebra; and the classification of modules over quiver algebras (tame modules exist over a few families of Dynkin diagrams; all others admit wild modules) is what forms the basis for persistence diagrams - a formalization (and visualization) of the idea that classes in the homology of a filtered simplicial complex carry lifetimes that end up telling us much about the topology of the things we used to construct the simplicial complex in the first place.

These ideas, again, are currently being applied in data analysis, in group cohomology and in computer graphics.

All of which is to say that classification in general is good, and vector bundles are interesting, thus classifying vector bundles is good.

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  • $\begingroup$ Can you say more or provide a reference as to how the classification of quiver representations is the basis for persistence diagrams? I am familiar with both of these objects (more with quivers than with persistent homology), but didn't know of such a connection. $\endgroup$ Nov 13, 2017 at 23:20
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I am not sure why you think the classification of vector bundles is a difficult problem. It quickly reduces to homotopy theoretic issues that are quite well understood (as well as anything in homotopy theory can be understood).

Vector bundles are important because they give natural invariants of manifolds (tangent bundle and its characteristic classes), and embeddings (normal bundle), as well as a natural place where various structures attached to a manifold live (e.g. metrics, connections, tensors).

Whether the classification of bundles matters depends on what you do. To make an analogy if you are studying PDE on $\mathbb R^n$, which is a huge subject, then you might not care about classifications of manifolds at all; except that PDE do come in handy in classifying 3-manifolds. As such classification of vector bundles isn't used in most of the mathematics, rather it represents a basic result in topology.

In fact I am curious how often do you use classification of vector bundles (be that identification of $\mathbb R^k$-bundles over $X$ with $[X, BO(k)]$ or the fact that complex line bundles are classified by first Chern class, or classification of bundles over complexes of dimension $\le 4$ in terms of characteristic classes)? In my own research I have used all this extensively but my impression that this is quite rare.

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  • $\begingroup$ If you were studying PDE on $\mathbb{R}^n$ you might also care about classifying manifolds because of the homotopy principle. By the same token, the topological K-theory led to the theory for C*-algebras, which in turn (for example) provided a complete classification for AF-algebras: see, e.g. tinyurl.com/yj99wpe (wiki article, the asterisk is troublesome here) $\endgroup$ Feb 24, 2010 at 19:33
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    $\begingroup$ With all respect to h-principle, my guess is that most PDE people do not care about it. Nor do I think that having a classification of say simply-connected manifolds help solve any PDE. $\endgroup$ Feb 24, 2010 at 20:04
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I suppose a simpler answer to your question is that the classification of vector bundles is to vector bundles as Whitney's embedding theorem is to manifolds.

Specifically, Whitney's weak embedding theorem says that all n-manifolds embed in $\mathbb R^{2n+1}$ and that any two embeddings of an $n$-manifold in $\mathbb R^{2n+2}$ are isotopic. So studying the "abstract" problem of $n$-manifolds up to diffeomorphism is equivalent to the "less abstract" problem of studying $n$-dimensional submanifolds of $\mathbb R^{2n+2}$ up to isotopy.

Whitney's proof of the above fact is almost exactly how the proof of the classification of vector bundles works. Moreover, they're philosophically almost identical proofs, as the Whitney theorem says abstract manifolds are submanifolds. The classification theorem says abstract vector bundles can be thought of as having their fibres in Euclidean space.

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