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in this question about characteristic classes it is answered this:

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 neighborhood.

I ask for some references where I can look at this more closely. To be concrete, I want further reference for:

  • knowing the vector bundles over that space amounts to knowing all of its tubular neighborhoods when you embed the space in another manifold

I think that tubular neighborhoods are always vector bundles (at least they are with the definition of Bott Tu). But I don't know if every vector bundle can be seen as a tubular neighborhood of some submanifold?

  • Relationships between the two manifolds given by the tubular neighborhoods and the relation to the tubular neighborhoods.

  • One classical application would be the proof that all smooth embeddings $S^n \to S^{n+2}$ 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 neighborhood.

Any help would be appreciated.

By the way, I have looked at:

  • Lee, Introduction to Smooth Manifolds, 2nd. ed.
  • Bott and Tu, Differential forms in Algebraic Topology
  • Tu, Introduction to Manifolds, 2nd ed.

I am looking at:

  • Hirsch, Differential Topology

but I haven't found what I was looking for.

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Tautologically every vector bundle can be realized as a tubular neighborhood (in some manifold). Namely if $V \to M$ is a vector bundle, the inclusion $M\to V$ coming from the 0-section gives the normal bundle of $M$ as (something isomorphic to) $V$.

If the ambient manifold is fixed (for instance if you want to know what vector bundles are tubular neighborhoods are for $M \to \Bbb R^n$) the problem becomes a lot more subtle. As the tangent bundle of $\Bbb R^n$ is trivial, we have $T\Bbb R^n$ restricted to $M$ is $TM \oplus \nu$, so for $\nu$ arise as a normal bundle to an embedding (or an immersion even) $\nu$ must become trivial after summing with $TM$. This is not the only assumption, but it can get you a long way.

I'll give you a way to see every embedding $S^n \to \Bbb R^{n+2}$ is "framed" (the normal bundle is trivial) for $n\ne 2$. In fact, I'll show that every 2-dim bundle over $S^n$ is trivial. Further reading can be found in Milnor-Stasheff and probably and in any regular algebraic topology book to fill in the details.

One very nice thing about isomorphism classes of n-dimensional real vector bundles over $M$ is that they are in bijection to the homotopy class of a map $M \to BO(k)$. Here $BO(n)$ is called the classifing space of the orthogonal group, and can be decomposed into a direct limit $Gr(k,n)$ (the space of k-planes in $\Bbb R^n$).

The fact that our map comes from an embedding $S^n \to \Bbb R^{n+2}$ tells us that our classifying map factors through the inclusion $Gr(n,n+2) \to BO(2)$. So we get a map $f:S^n \to Gr(n,n+2)$, and we'll show that any such map is nullhomotopic.

Now $O(n)$ fibers over $Gr(k,n)$ with fiber $O(k)\times O(n-k)$. The long exact sequence for a fibration gives $\pi_n(O(n)\times O(2)) \to \pi_n(O(n+2)) \to \pi_n(Gr(n,n+2)) \to \pi_{n-1}(O(n) \times O(2))\to\pi_{n-1}(O(n+2))$, and another long exact sequence argument (prop 2.2 here https://ncatlab.org/nlab/show/orthogonal+group) implies that the first and last arrows are surjective, so $\pi_n(Gr(n,n+2)$ is $0$.

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  • $\begingroup$ Could you elaborate a bit more on: "Namely if $V \to M$ is a vector bundle, the inclusion $M \to V$ coming from the 0-section gives the normal bundle of $M$ as (something isomorphic to) $V$ ." please? I have been trying to write that down but I am afraid I am missing something. What I have tried: Suppose $\sigma \colon M \to V$ is the zero section, $\sigma(p)=(p,0) \; \forall p \in M$. Then, is this the right exact sequence to obtain $V$ as the normal bundle? $$0 \to T_{\sigma(M)} \to T_{V|\sigma(M)}\to N \to 0$$ $\endgroup$ – D1811994 Apr 23 '17 at 11:59
  • $\begingroup$ Thanks in advance, I just want to formalize it a bit so I am sure I really understand it. And I feel there is still something that I don't see. $\endgroup$ – D1811994 Apr 23 '17 at 12:01

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