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Maybe someone could help me with a bit of alebraic topology.

Take $M$ a $n$-manifold with $n \geq 3$ , and $V$ a submanifold of codimension $2$ in $M$. Assume $H_{n-2}(V) = 0$. I've read that under those hypothesis, there exists a cyclic ramified covering $\tilde{M} \longrightarrow M$, where $V$ is the set of the ramification point.

I've been told that it was clearly a consequence of the fact that the normal bundle at $V$ in $M$was trivial.

1) I don't see why this implies the existence of such a covering

2) I don't understand why the normal bundle at $V$ in $M$ is trivial. I assume $H_{n-2}(V) = 0$ implies that $V$ bounds an hypersurface, and one can define a normal vector field at $V$ (puting a Riemmanian metric on $M$, and taking the orthonormal vector at $T_pV$ pointing towards such an hypersurface). I couldn't go any further.

Sorry if there are obvious property I have not seen, I've just begun learning homology.

Have a nice day, and thanks for future answers

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For (2), that frame is a trivialization of the normal bundle. For (1), cut $M$ along an orientable hypersurface and glue copies of it along the hypersurface. There is a nice decription of this in Lickorish, in the case $n=3$, where $V$ is a knot. –  Neal Nov 15 '12 at 13:47
    
Thank you for the answer. But for (2) the normal bundle is of order $2$ while the frame explicited is only a vector field. So how could this provide a trivialisation ? –  Selim Ghazouani Nov 15 '12 at 14:25
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For (2), what happens with $\mathbb{R}P^2\subseteq S^4$? This is a codimension $2$ embedding and $H_2(\mathbb{R}P^2) = 0$. The the Stiefel whitney class of $\mathbb{R}P^2$ is $1+a+a^2$ while for $S^4$ it's trivial, so that implies the normal bundle must have Stiefel-Whitney class $1+a$, so, is, in particular, nontrivial. Note that $\mathbb{R}P^2$ bounds a chain in $S^4$, but $\mathbb{R}P^2$ doesn't bound any manifold at all. –  Jason DeVito Nov 15 '12 at 14:38
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I think you meant $[V] = 0$ in $H_{n-2}(M)$ (the stronger hypothesis $H_{n-2}(V) = 0$ prevents $V$ from being closed and oriented). I'm assuming that everything in sight is smooth and oriented. In High-dimensional knot theory, Ranicki call such a pair $(M, V)$ a homology framed knot. The discussion at bottom of page XIX = 16 and at page XXI = 18 are particularly relevant to us.

Because of the Tubular Neighbourhood theorem, there is a neighbourhood of $V \subset M$ homeomorphic to the normal bundle $\nu_V$. In particular, as everything is oriented, there is a well-defined notion of meridian : for each $x \in V$, we have a positive generator $\mu_x$ of $H_1((\nu_V)_x \setminus \{x\}) = H_1(\mathbb R^2 \setminus \{0\}) = \mathbb Z)$ and all those $\mu_x$ are homologous in $M \setminus V$. I claim two things:

  1. There is a morphism $\theta : \pi_1(M \setminus V) \to \mathbb Z$ which sends the meridian to $1$ (note that morphisms from $\pi_1(X)$ to $\mathbb Z$, morphisms from $H_1(X)$ to $\mathbb Z$ and cohomology classes in $H^1(X)$ are always the same, due to Hurewicz theorem and Universal Coefficient theorem).
  2. This morphism is actually represented by a hypersurface : there exists an embedded hypersurface $F \subset M$ whose boundary is $\partial F = V$ and such that intersection with $F$ defines the morphism $\theta$.

Let's start with that second result: what's key here is the equality $H^1(X) = [X, S^1]$ : every cohomology class in $H^1(X)$ is the pull-back of $[d\theta] \in H^1(X)$ by a map $f : X \to S^1$. Obviously, two homotopic maps give the same cohomology class. (I guess it can be seen rather easily but it is a consequence of a much more general construction of cohomology, see e.g. Hatcher, theorem 4.57, p. 393). By transversality arguments, if $X$ is a smooth manifold, you can homotope $f$ into a smooth function and take a regular value $c \in S^1$ of $f$. The preimage $F = f^{-1}\{c\}$ is then a smooth oriented submanifold such that the intersection with $f$ is the cohomology class we started with. (BTW, this second result is superfluous to prove the existence of your ramified covering but I think it makes the situation much clearer).

Let's know prove the first statement. Essentially, the proof is that in a perfect world, the hypothesis $[V] = 0$ in $H_{n-2}(M)$ would give you an embedded Seifert surface $F$ and the intersection number with $F$ would in turn give you the morphism $\theta$ we're after. I do not see a way to make this into a proper proof, but the algebraic topology machinery is precisely here to say that everything works as if this idealised picture were true.

Let's consider the exterior $E = M \setminus \mathscr U(V)$ of your "knot". That's a compact oriented manifold with boundary (the boundary is the sphere normal bundle $S\nu_V$). Poincaré-Lefschetz duality gives you an isomorphism $$ H_{n-1}(E, \partial E) \simeq H^1(E).$$ (And we really should think of this morphism as given by some kind of intersection counting). But, clearly, $M \setminus V$ deformation retracts on $E$. So $H^1(E) = H^1(M \setminus V)$. In the same way, $\mathscr U(V) \setminus V$ deformation retracts on $\partial E(V)$. So we have a canonical isomorphism $$ H^1(E, \partial E) = H^1(M \setminus V, \mathscr U(V) \setminus V) = H^1(M, V)$$ (by deformation retraction and excision).

As $[V] = 0$, we have a chain whose boundary defines the homology class $[V]$. This chain so defines a homology class $\Phi \in H_{n-1}(M, V)$ (and we should think of this beast as a $[F]$, for some ethereal Seifert hypersurface $F$). Through the isomorphism $H_{n-1}(M, V) \simeq H^1(M \setminus V)$ we have discussed [a special case of Alexander duality], this gives our beloved $\theta$. It isn't too hard to see that the mere existence of $\theta$ gives you a framing of the normal bundle $\nu_V$. Of course, a moment of thought is necessary to understand why this morphism has the good properties...

You can now build a ramified covering space by gluing the true covering space given by this $\theta$ (or the reduction modulo $k$ thereof) away from $V$ and the “model” $\widetilde{\nu_V} \to \nu_V$ given fiberwise by the map from $D^2 = (D\nu_V)_x$ to itself by $z \mapsto z^k$.

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