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If we take a knot $K$ in $S^3$, there are several ways to construct the associated Seifert surface. One way, which I am not familiar with, I just came across in a paper I am reading. It goes like this:

Consider the regular neighborhood $n_K$ of $K$ in $S^3$, which is diffeomorphic to $K\times \mathbb{R}^2$. $S^3-n_K$ is the knot complement. By Alexander duality, $H^1(S^3-n_K)\cong H_1(K)\cong\mathbb{Z}$; but we also have $H^1(S^3-K)\cong \[S^3-n_K,S^1\]$, so we can take a (smooth) map $f:\ S^3-n_K\rightarrow S^1$ representing a generator of $H^1$. The preimage of a regular value of $f$ gives a codimension-1 submanifold of $S^3-n_K$; namely, a surface-with-boundary $S$, such that $\partial S\subset \partial (S^3-n_K)$. So far, so good. It is the next step that throws me:

$\partial S$ is cobordant to $K$.

I don't understand how this works, and am asking how one shows this. For example, it's not even clear that $S$ is connected, and each component could have boundary.

I should mention that this is of course trivial in this specific case, since all 1-manifolds are cobordant. But this should work for higher-dimensional knots. It is even used in this answer and the comments below, where $K$ is replaced by an orientable 3-manifold and $S^3$ is replaced by $S^5$ (or $S^6$).

Can someone enlighten me in how one shows this (in the general case)? Namely, without relying on the dimensions of the spaces involved, but rather their codimensions?

[It might be relevant to note that the "regular neighborhood" (normal bundle) of the two examples above - the knot and the 3-manifold - are trivial. I don't know if this matters or not.]

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1 Answer 1

up vote 7 down vote accepted

This is an old argument that essentially predates much modern knot theory, and goes back to Serre.

The basic idea goes like this: let $C$ be the complement of a co-dimension two knot in $S^n$. Apply Poincare/Alexander duality to deduce that $C$ is a homology $S^1 \times D^{n-1}$. So $H^1 C \simeq \mathbb Z$, and $H^1 \partial C$ is either $\mathbb Z^2$ or $\mathbb Z$ according to whether or not $n=3$ or $n>3$. In either case the restriction map is an injection.

Serre's theorem that $H^1 X = [X,S^1]$ gives you a map $C \to S^1$ which you make transverse to a point. Because the boundary is a product $\partial C \simeq S^1 \times S^{n-2}$, you can ensure the map $C \to S^1$ is not just homotopic to but actually projection onto the first factor (when restricted to $\partial C$).

This ensures the preimage of a regular value of $C \to S^1$ is a Seifert surface for the knot. By Seifert-surface I mean an orientable co-dimension one submanifold of $S^n$ whose boundary is the knot. If the surface is disconnected, you throw away any components that do not touch $\partial C$. Because $\partial C \to S^1$ is the projection onto the factor, only one path-component touches $\partial C$.

Serre and Thom used these ideas repeatedly in their early attacks on the Steenrod realization problem. This was the version where you're trying to realize the homology classes by embedded submanifolds.

The above is a relative version of their arguments. There's an old Springer Lecture Notes in Mathematics that outlines the history of this argument and when it was first brought to knot theory. Right, SLN in Math 685. If it's not in the essay by Cameron Gordon, it's in the other big essay, maybe it was Milgram?

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Ryan, thank you very much for your answer. To be clear, why can we ensure the map restricts to projection on the boundary? Also, this works for the 3-manifold case as well I think. If $M$ is a closed orientable 3-manifold, embedded in $S^6$, then similar arguments show that there is a codimension 2 submanifold in the complement $C=S^6-n_M$; if we can also ensure the map is projection on $\partial C\cong M\times S^2$, then the conclusion is the same. –  Steve D Jan 10 '13 at 20:31
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Think of it as a process. First homotope the map on the boundary -- since the boundary has the homotopy extension property (even a collar neighbourhood) this is perfectly doable. Since the map is smooth on the boundary, when you go to homotope the map $C \to S^1$ to be smooth, you do not have to change the map on $\partial C$ at all -- this is a standard smoothing argument that you can find in Hirsch's Differential Topology text. –  Ryan Budney Jan 10 '13 at 20:35
    
Sorry, that last part of my comment above makes no sense: it can not be projection, because it comes from a map into $\mathbb{CP}^k\subset \mathbb{CP}^\infty$. So I actually don't see how to make this work in the 3-manifold case. –  Steve D Jan 10 '13 at 20:36
    
OK, I feel confident I have worked this out, thanks to your answer and reading the relevant section (#8) in Gordon's paper. Again, thanks for the great answer! –  Steve D Jan 15 '13 at 6:08
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"Serre's theorem" that $H^1(X)=[X,S^1]$ was proved by Bruschlinsky, Math. Ann. 109 (1934), 525-537. –  Sergey Melikhov Mar 9 '13 at 7:55

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