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Consider in an $2n$ dimensional manifold $M$(compact, smooth), for embedded $n$ dimensional surfaces with boundary, denoted by $\Sigma$ and $\Sigma'$, in M, we can consider "intersection number"(counted with sign) of $\Sigma$ and $\Sigma'$, denoted as $I(\Sigma, \Sigma')$. However under some homotopy mapping of $\Sigma$ or(and) $\Sigma'$ without changing their boundaries, the "intersection number" of them may be changed. More explicitly, if we do some homotopy mapping $\phi: \Sigma \rightarrow \Sigma_1$, similarly $\phi': \Sigma' \rightarrow \Sigma_1'$, then under $\phi$ and $\phi'$, the "intersection number" may also change from $I(\Sigma, \Sigma')$ to $I(\Sigma_1, \Sigma_1')$.

My question is that whether it is possible to choose suitable above-mentioned $\phi:\Sigma \rightarrow \Sigma_1$ and $\phi':\Sigma' \rightarrow \Sigma_1'$($\phi, \phi'$ are homotopy mapping without changing their boundaries, namely $\partial \Sigma=\partial \Sigma_1$, $\partial \Sigma_1'=\partial \Sigma'$) so that the "intersection number"(counted with sign) of $\Sigma_1$ and $\Sigma_1'$ is exactly one? If for general $\Sigma$ and $\Sigma'$, it is not generally possible, how about the case where $M=S^4$, $\Sigma$ and $\Sigma'$ are some Seifert surfaces of two embedding closed lines $\gamma$ and $\gamma'$ in $S^4$, and $\gamma$, $\gamma'$ are homotopic?

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  • $\begingroup$ I still object to the phrasing "we can calculate intersection number..." since you have to move $\Sigma$ and $\Sigma'$ to do so! But this is a small point and I will leave it. $\endgroup$ – user98602 Jan 1 at 17:53
  • $\begingroup$ @Mike Just fix it to "we can consider intersection number...". $\endgroup$ – wln Jan 1 at 17:58
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You can make the intersection number whatever you want.

Some preliminaries on intersection numbers of manifolds with boundary (I know you read about this in my previous answer, but the goal is to make this answer self-contained).

$M$ is an ambient oriented manifold without boundary (this is inessential, but the boundary makes discussion less pleasant, and we're here to have a good time, no?), say $2n$-dimensional. $S$ and $S'$ are compact oriented submanifolds of complimentary dimension $\dim S + \dim S' = 2n$. If $S \cap \partial S' = \partial S \cap S' = \varnothing$, then one may isotope either $S$ or $S'$ (your choice), without modifying the boundary (this is where the assumption on $S \cap \partial S'$ etc is crucial) and so that $S$ never passes through $\partial S'$ and vice versa, so that $S \pitchfork S'$; then the signed count $\#(S \cap S')$ is independent of the choice of transverse representatives, up to homotopies $f_t$ with $f_t(S) \cap \partial S' = \varnothing$ for all $t$, and symmetrically with $f_t(S')$.

If $S \subset M$ is a compact submanifold with boundary, it still carries a vector bundle called $NS$, the normal bundle; and one still has a tubular neighborhood-type theorem. I will word it carefully in what follows.

Let $S \subset M$ be a compact submanifold with boundary of codimension $k$. Then $S$ carries a vector bundle of rank $k$ known as the normal bundle $NS$. There is a codimension-0 submanifold $D(NS)$ with boundary of $M$, neither closed nor open, containing $S$, with the following properties: 1) $D(NS)$ is diffeomorphic to the open unit disc bundle inside of $NS$ (hence the notation). 2) $\partial D(NS) = D(NS\big|_{\partial S})$, meaning that it the boundary of $D(NS)$ is the disc bundle of the normal bundle to $S$, restricted to $\partial S$. 3) One may take $D(NS)$ to be sufficiently small to be contained in any fixed neighborhood of $S$.

To visualize this it helps to have an example at hand; consider $S = [-1, 1] \subset \Bbb R^2$; then $NS$ is the bundle consisting of vertical lines, and one possible choice of tubular neighborhood $D(NS)$ is $[-1,1] \times (-\varepsilon, \varepsilon)$: it is "open on each fiber", but not open because of the boundary.

In this situation, suppose one has fixed ahead of time a nonvanishing section $X \in \Gamma(NS \big|_{\partial S})$: this reads "A nonvanishing normal vector field to $S$ along $\partial S$". Extend this however you like to a section $Z$ of $NS$ over all of $S$. Then the graph of $X$ in $D(NS)$ is a smooth submanifold, with boundary contained in $\partial D(NS)$; call this manifold $X(S)$, to indicate that I have pushed $S$ in the direction of $X$. Then the definition of $X(S)$ implies that $\partial X(S) \cap S = \varnothing$, and similarly $X(S) \cap \partial S = \varnothing$.

At this point, one may define the intersection number of $S$ and $X(S)$ precisely as in the beginning of this answer (with their fixed boundary values); fixing the boundary of $X(S)$ corresponds to fixing a nonvanishing section of $NS \big|_{\partial S}$.

What this tells us, altogether, is that if you're interested in studying self-intersection numbers of half-dimensional manifolds, you may as well restrict to discussion of the pair $(E, S)$, where $S$ is an $n$-dimensional oriented compact manifold with boundary and $E$ is the total space of an oriented rank $n$ vector bundle over $S$.


What we saw up above is that the only way for the intersection number to change is to move zeroes across the boundary. So that is what we'll do.

It amounts to the following; triangulate $S$ and let $E$ be its normal bundle (an equi-dimensional vector bundle over $S$). All you want is to say that on a compact manifold with boundary, one may find a nonvanishing section of any vector bundle of the same rank. Triangulate $S$ and equip $E$ with a section with finitely many nondegenerate zeroes, all on the interior of the top-dimensional simplices. By a homotopy, you can always move a zero to the adjacent cell (while having fewer zeroes on the original cell). Ultimately, one can ensure that all zeroes are contained in a simplex $\sigma$ with $\sigma \cap \partial S \neq \varnothing$. Assuming for convenience that the vector field along $\partial \sigma$ is a field of unit vectors, then what we have constructed is a map $(D^n, D^{n-1}) \to (D^n, S^{n-1})$. All such maps are null-homotopic because the domain is contractible (as a pair). So we may modify the vector field on this cell, without changing its values on $\partial \sigma - \partial S$, so that it has no zeroes. Thus the final vector field has no zeroes at all.

The picture is we are pushing zeroes off to "infinity", past the boundary.

Of course, though I told you how to construct something with intersection number zero, there was no reason I had to push all of the zeroes off of the boundary. I could have pushed off as many as I liked, or even brought some in from outside - all because every map $(D^n, S^{n-1}) \to (D^n, S^{n-1})$ (which are classified by their degree, aka, intersection number with $0$) is homotopic through maps of the form $(D^n, D^{n-1}) \to (D^n, S^{n-1})$. So what's happening is on $\partial \sigma \cap \partial S$, the "other" side of the disc, I allow zeroes to move in and out if I so desire.

This completes the proof of your question. Below, I'll write down a different approach.


What we learned above was: if you fix the (homotopy class of) nonvanishing section $\partial S \to NS$, then you have picked out a specific intersection number already. What's next to ask is: How does this change as we re-choose our nonvanishing section?

What follows is essentially a "codimension 1" version of the Hopf degree theorem.

Let $M$ be a compact $(n-1)$-dimensional manifold, equipped with an $n$-dimensional oriented vector bundle $E$. Let $\Gamma(E \setminus 0)$ denote the space of nonvanishing sections $M \to E \setminus 0$; write its set of connected components as $\pi_0 \Gamma(E \setminus 0)$. If one fixes ahead of time a specified $X \in \pi_0 \Gamma(E \setminus 0)$, then there is a map $I_X: \pi_0 \Gamma(E \setminus 0) \to \Bbb Z$, defined as follows: if $Y \in \pi_0 \Gamma(E \setminus 0)$, pick a homotopy $X_t$ through sections between $X$ and $Y$; the sections $X_t$ may vanish for $t \in (0,1)$. Then $I_X(Y) = \#(X_t \cap 0)$, the intersection number of the map $X_t: M \times [0,1] \to E$ with the zero-section of $E$. Because $X_0$ and $X_1$ are nonvanishing, this is well-defined as in the first section of this post. Then $I_X(X) = 0$ and $I_X$ is surjective.

The fact that $I_X(X) = 0$ is clear: take $X_t = X$ for all $t$!

To see that $I_X$ is surjective is only slightly more subtle. What we do now is as follows. Choose a small disc in $M$; over that disc, choose a trivialization of $E \cong \Bbb R^n$ so that in this trivialization, $X$ is just constant at the first basis vector.

What we want to do is exploit the Hopf degree theorem, which says that maps $(D^n, S^{n-1}) \to (D^n, S^{n-1})$ are classified by their degree (which, generically, is given by the signed count of zeroes); what we start with is a (rather trivial) map $(D^{n-1}, S^{n-2}) \to (D^n, S^{n-1})$. The thing to use here is that $D^{n-1} \times I \cong D^n$, so we will construct a vector field on $D^n$ which vanishes once on the interior and never on the boundary. To define such a map, we should construct a homotopy $X_t$ so that $X_0$ and $X_1$ are unit vector fields, as is the restriction of $X_t$ to $\partial D^{n-1}$.

Define the homotopy $X_t = X$ on $S^{n-2}$ so that we have constructed a map $D^{n-1} \cup S^{n-2} \times I \to S^{n-1}$. Construct an extension (using the Hopf degree theorem!) to a map $\partial D^n \to \partial D^n$ of degree $1$; then any choice of extension to all of $D^n$ (by definition of degree) must have signed count $\#(X_t \cap 0) = 1$.

We changed the degree of $X$ "on this disc" by $1$. To do this globally, remember that on the boundary of the disc, $X_t$ was constant; so we just take $X_t$ to be constant everywhere outside this disc as well and we have a continuously varying family of vector fields. (You may smooth them out, if you like.) In particular, they are never zero! So at this point we have proved the desired result: in the homotopy we constructed, there is only one zero, which is positive and cut out transversely.


What does this buy you? At the start, we had a continuous section $X: S \to E$ which was nonvanishing on the boundary; it gave us the invariant $\#(X \cap 0)$, which only depended on the fixed boundary value of $X$ (up to homotopy through nonvanishing sections). What we showed above is, given a fixed boundary value $X: \partial S \to E$, we may modify $X$ on the boundary so as to add $1$ to the intersection number $\#(X \cap 0)$. Whatever number we started with, we can now either add or subtract until it becomes any other integer. This is what you wanted.

I think this is all much more transparent in the language of the Euler class and characteristic classes, but I assume you prefer the more elementary differential topology language.

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