How to disjoint two cycles with zero intersection? - MathOverflow

How to disjoint two cycles with zero intersection?

2011-10-01T20:15:37Z

Suppose that $M^n$ is an orientable connected (thanks to Greg) manifold and $Z^k$ with $Z^{n-k}$ are two real cycles in $M^n$ with zero index of intersection $Z^k\cdot Z^{n-k}=0$ (for me these cylces are manifolds if this helps). Is it true that the cycles can be replaced by homologous ones that does not intersect at all? How one usually proves this?

Note that we don't put any restriction on cycles by which we replace $Z^k$ and $Z^{n-k}$ (they are not asked to be connected). I am mainly interested in positive statement, like the one given by Elisabeth in her comment. But if the answer is still "no" in general situation I would like to know this. One more detail, in my case $Z^k$ is two dimensional (if this helps).

Answer by Simon Rose for How to disjoint two cycles with zero intersection?

2011-10-01T21:39:56Z

Edited:

If you require the $Z_i$ to be connected, then this is not true.

Consider $M = \mathbb{R}^2$ less three points (say $-1, 0, 1$), and let $Z_1$ be a loop around $-1$ and $0$, while $Z_2$ is a loop around 0 and 1. Then their index of intersection is 0, but they cannot be replaced by cycles which do not intersect.

It is of course worth noting that $M$ is not compact in this case.

However, if the $Z_i$ are allowed to be disconnected, then the argument I just gave falls apart. Either of the $Z_i$ are homologous to a disjoint union of two circles around their respective centres, and so it is clear that they can be made to be disjoint.

Answer by Greg Friedman for How to disjoint two cycles with zero intersection?

2011-10-02T16:14:16Z

Let $M$ be the union of two oriented circles. Let $\Gamma$ be the orientation class of $M$. Let $Z$ be the $0$-dimensional homology class represented by a positively oriented point in one of the circles and a negatively oriented point in the other circle (so $Z$ representes the class $<1,-1>\in H_0(M)\cong \mathbb{Z}\oplus \mathbb{Z}$). Then the intersection number of $\Gamma$ with $Z$ should be $0$, but $\Gamma$ and $Z$ can't be made disjoint. In fact, the basic property of a fundamental class of a compact manifold is that it's supported at every point.

Answer by algori for How to disjoint two cycles with zero intersection?

2011-10-02T22:54:50Z

I think that the answer is yes and that we don't really need the full Whitney's trick for this since being homologous is a much coarser relation than being isotopic, which is what Whitney's trick gives. So, rather than using the full trick, one can use just a half of it.

Let $M$ be an oriented connected smooth manifold, and let $Z_1,Z_2$ be oriented pseudo-manifolds representing two cohomology classes (recall that a pseudo-manifold is a stratified space that has no codimension 1 strata and such that each connected component is the closure of a single connected codimension 0 stratum; any homology class can be represented by a pseudo-manifold). Assume $\dim Z_2\geq 1$ [upd: and $\mathop{\mathrm{codim}}_M Z_2>1$; this assumption excludes the cases when one of the cycles is 0-dimensional and the other is of the maximal dimension, in which case the statement we're after is clearly true, and when $\dim Z_1=\dim Z_2=1$; this case has to be considered separately].

First, let's make $Z_2$ connected by joining the connected components with tubes. [upd: as Bruno Martelli points out in the comments, some care is needed here. However, if we have a tube that induces the wrong orientations of one of the components it's supposed to connect, we can always twist the tube since we assume $\mathop{\mathrm{codim}}_M Z_2>1$.] While doing this we may introduce new intersection points, but after a small isotopy these will be all transversal and their signs will add up to 0. Second, take two intersection points $P,Q$ with opposite signs and join them with a non-self-intersecting path $\gamma\subset Z_2$ that does not pass through the singularities [upd: and through other intersecrtion points; some care is needed here as well when $\dim Z_2=1$: in this case we take $P$ and $Q$ to be neighbors on $Z_2$].

Now equip $M$ with a Riemannian metric and let's modify $Z_1$ by taking out two small balls around $P$ and $Q$ in $Z_1$ and inserting a thin tube $T$ instead where $T$ is obtained by exponentiating the sphere subbundle of $N_M Z_2|\gamma$ of sufficiently small radius. More precisely, some work is needed to identify the spheres in $N_MZ_2$ at $P$ and $Q$ with the boundaries of the balls, but this should be no problem.

The result will be homologous to $Z_1$: the fact that $P$ and $Q$ have different signs ensures that the small balls around them and the tube together form the boundary of the exponential of a a ball subbundle of $N_{M}Z_2|\gamma$. [I wish I could draw a picture here but don't know how to do that.] Notice that when $Z_1$ is a loop around $(0,0)$ and $(1,0)$ in $\mathbb{R}^2$ and $Z_2$ is a loop around $(-1,0)$ and $(0,0)$, as in Simon Rose's example, then this procedure cuts $Z_1$ into two loops, one around $(0,0)$, the other around $(1,0)$.

In this way one can eliminate every pair of intersection points with opposite signs. Notice that we haven't done anything to $Z_2$ in the process, apart from making it connected.

[upd: in the case when $M$ is a surface and $Z_1,Z_2$ are 1-dimensional, then one could try to adapt the above argument for possibly self-intersecting $Z_2$, but this would require some work. In a sense, this makes sense: if 4 dimensions are not enough to perform Whitney's trick, then it's not too surprising that 2 dimensions are not enough to perform half of it.]