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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).

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For an interesting "usual" argument, there is the Whitney trick when M is simply connected. There you find two points of intersection of opposite sign, find an arc between them along $Z^k$ and one along $Z^{n-k}$, find a disk between those arcs, then slide $Z^k$ along the disk so it becomes disjoint from $Z^{n-k}$. There are dimension constraints here, and if $Z^k$ and $Z^{n-k}$ are meant to be manifolds you want an embedded disk; Milnor writes well about it in the "h-cobordism" book. (pg. 76 of <a href="maths.ed.ac.uk/~aar/surgery/hcobord.pdf">this PDF</a> but it's all over). –  Elizabeth S. Q. Goodman Oct 1 '11 at 22:43

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up vote 12 down vote accepted

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.]

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Dear algori, thank you so much for this nice answer! –  aglearner Oct 3 '11 at 12:24
Dear aglearner -- welcome! –  algori Oct 3 '11 at 15:12
I think there is a little thing to fix in the argument: you cannot represent any homology class by a connected pseudomanifold. As an example, the homology class $(2,0)$ in the torus cannot be represented by a connected pseudo-manifold, which would be a simple closed curve. The point is that homology is in Z (not in Z2) and hence the "tubes" must be attached on the right side of each component. However, your argument probably works anyway: you might only need that $Z_1\cup Z_2$ is connected. –  Bruno Martelli Oct 3 '11 at 19:45
Bruno -- thanks! However, I think this problem arises only when the cycle is of codimension 1: otherwise there are no right and wrong sides and one can always attach the tubes correctly. –  algori Oct 3 '11 at 20:42
Dear algori, it is not true that every element of $\pi_1(\Sigma)$ can be represented by a closed simple loop (though for $H_1$ this is true for every *primitive* homology class). However a surgery argument like the one you describe in higher dimensions does go through for curves on surfaces, and so the result definitely holds there as well. –  Tom Church Oct 4 '11 at 5:48


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.

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Is it still false if $M$ is a projective complex variety (and $Z^k, Z^{n-k}$ are algebraic cycles)? –  Henri Oct 1 '11 at 22:19
In the algebro-geometric setting the "moving lemma" does everything you need. Note that Simon's example doesn't work, anyways, as he mentions in his edit: the cycles he mentions are homologous to unions of pairs of circles, and you can make these pairs disjoint. –  Jack Huizenga Oct 1 '11 at 22:25
Henri: I believe that situation can't happen. If two algebraic cycles, or rather two complex submanifolds of a complex manifold, have a point of intersection, then that point is counted positively. You can see "non-negative" because the complex plane defines a canonical orientation, so conjoining the orientations of the two tangent spaces at the intersection agrees with the ambient orientation. Actual positivity is trickier, but since this is complex algebraic geometry you should be able to do it by degrees of polynomials in local coordinates. (It is much harder in almost-complex geometry.) –  Elizabeth S. Q. Goodman Oct 1 '11 at 22:32
Sorry, I meant to say that positivity is perhaps trickier in the non-transverse case, where you have to show that you have an intersection of higher degree. –  Elizabeth S. Q. Goodman Oct 1 '11 at 22:33
The main issue is that the cycles could intersect in the wrong dimension. –  Jack Huizenga Oct 1 '11 at 22:35

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.

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Greg, great, thanks a lot! I will add the condition that the manifold M is connected. I do want either a non-trivial counterexample or a proof my statement. –  aglearner Oct 2 '11 at 21:10

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