When is a Homology Class Represented by a Submanifold? - MathOverflow [closed]

When is a Homology Class Represented by a Submanifold?

2010-04-13T01:15:48Z

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Cohomology and fundamental classes

Given an oriented manifold $M$ and an oriented submanifold $\phi:N\to M$ we can obtain a homology class $\phi_*[N]\in H_*(M)$ where $[N]$ is the fundamental class of $N$. In general, it is not true that every homology class of $M$ can be represented by a submanifold in this manner, however for some special cases it is.

For example, for $M$ an oriented (and closed maybe?) 4-manifold every homology class can be represented by a submanifold. Another example is when $M$ an Euclidean configuration space.

My questions are:

1) Under what circumstances can every homology class of $M$ be represented by a submanifold and

2) What are some examples of manifolds who have homology classes not representable in this manner?

Answer by Andy Putman for When is a Homology Class Represented by a Submanifold?

2010-04-13T01:51:28Z

A weaker question replaces "homology class of an embedded submanifold" with $f_{\ast}([M])$ for some compact smooth manifold $M$ and an arbitrary continuous map $f:M \rightarrow X$. Once you give up looking at embedded submanifolds, there is also no reason to restrict yourself to $X$ being a manifold.

A lot was proven about this by Thom in his classic paper "Quelques propriétés globales des variétés différentiables", which is more famous for containing his work on cobordism theory. A few of the results contained in that paper are as follows.

1) Every mod $2$ homology class can be so represented.

2) Integrally, it is true for every class in $H_k$ for $k \leq 6$.

3) For every $k \geq 7$, there exist polyhedra $X$ and classes in $H_k(X)$ that cannot be so represented.

EDIT : One should also remark that the above is germane to the original question too in many cases. Namely, if $X$ is a smooth $n$-manifold and $M$ is a compact smooth $k$-manifold and $f:M \rightarrow X$ is arbitrary, then $f$ is homotopic to an embedding as long as $k < n/2$.

Answer by Don Stanley for When is a Homology Class Represented by a Submanifold?

2010-04-13T06:36:46Z

If you really want a submanifold then I guess you can't always do it. For a closed manifold $M$ consider two times the fundamental class $2[M]$. It's easy to see you can't represent this class as a submanifold when $M=S^1$. Perhaps if you take any class in $a\in H_*(M)$ with nonzero self intersection, then $2a$ can't be represented as a submanifold?

Answer by Paul for When is a Homology Class Represented by a Submanifold?

2010-04-13T13:01:38Z

Here are a few simple answers to the question you asked:

Every class in $H_{n-1}(M;Z)$ for $M$ orientable is represented by a submanifold: choose a smooth map $f:M\to S^1$ representing the Poincare dual in $H^1(M;Z)=[M,S^1]$ and take the preimage of a point. In dimensions>2 it can be taken connected.
Similarly, every class in $H_{n-2}(M;Z)$ for $M$ orientable is represented by a submanifold: choose a smooth map $f:M\to CP^\infty$ representing the Poincare dual in $H^2(M;Z)=[M,CP^\infty]$, homotop $f$ into a finite skeleton, say $CP^N$, and take the preimage of $CP^{N-1}$.
Transversality says that if you can represent $x\in H_k(M)$ by a map of a smooth manifold (e.g. elements in the image of the Hurewicz map, or by Thom) , and $2k < n$, then you can represent it by an embedded submanifold (as Andy mentions above). For example, any class in $H_1(M)$ for $dim(M)\ge 3$. With care you can also make this work for $2k=n$, and there are techniques available in the "metastable" range (no triple points) involving generalizations of Whitney's trick and other ways to replace double points.