Integral homology classes that can be represented by immersed submanifolds but not embedded submanifolds

Question

Let $M$ be an $m$-dimensional compact closed smooth manifold and $z\in H_n(M,\mathbb{Z})$ an $n$-dimensional integral homology class, with $m>n.$ Does there exist a pair of $M$ and $z$ so that $z$ can be represented by an immersed submanifold while $z$ can never be represented by an embedded submanifold? Mod $2$ examples are also welcome.

I am aware of Thom's results in Chapters II and III of his classical 1954 paper. I have tried to use my limited knowledge of differential topology, e.g., generic immersions, stable mappings, etc., to see whether one can find an easy example directly from the results in that classical paper, but I cannot get anywhere. I am sorry if this question is trivial, as my field is very far away from topology. Many thanks.

The motivation comes from several different problems in geometric measure theory. Roughly speaking, I am concerned about whether there are topological obstructions to removing the self-intersection in generic immersions and finding homologous embeddings from possibly topologically different submanifolds.

Remark: I want to thank many colleagues for pointing out interesting examples that are relevant. I have to add that the intended question is essentially in the same setting as in Thom's paper. In other words, we care about only codimension $>0$, we do not care about the connectedness of the submanifold, and $M$ is orientable in the integral homology case and unorientable in the mod $2$ case, etc. However, one exception is that we can allow $M$ to be unorientable for the integral case and we can weaken the structure group of the normal bundle to $O(m-n)$ if this makes it easier to give an answer to the question.

Answer 1

This is a great question, and I don't have an answer but this is too long for a comment.

Working mod $2$, a codimension $k$ homology class $z\in H_{m-k}(M;\mathbb{Z}/2)$ is realizable by an embedding if and only if its Poincaré dual $u:=PD(z)\in H^k(M;\mathbb{Z}/2)$ is the pullback of the universal Thom class $t_k\in H^*(MO(k);\mathbb{Z}/2)$ under some map. That is, viewing $PD(z)$ as a map $u:M\to K(\mathbb{Z}/2,k)$, we need a map $\tilde{u}:M\to MO(k)$ such that $u=t_k\circ \tilde u$, where $t_k:MO(k)\to K(\mathbb{Z}/2,k)$. (I'm ignoring some stuff about basepoints here.)

So far, this is all in Thom's paper. The story for realizability by immersions follows from later work of Robert Wells. For a based space $X$ let $QX=\Omega^\infty\Sigma^\infty X$ (infinite loops on infinite suspensions of $X$, a.k.a the free infinite loopspace on $X$). This functor has the property that any map from $X$ into an infinite loopspace $Y$ extends uniquely to an infinite loop map $QX\to Y$. So the Thom class $t_k:MO(k)\to K(\mathbb{Z}/2,k)$ extends to a map $\tilde{t_k}:QMO(k)\to K(\mathbb{Z}/2,k)$. It follows from Wells' work on cobordism of immersions that $z\in H_{m-k}(M;\mathbb{Z}/2)$ is realized by an immersion if and only if $u=PD(z)\in H^k(M;\mathbb{Z}/2)$ is a pullback of $\tilde{t_k}\in H^k(QMO(k);\mathbb{Z}/2)$.

In these terms, we are looking for a map $u:M\to K(\mathbb{Z}/2,k)$ that lifts through $\tilde{t_k}:QMO(k)\to K(\mathbb{Z}/2,k)$, but does not lift through $t_k:MO(k)\to K(\mathbb{Z}/2,k)$. The second part is relatively easy, you just need to find a cohomology operation that vanishes on $t_k$ but not on $u$. The first part seems harder, because you need to understand enough of the Postnikov tower of $QMO(k)$ to show that the relevant lift exists. The primary obstruction to realizability in both cases is $\beta(x^2)$, where $\beta:H^{2k}(M;\mathbb{Z}/2)\to H^{2k+1}(M;\mathbb{Z})$ is the Bockstein. Beyond that, it gets a bit more complicated.

There is a similar story for integral (co)homology classes and (co)oriented embeddings and immersions, replacing $MO(k)$ by $MSO(k)$.

Answer 2

I don't know if you'll find this a satisfying example, but what about $2\in \mathbb{Z} \cong H_1(S^1)$? This can be represented by the immersion that wraps the circle twice around itself, but it can't be represented by an embedding because it must be represented by a degree 2 map (or a map from a disjoint union of circles, but in any case it won't be injective).