Realizing integral homology classes on non-orientable manifolds by embedded orientable submanifolds

Question

Let $M^m$ denote a compact, non-orientable smooth manifold and $\nu$ an integral homology class of dimension $n$. I am interested in understanding the representability of $\nu$ by embedded, orientable submanifolds in $M$. Specifically, my inquiries are twofold:

1. Can some multiple of $\nu$ be represented by an embedded, orientable submanifold of $M$?

2. For which smallest dimension and codimension do there exist $M$ and $\nu$ such that $\nu$ is not representable by any embedded submanifold?

I am aware of Thom's classical 1954 paper and the results therein. For integral homology on orientable manifolds, the first question has a positive answer and the second question has an answer $(7,3).$ However, to my limited knowledge, Thom's paper uses Poincare duality to transform the problem into a cohomological one. Thus for nonorientable manifolds his results concerned only $\mathbb{Z}/2\mathbb{Z}$ coefficient homology, not providing an answer to the above questions directly. I tried to solve the above questions using the corresponding results in the orientable double cover but did not get anywhere.

The motivation comes from a problem in geometric measure theory. I'm sorry if this question is trivial, as my field is very far from topology. Any insights or references that can shed light on these questions would be highly appreciated. If general results are not available, specific examples/partial results are also welcome. Many thanks!

Answer 1

Here are some comments that don't really answer the question, but are too long for the comment box.

Firstly, the Poincaré dual of $\nu\in H_n(M;\mathbb{Z})$ is a twisted integer class $D\nu\in H^{m-n}(M;\mathbb{Z}^{w_1(M)})$, where $w_1(M):\pi_1(M)\to \mathbb{Z}/2$ is the orientation, or first Stiefel-Whitney class, of $M$. There is a twisted Thom class $t\in \tilde{H}^{m-n}(MO(m-n);\mathbb{Z}^w)\cong H^{m-n}(BO(m-n),BO(m-n-1);\mathbb{Z}^w)$, where $w:\pi_1(BO(m-n))\to\mathbb{Z}/2$ is the universal first SW-class. I haven't seen this written down anywhere, but I'm fairly certain that $\nu$ is realisable by an embedded oriented submanifold if and only if there is a pointed map $\phi:M_+\to MO(m-n)$ such that $\phi^*(t)=D\nu$.

The usual way to rule out realizability is to find a cohomology operation which vanishes on the Thom class but not on the cohomology class to be realized. This makes the above criterion difficult to apply, unless one understands cohomology operations with local coefficients (which I'm sure some people do).

Another sufficient condition for realizability is that the Poincaré dual class $D\nu$ be the twisted Euler class $e(\xi)\in H^{m-n}(M;\mathbb{Z}^{w_1(M)})$ of some vector bundle with $w_1(\xi)=w_1(M)$. This is because then $\nu$ is realized by the zeroes of a section of $\xi$ transverse to the zero section. When the codimension $m-n$ equals $2$, every twisted integer class is a twisted Euler class. This follows from the representability of twisted cohomology by maps into generalized Eilenberg--MacLane spaces, and the fact that the generalized Eilenberg--MacLane space $L_{\mathbb{Z}/2}(\mathbb{Z})$ may be taken to be $BO(2)$. This has been discussed on MO before, see here and here. This gives a partial answer to your question 2.: the codimension must be at least $3$.