Integral homology classes of which no multiples admit embedded representatives with trivial normal bundle

Question

Let $M$ be a closed smooth manifold of dimension $n$ and $z\in H_l(M,\mathbb{Z})$ a $k$-dimensional integral homology class. Theorem II.4 of Thom's classical 1954 paper states that for $l< n/2$ or $n-l$ odd, some multiple of $z$ can be represented by a smoothly embedded submanifold with the trivial normal bundle.

I wonder if this conclusion is sharp. In other words, do there exist some $M$ and a non-torsion integral homology class $z$ on $M$, so that no multiples of $z$ can admit a smoothly embedded representative with a trivial normal bundle? (Non-torsion examples are what I need. Torsion class examples are also welcome.) I'm sorry if this is a trivial question, as my field is not adjacent to algebraic topology. Many thanks.

Answer 1

With the trivial normal bundle condition, it's fairly easy to produce non-realizable examples using Théorème II.2 of Thom's paper. Namely, a class $z\in H_l(M^n;\mathbb{Z})$ is realizable by an embedding with trivial normal bundle if and only if the Poincaré dual class $y\in H^{n-l}(M;\mathbb{Z})$ is spherical, meaning that there is a map $f:M\to S^{n-l}$ such that $y=f^*(s_{n-l})$, where $s_{n-l}\in H^{n-l}(S^{n-l};\mathbb{Z})$ is the generator.

For $y$ to be spherical, any cohomology operation which vanishes on $s_{n-l}$ must vanish on $y$. In particular, $y^2$ must be zero.

To give an explicit torsion-free example, no nonzero element $z\in H_2(\mathbb{C}P^2;\mathbb{Z})\cong\mathbb{Z}$ is realized by an embedding with trivial normal bundle.