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

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.

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.