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.



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


