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.


1 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.