Question

Let X be a real orientable compact differentiable manifold. Is the (co)homology of X generated by the fundamental classes of oriented subvarieties? And if not, what is known about the subgroup generated?

Answer 1

Rene Thom answered this in section II of "Quelques propriétés globales des variétés différentiables." Every class $x$ in $H_r(X; \mathbb Z)$ has some integral multiple $nx$ which is the fundamental class of a submanifold, so the homology is at least rationally generated by these fundamental classes.

Section II.11 works out some specific cases: for example, every homology class of a manifold of dimension at most 8 is realizable this way, but this is not true for higher dimensional manifolds and the answer in general has to do with Steenrod operations.

Answer 2

This is a reply to Alon's comment, but it's too long to be a comment and is probably interesting enough to be an answer.

Here's an example Thom gives of a homology class that is not realized by a submanifold: let $X=S^7/\mathbb Z_3$, with $\mathbb Z_3$ acting by rotations, and $Y=X \times X$.
Then $H^1(X;\mathbb Z_3)=H^2(X;\mathbb Z_3)=\mathbb Z_3$ (and they are related by a Bockstein); let $u$ generate $H^1$ and $v=\beta u$ be the corresponding generator of $H^2$. Then it can be shown that the class $u \otimes vu^2 - v \otimes u^3 \in H^7(Y;Z_3)$ is actually integral (i.e., in $H^7(Y;Z)$), and its Poincare dual in $H_7$ cannot be realized by a submanifold (in fact, it can't be realized by any map from a closed manifold to $Y$, which need not be the inclusion of a submanifold). This is a natural example to consider because the first obstruction to classes being realized by submanifolds comes from a mod 3 Steenrod operation, and these are easy to compute on $Y$ because $X$ is the 7-skeleton of a $K(\mathbb Z_3,1)$. Note that the class in question is 3-torsion, so trivially 3 times it is realized by a submanifold.