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?

1$\begingroup$ Do you mean submanifolds, not subvarieties? $\endgroup$ – Kevin H. Lin Feb 20 '10 at 5:55

5$\begingroup$ Yes, I meant submanifolds. The two words are interchangeable in italian, and I sometimes mix them up :) $\endgroup$ – Andrea Ferretti Feb 20 '10 at 10:38
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.

1$\begingroup$ Are you aware of an explicit example where n>1 is needed? that is, an example of a cohomology class which is not the fundamental class of a manifold, but some multiple of it is? $\endgroup$ – Alon Amit Oct 20 '09 at 21:39

8$\begingroup$ Every class in H<sub>k</sub>(X) is realizable for k <= 6 or k >= n2, so the first possible example is H<sub>7</sub>(X) for X a 10manifold. Apparently the 10dimensional Lie group SP(2) provides such a class; this is constructed in "Cycles, submanifolds, and structures on normal bundles" by Bohr, Hanke and Kotschick, arXiv:0011178. $\endgroup$ – Steven Sivek Oct 20 '09 at 22:18

7$\begingroup$ Despite the genericsounding name, this is a landmark paper in algebraic topology. Besides answering this question, it also computes (for the first time?) the cobordism ring of unoriented manifolds. $\endgroup$ – Reid Barton Oct 21 '09 at 0:16

5$\begingroup$ As discussed here math.stackexchange.com/questions/281931/… and here mathoverflow.net/questions/21171/… this answer doesn't sound quite right; for instance, no integral multiple of $2[S^1]$ is the fundamental class of a submanifold of $S^1$). The relevant result in Thom's paper is Theorem II.4; one needs some hypotheses on the dimensions of the homology class $x$ and of the manifold $X$. $\endgroup$ – Dan Ramras Jan 22 '13 at 7:54

$\begingroup$ One should note that some multiple of any given homology class can be represented as the image of some fundamental class of some manifold under some continuous map into the space; see, e.g., the following article of Gaifullin: arxiv.org/abs/1201.4823 $\endgroup$ – AlexE Jan 5 '17 at 16:21
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 7skeleton of a $K(\mathbb Z_3,1)$. Note that the class in question is 3torsion, so trivially 3 times it is realized by a submanifold.