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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?

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Do you mean submanifolds, not subvarieties? – Kevin H. Lin Feb 20 '10 at 5:55
Yes, I meant submanifolds. The two words are interchangeable in italian, and I sometimes mix them up :-) – Andrea Ferretti Feb 20 '10 at 10:38

2 Answers 2

up vote 47 down vote accepted

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.

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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? – Alon Amit Oct 20 '09 at 21:39
Every class in H<sub>k</sub>(X) is realizable for k <= 6 or k >= n-2, so the first possible example is H<sub>7</sub>(X) for X a 10-manifold. Apparently the 10-dimensional 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. – Steven Sivek Oct 20 '09 at 22:18
Despite the generic-sounding 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. – Reid Barton Oct 21 '09 at 0:16
As discussed here… and here… 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 $X$. – Dan Ramras Jan 22 '13 at 7:54

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.

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Amazing - thank you (and Steven). – Alon Amit Oct 20 '09 at 23:05

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