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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asked Oct 20, 2009 at 20:57
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3$\begingroup$ Do you mean submanifolds, not subvarieties? $\endgroup$– Kevin H. LinCommented Feb 20, 2010 at 5:55
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7$\begingroup$ Yes, I meant submanifolds. The two words are interchangeable in italian, and I sometimes mix them up :-) $\endgroup$– Andrea FerrettiCommented Feb 20, 2010 at 10:38
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3 Answers
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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.
answered Oct 20, 2009 at 21:13
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2$\begingroup$ 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? $\endgroup$ Commented Oct 20, 2009 at 21:39
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9$\begingroup$ 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. $\endgroup$ Commented Oct 20, 2009 at 22:18
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12$\begingroup$ 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. $\endgroup$ Commented Oct 21, 2009 at 0:16
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10$\begingroup$ 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 $X$. $\endgroup$ Commented Jan 22, 2013 at 7:54
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$\begingroup$ 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 $\endgroup$– AlexECommented Jan 5, 2017 at 16:21
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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 freely 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 $v$ generate $H^1$ and $u=\beta v$ be the corresponding generator of $H^2$. Then it can be shown that the class $$u \otimes vu^2 - v \otimes u^3 \in H^7(Y;\mathbb{Z}_3)$$ is actually integral (i.e., in $H^7(Y;\mathbb{Z})$), 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.
answered Oct 20, 2009 at 22:25
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$\begingroup$ Is it possible that u and v are switched in the formula for the class in question? The formula here seems to be a class in $H^5$, not $H^7$ and, moreover, it appears to be $0$ (because $u^2=0$.) $\endgroup$ Commented May 11, 2021 at 13:51
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$\begingroup$ Good catch; that's exactly what happened (or rather the dual--I switched them when first defining them)! For anyone who wants to follow along in Thom's paper, this example is on the bottom of page 62 and the top of page 63. $\endgroup$ Commented May 11, 2021 at 15:32
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According to an article in The Geometry and Topology of Submanifolds and Currents, edited by Weiping Li and Shinshu Walter Wei
The lack of suitable representatives in smooth homology by submanifolds was 'partial motivation' for Federer to introduce rectifiable homology by defining
• the chain groups as:
$Z_{i}(A,B):=\{T\in R_{m}(\mathbb{R}^{n},\mathbb{Z}): \partial T\in R_{m-1}(\mathbb{R}^{n},\mathbb{Z}),spt(T)\subset A$ and $spt(\partial T)\subset B\}$
• the boundary groups as
$B_{i}(A,B):=\{\partial T:T\in Z_{i+1}(A,A)\}$
And the homology groups as usual by $H_{i}(A,B):=Z_{i}(A,B)/B_{i}(A,B)$
(Here, $A$ & $B$ are both compact Lipschitz neghbourhood retracts (of $\mathbb{R}^{n}$) and we have $B\subset A$).
He showed that it satisfied the Eilenberg-Steenrod axioms, and hence is isomorphic to singular homology; and that every homology class in this homology contains a mass-minimising rectifiable representative. Thus we should see Federers notion of rectifiable currents as an appropriate generalisation of submanifold which allows us to identify homology classes with special sub-manifolds.
