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Pick an integral basis $a_i,\ i=1,\ldots,n=b_1(X)$, for $H^1(X,\mathbb Z)$, then form the element $$a:=a_1\cup\ldots\cup a_n\in H^{b_1}(X,\mathbb Z).$$

This is canonical up to sign; any other choice of basis differs by some $g\in GL(H^1(X,\mathbb Z))$, which changes $a$ by $\det g=\pm1$. There is no sign ambiguity on a Kähler manifold because we can choose the complex orientation on $H^1(X)$.

Has anyone seen this thing before ? Does it even have a name ?

It depends on $X$. For instance examples with $b_1(X)=4$ include $T^4$, where it is the volume form, and $\Sigma_2\times S^2$, where it is zero. ($\Sigma_2:=$ genus $2$ Riemann surface.)

I'm not sure what kind of answer I'm after. I have some curve counting invariants on a projective variety $X$ that depend only on the Chern numbers of $X$, and Chern classes evaluated against this strange canonical class $a$. Is that the best I can say ?

edited Sep 16 2011 at 21:40

asked Sep 16 2011 at 20:31

Richard Thomas
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You seem to assume that $H^1(X,\mathbb{Z})$ is torsion-free: is that so? – MP Sep 16 2011 at 20:42

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@MP: $H_0(X)$ is free so there is no $Ext$ term and $H^1(X)=Hom(H_1(X),\mathbb{Z})$, which is torsion-free. – Neil Strickland Sep 16 2011 at 20:51

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If $X$ is an algebraic variety and $Alb(X)$ denotes its Albanese variety, then is it correct that the class you want is the pull-back of the fundamental class of $Alb(X)$? In particular, your class will be trivial whenever the albanese map is not surjective. For a general topological space, the albanese variety should be replaced by the appropriate Eilenberg-Maclane space. Not sure if this helps, though! – MP Sep 16 2011 at 21:06

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That's certainly a slightly nicer way of looking at it, thanks. I still feel like this thing should have a name, simple though it is. Anyone for Albanese class ? – Richard Thomas Sep 16 2011 at 21:38

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For links to another topic, the Novikov conjecture is the assertion that the L-polynomial in the Pontryagin classes evaluated on classes of this type (i.e puled back to $H^*X$ from $H^*(K(\pi,1))$ is a homotopy invariant. For $\pi=\ZZ^n$ NC is proved, and for $X$ projective the Chern classes and Pontryagin classes are related. – Paul Sep 18 2011 at 3:41

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Ryan Budney · Answer 1 · 2011-09-16 22:34:33Z UTC

Given a space $X$ you can ask for a universal

$$ X \to (S^1)^n$$

such that any other map

$$ X \to (S^1)^j$$

factors (up to homotopy) through a map/homomorphism of groups $(S^1)^n \to (S^1)^j$

One way to construct this map is to abelianize the fundamental group of $X$ by attaching 2-cells that are commutators. Then kill $\pi_2$ of this space by attaching 3-cells, and so on. So you construct a $K(\pi,1)$ from $X$ by attaching only 2-cells and higher dimensional cells, and the 2-cells are commutators. So this constructed spaces is a $K(AB(\pi_1 X),1)$-space, i.e. its fundamental group is the abelianization $AB(\pi_1 X)$ of $\pi_1 X$.

The torsion part of the abelianization factors off (some lens space product factors) and you're left with just a product of circles.

I'm not sure if that has a name, but that's one way to think of it.

In low-dimensional topology, this is relevant because this is the map whose homotopy-fibre has the homology of the universal (multi-variable) Alexander module.

edit: of course, if $H_1 X$ isn't finitely-generated, this universal map may not be to a product of circles. But there still is a universal map.

Strange canonical cohomology class on topological spaces

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