Strange canonical cohomology class on topological spaces 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 ?
 A: 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. 
