Let $X$ be a compact manifold of dimension $\geq k$. Denote by
\begin{equation}
h: \pi _k(X) \rightarrow H_k(X,\mathbb{Z})
\end{equation}
be Hurewicz homomorphism and by $\Gamma _k(X)\subset H_k(X,\mathbb{Z})$ its image. I want to look at the pairing
\begin{equation}
\langle \ . \ ,\ . \ \rangle : \ \Gamma _k(X)_{\text{free}} \times H^k(X,\mathbb{Z})_{\text{free}} \rightarrow \mathbb{Z}
\end{equation}
obtained by restricting the the Kronecker pairing between the free subgroups of homology and cohomology, which is non-degenerate. In general the restriction above will be degenerate. Indeed given a non-vanishing class $\omega \in H^k (X,\mathbb{Z})_{\text{free}}$, it is possible that for any $f: S^k \rightarrow X$ the pull-back $f^* \omega$ is zero. I'm looking for some cases in which I can say something more on this degeneracy.
In particular if $\omega \in H^k(X,\mathbb{Z})_{\text{free}}$ is also a generator of the cohomology ring (so that it cannot be written as the product of two classes of lower degree) I have some intuition that maybe $\langle \Sigma , \omega \rangle \neq 0$ for at least one $\Sigma \in \Gamma _k (X)$. I am not sure whether this is true, and if it is how to prove it. Or if it is false how to modify a bit statement to get some true fact connecting the generators of the cohomology ring with the image of the Hurewicz map. Can somebody say something about this?
Best Answer
The answer is no, by the existence of nontrivial Massey products.
Call a cohomology class decomposable if it is in the subgroup generated by cup products of pairs of classes of positive dimension.
As you say, any decomposable class pairs to zero with all spherical homology classes (i.e. those in the image of the Hurewicz map). That's because in the cohomology of a sphere all decomposable classes are zero.
But more generally in the cohomology of a sphere the Massey product of three positive-degree cohomology classes is always zero, and this implies that in general the Massey product of three positive-degree cohomology classes (when it is defined) must pair to zero with all spherical classes.
Here the Massey product of $a\in H^i(X)$, $b\in H^j(X)$, and $c\in H^k(X)$ is defined when $a\cup b=0$ and $b\cup c=0$, and it belongs to the quotient of $H^{i+j+k+1}(X)$ by its decomposable part. So any Massey product that is non-zero (in that quotient) gives us a cohomology class that is not decomposable but does pair to zero with spherical homology classes.