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) \times H^k(X,\mathbb{Z})_{\text{free}} \rightarrow \mathbb{Z} \end{equation} obtained by restricting the the natural pairing between homology and cohomology, which is non-degenerate. Here $H^k(X,\mathbb{Z})_{\text{free}}$ denotes the torsion-free part of the cohomology. In general the restriction above will be degenerate, and I'm looking for some cases in which I can say something more on this degeneracy. Indeed given a non-vanishing class $\omega \in H^k (X,\mathbb{Z})_{\text{free}}$, of course it is possible that for any $f: S^k \rightarrow X$ the pull-back $f^* \omega$ is exact. However 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 the intuition that $\langle \Sigma , \omega \rangle \neq 0$ for at least one $\Sigma \in \Gamma _k (X)$. However 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?

$\textbf{EDIT:}$ As @ConnorMalin correctly pointed out it is not hard to cook up counterexamples with torsion classes. For this reason I edited by clarifying that my guess is only for non-torsion classes.


2 Answers 2


Consider $\mathbb{R}P^{2n-1}$. By covering space theory $\Gamma_k(\mathbb{R}P^{2n-1})$ is $0$ if $k\neq 1,2n-1$, so if we take a cohomology class in degree $2$ this will provide a counterexample ( it is indecomposable since first cohomology with $\mathbb{Z}$ coefficients vanishes). In general, there is little we can say about $\Gamma$, so I would not expect any statement like what you are asking for to be true. A related question is when we can take representing classes to be manifolds rather than general simplicial complexes. Thom proved you always can for $\mathbb{Z}/2$ coefficients, but not for $\mathbb{Z}$ coefficients.

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    $\begingroup$ Do you know of any non-torsion counterexamples? $\endgroup$ Jan 9, 2023 at 17:40
  • $\begingroup$ @JasonDeVito Complex projective space, for similar reasons $\endgroup$ Jan 9, 2023 at 21:45
  • $\begingroup$ In this case, the only generator of the cohomology ring is dual to the image of the Hurewicz map $S^2\cong \mathbb{C}P^1\subseteq \mathbb{C}P^n$. $\endgroup$ Jan 10, 2023 at 0:51
  • $\begingroup$ @JasonDeVito Sorry, for whatever reason I was thinking because the $2n-2$ class was not spherical this was an example, but of course that is not indecomposable. To get a nontorsion example, I think you need something that is multiplicatively generated by at least two elements. I don't think any Lie groups will do because they will split rationally into Eilenberg-MacLane spaces, so the indecomposables should all be hit by the Hurewicz map. $\endgroup$ Jan 10, 2023 at 1:14
  • $\begingroup$ Yeah, I tried many examples (Lie groups, homogeneous spaces, products, connect sums, etc) and couldn't find any. I can prove that a non-torsion counterexample requires at least two non-torsion multiplicative generators using some easy rational homotopy theory. I wonder if, at least for 1-connected spaces there are no non-torsion counterexmaples.... $\endgroup$ Jan 10, 2023 at 2:19

Here's a simply connected counterexample where the element $\omega \in H^k(X)$ is not torsion.

We construct $X$ via rational homotopy theory. (This is Example 1 on pg 157 of Felix, Halperin, and Thomas book Rational Homotopy Theory).

Consider the differential graded $\mathbb{Q}$-algebra $(\Lambda V, d)$ with $V$ spanned by $\{x,y,z\}$ with $\deg x = 3, \deg y = 5,$ and $\deg z = 7$, and where $dx = dy =0$ and $dz = xy$.

Claim: The cohomology ring $H^\ast(\Lambda V;\mathbb{Q})$ is trivial except in degrees $3$, $5$, $10$, $12$, and $15$ where it is isomorphic to $\mathbb{Q}$. Writing $v_i$ with $i\in \{3,5,10,12,15\}$ as a graded basis, we have $v_3 v_{12} = v_{15} = -v_{5}v_{10}$ with all other products trivial.

Proof: Because each of $x,y,$ and $z$ has odd degree, their squares vanish, so the only elements in $\Lambda V$ are spanned by $S:=\{1,x,y,z,xy,xz,yz,xyz\}$.

Of course, $dx = dy=0$ and $dz = xy$ are given. The rest are determined by the product rule: \begin{align*} d(xy) &= (dx)y - x(dy) &= 0\\d(xz) &= (dx)z - x(dz) &= 0\\d(yz) &= (dy)z-y(dz) &= 0\\d(xyz) &= d(xy)z + xy(dz) &= 0\end{align*}

So the cohomology groups are easy: each element of $S$ is closed, so represents something in cohomology. The only possible non-trivial image of a differential is $dz = xy$, so $v_3v_5 = 0$. But, e.g., $v_3v_{12} = [x][yz] = [xyz] = v_{15}$, etc. $\square$

Note that all elements are indecomposable, except for those in $H^{15}(X)$. We also observe that the cohomology of $(\Lambda V,d)$ satisfies Poincare duality. From the Barge-Sullivan Theorem see this MO post, there is a closed simply connected manifold $X$ with $H^\ast(X;\mathbb{Q})\cong H^\ast(\Lambda V)$.

Suppose $k\in \{10,12\}$. Because $H^k(X;\mathbb{Q})\neq 0$, it follows that $H^k(X;\mathbb{Z})$ contains a $\mathbb{Z}$-summand. Let $\omega\in H^k(X;\mathbb{Z}$ be a non-zero element of this $\mathbb{Z}$-summand. Then $\omega$ is indecomposable since its image in $H^k(X;\mathbb{Z})$ is indecomposable.

We claim that for any $\Sigma \in \Gamma_k(X)$, that $\langle \Sigma, \omega\rangle = 0$. To see this, simply note that since $V$ is a minimal Sullivan algebra, it follows from rational homotopy theory that $\pi_s(X)$ is finite except when $s = 3,5,7$. In particular, $\Gamma_k(X)$ is a finite group. But then the map $\Gamma_k(X)\rightarrow \mathbb{Z}$ given by $\Sigma \mapsto \langle \Sigma, \omega\rangle$ is a homomorphism from a finite group to $\mathbb{Z}$, so is necessarily trivial.


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