MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

As we know, by universal coefficient theorem, $H^{1}(X,\mathbb{Z})$ is torsion-free. My question is: for cup product $H^{1}(X,\mathbb{Z})\otimes H^{1}(X,\mathbb{Z})\rightarrow H^{2}(X,\mathbb{Z})$ could $a\cup b$ be a torsion element in $H^{2}(X,\mathbb{Z})$.

share|cite|improve this question
This is more appropriate for this forum: (anyway: take $X=S^1$). – Chris Gerig Jun 6 '13 at 2:29
Sorry, I always mean nonzero torsion element here. – Allen Jun 6 '13 at 2:45
up vote 14 down vote accepted

An example is given by a 3-manifold. Specifically, for any $n$ other than $0$ or $\pm 1$ we can take the real Heisenberg group $G$ of $3 \times 3$ real matrices of the form $$ \begin{bmatrix} 1 & a & b \\\\ 0 & 1 & c \\\\ 0 & 0 & 1 \end{bmatrix} $$ where $a,b,c \in \mathbb{R}$. We then take the quotient $G/H$ by the subgroup $H$ of elements where $a,b,(nc) \in \mathbb{Z}$. These manifolds are $K(H,1)$ manifolds and the group $H$ is a nilpotent group with cohomology $$ \mathbb{Z}, \mathbb{Z}^2, \mathbb{Z}/n \times \mathbb{Z}^2, \mathbb{Z}. $$ The cup product of the two generators in degree one is the generator of the $\mathbb{Z}/n$ in degree two. This can be checked by using the Lyndon-Hochschild-Serre spectral sequence associated to the central extension $1 \to \mathbb{Z} \to H \to \mathbb{Z}^2 \to 1$, with a useful intermediate step being the determination of the abelianization of $H$.

In some sense, these examples are universal, which can be seen using a little bit of homotopy theory. If you have two cohomology classes whose cup product is $n$-torsion, then (by the correspondence between cohomology classes and maps to Eilenberg-Mac Lane spaces) we have two cohomology classes $X \to K(\mathbb{Z},1)$ such that the composite $$ X \to K(\mathbb{Z},1) \times K(\mathbb{Z},1) \stackrel{cup}{\longrightarrow} K(\mathbb{Z},2) \stackrel{n}{\longrightarrow} K(\mathbb{Z},2) $$ is nullhomotopic. This means that it lifts to the homotopy fiber $$ F \to K(\mathbb{Z},1) \times K(\mathbb{Z},1) \stackrel{n \cdot cup}{\longrightarrow} K(\mathbb{Z},2).$$ This homotopy fiber $F$, in fact, is precisely the classifying space of the group $H$ (which is at least plausible from the long exact sequence of homotopy groups).

share|cite|improve this answer
Thanks Tyler, I really appreciate your answer. – Allen Jun 6 '13 at 6:55

Here's an example that's a 2-dimensional CW complex. Start with a 0-cell, then attach three 1-cells labeled $a$, $b$, $c$ to get a wedge of three circles, then attach a 2-cell via the word $aba^{-1}b^{-1}c^n$ for a fixed integer $n>1$. From the cellular cochain complex one then reads off that the resulting complex $X$ has $H^1X={\mathbb Z}\times{\mathbb Z}$ with generators $a$ and $b$ (by abuse of notation) and $H^2X={\mathbb Z}_n$. The claim is that $a\cup b$ is a generator of $H^2X$. To see this consider the quotient space of $X$ obtained by collapsing the 1-cell $c$ to a point. This is a torus $T$ and the quotient map $X\to T$ induces an isomorphism on $H^1$ and a surjection on $H^2$, as one can see by looking at the induced map on cellular cochain complexes. In $T$ the cup product $a\cup b$ generates $H^2$ so the same is true for $X$ by naturality of cup product.

When $n=2$ the complex $X$ is a closed surface since it's a hexagonal 2-disk with edges identified in pairs. Its Euler characteristic is $-1$ so it's the connected sum of a torus and the real projective plane.

share|cite|improve this answer
Thanks Allen. Your example is really helpful. – Allen Jun 6 '13 at 11:29

You can also make an example involving a closed surface. Let $X$ be the connected sum of a torus $T$ and a projective plane, and let $f:X\to T$ be nontrivial on $H^2$. Two elements of $H^1(T)$ whose cup product is nonzero mod $2$ will pull back by $f$ to two elements of $H^1(X)$ whose cup product generates $H^2(X;\mathbb Z/2)=\mathbb Z/2$.

share|cite|improve this answer
Oh, now I see that Allen mentions this as a special case of his example. – Tom Goodwillie Jun 7 '13 at 14:51

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.