Wanted: Odd-dimensional integral cohomology class whose square is nonzero - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T03:34:50Z http://mathoverflow.net/feeds/question/105124 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105124/wanted-odd-dimensional-integral-cohomology-class-whose-square-is-nonzero Wanted: Odd-dimensional integral cohomology class whose square is nonzero Mark Grant 2012-08-20T22:01:15Z 2012-08-29T11:49:01Z <p>Does anyone know of a nice simple example of a space $X$ with an odd-dimensional integral cohomology class $a\in H^{2k+1}(X;\mathbb{Z})$ whose square is nonzero?</p> <p>I once thought that the one-dimensional generator $a\in H^1(K;\mathbb{Z})\cong\mathbb{Z}$ in the cohomology of the Klein bottle had $a^2\in H^2(K;\mathbb{Z})\cong\mathbb{Z}/2$ nonzero, but it appears this is not the case. </p> http://mathoverflow.net/questions/105124/wanted-odd-dimensional-integral-cohomology-class-whose-square-is-nonzero/105817#105817 Answer by Mark Grant for Wanted: Odd-dimensional integral cohomology class whose square is nonzero Mark Grant 2012-08-29T11:49:01Z 2012-08-29T11:49:01Z <p>As Ralph is being modest, I have decided to make his comment into a CW answer.</p> <p>Recall that the short exact coefficient sequence $0\to \mathbb{Z}\to \mathbb{Z}\to \mathbb{Z}/2\to 0$ leads to a long exact sequence $$\cdots \to H^\ast(X;\mathbb{Z})\to H^\ast(X;\mathbb{Z})\stackrel{\rho}{\to} H^\ast(X;\mathbb{Z}/2)\stackrel{\beta}{\to} H^{\ast+1}(X;\mathbb{Z})\to\cdots$$ for any space $X$, where $\rho$ denotes reduction mod 2 and $\beta$ is the Bockstein homomorphism. Recall also that $\rho$ is a ring homomorphism, and that $\rho\circ\beta = Sq^1$, the first <a href="http://en.wikipedia.org/wiki/Steenrod_algebra" rel="nofollow">Steenrod square</a>. </p> <p>Let $X=\mathbb{R}P^\infty\times\mathbb{R}P^\infty$, and let $x,y\in H^1(X;\mathbb{Z}/2)$ denote the generators over each axis. Note that $c=\beta(xy) \in H^3(X;\mathbb{Z})$ is nonzero, since $$\rho(c) = Sq^1(xy) = Sq^1(x)y + x Sq^1(y) = x^2y + xy^2 \neq 0.$$ </p> <p>Similarly $c^2 \in H^6(X;\mathbb{Z})$ is nonzero, since we have $$\rho(c^2) = \rho(c)^2 = (x^2y+xy^2)^2 = x^4y^2 + x^2 y^4 \neq 0.$$</p> <p>As Will Sawin noted in his comment, we can get a finite-dimensional example by taking $Y=X^{(6)}$, the $6$-skeleton of $X$ (this is the smallest possible dimension, by Tom Goodwillie's comment). We could also take $Y=\mathbb{R}P^4\times\mathbb{R}P^4$ (but $\mathbb{R}P^3\times\mathbb{R}P^3$ won't work, since its six-dimensional cohomology is torsion-free).</p>