6
$\begingroup$

$\newcommand\pt{\mathrm{pt}}$Let $(X,\pt)$ be a connected, pointed, finite CW complex and let $h$ be a generalized cohomology theory. Let $\smash{\tilde{h}}^*(X)$ denote the kernel of restriction $h^*(X)\to h^*(\pt)$. In Corollary 3.1.6 of Atiyah's book on K-theory, he proves that for complex K-theory, $\smash{\tilde{K}}^*(X)$ consists of nilpotent elements. Is there an analogue of this statement for other generalized cohomology theories?

$\endgroup$
2
  • 3
    $\begingroup$ TeX note: a tilde as an accent increases the height of a letter, so a following star is too high. This can be fixed with \smash: $\tilde h^*(X)h^*(X)\smash{\tilde h}^*(X)$ \tilde h^*(X)h^*(X)\smash{\tilde h}^*(X). I edited accordingly. $\endgroup$
    – LSpice
    Commented Jul 9 at 21:19
  • 2
    $\begingroup$ OK!............. $\endgroup$ Commented Jul 9 at 22:13

1 Answer 1

9
$\begingroup$

$\newcommand\smashtilde[1]{\smash{\tilde{#1}}}$Yes, by the Atiyah–Hirzebruch spectral sequence applied to the identity map: $$E^{pq}_2=\tilde H^p(X,h^q(pt))\Rightarrow \smashtilde h^{p+q}(X).$$ Here $\smashtilde H^*$ is the usual reduced Eilenberg–MacLane cohomology. Assume $X$ is path connected and finite dimensional, and $h$ multiplicative. Take an $x\in \smashtilde h^*(X)$ and represent it by $\tilde x\in E_\infty$. A sufficiently large power $\tilde x^N$ ends up in a cell $E^{p,q}_\infty=0$ such that there are only zeros to the right of it in the same diagonal: $E^{p',q'}_\infty=0$ if $p'\geq p$ and $p'+q'=p+q$. This means that $x^N=0$. In fact, one can take $N=\dim X+1$.

$\endgroup$

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