Let $E$ be a spectrum, $X$ a CW-complex and associate a graded abelian group $$ E^*(X)=\bigoplus_{k\in\mathbb{Z}}[X_+,S^k\wedge E] $$ to it. The brackets denote stable homotopy classes. Please let me ignore the suspension spectrum symbol $\Sigma^\infty$ and problems with the smash product in what follows and take symmetric spectra if you feel uncomfortable. This is not essential to the question.

Let $E$ be a ring spectrum with multiplication $\mu:E\wedge E\to E$.

Take two (representatives of) elements $a:X_+\to S^k\wedge E$ and $b:X_+\to S^l\wedge E$ and consider the map $a\star b$ $$ X_+\xrightarrow{\Delta} X_+\wedge X_+\xrightarrow{a\wedge b} (S^k\wedge E)\wedge (S^l\wedge E)\cong S^{k+l}\wedge (E\wedge E)\xrightarrow{id\wedge \mu} S^{k+l}\wedge E. $$

Is $\star$ the way $E^*(X)$ usually gets into a graded ring? In particular if $E$ is the Eilenberg-Mac Lane spectrum, is $a\star b$ the cup-product $\cup:E^k(X)\otimes E^l(X)\to E^{k+l}(X)$?

  • $\begingroup$ I don't think there's any overwhelming consensus one way or the other, but I find it's nice to use \smallsmile and \smallfrown for cup and cap product respectively, just to have different symbols. To ensure proper spacing, it may be necessary to write \mathbin{\smallsmile} or \mathbin{\smallfrown} (of course in a TeX document one could just define \cupprod and \capprod to be these things and forget about it). $\endgroup$ Commented Apr 11, 2013 at 3:09

1 Answer 1


Absolutely. A reference for this is Adams' book ``Stable homotopy and generalized homology''


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