do spectra have diagonal maps? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T15:44:03Z http://mathoverflow.net/feeds/question/84458 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84458/do-spectra-have-diagonal-maps do spectra have diagonal maps? Cary 2011-12-28T18:00:17Z 2012-01-02T00:34:59Z <p>Topological spaces have diagonal maps $X \rightarrow X \times X$ and $X \rightarrow X \wedge X$, and suspension spectra also have diagonal maps $\Sigma^\infty X \rightarrow \Sigma^\infty(X \wedge X) \cong (\Sigma^\infty X) \wedge (\Sigma^\infty X)$. What about general spectra? (i.e. symmetric spectra, S-modules, or any other convenient definition.) I always assumed you could, but I haven't thought through it carefully. And if not, can we still get a cup product on $E^*(X)$ when $E$ and $X$ are spectra?</p> http://mathoverflow.net/questions/84458/do-spectra-have-diagonal-maps/84459#84459 Answer by Fernando Muro for do spectra have diagonal maps? Fernando Muro 2011-12-28T18:17:12Z 2011-12-28T18:17:12Z <p>The smash product is not a categorical product, so you can't speak of diagonal map, in the same way as you don't have a natural diagonal map $M\rightarrow M\otimes M$, for $M$ an abelian group or vector space. The analogy is very pertinent since the homotopy category of spectra with homotopy concentrated in dimension $0$ is equivalent to the category of abelian groups, and if we restric to abelian groups which are $\mathbb{Q}$-vector spaces, the smash product corresponds to the tensor product.</p> <p>BTW, suspension spectra of base spaces (as you seem to consider) do not have a diagonal map either. You have a diagonal map if you consider suspension spectra of unbased spaces, since you need to add an outer base point first, and this operation takes products to smash products.</p> http://mathoverflow.net/questions/84458/do-spectra-have-diagonal-maps/84462#84462 Answer by Tom Goodwillie for do spectra have diagonal maps? Tom Goodwillie 2011-12-28T18:30:38Z 2012-01-01T15:41:42Z <p>No. Let $[X,A]$ be the set (in fact abelian group) of homotopy classes of maps from one spectrum to another. If $A$, $B$, and $C$ are spectra, any natural map of sets $[X,A]\times [X,B]\to [X,C]$ is induced by a map $A\times B\to C$. Since $A\times B=A\coprod B$, this amounts to two maps $A\to C$ and $B\to C$ inducing two homomorphisms $[X,A]\to [X,C]$ and $[X,B]\to [X,C]$ which are then added to give one homomorphism $[X,A]\times [X,B]=[X,A]\oplus [X,B] \to [X,C]$. This map cannot be distributive over addition except by being identically zero. </p> <p>EDIT Taking $A$, $B$, and $C$ to be Eilenberg-MacLane spectra, this rules out nontrivial natural bilinear products on ordinary cohomology of spectra. More generally it rules out such products on generalized cohomology of spectra. And it also rules out any nontrivial natural map $X\to X\wedge X$ because if $A\to A\wedge A$ were nontrivial then this would lead to a natural bilinear map $[X,A]\times [X,A]\to [X,A\wedge A]$ that (for example when $X=A$) is nontrivial.</p> http://mathoverflow.net/questions/84458/do-spectra-have-diagonal-maps/84466#84466 Answer by Marc Hoyois for do spectra have diagonal maps? Marc Hoyois 2011-12-28T18:45:33Z 2011-12-28T18:45:33Z <p>The answer is no in general. So if $E$ is a ring spectrum, $E^\ast(X)$ need not be a ring, unless $X$ is a suspension spectrum. It is only a module over $E^\ast(pt)$. The ring structure on $E$ only gives external multiplications:</p> <p>$E^\ast(X)\otimes_{E^\ast} E^\ast(Y)\to E^\ast(X\wedge Y)$</p> <p>$E_\ast(X)\otimes_{E_\ast} E_\ast(Y)\to E_\ast(X\wedge Y)$</p> <p>In the case that $X$ is also a ring spectrum, then the homology $E_\ast(X)$ becomes a ring, while the cohomology $E^\ast(X)$ will be a coalgebra under the assumption that $E^\ast(X\wedge X)\cong E^\ast(X)\otimes_{E^\ast}E^\ast(X)$.</p> http://mathoverflow.net/questions/84458/do-spectra-have-diagonal-maps/84686#84686 Answer by John Klein for do spectra have diagonal maps? John Klein 2012-01-01T16:20:16Z 2012-01-02T00:34:59Z <p>The existence of an $E_\infty$-diagonal is an obstruction for equipping a spectrum $E$ with the structure of a suspension spectrum. Conversely, in</p> <p>Klein, J.R.: Moduli of suspension spectra. <em>Trans. Amer. Math. Soc.</em> <strong>357</strong> (2005), 489–507</p> <p>I showed that the existence of a suitably defined notion of $A_\infty$-diagonal on $E$ is equivalent to equipping $E$ with the structure of a suspension spectrum provided we are in the metastable range. Here "metastable" means $E$ is $r$-connected (for $r \ge 1$) and is weak equivalent to a cell spectrum of dimension $\le 3r+2$. </p> <p>There are various elementary ways of defining the notion of $A_\infty$-diagonal, but they in the end amount to the existence of a map $\delta: E \to (E\wedge E)^{\Bbb Z_2}$ (for a suitably defined version of the smash product), which is a homotopy section to the map $(E\wedge E)^{\Bbb Z_2} \to E$ which is given by passing from categorical to geometric fixed points. The way I do this in the paper is the use the second stage of the Taylor tower of the functor $E \mapsto \Sigma^\infty \Omega^\infty E$; this second stage turns out to be a model for $(E\wedge E)^{\Bbb Z_2}$.</p>