## do spectra have diagonal maps?

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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?

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DGA's also don't have diagonal maps, there is "an obvious" (naive) choice, but this is not a map in the appropriate category. It doesn't preserve the grading for example. – Sean Tilson Dec 28 2011 at 19:40

## 4 Answers

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.

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.

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So if I had a natural "diagonal" $X \rightarrow X \wedge X$ then it would give a bilinear pairing $[X,A] \times [X,B] \rightarrow [X,A \wedge B]$. By your argument, such a pairing must be zero. So the only candidate for $X \rightarrow X \wedge X$ is the zero map. Did I understand that correctly? – Cary Jan 1 2012 at 15:28
Yes. I have edited to clarify this. – Tom Goodwillie Jan 1 2012 at 15:42
Thank you, this was very helpful. – Cary Jan 7 2012 at 16:21

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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.

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.

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In based spaces, $X\sma X$ is a quotient of $X\times X$, so we do have a diagonal, namely the composite of the diagonal $X\to X\times X$ and the quotient map $X\times X\to X\sma X$. Of course, it is not a categorical diagonal, but it is often useful. Since the suspension spectrum functor commutes with smash products, this does give suspension spectra a diagonal. It is used all the time in duality theory. A relevant conceptual point is that, in spectra, the canonical map $X\wedge X\rtarr X\times X$ is an equivalence, just as in Abelian categories. – Peter May Dec 28 2011 at 18:26

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By $X\wedge X\to X\times X$ Peter presumably meant $X\vee X\to X\times X$. – Tom Goodwillie Dec 28 2011 at 19:51

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Thanks, Tom, of course that is another typo. It would be nice if comments could also be edited. – Peter May Dec 28 2011 at 20:54

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So, suspension spectra are commutative coalgebras in spectra. Is there a converse? Is the homotopy theory of $E_\infty$ coalgebras in spectra the homotopy theory of $HZ$ local spaces. – Jeff Smith Dec 29 2011 at 1:09

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Jeff, there certainly seem to be coalgebras that are not bounded below: start with a commutative differential graded coalgebra over $\mathbb Q$. – Tom Goodwillie Jan 2 2012 at 1:12
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The existence of an $E_\infty$-diagonal is an obstruction for equipping a spectrum $E$ with the structure of a suspension spectrum. Conversely, in

Klein, J.R.: Moduli of suspension spectra. Trans. Amer. Math. Soc. 357 (2005), 489–507

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 $\le 3r+2$.

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 $(E\wedge E)^{\Bbb Z_2}$.

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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:

$E^\ast(X)\otimes_{E^\ast} E^\ast(Y)\to E^\ast(X\wedge Y)$

$E_\ast(X)\otimes_{E_\ast} E_\ast(Y)\to E_\ast(X\wedge Y)$

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 $E^\ast(X\wedge X)\cong E^\ast(X)\otimes_{E^\ast}E^\ast(X)$.

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