One can not define a symmetric monoidal smash product for ordinary spectra because of the twist isomorphism wanting to be involved.

But still, the stable homotopy category (viewed as the homotopy category of the stable model structure on spectra) can be equipped with a symmetric monoidal smash product, I've read.

My first question is: Can I define this smash product on the homotopy category of the stable model structure on spectra by setting $$ (X\wedge Y)_n=\bigvee_{p+q=n} X_p\wedge Y_q $$ with structure maps induced by (only) the structure maps of $Y$ defined by a coequalizer $$ \bigvee_{p+q=n-1} X_p\wedge Y_q \wedge S^1\stackrel{\to}{\to} \bigvee_{p+q=n} X_p\wedge Y_q $$ where one map uses the structure map of $X$ and the other one switches the factors and uses the structure map of $X$. If yes, what is the reason that this defines a monoidal operation on the homotopy category? Probably this is a consequence of the fact (?) that applying the twist $\tau$ is homotopic to the the identity but I neither see why this should be true nor how it implies the existence of the smash product.

There are two ways to extend the functor $-\wedge S^1$ on spaces to spectra, e.g. one by setting $(X\wedge S^1)_n=X_n\wedge S^1$ and structure maps $$ X_n\wedge S^1\wedge S^1\xrightarrow{id\wedge\tau}X_n\wedge S^1\wedge S^1\to X_{n+1}\wedge S^1. $$ The other one does not involve the twist.These functors should correspond on the stable homotopy category to smashing with the sphere spectrum $S$.

But evaluating $X\wedge S$ with the smash product (if it actually is the right definition) defined above gives $(X_0\wedge S^0)$, $(X_1\wedge S^0\vee X_0\wedge S^1)$, $\ldots$ which doesn't look right. My second question is: What am I missing here?

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    $\begingroup$ You know, for the second time today you just delete your question after I indicated something in a comment. This time I gave you a link, a definition, an outline of what to look for and so on. Not even a little thank you as a comment from your part. $\endgroup$
    – t.b.
    Aug 23 '11 at 22:11

Unfortunately I don't see an "obvious" structure map on the wedge that you describe. It looks more like the smash product for symmetric spectra and doesn't work without the symmetric group action. You'd need to describe maps $$ S^1 \wedge X_p \wedge Y_q \to \bigvee_{r+s=p+q+1} X_r \wedge Y_s $$ and it's not clear whether one should use the structure map of $X$ or $Y$ (possibly first applying the twist), or perhaps you had an application of the pinch-multiplication in mind?

So far as the two definitions of smashing with $S^1$, you need to be careful; I'm not sure that both of these definitions are functorial.

In case you hadn't already investigated: for questions about constructing a smash product on these types of spectra, Adams' "Stable homotopy and generalized homology" describes in detail a construction of "handicrafted" smash products by starting with $X_0 \wedge Y_0$ and making a sequence of choices about whether to apply the structure map to the left- or right-hand factor. The proof that it is independent of these choices and gives rise to a symmetric monoidal structure (all after passing to the homotopy category) has been superseded technologically and I wouldn't recommend reading it in detail.

  • $\begingroup$ Thanks for the answer, Tyler. The term "obvious" was really bad. I meant "defined by the structure maps of $Y$". In the symmetric spectra case, this gives the structure map and one may equivalently do this with the structure map of $X$ there, I've read. I thought, that on the homotopy category this (i.e. using the structure maps of $Y$ only) would be somehow also possible for ordinary spectra. $\endgroup$
    – John Zhao
    Aug 15 '11 at 20:05
  • $\begingroup$ John, you could certainly define something like that. However, the resulting homotopy type that you got wouldn't be using any data about the spectrum structure on $X$, and you get a lack of symmetry. In symmetric spectra, you actually take a coequalizer to force both structure maps to have the same target! $\endgroup$ Aug 16 '11 at 5:13
  • $\begingroup$ I did get mixed up with symmetric sequences and now the structure maps in the question are "defined" similar to symmetric spectra. This, i.e. "like for symmetric spectra but without the action" is, what I initially wanted to ask. I am sorry for the imprecision and the confusion. $\endgroup$
    – John Zhao
    Aug 16 '11 at 6:28

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