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I have been informed that there is a reference out there which specifically details what goes wrong with the mod 2 Moore spectrum, i.e. why it is not $A_\infty$ or something? I do not know the details, but I am interested in this from the point of view of trying to understand how to come up with the correct notion of ideals of spectra (in the sense of Smith and others), especially the ideal generated by multiplication by 2 on the sphere spectrum. I think it is a paper by Neeman. Does anyone know of this paper, or of other papers which might detail this situation carefully?

Thanks!

PS Is this question appropriate, as it is only a reference request, in the strongest sense of the phrase?

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2 Answers 2

up vote 19 down vote accepted

The actual statement is much stronger than you suggest, namely: the mod 2 Moore spectrum does not admit a unital multiplication (even if it is non-associative). I don't know a reference so I'll sketch the proof:

Let $R$ be a spectrum with unital product, with unit map $\eta\colon S^0\to R$ and product map $\mu: R\wedge R\to R$. Then it is straightforward to show that if $n\eta=0$ in $\pi_0R$ for some integer $n$, then $n\cdot\mathrm{id}_R: R\to R$ is homotopic to the null map as well. (The key point is that proving this uses the existence of $\mu:R\wedge R\to R$ such that $\mu\circ (\eta\wedge \mathrm{id}_R) = \mathrm{id}_R$, but nothing about associativity of such $\mu$.)

If $R$ is the mod $2$ Moore spectrum, with $\eta: S^0\to R$ the generator of $\pi_0R$, then you calculate that:

  • $\pi_0R = Z/2$, but
  • $\pi_2R = Z/4$,

from which it follows that

  • $2\eta=0$,
  • $2\mathrm{id}_R\neq 0$.

Therefore no such unital multiplication $\mu$ on $R$ can exist.

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Thankyou, this is a very clear explanation! –  Jon Beardsley Feb 8 '12 at 19:37
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Yes, very nice. An alternative reason is that for the mod $2$ Moore spectrum any map $R\wedge R\to R$ must induce the zero map $\pi_0(R)\otimes \pi_0(R)\to\pi_0(R)=Z/2$ because $\pi_0(R\wedge R)=Z/4$. –  Tom Goodwillie Feb 9 '12 at 2:19
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Tom, I think your proof is better. –  Charles Rezk Feb 9 '12 at 6:35
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@Elden: You are right, of course. I was a little confused. What I should have said is that $\pi_0Hom(R,R)=\pi_1(R\wedge R$ is $\mathbb Z/4$. –  Tom Goodwillie Jul 12 at 2:25
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Ah thanks for the clarification! So we just note that a map $R \wedge R \rightarrow R$ will induce a map $R \rightarrow Hom(R,R)$ by adjunction and if $R$ is a ring this map has to induce a map of rings $Z/2 \rightarrow Z/4$ which is absurd. –  Elden Elmanto Jul 12 at 18:32

An argument from the old days. A unital multiplication gives a non-trivial splitting of $R\wedge R$ whereas its mod 2 cohomology is indecomposable as a module over the steenrod algenbra.

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