Homotopy type of tensors of Moore spectra - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T02:55:26Z http://mathoverflow.net/feeds/question/85313 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/85313/homotopy-type-of-tensors-of-moore-spectra Homotopy type of tensors of Moore spectra Eric Peterson 2012-01-10T06:36:49Z 2012-01-13T13:00:07Z <p>I would like to hear what's known about the homotopy type of smash products of mod-$p^j$ Moore spectra, for $p$ an odd prime.</p> <p>First, here is what I'm specifically interested in: there is a short exact sequence <code>$0 \to \mathbb{Z} \xrightarrow{p^j} \mathbb{Z} \to \mathbb{Z}/p^j \to 0.$</code> Tensoring this short exact sequence against your favorite group $G$ yields an exact sequence <code>$\cdots \to G \xrightarrow{p^j} G \to G \otimes \mathbb{Z}/p^j \to 0,$</code> which exhibits $G \otimes \mathbb{Z}/p^j \cong G / p^j G$ as $G$ with its $p^j$-divisible part stripped out.</p> <p>Moore spectra play a related role in homotopy theory: they are defined by the cofiber sequence <code>$S \xrightarrow{p^j} S \to M(p^j).$</code> Smashing through with any spectrum $X$ gives the new cofiber sequence <code>$X \xrightarrow{p^j} X \to X \wedge M(p^j),$</code> and chasing this around shows that the homotopy group $\pi_n X \wedge M(p^j)$ is a mix of $\pi_n X / p^j(\pi_n X)$, as one would expect, together with the $p^j$-torsion of $\pi_{n-1} X$, which is new and different. So, though $X \wedge M(p^j)$ is sometimes written $X / p^j$, and though this notation suggests a useful analogy, this isn't exactly true, and we have to be careful about things we expect to follow from the algebraic setting.</p> <p>The specific algebraic fact I'm interested in is that the composition of the tensor functors $- \otimes \mathbb{Z}/p^j$ and $- \otimes \mathbb{Z}/p^i$ for $j > i$ has a reduction: <code>$- \otimes \mathbb{Z}/p^j \otimes \mathbb{Z}/p^i \cong - \otimes \mathbb{Z}/p^i.$</code> The exact translation of this statement to Moore spectra and the smash product is not true --- one can, for instance, compute the reduced integral homology of $M(p^i) \wedge M(p^j)$ to see that there are too many cells around for it to be equivalent to $M(p^i)$ alone. However, the same homology calculation suggests something related: there is an abstract isomorphism between the reduced homology groups of $M(p^j) \wedge M(p^i)$ and those of $M(p^i) \wedge M(p^i)$. This is, of course, also true for groups; it is indeed the case that $\mathbb{Z}/p^i \otimes \mathbb{Z}/p^j \cong \mathbb{Z}/p^i \otimes \mathbb{Z}/p^i$. This is what I want to know:</p> <blockquote> <p>For $j > i$, is $M(p^j) \wedge M(p^i)$ homotopy equivalent to $M(p^i) \wedge M(p^i)$?</p> </blockquote> <p>If this is not true, I'm willing to throw in some extra qualifiers. For instance, is the situation improved if we work $K(n)$-locally? Is it true only when $j \gg i$? What if we additionally restrict to $j \gg i \gg 0$?</p> <p>This specific question aside, I am also interested in any and all known features of the homotopy type of $M(p^i) \wedge M(p^j)$ --- any favorite fact you have that would help me get a grip on them. I'm also specifically interested in variants of the above question for generalized Moore spectra: can anything similar be said about those?</p> http://mathoverflow.net/questions/85313/homotopy-type-of-tensors-of-moore-spectra/85321#85321 Answer by Fernando Muro for Homotopy type of tensors of Moore spectra Fernando Muro 2012-01-10T08:44:50Z 2012-01-10T08:44:50Z <p>For $p\neq 2$ the answer is yes. It's an easy computation usinghomology decompositions in the sense of Eckmannâ€“Hilton the well known exact sequence</p> <p>$$\operatorname{Ext}(A,\pi_{n+1}(X))\hookrightarrow [M(A,n),X]\twoheadrightarrow \operatorname{Hom}(A,\pi_n(X)),$$</p> <p>and the computation</p> <p>$$\pi_{n+1}(M(A,n))=A\otimes\mathbb{Z}/2.$$</p> <p>Here $M(A,n)$ is the Moore spectrum whose homology is the abelian group $A$ concentrated in degree $n$.</p> <p>As you point out, it is easy to check that</p> <p>$$H_n(M(A,s)\wedge M(B,t))= \left\{\begin{array}{ll} A\otimes B,&amp;n=s+t,\\ \operatorname{Tor}_1(A,B),&amp;n=s+t+1,\\ 0,&amp;\text{otherwise}. \end{array}\right.$$</p> <p>Therefore, $M(A,s)\wedge M(B,t)$ can be obtained as the homotopy cofiber of a map $$f\colon M(\operatorname{Tor}_1(A,B),s+t)\longrightarrow M(A\otimes B,s+t)$$ which is trivial in homology $H _{*}(f)=0$.</p> <p>Suppose for instance that $A$ and $B$ are finite and do not have $2$-torsion. Then the previous short exact sequence shows that homology induces an isomorphism</p> <p>$$H _{s+t}\colon [M(\operatorname{Tor}_1(A,B),s+t), M(A\otimes B,s+t)]\cong \operatorname{Hom}(\operatorname{Tor}_1(A,B),A\otimes B).$$</p> <p>Therefore $f$ must be null-homotopic, so</p> <p>$$M(A,s)\wedge M(B,t) \simeq M(A\otimes B,s+t)\vee M(\operatorname{Tor}_1(A,B),s+t+1).$$</p> <p>If you take $A=\mathbb{Z}/p^i$ and either $B=\mathbb{Z}/p^j$ or $B=\mathbb{Z}/p^i$ you always obtain the same thing on the right.</p> <p>For $p= 2$ the answer is no. The spectrum $M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/4,0)$ is the mapping cone of $$4\colon M(\mathbb{Z}/2,0)\longrightarrow M(\mathbb{Z}/2,0)$$ which is knonw to be null-homotopic, actually $[M(\mathbb{Z}/2,0),M(\mathbb{Z}/2,0)]\cong\mathbb{Z}/4$ generated by the identity. Hence</p> <p>$$M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/4,0)\simeq M(\mathbb{Z}/2,0)\vee M(\mathbb{Z}/2,1).$$</p> <p>In particular the action of the Steenrod algebra on the mod 2 homology of $M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/4,0)$ is trivial.</p> <p>On the other hand, it is known that the Steenrod algebra mode 2 acts on the mod 2 homology of $M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/2,0)$ in a non-trivial way, i.e. the first Steenrod operation sends the 0-dimensional generator to the 1-dimensional generator. Therefore $M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/4,0)$ and $M(\mathbb{Z}/2,0)\wedge M(\mathbb{Z}/2,0)$ cannot be homotopy equivalent.</p> <p>As for references, see Hatcher's book. This book doesn't deal with spectra but since we are working with finite-dimensional CW-complexes you can assume that we are in the stable range.</p> http://mathoverflow.net/questions/85313/homotopy-type-of-tensors-of-moore-spectra/85575#85575 Answer by Neil Strickland for Homotopy type of tensors of Moore spectra Neil Strickland 2012-01-13T13:00:07Z 2012-01-13T13:00:07Z <p>You might find this paper useful:</p> <pre><code>\bib{MR760188}{article}{ author={Oka, Shichir{\^o}}, title={Multiplications on the Moore spectrum}, journal={Mem. Fac. Sci. Kyushu Univ. Ser. A}, volume={38}, date={1984}, number={2}, pages={257--276}, issn={0373-6385}, review={\MR{760188 (85j:55019)}}, doi={10.2206/kyushumfs.38.257}, } </code></pre>