It is well-known that for the sphere spectrum $S$ in the ('topological') stable homotopy category the object $S/2S$ i.e. the cone of $S\stackrel{\times 2}{\to}S$, is not $2$-torsion.

So I wonder where there exists an object $X$ in a (topological?) triangulated category such that

  1. $2End(X/2X)\neq 0$.

  2. $End(X,X)$ is torsion ($\cong \mathbb{Z}/4 \mathbb{Z}$?).

  3. $Hom(X,X[i])=0$ for any $i\neq 0$ (or at least for 'small' $i$).

I would be grateful for any hints or references concerning this question! I believe that I have proved that condition 3 contradicts 2 if $End(X,X)\cong \mathbb{Z}$ (since in this case the triangulated subcategory 'strongly' generated by $X$ is isomorphic to $K^b(B)$, where $B$ is the category of finitely generated free $\mathbb{Z}$-modules); yet I cannot prove anything like that if condition 2 is fulfilled.

  • 1
    $\begingroup$ Whatever your category is, $\hom(X,X/2X)$ is a right $\hom(X,X)$-module where the action of $2\cdot 1_X$ is trivial, hence $2\cdot\hom(X,X/2X)=0$. $\endgroup$ – Fernando Muro Nov 14 '12 at 9:43
  • $\begingroup$ Yes, you are right! I updated the question. $\endgroup$ – Mikhail Bondarko Nov 14 '12 at 9:49

If $\text{Hom}(S,S/2)$ refers to maps of degree zero, then that group is $\mathbb{Z}/2$. However, $\text{Hom}(S[2],S/2)$ is $\mathbb{Z}/4$, as is $\text{Hom}(S/2,S/2)$. (My sign convention for the shift is such that $S[n]$ is the sphere $S^n$.) On the other hand, $\text{Hom}(S/2,S/2[i])$ will be zero for $i>1$ but nonzero for most $i\leq 1$.

  • $\begingroup$ Thank you!! Now I understand the situation with $S$ better; I probably cannot use it for my purposes. Yet do you think that there exists a spectrum that satisfies my conditions 1-3? $\endgroup$ – Mikhail Bondarko Nov 14 '12 at 9:37
  • $\begingroup$ Sorry, my first version of the question was self-contradictory; I made an update. The information contained in your answer is very interesting for me still! $\endgroup$ – Mikhail Bondarko Nov 14 '12 at 9:52

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