Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
1  
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 $2\cdot\hom(X,X/2X)=0$. –  Fernando Muro Nov 14 '12 at 9:43
    
Yes, you are right! I updated the question. –  Mikhail Bondarko Nov 14 '12 at 9:49
add comment

1 Answer

up vote 3 down vote accepted

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 $i\leq 1$.

share|improve this answer
    
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? –  Mikhail Bondarko Nov 14 '12 at 9:37
    
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! –  Mikhail Bondarko Nov 14 '12 at 9:52
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.