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Let $\mathcal{A}$ be an abelian category with enough projectives and let $\underline{\mathcal{A}}$ be its stable category with loop functor $\Omega: \underline{\mathcal{A}} \to \underline{\mathcal{A}}$ (for definitions see remark 3 below). Define $\Omega^0 := id_{\underline{\mathcal{A}}}$ and $\Omega^{n+1} := \Omega \circ \Omega^n$ and denote the hom-groups of $\underline{\mathcal{A}}$ by $[-,-]$. Then, if

$$ 0 \to B' \to B \to B'' \to 0$$

is a short exact sequence in $\mathcal{A}$, there is a sequence (starting with $n=0$ from the left):

$$ ... \to [\Omega^n(A),B'] \to > [\Omega^n(A),B] \to > [\Omega^n(A),B''] \to > [\Omega^{n+1}(A),B'] \to ...$$

and the composition of two consecutive maps is zero.

Is this sequence exact, or equivalently, is $[\Omega^n(A),-]_{n > \ge 0}$ a delta-functor ?

Remark 1: I know that the following is a long exact sequence (ending with $n=0$ at the right):

$$ ... \to [A, \Omega^n(B')] \to [A, \Omega^n(B)] \to [A, \Omega^n(B'')], \to [A, \Omega^{n-1}(B')] \to ...$$ Therefore is guess that the sequence above is also exact.

Remark 2: There is a natural epimorphism $$Ext_\mathcal{A}^n(A,B) \to [\Omega^n(A),B].$$ If $\mathcal{A}$ satisfies $Ext_\mathcal{A}^n(-,P)=0$ for all projectives $P$ and all $n > 0$ then the epimorphism is actually an isomorphism (for $n >0$) and the exactness of the sequence follows from the long exact $Ext$-sequence.

Remark 3: The stable category $\underline{\mathcal{A}}$ is defined as follows: It has the same objects as $\mathcal{A}$ and the hom's are given by $$[A,B] := Hom_{\underline{\mathcal{A}}}(A,B) := Hom_\mathcal{A}(A,B) / P(A,B)$$ where $P(A,B)$ is the subgroup of homomorphisms that factor through a projective. The endo-functor $\Omega$ is obtained by taking fixed projective presentations in $\mathcal{A}$: $$0 \to \Omega(A) \to P \to A \to 0.$$

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+1 for write-up and details. –  Theo Johnson-Freyd Mar 30 '11 at 0:53
    
I think (but am not totally sure) that we are dealing with the homotopy category of a model category with $\Omega$ being the loop space operation in which case the exactness follows from general considerations. –  Torsten Ekedahl Mar 30 '11 at 4:24
    
Sounds interesting. I'll have a look at it. Thanks, Torsten. –  Ralph Mar 30 '11 at 9:16

3 Answers 3

It's not exact in general. Take $\mathcal{A}$ to be the category of finitely generated abelian groups, so that $\underline{\mathcal{A}}$ is the category of finite abelian groups and $\Omega=0$. Take the short exact sequence $$0\rightarrow \mathbb{Z}/2 \rightarrow \mathbb{Z}/4 \rightarrow \mathbb{Z}/2\rightarrow 0$$ and $A=\mathbb{Z}/2$. Then the long sequence is $$\mathbb{Z}/2\stackrel{1}\rightarrow \mathbb{Z}/2\stackrel{0}\rightarrow \mathbb{Z}/2\rightarrow 0\rightarrow \cdots,$$ which is not exact.

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Nice example. Thank you very much, Fernando. –  Ralph Mar 30 '11 at 10:58

If the category ${\mathcal{A}}$ is Frobenius (that is it has also enough injectives and the classes of injective and projective objects coincide) then $\underline{\mathcal{A}}$ is triangulated with $\Omega$ being the shift $[-1]$ functor. Then the sequence you write down is the usual long exact sequence of a distinguished triangle, so it is exact.

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Isn't this case already contained in the one the OP mentions in Remark 2? –  Daniel Litt Mar 30 '11 at 4:27
    
I agree with Daniel: $Ext_\mathcal{A}^n(A,-)$ is the right derived functor of $Hom_\mathcal{A}^n(A,-)$, hence $Ext_\mathcal{A}^n(A,I) = 0$ for $I$ injective and all $n>0$. In the Frobenius situation this implies that the condition of remark 2 is satisfied. –  Ralph Mar 30 '11 at 6:56
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I agree that the exactness in the Frobenius case follows from Remark 2. However, I think that it worth emphasizing the relevance of the Frobenius property and the triangulated structure of the resulting category. –  Sasha Mar 30 '11 at 7:59
    
@Sasha: I totally agree. I merely used the formulation in remark 2 for technical reasons since it sometimes yields an isomorphism even if $\mathcal{A}$ isn't Frobenius. E.g. let $\mathcal{A}$ be the category of $RG$-modules ($G$ a finite group, $R$ a commutative ring). If $A$ is $R$-projective and $B$ is finitely generated over $RG$ then there is an isomorphism in remark 2 (those assumption is fulfilled by Shapiro's lemma). But of course there is also some kind of Frobenius structure on $\mathcal{A}$ since relative injectives and projectives coincide. –  Ralph Mar 30 '11 at 8:49

Due to better editing facilities I post this as an answer, though it's rather a comment. My motivation for beliving the long sequence might be exact was the (unconditional) exactness of the sequence in remark 1. But when applied to the short exact sequence $$ 0 \to \Omega(B) \to P \to B \to 0$$ with $P$ projective, the two long sequences show a significant difference. For, since $\Omega^n(P)$ is again projective we have $[-,\Omega^n(P)] = 0$ and the sequence in remark 1 just yields the tautological $$[A,\Omega^n(B)] \cong [A, \Omega^{n-1}(\Omega(B))],$$ while the sequence in question yields the sequence $$0 \to [\Omega^n(A),B] \to [\Omega(\Omega^n(A)), \Omega(B)] \to 0.$$ I see no reason, why this sequence should be exact in general. Therefore, I guess, one needs further assumptions, like $\mathcal{A}$ being Frobenius as observed by Sasha, in order to make the questionable long sequence exact.

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