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Let $\mathcal{A}$ be an abelian category and $Kom(\mathcal{A})$ the abelian category of complexes in $\mathcal{A}$. In his book Fourier-Mukai transform in algebraic geometry, Huybrechts states that $Kom(\mathcal{A})$ is not a triangulated category because the natural choices of distinguished triangles do not work.

These natural choices are: short exact sequences; mapping cones.

The first choice doesn't work because it breaks, for example, the axiom TR2 (using Huybrechts notation).

The second choice doesn't work because the mapping cone construction is not even a complex in $Kom(\mathcal{A})$ (we need to pass to the homotopy category).

I am happy with that, but why does it imply that we cannot find another "non-nutural" set of distinguished triangles which works?

I have tried to use the axioms of triangulated category to arrive to a contradiction or a constraint on the shape of this distinguished triangles, but I failed (I should say that I have not worked out the actahedral axiom so far). So I am starting to think that maybe $Kom(\mathcal{A})$ can be triangulated in some crazy way, but that this structure is useless because it has not a geometric meaning..

Thank you for any comments about that!

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The category of complexes over an abelian category is also abelian, and every abelian triangulated category is necessarily semisimple (= any exact triple splits). See p. 250 of the book by Gelfand and Manin (Exercise 1 at the end of §IV.1). Usually the category of complexes is not semisimple, and this is a contradiction you are looking for.

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  • $\begingroup$ Dear Alejo, thank you for this answer. I suspected something deeper was hidden here. I have a further question: does the word "usually" means that there are examples of semisimple categories of complexes? For example I was reading on the reference you pointed out that the full subcategory of acyclic complexes is always (trivially) semisimple. Maybe there are cases in which this subcategory is the whole category..? $\endgroup$ – Cla Oct 7 '16 at 14:59
  • $\begingroup$ For my main application $\mathcal{A}$ will be the category of coherent sheaves on smooth projective scheme, so the corresponding category of complexes is not semisimple: I am happy with that. My previous question is just for curiosity. Thank you! $\endgroup$ – Cla Oct 7 '16 at 15:02
  • $\begingroup$ Note that to prove that $Kom (\mathcal{A})$ is not semisimple, it will be enough to find a non-split short exact sequence in $\mathcal{A}$. $\endgroup$ – user144221 Oct 7 '16 at 16:06
  • $\begingroup$ yep, this is why I said that $Kom(Coh(X))$ is not semisimple. If $\mathcal{A}$ is the category of finite dimensional vector spaces over a field $k$, $\mathcal{A}$ is abelian and so is $Kom(\mathcal{A})$. I know $\mathcal{A}$ is semisimple and I think $Kom(\mathcal{A})$ is also semisimple (is it right?). Then this should be an example in which the category of complexes is triangulated. Anyway, it looks like the natural choice of s.e.s. as distinguished triangles do not work... $\endgroup$ – Cla Oct 7 '16 at 18:14
  • $\begingroup$ @Cla: Kom(A) is never semisimple (even if A=Vect). Take a nonzero object X in A viewed as a complex concentrated in degree zero and C the complex with X in degree 0 and -1 and differential = $Id_X$. Then, the identity of $X$ gives you a map of complexes between $X$ and $C$. It is clearly an injective map. If $X$ were an injective object (as complex), this map would be split as map of complexes, but the homology of X is X, while the homology of C is zero, so, it cannot have X as a direct summand. $\endgroup$ – Marco Farinati Sep 4 at 15:53

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