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!
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.