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Two (finite-dimensional) $k$-algebras $A$ and $B$ are said to be stable equivalent if their stable module categories $\underline{\rm mod}(A)$ and $\underline{\rm mod}(B)$ are equivalent as $k$-linear categories. If these algebras are self-injective, the stable module categories are triangulated categories.

So the question is: if $A$ and $B$ are stable equivalent, are the categories $\underline{\rm mod}(A)$ and $\underline{\rm mod}(B)$ equivalent as triangulated categories?

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There are some dull counterexamples.

If $A$ is a self-injective algebra such that the square of the radical is zero, then the stable module category is a semi-simple $k$-linear category with one simple object for each non-projective simple $A$-module.

So, for example, if $A = k[x]/(x^2) \times k[y]/(y^2)$ and $B$ is the path algebra of a quiver with two vertices and two arrows $a$ and $b$ between the vertices in opposite directions, modulo the relations $ab=0=ba$, then the stable module categories of $A$ and $B$ are equivalent as $k$-linear categories (both being semisimple categories with two simple objects). However, they are not equivalent as triangulated categories, as $\Omega$ acts differently on the simple objects.

I don't think I know of any more interesting kinds of example.

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  • $\begingroup$ Interesting! What about an example where the categories are equivalent as $k$-linear categories in a way compatible with $\Omega$ but incompatible with exact triangles? $\endgroup$
    – Fernando Muro
    Commented Nov 21, 2012 at 16:04
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    $\begingroup$ It is a very interesting and simple example! But what if both of them are connected? $\endgroup$
    – Sergei Ivanov
    Commented Nov 21, 2012 at 18:43

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