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though classifying pairs of matrices up to simultaneous conjugation is known to be wild, it seems to me a folklore that classifying pairs of idempotent matrices (up to simultaneous conjugation) is possible -- or in other words, tame. However I could not find a proper reference for this. Putting it differently, I would like to know a classification of the indecomposable modules of $k\langle x, y\rangle/(x^2-x,y^2-y)$.

My motivation is to understand the paper of W. W. Crawley-Boevey with the title Functorial filtrations III: Semidihedral algebras, where this is used implicitly.

I understand that this must be common knowledge for experts in representation theory of associative algebras, but I tried and could not find a reference.

Thank you, and regards, Jimmy

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    $\begingroup$ If the characteristic is not 2, then A is an idempotent iff I-2A is an involution. So you are essentially looking at the representation type of the infinite dihedral group. Quick googling of representation type infinite dihedral group gave me a chapter of a book by Benson giving all indecomposables from which you can deduce tame $\endgroup$
    – Benjamin Steinberg
    Commented Apr 13, 2015 at 2:56
  • $\begingroup$ @BenjaminSteinberg Thank you for the reference. Yes that solves the case for fields with characteristic not 2. However I am actually interested in the characteristic 2 case -- in particular the GF(2). What can we say about this case? Thanks. $\endgroup$
    – Jimmy
    Commented Apr 13, 2015 at 3:49
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    $\begingroup$ So you want to characterize $k$-vector spaces with two projections. A projection is characterized by its image and its kernel, which have to be complementary subspaces. So a solution will follow from the solution to the four-subspaces problem (e.g., sciencedirect.com/science/article/pii/S0024379504002575 ) after throwing away the indecomposables where the appropriate pairs of subspaces are not complementary. Or am I off here? $\endgroup$
    – darij grinberg
    Commented Apr 13, 2015 at 5:38
  • $\begingroup$ @darijgrinberg I thought so yesterday, but was not sure whether this correspondence is correct. Now I verified it and I think you are right. Thank you! I will go ahead to read about the 4-subspace problem. $\endgroup$
    – Jimmy
    Commented Apr 13, 2015 at 6:13
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    $\begingroup$ @darijgrinberg And the $4$-subspace problem itself is a subproblem of classifying representations of the $\tilde{D}_4$ quiver, with all arrows directed inward. (Namely, the subproblem where the maps associated to the $4$ arrows are injective.) $\tilde{D}_4$ is affine, so this is a tame problem. $\endgroup$
    – David E Speyer
    Commented Apr 13, 2015 at 12:32

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