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What is known about the matrix factorization categories of singularities of type ADE? Any references on this would be greatly appreciated.

Background: For ADE singularities, see for example this. For matrix factorizations, see for example this.

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up vote 2 down vote accepted

See: "Matrix Factorizations and Representations of Quivers II: type ADE case" (math/0511155) by Kajiura, Saito, and Takahashi for a recent account.

Older references include: "Construction geometrique de la correspondance de McKay" Gonzalez-Sprinberg,and Verdier (1983) Y. Yoshino, Cohen-Macaulay modules over Cohen-Macaulay rings (1990)

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Matrix factorization categories for these singularities depend on a grading that you consider. If you consider the maximal grading for ADE singularities in a standard form like

$X^{l+1}+ Y^2+\cdots$(sum of squares) for $A_l$ and so on till $X^3+Y^5+\cdots$(sum of squares) for $E_8$, then the category will be equivalent to the derived category of representations of the corresponding Dynkin quiver. (see paper math/0511155 and especially Appendix A for a short proof)

If you consider non-graded case then for A-type singularities the category is described in the end of the paper math/0302304. I am sure that these non-graded categories can be obtained from the graded versions as orbit categories with respect to a related autoequivalence in Definition of Bernhard Keller math/0503240. But it seems that this fact is not written yet.

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Sorry for the stupid question, but I don't understand what is meant by grading here. – Kevin H. Lin Apr 7 '10 at 21:01
In non graded case it is known that the matrix factorization category is equivalent to a triangulated category of singularities of the singular fiber. There is a so called a graded version of triangulated category of singularities. By definition a triangulated category of singularities is the quotient of the bounded derived category of f.g modules by the subcategory of perfect complexes. Now if we have an algebra with a grading then we can consider a graded version, i.e. the quotient of bounded derived category of graded modules by the subcategory of perfect complexes. – user2464 Apr 8 '10 at 10:42

@ploughshare: If M is a CM A-module, then M and M(n) are isomorphic in the completion \hat{A}. this is how one goes from the graded to the ungraded.

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Kevin: C[x,y] is naturally Z-graded by deg x = 1, deg y = 1. this induces a Z-grading on the ring of invariants A = C[x,y]^G. If G is not cyclic of odd order, then A is supported in even degrees, i.e. A_n = 0 for n odd. this is the natural grading on A. One often changes the grading by A_n = A_{n/2}; this is called the reduced grading.

There are also fractional gradings: Write C[x,y] = Sym V, where V is a two dim'l vector space. Let R = C[x^2,xy,y^2] = C[u,v,w]/uv-w^2. If we set deg (u,v,w) = (1,1,1), then uv-w^2 is homogeneous. then x and y have degree "1/2".

Incidentally, this is what all the "Spin 1/2" business is about. V is called D_{1/2}. V^{otimes 2j} is called D^j. The Clebsch-Gordan formula tells one how to decompose tensor powers of V. It says D^j otimes D^k = D^{abs j -k} oplus D^{abs {j - k}+1} oplus ... oplus D^{j+k}. see Varadarajan Supersymmetry chapt 1.

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Thanks, Brian. By the way, it looks like you have a bunch of user accounts on MO. You should contact one of the admins (by email or on the meta board), so they can merge your accounts. – Kevin H. Lin Apr 7 '10 at 22:57
thanks but i like not having to log in. it's so liberating. – Brian Jurgelewicz Apr 7 '10 at 23:01
Sure; but if you log in, you can make comments. – Kevin H. Lin Apr 7 '10 at 23:03
I registered with some kind of universal ID from having used Google, now it always logs me in if I visit the site from the same computer. I don't expect I would know how to log in from elsewhere. – Will Jagy Apr 8 '10 at 0:42

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