Matrix factorization categories for ADE singularities

Question

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.

Answer 1

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)

Answer 2

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.

Answer 3

@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.

Answer 4

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.