The space of Penrose tilings and the non-commutative curve with homogeneous coordinate ring k<x,y>(y^2)

Details The space of Penrose tilings and the non-commutative curve with homogeneous coordinate ring k<x,y>(y^2) The space of Penrose tilings and the non-commutative curve with homogeneous coordinate ring k<x,y>(y^2)

2011-04-19

arxiv.org/abs/1104.3811v1

Mathematics

Rings and Algebras

Scientific paper

We construct a non-commutative scheme that behaves as if it is the space of Penrose tilings of the plane. Let k be a field and B=k(y^2). We consider B as the homogeneous coordinate ring of a non-commutative projective scheme. The category of "quasi-coherent sheaves" on it is, by fiat, the quotient category QGr(B):=Gr(B)/Fdim(B) and the category of coherent sheaves on it is qgr(B):=gr(B)/fdim(B), where gr(B) is the category of finitely presented graded modules and fdim(B) is the full subcategory of finite dimensional graded modules. We show that QGr B is equivalent to Mod S, the category of left modules over the ring S that is the direct limit of the directed system of finite dimensional semisimple algebras S_n=M_{f_n}(k) + M_{f_{n-1}}(k) where f_{n-1} and f_n$ are adjacent Fibonacci numbers and the maps S_n \to S_{n+1} are (a,b)--->(diag(a,b),a). When k is the complex numbers, the norm closure of S is the C^*-algebra Connes uses to view the space of Penrose tilings as a non-commutative space. Objects in QGr B have projective resolutions of length at most one so the non-commutative scheme is, in a certain sense, a smooth non-commutative curve. Penrose tilings of the plane are in bijection with infinite sequences z=z_0z_1 ... of 0s and 1s with no consecutive 1s. We associate to each such sequence a graded B-module, a "point module", that becomes a simple object O_z in QGr B that we think of as a "skyscraper sheaf" at a "point" on this non-commutative curve. Tilings T_z and T_{z'} determined by two such sequences are equivalent, i.e., the same up to a translation on R^2, if and only if O_z is isomorphic to O_{z'}. A result of Herbera shows that Ext^1(O_z,O_{z'}) is non-zero for all z and z'. This as an algebraic analogue of the fact that every equivalence class of tilings is dense in the set of all Penrose tilings.

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