the space of penrose tilings as a non-commutative schemeof x.buzzword: noncommutative topology....

The space of Penrose tilings as a non-commutativescheme

S. Paul Smith

University of WashingtonSeattle, WA 98195.

[email protected]

May 8, 2012

Linking representation theory, singularity theoryand non-commutative algebraic geometry

S. Paul Smith The space of Penrose tilings as a non-commutative scheme

A non-commutative homogeneous coordinate ring

R =C〈x , y〉

= the free algebra with relation y 2 = 0

Basis = words in x and y not having yy as a subword

= finite paths in

%%x 99 1

ee starting at 0.

The non-commutative scheme X = Projnc R

R is the homogeneous coordinate ring of a

“non-commutative scheme”

X = Projnc R

defined implicitly by declaring that the category of “quasi-coherentsheaves on X ” is

Qcoh(X ) :=Gr(R)

Fdim(R)=

Z-graded R-modules

sums of finite diml mods

Philosophy of nc projective algebraic geometry: Certain abelianand triangulated categories are “quasi-coherent sheaves” on“non-commutative schemes”.

We will argue that Projnc R is “the space of Penrose tilings”.

X = Projnc R

Qcoh(X ) :=Gr(R)

Fdim(R)=

Z-graded R-modules

X = Projnc R

Qcoh(X ) :=Gr(R)

Fdim(R)=

Z-graded R-modules

Penrose tilings and Penrose sequences

PT = {Penrose tilings of R2}

PS = {z ∈ {0, 1}N | znzn+1 6= 11}

= semi-infinite paths in •1

��0 99 •

Theorem (R. Robinson? N. de Bruijn?)

There is a bijective map z : PT → PS.

z ∈ PS The module Mz and the skyscraper sheaf Oz

Let z ∈ PS . Define the graded left R-module

Mz = Ce0 ⊕ Ce1 ⊕ · · · degree(en)=n

x · en = (1− zn)en+1

y · en = znen+1

Note: y 2 · en = zn+1znen+2 = 0 =⇒ Mz ∈ Gr(R)

Quotient functor π∗ : Gr(R) −→ Qcoh(X ) =Gr(R)

Fdim(R)

Oz := π∗Mz

Oz is a simple object in Qcoh(X ).

π∗ is analogous to the functor M 7→ M in algebraic geometry

M(f 6= 0) := M[f −1]0S. Paul Smith The space of Penrose tilings as a non-commutative scheme

x · en = (1− zn)en+1

y · en = znen+1

Fdim(R)

Oz := π∗Mz

x · en = (1− zn)en+1

y · en = znen+1

Fdim(R)

Oz := π∗Mz

x · en = (1− zn)en+1

y · en = znen+1

Fdim(R)

Oz := π∗Mz

Serre (1955) Faisceaux Algebriques Coherents

Z ⊂ Pn a closed subvariety

A = C[x0, . . . , xn]/(homogeneous f vanishing on Z )

Theorem: Qcoh(Z ) ≡ Gr(A)Fdim(A)

π∗←− Gr(A)

coh(Z ) ≡ gr(A)fdim(A)

= finitely generated graded A-modulesfinite dimensional modules

OZ = π∗A = A= the structure sheaf for Z= the sheaf of holomorphic fns on Z

Z ⊂ Pn a closed subvariety

A = C[x0, . . . , xn]/(homogeneous f vanishing on Z )

Theorem: Qcoh(Z ) ≡ Gr(A)Fdim(A)

π∗←− Gr(A)

coh(Z ) ≡ gr(A)fdim(A)

= finitely generated graded A-modulesfinite dimensional modules

OZ = π∗A = A= the structure sheaf for Z= the sheaf of holomorphic fns on Z

If z ∈ Z and

Mz = A/(homogeneous f vanishing at z)

π∗Mz = Oz

= the skyscraper sheaf at z

Oz(U) =

{C if z ∈ U

0 if z /∈ U

{simple objects in QcohZ} = {Oz | z ∈ Z}points in Z ←→ simple objects in QcohZ

Tilings up to isometry

Recall: ∃ a bijection z : PT −→ PS ⊂ {0, 1}N.

Proposition

Let T and T ′ be Penrose tilings of R2. Let z = z(T ) = z0z1 . . .and z ′ = z(T ′) = z ′0z ′1 . . . be the corresponding sequences in{0, 1}N. The following are equivalent

1 Oz∼= Oz ′

2 there is an isometry σ such that σ(T ) = T ′

3 z≥n = z ′≥n for some n

4 (Mz)≥n∼= (Mz ′)≥n for some n.

(2) ⇔ (3) is due to R. Robinson or N. de Bruijn. See Chapter 10of Grunbaum and Shepard’s masterpiece, Patterns and Tilings.

Tilings up to isometry

Recall: ∃ a bijection z : PT −→ PS ⊂ {0, 1}N.

Proposition

Let T and T ′ be Penrose tilings of R2. Let z = z(T ) = z0z1 . . .and z ′ = z(T ′) = z ′0z ′1 . . . be the corresponding sequences in{0, 1}N. The following are equivalent

1 Oz∼= Oz ′

2 there is an isometry σ such that σ(T ) = T ′

3 z≥n = z ′≥n for some n

4 (Mz)≥n∼= (Mz ′)≥n for some n.

(2) ⇔ (3) is due to R. Robinson or N. de Bruijn. See Chapter 10of Grunbaum and Shepard’s masterpiece, Patterns and Tilings.

Topology

Equivalences: T ∼ T ′ if σ(T ) = T ′ for some isometry σz ∼ z ′ if z≥n = z ′≥n for some n

Topologies:{0, 1} discrete topology{0, 1}N product topology

PT1:1←→ PS ⊂ {0, 1}N subspace topology

PT/∼ and PS/∼ quotient topology

PT/∼ is homeomorphic to PS/∼

Equivalent metric topology: If z , z ′ ∈ PS define d(z , z ′) := 2−n

where n = min{i | zi 6= z ′i }.

Every equivalence class in PS is dense. Every point in PS/∼ isdense.

Topology

Enter non-commutative geometry

Gelfand-Naimark:commutative C∗-algebras with 1 ←→ compact Hausdorff spaces

X C (X ) = {continuous maps X → C}

X ∼= Max C (X )

Philosophy: if X is a non-Hausdorff topological space there is anon-commutative C∗-algebra that “captures” the essential featuresof X . Buzzword: noncommutative topology.

Example 1: if X = Y /G , use C (Y ) o G

Example 2 (Bratteli): if X is a finite T0 topological space there isa C∗-algebra AF (X ) such that Prim AF (X ) ∼= X .

X ∼= Max C (X )

Every equivalence class in PT is dense. Every point in PT/∼ isdense.

PT/∼ is not Hausdorff.

Compare the lemma to

Proposition (?)

If P is a finite patch in a Penrose tiling T , P appears infinitelyoften in every Penrose tiling.

The AF algebra Connes’s associates to PT/∼General principles...

>>>>>>>> 1

��

>>>>>>>> 1

��

>>>>>>>> 2

��

5...3...

f0 = f1 = 1

fn+1 = fn + fn−1

��

= Mfn(C)⊕Mfn−1(C)

Sn+1 = Mfn+1(C)⊕Mfn(C)

Sn ↘ Sn+1 (a, b) 7→((

a 00 b

)S := lim−→ Sn in the category of C∗-algebras

>>>>>>>> 1

��

>>>>>>>> 1

��

>>>>>>>> 2

��

5...3...

f0 = f1 = 1

fn+1 = fn + fn−1

��

Sn ↘ Sn+1 (a, b) 7→((

a 00 b

>>>>>>>> 1

��

>>>>>>>> 1

��

>>>>>>>> 2

��

5...3...

f0 = f1 = 1

fn+1 = fn + fn−1

��

Sn ↘ Sn+1 (a, b) 7→((

a 00 b

>>>>>>>> 1

��

>>>>>>>> 1

��

>>>>>>>> 2

��

5...3...

f0 = f1 = 1

fn+1 = fn + fn−1

��

Sn ↘ Sn+1 (a, b) 7→((

a 00 b

The Grothendieck group

As an ordered abelian group with order unit

K0(S) = Z[τ ]

τ =1 +√

2= the Golden ratio

=the area of a kite

the area of a dart

= limP→∞

the number of kites in a finite patch Pthe number of darts in P

The non-commutative scheme X = Projnc(R)

Theorem

Let R = C〈x , y〉/(y 2) and define X = Projnc(R) by declaring

Qcoh(X ) :=Gr(R)

Fdim(R)π∗←− Gr(R).

There is an equivalence of categories

Qcoh(X ) ≡ Mod(S)

where S = lim−→ Sn in the category of C-algebras and

Connes’s algebra S = the norm closure of S.

Inspired by Serre, define OX := π∗(R).The equivalence is given by HomQcoh(X )(OX ,−) because

OX is projectiveOX generates Qcoh(X )End(OX ) ∼= S .

Theorem

Qcoh(X ) :=Gr(R)

Qcoh(X ) ≡ Mod(S)

Theorem

Qcoh(X ) :=Gr(R)

Qcoh(X ) ≡ Mod(S)

Theorem

Qcoh(X ) :=Gr(R)

Qcoh(X ) ≡ Mod(S)

Computing End(OX )

Definition:HomQGr(R)(π∗R, π∗R) = lim−→HomGrR(R≥n,R≥n) = lim−→Tn.

Each Tn is Morita equivalent to

(C 0C C

��

Sn+1oo

��

= Mfn+1(C)⊕Mfn(C)

and End(OX ) = lim−→Tn∼= lim−→ Sn = S .

dim X = 0?

Theorem

The equivalence of categories restricts to an equivalence

coh(X ) :=gr(R)

fdim(R)≡ mod(S) = finitely presented S-modules.

Every finitely presented S-module is projective soevery short exact sequence in coh(X ) splits.The only irreducible variety with this property is a point.Thus X behaves like a 0-dimensional scheme but

topological dimension(PT/∼) = 0. :)

dim X = 0?

Theorem

coh(X ) :=gr(R)

dim X = 0?

Theorem

coh(X ) :=gr(R)

dim X = 1?

Qcoh(X ) = ModS =⇒ X is an affine non-commutative scheme.

S is hereditary =⇒ dim X = 1.

Should the dimension of X lie between 0 and 1?

Dynamics=degree shift/twist

Define

dynamical shift s : PS → PS ,s(z0z1 . . .) = z1z2 . . .

degree shift (1) : Gr(R)→ Gr(R)N(1) = N as an R-module butN(1)n = Nn+1

compare with L ⊗− where L =an ample line bundle.

(1) : fdim(R)→ fdim(R) so we obtain an auto-equivalence(1) : Qcoh(X )→ Qcoh(X ).

Proposition

If z ∈ PS, then Oz(1) ∼= Os(z).

What does this result mean in terms of tilings?

Define

Proposition

Define

Proposition

Define

Proposition

Define

Proposition

Inflation/composition/deleting edges

z = z0z1 . . .

s(z) = z1z2 . . .

Oz(1) ∼= Os(z)

Question: If T is the tiling corresponding to z what is the tiling T ′that corresponds to s(z)?

Answer: T ′ is obtained by “inflating” T .

Explanation:

Start with a tiling T .

Insert the axis of symmetry for each kite and dart to obtain atiling by isoceles triangles.

Delete certain edges to get a new tiling by triangles inflatedby a factor of τ .

Delete appropriate edges to get a tiling by kites and darts, T ′.

z = z0z1 . . .

s(z) = z1z2 . . .

Oz(1) ∼= Os(z)

Explanation:

z = z0z1 . . .

s(z) = z1z2 . . .

Oz(1) ∼= Os(z)

Explanation:

The 01-sequence associated to T

4/3/12 12:23 AMGmail - favor to ask

of 2https://mail.google.com/mail/?ui=2&ik=bc3a51fad3&view=pt&search=inbox&th=1367327e147823f4

I used screen shots at the various stages of composition (or perhaps decomposition and then switched the order ofthe pictures); I then used MS Paint to make certain lines dotted and to color the triangles.

Today, I used a simple mac program called Seashore to color the big triangles (containing the point P) red.

Let me know if you would prefer something else. It might take me a bit, because I'd like to finish this draft first.

Best wishes,

Pick p ∈ T . If p is in a big/small triangle write 0/1.Delete certain edges and repeat. Here z = 010100 . . .

Closed subspaces of X = Projnc R

A closed subspace Z of a nc-scheme Y is a full subcategoryQcoh(Z ) ⊂ Qcoh(Y ) that is closed under subquotients and forwhich the inclusion functor i∗ : Qcoh(Z )→ Qcoh(Y ) has a leftadjoint i∗ and a right adjoint i !.

Proposition

If Y is an affine non-commutative scheme, i.e.,Qcoh(Y ) = Mod(A) for some ring A, there is a bijection

closed subspaces of Y ←→ two-sided ideals in A

Mod(A/I )←→ I .

Closed subspaces of X = Projnc R

A closed subspace Z of a nc-scheme Y is a full subcategoryQcoh(Z ) ⊂ Qcoh(Y ) that is closed under subquotients and forwhich the inclusion functor i∗ : Qcoh(Z )→ Qcoh(Y ) has a leftadjoint i∗ and a right adjoint i !.

Proposition

If Y is an affine non-commutative scheme, i.e.,Qcoh(Y ) = Mod(A) for some ring A, there is a bijection

closed subspaces of Y ←→ two-sided ideals in A

Mod(A/I )←→ I .

This definition is compatible with the notion of closed subscheme:

Proposition

If Y is a noetherian scheme having an ample line bundle (e.g., aquasi-projective scheme over a field), then

closed subspaces = closed subschemes.

Closed subspaces and density

S is a simple ring so the lemma saysthe only closed subspaces of X = Projnc R are ∅ and X .

Compare: the only closed subspaces of PT/∼ are ∅ and PT/∼because every point is dense.Compare: every finite patch of one tiling appears infinitely often inevery tiling.

Proposition (Herbera)

If z , z ′ ∈ PS, then Ext1S(Oz ,Oz ′) 6= 0.

Translation: Ext1X (OT ,OT ′) 6= 0 for all tilings T and T ′.

Philosophy: simple modules M and M ′ are “close” to each other ifExt1(M,M ′) 6= 0.

Evidence: points are “close” if Ext1(Ox ,Oy ) 6= 0

Proposition

Let x and y be closed points on an irreducible variety Y ofdimension ≥ 1. Then Ext1

Y (Ox ,Oy ) 6= 0 if and only if x = y.

Let X be a finite T0 topological space. Make X a poset: y ≤ x ifx ∈ {y}.

Ladkani: the category of sheaves of vector spaces on X isequivalent to ModI(X ) where I(X ) = the incidence algebra;

{simple I(X )-modules} = {skyscraper sheaves Ox | x ∈ X};Ext1

I(X )(Ox ,Oy ) 6= 0 ⇐⇒ x covers y .

Compatible facts:

topology: the isometry class of every tiling in PT is dense

algebra: Ext1S(OT ,OT ′) 6= 0 for all tilings T and T ′

Proposition

Compatible facts:

Proposition

Compatible facts:

Proposition

Compatible facts:

The cartwheel tiling

If P is a finite patch of the cartwheel tiling, there is a patchP ′ ⊃ P having D10 symmetry.

Ocartwheel = O00... = π∗M00...

M00... =R

C〈x , y〉(y 2, y)

∼= C[x ].

The cartwheel tiling

If P is a finite patch of the cartwheel tiling, there is a patchP ′ ⊃ P having D10 symmetry.

Ocartwheel = O00... = π∗M00...

M00... =R

C〈x , y〉(y 2, y)

∼= C[x ].

General phenomena I: monomial algebras

A monomial algebra is an algebra of the form

A =k〈x1, . . . , xg 〉(w1, . . . ,wr )

where w1, . . . ,wr are words in the letters x1, . . . , xg .

Theorem (Holdaway-S)

Let A be a monomial algebra and Q its Ufnarovskii graph. There isa homomorphism of graded algebras f : A→ kQ such thatkQ ⊗A − induces an equivalence of categories

QGr A ≡ QGr kQ.

Example: A = k〈x ,y〉(y2)

and Q = • ((99 •hh

General phenomena I: monomial algebras

A monomial algebra is an algebra of the form

A =k〈x1, . . . , xg 〉(w1, . . . ,wr )

where w1, . . . ,wr are words in the letters x1, . . . , xg .

Theorem (Holdaway-S)

Let A be a monomial algebra and Q its Ufnarovskii graph. There isa homomorphism of graded algebras f : A→ kQ such thatkQ ⊗A − induces an equivalence of categories

QGr A ≡ QGr kQ.

Example: A = k〈x ,y〉(y2)

and Q = • ((99 •hh

General phenomena II: singularity category

Q = a finite quiver

S(Q) = lim−→EndkI (kQ⊗n1 )

= lim−→(finite dimensional semisimple algebras)

Λ := kQ/kQ≥2

Theorem (Chen + S)

(qgr(kQ), (−1))4≡ (modS(Q),−⊗ B)

4≡ Dsing(Λ) := Db(modΛ)

Dperf (modΛ)

For Q = • ))99 •ii the category Dsing(kQ/kQ≥2) is

at least as complicated as the space of Penrose tilings.

General phenomena II: singularity category

Q = a finite quiver

S(Q) = lim−→EndkI (kQ⊗n1 )

= lim−→(finite dimensional semisimple algebras)

Λ := kQ/kQ≥2

Theorem (Chen + S)

(qgr(kQ), (−1))4≡ (modS(Q),−⊗ B)

4≡ Dsing(Λ) := Db(modΛ)

Dperf (modΛ)

For Q = • ))99 •ii the category Dsing(kQ/kQ≥2) is

at least as complicated as the space of Penrose tilings.

THE END

the space of penrose tilings as a non-commutative schemeof x.buzzword: noncommutative topology....

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