rigidity of artin-schelter regular algebras · for an artin-schelter regular algebra a, when is ag...

Rigidity of Artin-Schelter Regular Algebras

Ellen Kirkman and James KuzmanovichWake Forest University

James ZhangUniversity of Washington


Shephard-Todd-Chevalley Theorem

Theorem. The ring of invariants C[x1, · · · , xn]G under a finitegroup G is a polynomial ring if and only if G is generated byreflections.

Question. For an Artin-Schelter regular algebra A, when is AG

isomorphic to A for a finite group of graded automorphisms G ?

Question. For an Artin-Schelter regular algebra A, when is AG

Artin-Schelter regular?


Results in Another Direction

Theorem. (S. P. Smith, 1989) The Weyl algebra A1(k) is not thefixed subring SG under a finite solvable group of automorphisms ofa domain S.Theorem. (J. Alev and P. Polo, 1995)

1. Let g and g′ be two semisimple Lie algebras. Let G be a finitegroup of algebra automorphisms of U(g) such thatU(g)G ∼= U(g′). Then G is trivial and g ∼= g′.

2. If G is a finite group of algebra automorphisms of An(k), thenthe fixed subring An(k)G is isomorphic to An(k) only when Gis trivial.


Questions in this Other Direction

Let A be Artin-Schelter regular.

Question. When is it the case that AG is never isomorphic to Afor a finite group G of graded automorphisms?

Call such algebras rigid.

Question. When is it the case that AG is never Artin-Schelterregular for a finite group G of graded automorphisms?


Example

Let A = C−1[x , y ] and let 〈g〉 where g =

[ξ 00 1

]for a primitive

nth root of unity ξ.

I If n is odd, then A〈g〉 = C−1[xn, y ] which is isomorphic to A.

I If n is even, then A〈g〉 = C[xn, y ], a commutative polynomialring, which is not isomorphic to A.


Hilbert Series of Regular Algebras

Let B be a graded algebra. The Hilbert series of B is defined by

HB(t) =∞∑k=0

dimBktk .

Proposition. (Stephenson-Zhang, Jing-Zhang, ATV) Let B be anArtin-Schelter regular algebra and let

HB(t) =1

(1− t)np(t)

where p(1) 6= 0. Furhtermore n = GKdim(B) and p(t) is a productof cyclotomic polynomials.


Traces of Graded Automorphisms

Let g be a graded automorphism of a graded algebra A. Thetrace of g is defined by

Tr(g , t) =∞∑n=0

tr(g |An)tn.

Note HA(t) = Tr(Id , t).


Molien’s Theorem

Theorem. (Jørgensen-Zhang) Let B be a connected gradedK -algebra and let G be a finite group of graded automorphisms ofB with |G |−1 ∈ K . Then

HBG (t) =1

|G |∑g∈G

TrB(g , t).


Quasi-Reflections

Let G be an automorphism of an AS-regular algebra A withGKdim(A) = n. We call g a quasi-reflection if

Tr(g , t) =1

(1− t)n−1p(t)

with p(1) 6= 0.


You Need Quasi-Reflections

Theorem. Let G be a finite group of graded automorphisms of aNoetherian AS-regular algebra A. If AG is AS-regular, then G mustcontain a quasi-reflection.

Lemma. Let f (t) = a0 + a1t + · · ·+ antn be a palindrome

polynomial; that is, an−i = ai for all i . Then f ′(1) = nf (1)

2.


A Sketch of the Proof

Proof. Assume that G does not contain a quasi-reflection.Let

HA(t) =1

(1− t)np(t).

Suppose that AG is regular. Then

HAG (t) =1

(1− t)nq(t),

where q(t) is a product of cyclotomic polynomials. Since G isnontrivial, ` = deg(q(t)) > deg(p(t)) = k .


Expand HA(t) and HAG (t) into a Laurent series about t = 1.

HA(t) =1

(1− t)n1

p(1)+

1

(1− t)n−1p′(1)

p(1)2+ · · ·

HAG (t) =1

(1− t)n1

q(1)+

1

(1− t)n−1q′(1)

q(1)2+ · · ·


If we expand HAG (t) =1

|G |∑g∈G

Tr(g , t) into a Laurent series

around t = 1, the first terms come entirely from the trace of theidentity.

HAG (t) =1

|G |

[1

(1− t)n1

p(1)+

1

(1− t)n−1p′(1)

p(1)2+ · · ·

HAG (t) =1

(1− t)n1

q(1)+

1

(1− t)n−1q′(1)

q(1)2+ · · ·

Equating coefficients q(1) = |G |p(1), andq′(1)

q(1)2=

1

|G |p′(1)

p(1)2.


Since p(t) and q(t) are products of cyclotomic polynomials, theyare palindrome polynomials, and hence by the Lemma

q′(1) = `q(1)

2and p′(1) = k

p(1)

2.

Substituting inq′(1)

q(1)2=

1

|G |p′(1)

p(1)2we have

`

2q(1)=

1

|G |k

2p(1).

Since q(1) = |G |p(1), it follows that ` = k , which is acontradiction. �


Jordan Plane

The Jordan Plane CJ [x , y ] is defined by yx − xy = x2.

All graded automorphisms are of the form g = ξId with trace

Tr(g , t) =1

(1− ξt)2.

Hence the Jordan Plane is rigid.


Graded Down-Up Algebras

Let α, β ∈ C with β 6= 0. Then the down-up algebra A(α, β, 0) isthe algebra generated by two elements u, d subject to the relations

d2u = αdud + βud2

du2 = αudu + βu2d .

Then A is a Noetherian Artin-Schelter domain with gldim(A) = 3.Benkart and Roby have shown that A has a vector space basisconsisting of all monomials of the form ui (du)jdk . Hence

HA(t) =1

(1− t)2(1− t2).


Proposition. The graded automorphisms of a graded down-upalgebra A = A(α, β, 0) are given by

1.

[w 00 z

]for any A(α, β, 0).

2.

[0 xy 0

]when A = A(0, 1, 0) or A(α,−1, 0) for any α.

3.

[w xy z

], when A = A(0, 1, 0) or A(2,−1, 0).

Furthermore if the eigenvalues of the defining matrix for g are λand µ, then

Tr(g , t) =1

(1− λt)(1− µt)(1− λµt2).


Graded Down-Up Algebras are Rigid

If g is a graded automorphism, then

Tr(g , t) =1

(1− λt)(1− µt)(1− λµt2).

Hence if g is a quasi-reflection

Tr(g , t) =1

(1− t)(1− µt)(1− µt2).

Thus µ = 1 and we have a pole of order 3, not 2.


A More General Result

Theorem. Let A be Noetherian regular with gldim(A) = 3. If A isgenerated by two elements of degree 1, then A is rigid.

Nonproof. If g is a quasi-reflection, then

Tr(g , t) =1

(1− t)2(1− ξ1t)(1− ξ2t)

for roots of unity ξ1, ξ2.


Quantum Polynomial Rings

A quantum polynomial ring of dimension n is a NoetherianAS-regular domain of global dimension n with Hilbert series

HA(t) =1

(1− t)n.

Hence quasi-reflections have traces of the form

TrA(g , t) =1

(1− t)n−1(1− ξt).


The Quasi-Reflections

Theorem. The quasi-reflections of a quantum polynomial ring Aare of two types. If g is a quasi-reflection, then either

I (Reflections) there is a basis {b1, · · · , bn} of A1, such thatg(b1) = ξb1 and g(bj) = bj for j ≥ 2, or

I (Mystic Reflections) the order of g is 4, and there is a basis{b1, · · · , bn} of A1, such that g(b1) = ib1, g(b2) = −ib2, andg(bj) = bj for j ≥ 3.

I In each case A〈g〉 is regular.


Example

Let A = C−1[x , y ].

I Let g be given by

[ξ 00 1

]where ξ is an nth root of unity.

Then g is a reflection and

Tr(g , t) =1

(1− t)(1− ξt).

A〈g〉 ∼= C±1[xn, y ].

I Let g be given by

[0 −11 0

].


Then g is a mystic reflection with

b1 = x − iy , b2 = x + iy .

The trace of g is given by

Tr(g , t) =1

(1− t)(1 + t).

The invariant ring is given by A〈g〉 ∼= C[x2 + y2, xy ].


Consequences

Theorem. Let A be a quantum polynomial ring.

I If A has a reflection, then A has a normal element of degree 1.I If A has a mystic reflection, then there is a basis{b1, b2, . . . bn} of A1 such that

I b21 = b2

2 is a normal element, andI C〈b1, b2〉 ∼= C−1[x , y ].


Sklyanin Algebras

I Let A be a non-PI Skylanin algebra of global dimension n ≥ 3.Then A has no element b of degree 1 with b2 normal. HenceA is rigid.

I If A has dimension 4, then the graded automorphisms werefound by Smith and Staniszkis, and their traces, none ofwhich have a pole of order 3, were found by Jing and Zhang.

I Dimension 3 PI calculations?


Theorem. Let A be a quantum polynomial ring and G a finitegroup of graded automorphisms such that AG is regular with

HAG (t) =1

(1− t)nq(t)

having q(1) 6= 0. Then q(1) = |G | and degree(q(t)) is the numberof quasi-reflections in G .


Rees Ring of the Weyl Algebra An

Let A the algebra generated by x1, . . . , xn, y1, . . . yn, z withrelations xiyi − yixi = z2 for i = 1, 2, . . . , n.Let all other pairs of generators commute.

The algebra A is AS-regular of dimension 2n + 1.

Proposition. The algebra A is rigid.


Suppose AG is regular.

I If g is a quasi-reflection, then g is given by matrix of the form[I 0v −1

]where I is a 2n × 2n identity matrix. All have

order 2.

I Any group containing two quasi-reflections is infinite.[I 0v −1

] [I 0u −1

]=

[I 0

v − u 1

].


I If AG is regular, then G = {e, g} where g is a quasi-reflection.If HAG (t) = 1

(1−t)2n+1q(t), then degree(q) is the number of

quasi-reflections and q(1) = |G |.I In this case AG is isomorphic to the algebra generated by

X1,X2, . . . ,Xn,Y1,Y2, . . . ,Yn, z2 subject to the relations

XiYi − YiXi = z2 with all other pairs of generatorscommuting.

I The algebra AG is regular but cannot be isomorphic to A,since AG can be generated by 2n elements over C whereas Arequires 2n + 1.


Let g be a finite dimensional Lie algebra over K with bracketoperation [, ]. If b1, b2, . . . , bn is a basis for g over K , then theenveloping algebra U(g) is the associative algebra generated byb1, b2, . . . , bn subject to the relations bibj − bjbi = [bi , bj ].

The homogenization of U(g),H(g) is the associative algebragenerated by , b1, b2, . . . , bn, z subject to the relations biz = zbi

and bibj − bjbi = [bi , bj ]z .Then H(g) is regular of dimension n + 1.


Proposition. Let g be a finite dimensional Lie algebra with no1-dimensional Lie ideal. Then H(g) is rigid.

The proof consists of showing that there are no quasi-reflections.


Reference

E. Kirkman, J. Kuzmanovich and J. J. Zhang, Rigidity of GradedRegular Algebras, Trans. Amer. Math. Soc. 360 (2008), No. 12,pages 6331 - 6369.


rigidity of artin-schelter regular algebras · for an artin-schelter regular algebra a, when is ag...

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