ON THE TRACE OF GRADED AUTOMORPHISMS

Naihuan Jing and James J. Zhang

Abstract. Let A = ⊕_{n≥0}A_{n} be a connected graded algebra with a graded algebra automorphism σ. The trace of σ is defined to be a power series T r(σ, t) = ∑_{n≥0} tr(σ|_{A_{n}})t^{n}. In particular, the trace of the identity map of A is the Hilbert series of A. We prove that T r(σ, t) is a rational function if either

1. A is a finitely generated commutative algebra, or

2. A is a right noetherian algebra with finite global dimension, or

3. A is a regular (but not necessarily noetherian) algebra with finite global dimension. Using this we show that the Hilbert series of subrings of invariants are rational functions in these three cases. Some basic properties of the trace are discussed and some applications are given. We also study the Gorenstein property for some special subrings of invariants.

0. Introduction

Throughout let k denote an algebraically closed field of characteristic zero. Notions such as vector space, algebra, homomorphism and so on are defined over the base field k.

Let M = ⊕_{n}M_{n} be a ℤ-graded vector space. We say M is locally finite if dim(M_{n}) < ∞ for all n, where dim is always the dimension of a vector space. Let A = ⊕_{n≥0}A_{n} be an ℕ-graded algebra. We say A is connected graded if A_{0} = k. A connected graded right noetherian algebra is locally finite. The Hilbert series (or the Poincarè series) of M = ⨁_{n} M_{n} is defined to be a power series

H_{M}(t) = ∑_{n∈ℤ} dim(M_{n})t^{n}.

The Hilbert series is an efficient tool to study commutative and noncommutative graded algebras (see [Be], [St], [KL], and [SpZ]). For example, the Hilbert series provides a measurement of the size of a graded module by giving its Gelfand-Kirillov dimension [KL].

If M is a graded module over an infinite dimensional Kac-Moody Lie algebra 𝔤 with respect to a grading of 𝔤, then the series H_{M}(q) is also called the q-dimension or character in the representation theory of Kac-Moody Lie algebras [Ka]. In this case the series H_{M}(q) is a modular function of τ with q = e^{2πiτ}. Moreover there is a more general notion–the trace of endomorphisms used widely in mathematical physics, for instance the trace of certain vertex operators are n-point functions in the statistical mechanics models in connection with affine Lie algebras [Ba, JM].

Typeset by AMS-TEX

1980 Mathematics Subject Classification (1985 Revision). 13D40, 16W50 Secondary 16W20.

Key words and phrases. Trace, graded algebra, graded automorphism.

The first author was supported in part by the NSA and the second author by the NSF..

Let us introduce the notion of the trace as follows. A linear map σ of M = ⊕_{n}M_{n} is called graded if σ(M_{i}) ⊂ M_{i} for all i ∈ ℤ. The trace of a graded map σ is defined to be

T r_{M}(σ, t) = ∑_{n∈ℤ} tr(σ|_{M_{n}})t^{n}.

We will omit M from T r_{M}(σ, t) if no confusion occurs. Clearly T r(id_{M}, t) is the Hilbert series H_{M}(t). We say M is left bounded if M_{n} = 0 for all n ≪ 0. If M is a left bounded and locally finite, then T r(σ, t) is in the power series ring k[[t, t^{−1}]].

The trace has been used to study subrings of invariants of commutative polynomial algebras since last century (see [Be] for more information). Recently in ring theory there have been increased activities to study noncommutative graded rings coming from quantum groups [Ma, PW] and noncommutative projective geometry [AZ, SSf]. One important class is graded regular algebras, which are viewed as noncommutative analogues of commutative polynomial algebras. A connected graded algebra A is called regular if (i) A has finite global dimension d and (ii) Ext_{A}^{i}(k_{A}, A) is zero if i ≠ d and is isomorphic to k if i = d, where k_{A} is the trivial A-module A/(⊕_{n>0}A_{n}). There are two subtle points in studying subrings of invariants in noncommutative case. One is that the Hilbert basis theorem doesn’t hold for finitely generated noncommutative algebras, for example there are finitely presented but non-noetherian regular algebras [SpZ, 3.4]. The other is that the graded algebra automorphism group is not easy to determined, even for the Sklyanin algebra [SSn]. Examining the trace of the graded automorphisms provides a way to study the automorphism group and subrings of invariants. One aim of this paper is to extend some ideas in [Be] to noncommutative algebras ($ 5 and $ 6). In particular, we partially generalize a theorem of Molien [Be, 2.5.2] and a theorem of Watanabe [Be, 5.3.2] to the noncommutative case.

Another aim of this paper is to prove that the trace T r(σ, t) has nice properties, and under some finiteness conditions it has a simple form. The Hilbert series of a connected graded algebra A is a rational function if either (a) A is commutative [Be, 2.1.1] or (b) A is a right noetherian algebra with finite global dimension [SpZ, 2.4] or (c) A is regular [SpZ, 3.1]. A similar property holds for the trace (§1, 2 and 3). As a consequence, the trace T r(σ, t) is determined by finitely many terms tr(σ|_{A_{n}}). This property allows us to compute the trace for graded automorphisms of many familiar graded algebras. For regular algebras, Koszul algebras and Frobenius algebras, we prove that there is a reciprocity formula for the trace (§3 and 4). We also list the trace for all graded algebra automorphisms of the Sklyanin algebra (§4).

In this paper every graded vector space is assumed to be locally finite, a graded algebra is always connected graded and an endomorphism or automorphism of an algebra means algebraic.

1. Commutative polynomial algebras

In this section we will compute the trace for graded automorphisms of polynomial algebras by using elementary linear algebra. The result is quite obvious and the proof given below is essentially from that of [Be, 2.5.1]. Let A = k[x_{1}, ⋯, x_{d}] be the commutative polynomial algebra generated by {x_{1}, ⋯, x_{d}} with deg(x_{i}) > 0. Then the Hilbert series of A is

H_{A}(t) = \frac{1}{∏_{i=1}^{d}(1 − t^{deg(x_{i})})}.  (1-1)

After rearranging the indices, we may assume that

deg(x_{1}) = ⋯ = deg(x_{d_{1}}) = s_{1} < deg(x_{d_{1}+1}) = ⋯ = deg(x_{d_{2}}) = s_{2} < ⋯ deg(x_{d_{l}}) = s_{l}

where d_{l} = d. Every graded endomorphism σ of A is determined by σ(x_{i}) and we can write σ as follows:

σ( x_{1} ; ⋮ ; x_{d_{1}} )  = C_{1}( x_{1} ; ⋮ ; x_{d_{1}} ),  C_{1} ∈ M_{d_{1}}(k)

σ( x_{d_{1}+1} ; ⋮ ; x_{d_{2}} ) = C_{2}( x_{d_{1}+1} ; ⋮ ; x_{d_{2}} ) + ( f_{d_{1}+1}^{2} ; ⋮ ; f_{d_{2}}^{2} ), C_{2} ∈ M_{d_{2}−d_{1}}(k), f_{i}^{2} ∈ k[x_{1}, ⋯, x_{d_{1}}]
⋮
⋮
σ( x_{d_{l−1}+1} ; ⋮ ; x_{d} ) = C_{l}( x_{d_{l−1}+1} ; ⋮ ; x_{d} ) + ( f_{d_{l−1}+1}^{l} ; ⋮ ; f_{d}^{l} ), C_{l} ∈ M_{d_{l}−d_{l−1}}(k), f_{i}^{l} ∈ k[x_{1}, ⋯, x_{d_{l−1}}]

Associated with σ we have a sequence of square matrices C_{w} for w = 1, ⋯, l. Let I_{n} denote the identity n × n-matrix.

Proposition 1.1. Let σ be a graded endomorphism of k[x_{1}, ⋯ x_{d}] and C_{w} (w = 1, ⋯, l) as above. Then

T r(σ, t) = \frac{1}{∏_{w=1}^{l} det(I_{d_{w}−d_{w−1}} − C_{w} t^{s_{w}})}.

Remark: If σ is the identity map of k[x_{1}, ⋯ x_{d}] in Proposition 1.1, we obtain (1-1).

Proof. By changing basis in the linear space ⊕_{i=1}^{d}kx_{i}, we can assume that each C_{w} is an upper triangular matrix and further assume that

σ(x_{j}) = λ_{j}x_{j} + f_{j}(x_{1}, ⋯ x_{j−1}), j = 1, ⋯, d

where λ_{j}’s are eigenvalues of the matrices C_{w}’s and f_{j}’s are homogeneous polynomials.

We take an order x_{1} < x_{2} < ⋯ < x_{d} and order lexicographically the monomial basis {x_{1}^{n_{1}} ⋯ x_{d}^{n_{d}} | n_{i} ≥ 0}. To compute the trace we apply σ to this monomial basis. It follows that

σ(x_{1}^{n_{1}} ⋯ x_{d}^{n_{d}}) = λ_{1}^{n_{1}} ⋯ λ_{d}^{n_{d}}x_{1}^{n_{1}} ⋯ x_{d}^{n_{d}} + lower terms.

Then we have

T r(σ, t) = ∑_{n≥0} (  ∑ ; ∑ n_{i} deg(x_{i})=n λ_{1}^{n_{1}} … λ_{d}^{n_{d}} ) t^{n}

= \frac{1}{∏_{i≥1}(1 − λ_{i}t^{deg(x_{i})})} = \frac{1}{∏_{w=1}^{l} det(I_{d_{w}−d_{w−1}} − C_{w}t^{s_{w}})}.

□

As an immediate consequence, we have the following corollary [Be, 2.5.1].

Corollary 1.2. Let deg(x_{i}) = 1, i = 1, ⋯, d for the polynomial algebra A = k[x_{1}, ⋯ x_{d}] with a graded endomorphism σ. Then

T r(σ, t) = \frac{1}{det(1 − σ|_{A_{1}} t)}.

□

The reason why the proof of Proposition 1.1 works is that A has a monomial basis such that σ is an upper triangular matrix. In noncommutative case, there is no such monomial basis which we can work on. Starting from next section we will introduce some methods to calculate the trace for noncommutative algebras.

2. Properties

In this section we will investigate some elementary properties of T r(σ, t) and prove that if a graded right noetherian algebra has finite global dimension, then the trace of any graded endomorphism has a form of 1/p(t) for some p(t) ∈ k[t]. Similar to Euler characteristic T r is additive with respect to exact sequences.

Lemma 2.1. Let M^{i} = ⨁_{d} M_{d}^{i} be graded vector spaces for i ≥ 0 with the minimal degree of M^{i} going to the positive infinity when i goes to infinity. Suppose that the complex

⋯ ⟶ M^{n} ⟶ ⋯ ⟶ M^{1} ⟶ M^{0} ⟶ 0  (2-1)

is exact and that ϕ_{i} (i ≥ 0) are k-linear graded endomorphisms of M^{i} commuting with the boundary maps of (2-1). Then ∑_{i≥0}(−1)^{i}T r(ϕ_{i}, t) = 0.

Proof. First we assume that the complex (2-1) is finite, i.e., there is an n such that M^{i} = 0 for all i > n. Using the commutativity between ϕ_{i}’s and boundary maps of (2-1) we can break (2-1) into two exact sequences:

0 ⟶ M^{n} ⟶ ⋯ ⟶ M^{2} ⟶ L ⟶ 0  (2-2)

and

0 ⟶ L ⟶ M^{1} ⟶ M^{0} ⟶ 0  (2-3)

such that {ϕ_{i}|i = 2, ⋯, n} ∪ {ϕ_{1}|_{L}} and {ϕ_{1}|_{L}, ϕ_{1}, ϕ_{0}} commute with the boundary maps of (2-2) and (2-3) respectively. By induction, we only need to consider the case when n = 2. By the definition of T r it suffices to prove ∑_{i=0}^{2}(−1)^{i}tr(ϕ_{i}|_{M_{d}^{i}}) = 0 for the exact sequence

0 ⟶ M_{d}^{2} ⟶ M_{d}^{1} ⟶ M_{d}^{0} ⟶ 0

for every degree d ∈ ℤ. But this follows immediately from linear algebra.

In general we consider the subcomplex of degree less than m. Since the minimal degree of M^{i} goes to the positive infinity, such subcomplex is finite and the formula holds. The general case follows by letting m go to the positive infinity. □

The tensor product over a ring A is denoted by ⊗_{A} and if A = k, then we simply write it as ⊗. Let M^{1} and M^{2} be graded vector spaces. Then M^{1} ⊗ M^{2} is a graded vector space with (M^{1} ⊗ M^{2})_{d} = ⊕_{s}M_{s}^{1} ⊗ M_{d−s}^{2}. If M^{1} and M^{2} are left bounded and locally finite, then so is M^{1} ⊗ M^{2}. Let σ_{i} be a graded endomorphism of M^{i} for i = 1, 2 respectively. Then σ = σ_{1} ⊗ σ_{2} is a graded endomorphism of M^{1} ⊗ M^{2} and its trace is the product of those of its factors:

T r_{M^{1}⊗M^{2}}(σ, t) = T r_{M^{1}}(σ_{1}, t)T r_{M^{2}}(σ_{2}, t).  (2-4)

Let A be a graded algebra. We denote by Gr– A the category of graded right A-modules with morphisms being graded A-homomorphisms of degree 0. Let σ be a graded endomorphism of A. For M ∈ Gr– A we define the σ-twisted module M^{σ} such that M^{σ} = M as a graded vector space and the action is given by

m ∗ a = mσ(a), m ∈ M, a ∈ A.

We define a functor on Gr– A associated to σ by F_{σ}(M) = M^{σ} and F_{σ}(f) = f for any f ∈ Hom_{Gr– A}(M, N). It is easy to see that F_{σ} is indeed a functor from the category Gr– A to itself. If σ is an automorphism, then F_{σ} is an invertible functor and A^{σ} is isomorphic to A as a graded right R-module. Let M and N be graded A-modules and ϕ a graded k-linear map from M to N. We say ϕ is σ-linear if

ϕ(ma) = ϕ(m)σ(a), m ∈ M, a ∈ A.

For example, σ itself is a σ-linear endomorphism of A. A graded map ϕ : M ⟶ N is σ-linear if and only if it is a graded A-homomorphism from M to N^{σ}.

For a graded A-module M = ⊕_{i}M_{i} and an integer s ∈ ℤ we define the shift module M(s) = ⊕_{i}M(s)_{i} such that M(s)_{i} = M_{s+i}. Every left bounded and locally finite graded projective right A-module P is free, namely it has a form of ⨁_{i} A(−s_{i}). We can write it as P = V ⊗ A where V is a left bounded and locally finite graded vector space ⊕_{i}kx_{i} with deg(x_{i}) = s_{i}. It is clear that V = P ⊗_{A} k_{A} where k_{A} (or simply k if no confusion occurs) is the trivial A-bimodule A/(⊕_{n>0}A_{n}). Given a graded endomorphism ϕ′ : V ⟶ V, ϕ′ ⊗ σ is a σ-linear endomorphism of V ⊗ A for a graded endomorphism σ of A. Given a σ-linear endomorphism ϕ : V ⊗ A, we can define ϕ′ to be ϕ ⊗_{A} id_{k_{A}}. Then ϕ′ is a graded endomorphism of V. In general ϕ is not equal to ϕ′ ⊗ σ. Next lemma shows that we still have a formula similar to (2-4).

Lemma 2.2. Let A be a graded algebra with a graded endomorphism σ. Let P = V ⊗ A be a left bounded and locally finite free A-module. If ϕ is a σ-linear graded endomorphism of P, then

T r_{P} (ϕ, t) = T r_{V} (ϕ′, t)T r_{A}(σ, t)

where ϕ′ = ϕ ⊗_{A} id_{k_{A}}.

Proof. First we assume that dim(V) is finite and use induction on dim(V). If dim(V) = 1, then P = A(−s) and every σ-linear endomorphism of P is of form

ϕ(a) = ϕ(1)σ(a), ϕ(1) ∈ k.

Hence

T r(ϕ, t) = ϕ(1)t^{s}T r(σ, t) = T r(ϕ′, t)T r(σ, t).

If dim(V) > 1, let s be the lowest degree in V. Since A is connected graded, P_{s} = V_{s} and ϕ has to preserve the summand V_{s} ⊗ A. Let x ∈ V_{s} be an eigenvector of ϕ (and hence, of ϕ′) such that ϕ(x) = λx for λ ∈ k. Then the map ϕ restricted on kx ⊗ A is a σ-linear endomorphism of kx ⊗ A and the map induced on (V/kx) ⊗ A, which is still denoted by ϕ, is a σ-linear endomorphism of (V/kx) ⊗ A. Hence we have a short exact sequence

0 ⟶ kx ⊗ A ⟶ V ⊗ A ⟶ (V/kx) ⊗ A ⟶ 0

such that {ϕ|_{kx⊗A}, ϕ, ϕ|_{(V/kx)⊗A}} commutes with the boundary maps of the above complex. By Lemma 2.1,

T r_{V ⊗A}(ϕ, t) = T r_{kx⊗A}(ϕ, t) + T r_{(V/kx)⊗A}(ϕ, t)

and

T r_{V} (ϕ′, t) = T r_{kx}(ϕ′, t) + T r_{V/kx}(ϕ′, t).

Hence the statement follows by induction.

In general case we let V_{≤n} = ⊕_{i≤n}V_{i}. Consider V_{≤n} ⊗ A as a submodule of V ⊗ A, it is ϕ-stable. Since V is left bounded and locally finite, V_{≤n} is finite dimension and the formula holds for V_{≤n} ⊗ A. It is easy to see that (V ⊗ A)_{≤n} = (V_{≤n} ⊗ A)_{≤n}. Therefore

T r(ϕ, t) = lim_{n→∞} T r(ϕ|_{V_{≤n}⊗A}, t) = lim_{n→∞} T r((ϕ|_{V_{≤n}⊗A})′, t)T r(σ, t) = T r(ϕ′, t)T r(σ, t).

□

Let A be a graded algebra and M a graded right A-module. We say M_{A} has a finite minimal free resolution (FMFR) if (i) the minimal free resolution of M has finite length and (ii) every free module in the minimal free resolution is finitely generated. If A is right noetherian and has finite global dimension, then every finitely generated right A-module has an FMFR. By [StZ, 3.1] if A is regular then k_{A} has an FMFR.

Theorem 2.3. Let A be a graded algebra with a graded endomorphism σ.

1. Let M be a graded right A-module which has an FMFR. If ϕ is a σ-linear graded endomorphism of M, then

T r_{M}(ϕ, t) = f(t)T r_{A}(σ, t)

for some f(t) in k[t, t^{−1}].

2. If B = A/I is a graded factor algebra of A, σ preserves the ideal I and B_{A} has an FMFR, then

T r_{B}(σ, t) = g(t)T r_{A}(σ, t)

for some g(t) in k[t] with constant coefficient 1.

3. Suppose the trivial module k_{A} has an FMFR, then

T r_{A}(σ, t) = \frac{1}{p(t)}

for some p(t) in k[t] with constant coefficient 1.

Remark: Let p(t) = 1 + ∑_{i=1}^{l} c_{i}t^{i}. Then 1/p(t) = ∑_{j≥0}(−1)^{j}(∑_{i=1}^{l} c_{i}t^{i})^{j}. If 1/p(t) = ∑_{i≥0} d_{i}t^{i}, then d_{i} is an integer polynomial of c_{1}, ⋯, c_{i} of degree i. Conversely, every c_{i} is an integer polynomial of d_{1}, ⋯, d_{i} of degree i. Thus d_{i} for i > l is an integer polynomial of d_{1}, ⋯, d_{l}. In the case of T r(σ, t) = 1/p(t), every tr(σ|_{A_{i}}) is uniquely determined by tr(σ|_{A_{1}}), ⋯, tr(σ|_{A_{l}}) where l = deg(p(t)).

Proof. 1. Suppose the minimal free resolution of M is

P^{·} := 0 ⟶ P^{n} ⟶ P^{n−1} ⟶ ⋯ ⟶ P^{0} ⟶ M ⟶ 0

where P^{i} = V^{i} ⊗ A for some finite dimensional graded vector spaces V^{i} (i = 0, 1, ⋯, n). The boundary map from P^{i} to P^{i−1} is denoted by δ_{i}. Applying F_{σ} to P^{·}, (P^{·})^{σ} is a resolution of graded right A-module M^{σ}. By definition ϕ can be viewed as an A-homomorphism from M to M^{σ}. Hence there exists a sequence of A-homomorphisms ϕ_{i} : P^{i} ⟼ P^{i}^{σ} such that ϕ_{i−1}δ_{i} = δ_{i}ϕ_{i} for i ≥ 0 ([Ja, 6.3]), where ϕ_{−1} = ϕ. Consider ϕ_{i} as an endomorphism of P^{i}, {ϕ_{i}} are σ-linear endomorphisms commuting with the boundary maps. By Lemmas 2.1 and 2.2, we obtain

T r_{M}(ϕ, t) = ∑_{i=0}^{n}(−1)^{i}T r(ϕ_{i}, t) = ∑_{i=0}^{n}(−1)^{i}T r_{V^{i}}(ϕ′_{i}, t)T r_{A}(σ, t) = f(t)T r_{A}(σ, t)  (2-5)

where f(t) = ∑_{i=0}^{n}(−1)^{i}T r_{V^{i}}(ϕ′_{i}, t) ∈ k[t, t^{−1}].

2. Let M = B and ϕ the induced map σ|_{B}. Then ϕ is a σ-linear endomorphism as a right A-module. By part 1 and (2-5), T r_{B}(σ, t) = g(t)T r_{A}(σ, t) where g(t) = ∑_{i=0}^{n}(−1)^{i}T r_{V^{i}}(ϕ′_{i}, t) for some V^{i} and ϕ′_{i}. But every free A-module in the minimal projective resolution of B_{A} is ℕ-graded, whence the minimal degree of nonzero elements in V^{i} is non-negative. Therefore g(t) ∈ k[t]. Since A and B are connected graded, the constant terms of T r_{B}(σ, t) and T r_{A}(σ, t) are 1 and hence the constant term of g(t) is 1.

3. Let B = k_{A} and ϕ = id_{k_{A}}. By part 2, T r(id_{k_{A}}, t) = g(t)T r(σ, t) for some g(t) ∈ k[t]. Hence T r(id_{k_{A}}, t) = 1 implies that T r(σ, t) = 1/g(t). □

We have two immediate consequences. First let B be a commutative graded algebra with a minimal generating set {x_{1}, ⋯, x_{n}} and σ a graded endomorphism of B. Since A = k[x_{1}, ⋯, x_{n}] is a free commutative algebra, we can lift σ to a graded algebra endomorphism of A. Since A is noetherian and has finite global dimension, B_{A} has an FMFR. By Theorem 2.3.2, T r_{B}(σ, t) = g(t)T r_{A}(σ, t). By Proposition 1.1, T r_{A}(σ, t) = 1/p(t), whence T r_{B}(σ, t) is a rational function.

Corollary 2.4. Let B be a graded finitely generated commutative algebra and σ a graded endomorphism. Then T r(σ, t) is a rational function. □

Second we consider a usual graded homomorphism ϕ of a graded A-module M. In this case we choose σ = Id_{A}. If A is finitely generated connected graded algebra, then the Gelfand-Kirillov dimension (GK-dimension) is defined to be

GKdim(A) = \overline{lim}_{n→∞} log_{n}(∑_{i≥n} dim(A_{i})).

If A has finite GK-dimension then there are c_{1}, c_{2} > 0 such that dim(A_{n}) ≤ c_{1}n^{c_{2}} for all n > 0 (see [KL] for more details).

Corollary 2.5. Let M be a graded A-module and ϕ a graded A-endomorphism of M. Suppose M has an FMFR. Then

T r_{M}(ϕ, t) = g(t)H_{A}(t)

where g(t) is in k[t, t^{−1}] and H_{A}(t) is the Hilbert series of A. If, moreover, A has finite GKdimension, then there are constants d_{1}, d_{2} > 0 such that |tr(ϕ|_{M_{n}})| ≤ d_{1}n^{d_{2}} for n > 0.

Proof. The first part follows from Theorem 2.3.1. Now suppose A has finite GK-dimension. Then there are constants c_{1}, c_{2} > 0 such that dim(A_{n}) ≤ c_{1}n^{c_{2}} for all n > 0. If g(t) = ∑_{i=l}^{m} a_{i}t^{i}, then tr(ϕ|_{M_{n}}) = ∑_{i=l}^{m} a_{i} dim(A_{n−i}) and

|tr(ϕ|_{M_{n}})| ≤ ∑_{i=l}^{m} |a_{i}|c_{1}(n − i)^{c_{2}} ≤ d_{1}n^{d_{2}}

for all n > 0 where d_{1}, d_{2} are some positive constants. □

Next simple and well-known example illustrates how to use Theorem 2.3 to compute the trace.

Example 2.6. The trace of graded endomorphisms of a free algebra.

Let A = k⟨x_{1}, ⋯, x_{d}⟩ be the free algebra generated by x_{1}, ⋯, x_{d} with deg(x_{i}) = 1. Then the trivial module k_{A} has an FMFR:

0 ⟶ V ⊗ A ⟶ A ⟶ k_{A} ⟶ 0

where V = ⊕_{i=1}^{d}kx_{i}. Every graded endomorphism σ of A is determined by its action on the linear space V, or given by a matrix C = (c_{ij})_{n×n} where σ(x_{i}) = ∑_{j} c_{ij}x_{j}. By Theorem 2.3, T r(σ, t) = 1/p(t) and p(t) = 1 − T r_{V} (ϕ′_{1}, t) = 1 − rt for r ∈ k. But T r(σ, t) = 1 + tr(C)t + · · ·. Hence r = tr(C) and T r(σ, t) = (1 − tr(C)t)^{−1}. □

To conclude this section we point out a relationship between graded automorphisms and their traces.

Proposition 2.7. 1. Let M be a graded vector space and τ a graded automorphism of M. If σ is a graded endomorphism of M, then T r(σ, t) = T r(τστ^{−1}, t).

2. Let G_{1} ⊂ G_{2} be two subgroups of the graded automorphism group of M. Suppose that, for every two distinct g_{1}, g_{2} ∈ G_{1}, T r(g_{1}, t) ≠ T r(g_{2}, t). Then G_{1} is abelian. If, moreover, G_{1} is a normal subgroup of G_{2}, then G_{1} is in the center of G_{2}.

Proof. 1. It is obvious from linear algebra.

2. Let g_{1}, g_{2} ∈ G_{1}. By part 1, T r(g_{1}, t) = T r(g_{2}g_{1}g_{2}^{−1}, t). By the hypothesis, g_{1} = g_{2}g_{1}g_{2}^{−1} and hence g_{1}g_{2} = g_{2}g_{1}. Thus G_{1} is abelian. If G_{1} is normal in G_{2}, we may also assume g_{2} ∈ G_{2}, then g_{1}g_{2} = g_{2}g_{1} implies that G_{1} is in the center of G_{2}. □

3. Reciprocity

One elegant result of Stanley [St, 12.7] states that if A is a connected graded commutative Cohen-Macaulay ring, then A is Gorenstein if and only if H_{A}(t^{−1}) = ±t^{l}H_{A}(t) for some l ∈ ℤ. In noncommutative case it was proved that if A is a connected graded either right noetherian or regular algebra with finite global dimension, then H_{A}(t^{−1}) = ±t^{l}H_{A}(t) for some l [SpZ, 2.4 and 3.1]. In this section we will show that such a reciprocity formula holds for the trace on regular algebras.

Let A be a connected graded algebra. We say A is Gorenstein if (i) A has left and right injective dimension d and (ii) Ext^{i}(k_{A}, A) = 0 for all i ≠ d and Ext^{d}(k_{A}, A_{A}) ≅ Ext^{d}(k_{A},_{A}A) ≅ k (this is equivalent to the classical definition of Gorenstein if A is commutative). If, moreover, A has finite global dimension, then A is called regular. If A is connected graded noetherian PI (including commutative) algebra, then A having finite left and right injective (or global) dimension implies that A is Gorenstein (or regular) [SfZ, 1.1]. If A is regular (noetherian or not), then k_{A} has an FMFR (the proof is similar to the proof of [SpZ, 3.1]). Hence we can apply Theorem 2.3.3 to regular algebras. We don’t have any example that a connected graded noetherian algebra of finite global dimension is not regular. The Sklyanin algebra (see the end of section 4) is noetherian and regular [SSf, 0.3]. There are non-noetherian regular algebras [SpZ, 3.4]. Also note that a connected graded algebra of injective dimension 0 is finite dimensional, Gorenstein and Frobenius [SZ].

If M and N are graded right A-modules, we use Hom_{A}^{d}(M, N) to denote the set of all Amodule homomorphisms h : M → N such that h(M_{i}) ⊆ N_{i+d}. We set \underline{Hom}_{A}(M, N) = ⊕_{d∈ℤ} Hom_{A}^{d}(M, N), and we denote the corresponding derived functors by \underline{Ext}_{A}^{i}(M, N). If M has an FMFR, \underline{Ext}_{A}^{i}(M, N) = Ext_{A}^{i}(M, N), but this is not true in general. If M is a graded right module and N is a graded left module, the usual Tor-groups Tor_{i}^{A}(M, N) have a graded structure, which we denote by \underline{Tor}_{i}^{A}(M, N). If A is a graded regular or noetherian Gorenstein algebra of injective dimension d, then, by definition, \underline{Ext}^{i}(k_{A}, A) = 0 for all i ≠ d and by [SpZ, 3.1.2.] and [Zh, 1.2.1], there is an integer l such that \underline{Ext}^{d}(k_{A}, A_{A}) ≅ \underline{Ext}^{d}(k_{A},_{A}A) ≅ k(l). Theorem 3.1. Let A be a graded regular algebra of global dimension d and σ a graded automorphism of A. Suppose T r(σ, t) = 1/p_{σ}(t) as in Theorem 2.3.3. Then p_{σ^{−1}}(t^{−1}) = a^{−1}t^{−l}p_{σ}(t) or equivalently T r(σ^{−1}, t^{−1}) = at^{l}T r(σ, t) where a is the leading coefficient of p_{σ}(t). Proof. By [SpZ, 3.1], k_{A} has an FMFR

P^{·} := 0 → A(−l) → ⨁_{i} A(−s_{i}^{d−1}) → ⋯ → ⨁_{i} A(−s_{i}^{1}) → A → k → 0

where l > s_{i}^{j} > 0. By Lemma 3.2.1 next, there is a σ-linear automorphism ϕ^{·} := {ϕ_{j}} of the complex P^{·}. Then by the proof of Theorem 2.3, T r(σ, t) = 1/p_{σ}(t) and p_{σ}(t) = ∑_{j=0}^{d}(−1)^{j}T r(ϕ′_{j}, t) where ϕ′_{j} = ϕ_{j} ⊗_{A} k_{A}. Since A is regular, the dual complex P^{·}^{∨} := Hom(P^{·}, A) is the minimal free resolution of the left A-module k(l). Applying (−)^{∨} := Hom(−, A) to {ϕ_{j}}, we obtain a σ^{−1}-linear automorphism {ϕ_{j}^{∨}} of the exact sequence of left A-modules

0 ← k(l) ← A(l) ← ⨁_{i} A(s_{i}^{d−1}) ← ⋯ ← ⨁_{i} A(s_{i}^{1}) ← A ← 0.

By Lemmas 2.1 and 2.2,

T r(ϕ_{d}^{∨}|_{k(l)}, t) = ∑_{j}(−1)^{(d−j)}T r(ϕ_{j}^{∨}, t) = ∑_{j}(−1)^{(d−j)}T r((ϕ_{j}^{∨})′, t)T r(σ^{−1}, t).

Since k(l) is one-dimensional, T r(ϕ_{d}^{∨}|_{k(l)}, t) = bt^{−l} where b is a nonzero scalar. If we write P^{j} = ⨁_{i} A(−s_{i}^{j}) as V ⊗ A, then (P^{j})^{∨} = A ⊗ V^{∗}. By the duality between V and V^{∗}, we have T r((ϕ_{j}^{∨})′, t) = T r(ϕ′_{j}, t^{−1}). Therefore

T r(σ^{−1}, t) = \frac{bt^{−l}(−1)^{d}}{∑(−1)^{j}T r((ϕ_{j}^{∨})′, t)} = \frac{bt^{−l}(−1)^{d}}{∑(−1)^{j}T r((ϕ_{j})′, t^{−1})} = \frac{at^{−1}}{p_{σ}(t^{−1})} = at^{−l}T r(σ, t^{−1})

or equivalently, T r(σ^{−1}, t^{−1}) = at^{l}T r(σ, t) and p_{σ^{−1}}(t^{−1}) = a^{−1}t^{−l}p_{σ}(t). □ Lemma 3.2. Let A be a graded algebra, σ a graded automorphism of A, and M a left bounded locally finite right A-module. Suppose P^{·} → M → 0 is the minimal free resolution of M.

1. If ϕ_{−1} is a graded isomorphism from M to M^{σ}, then there is a complex isomorphism ϕ^{·} = {ϕ_{i}} from P^{·} → M → 0 to P^{σ} → M^{σ} → 0.

2. If ϕ^{·} = {ϕ_{i}} is a complex morphism from (P^{·}) → M → 0 to (P^{·})^{σ} → M^{σ} → 0, then ϕ^{·} is a complex morphism from (P^{·})^{σ^{s}} → M^{σ^{s}} → 0 to (P^{·})^{σ^{s+1}} → M^{σ^{s+1}} → 0 for all s.

3. If σ^{n} = id_{A}, then (ϕ^{·})^{n} is a complex automorphism of P^{·} → M → 0, where ϕ^{·} is as in part 1. Proof. 1. By induction and the construction of the minimal free resolution, it suffices to show that if P^{0} is a projective cover of M, then there is an A-homomorphism ϕ_{0} such that

\begin{matrix}P^{0} & ⟶^{δ_{0}} & M & ⟶ & 0 \\ ^{ϕ_{0}}↓ & ^{ϕ_{−1}}↓ \\ P^{0}^{σ} & ⟶^{δ_{0}} & M^{σ} & ⟶ & 0 \end{matrix}  (3-1)

commutes and ϕ_{0} is an isomorphism. Since P^{0} is projective and δ_{0} : P^{0}^{σ} → M^{σ} is surjective, there is a morphism ϕ_{0} such that (3-1) commutes. Tensoring with the trivial A-module k_{A} we see that ϕ_{−1} maps a generating set of M to a generating set of M^{σ}. Since P^{0} is a projective cover, which is also free, δ_{0} maps a generating set of P^{0} to a generating set of M. Hence ϕ_{0} ⊗_{A} k is an isomorphism. By the Nakayama’s lemma for graded modules, ϕ_{0} is an isomorphism.

Part 2 is obtained by applying F_{σ^{s}} to ϕ^{·} and part 3 follows from part 1 and part 2. □

If A has a graded regular algebra with finite GK-dimension, then H_{A}(t) = 1/p(t) and every root of p(t) is a root of 1 (combining [SpZ 2.2] and [SpZ, 3.1]). Next we will show a similar property for the trace.

Proposition 3.3. Suppose A is a graded regular algebra with finite global dimension and finite GK-dimension, and suppose σ is a graded automorphism of A of finite order. Let T r(σ, t) = 1/p_{σ}(t) be as in Theorem 3.1. Then every root of p_{σ}(t) has absolute value 1.

If, moreover, p_{σ}(t) is a polynomial with integer coefficients, then every root of p_{σ}(t) is a root of 1

Proof. Consider the minimal free resolution of k_{A},

P^{·} := 0 → A(−l) → ⊕_{i}A(−s_{i}^{d−1}) → ⋯ → ⊕_{i}A(−s_{i}^{1}) → A → k → 0

where l > s_{i}^{j} > 0. By the proof of Theorem 2.3 or 3.1, p_{σ}(t) = ∑_{i=0}^{d}(−1)^{i}T r(ϕ′_{i}, t). Hence the leading coefficient of p_{σ}(t) is equal to (−1)^{d}a if ϕ_{d}(x) = aσ(x) for all x ∈ A(−l). By Lemma 3.2.3, {ϕ_{i}^{n}} (where n is the order of σ) is a morphism of P^{·}, then H_{A}(t) = 1/p(t) where p(t) = ∑_{i=0}^{d}(−1)^{i}T r((ϕ_{i}^{n})′, t). Thus the leading coefficient of p(t) is (−1)^{d}a^{n}. By [SpZ, 3.1.4], (−1)^{d}a^{n} is either 1 or −1. As a consequence, a is a root of 1.

By definition, T r(σ, t) = ∑_{n≥0} tr(σ|_{A_{n}})t^{n}. Since σ has finite order, every eigenvalue of σ|_{A_{n}} is a root of 1 and tr(σ|_{A_{n}}) has absolute value less than or equal to dim(A_{n}). Hence \overline{lim}_{n→∞} |tr(σ|_{A_{n}})|^{\frac{1}{n}} ≤ \overline{lim}_{n→∞} dim(A_{n}) ≤ 1 (because A has finite GK-dimension). Now suppose p_{σ}(t) = ∏_{i=1}^{l}(1 − r_{i}t). By [SpZ, 2.1], |r_{i}| ≤ 1. But the product of r_{i} is a which is a root of 1. Hence r_{i} has absolute value 1.

If p_{σ}(t) is a polynomial with integer coefficients, then the leading coefficient of p_{σ}(t) is a = ±1 and every root of p_{σ}(t) is an algebraic integer. But every algebraic integer with absolute value 1 is a root of 1 [CF, Lemma 2, p.72]. □

Next we consider the trace for Frobenius algebras. Recall that an algebra A is called Frobenius if it is finite dimensional and Hom_{k}(A, k) is isomorphic to A as left and right A-modules. By [Sm, 3.2], a finite dimensional connected graded algebra A = ⊕_{i=0}^{n}A_{i} is Frobenius if and only if (i) dim(A_{n}) = 1 and (ii) for every x ∈ A_{i} − {0}, xA_{n−i} = A_{n−i}x = A_{n}. Suppose σ is a graded automorphism of A. Pick a basis {x_{1}^{i}, ⋯ x_{w_{i}}^{i}} of A_{i}, we write σ(x_{j}^{i}) = ∑_{j′} c_{j′j}^{i}x_{j′}^{i} and C_{i} = (c_{jj′}^{i}). Then

T r(σ, t) = ∑_{i} tr(C_{i})t^{i}.

If A is Frobenius, we can assume that x_{j}^{i}x_{j′}^{n−i} = δ_{ij}x^{n} [Sm, 3.2]. Then we have C_{i}(C_{n−i})^{τ} = c_{11}^{n}I and hence c_{11}^{n}(C_{i})^{−1} = (C_{n−i})^{τ}. This implies the following result.

Theorem 3.4. Let A = ⊕_{i=0}^{n}A_{i} be a graded Frobenius algebra with A_{n} = kx^{n} and σ a graded automorphism of A. Suppose σ(x^{n}) = cx^{n}. Then

T r(σ, t) = ct^{n}T r(σ^{−1}, t^{−1}).
□

The following immediate consequence will be used in next section.

Corollary 3.5. Let A and σ be as in Theorem 3.4. Then, for all 1 ≤ i ≤ n − 1, tr(σ^{−1}|_{A_{i}}) = 0 if and only if tr(σ|_{A_{n−i}}) = 0. □

4. Induced automorphism on Ext

In this section we will make some ideas in section 2 more precise and then re-prove Theorem 3.1 by using Theorem 3.4. At the end we will calculate the trace for graded automorphisms of the Sklyanin algebra.

Let σ be a graded automorphism of a graded algebra A. Suppose M is a graded right A-module with an FMFR

P^{·} := 0 → P^{d} → ⋯ → P^{1} → P^{0} → M → 0.

Let ϕ_{−1} be a σ-linear automorphism of M. By Lemma 3.2.1, there is a σ-linear automorphism ϕ^{·} = {ϕ_{i}} of the complex P^{·}. Note that ϕ^{·} is not unique, but two different ϕ^{·}’s are always homotopic. Then ϕ^{·} induces a k-linear automorphism \overline{ϕ}_{i} of \underline{Ext}^{i}(M, k) = \underline{Hom}(P^{i}, k) by sending f ∈ \underline{Hom}(P^{i}, k) to fϕ_{i}. It is easy to see that fϕ_{i} ∈ \underline{Hom}(P^{i}, k) and it is independent of the choices of ϕ^{·}. Similarly ϕ^{·} induces a k-linear automorphism \underline{ϕ}_{i} of \underline{Tor}_{i}(M, k) = P^{i} ⊗_{A} k. We define the following double power series:

T r_{\underline{Ext}^{·}(M,k)}(ϕ_{−1}, t, u) = ∑_{i≥0}(−1)^{i}T r_{\underline{Ext}^{i}(M,k)}(\overline{ϕ}_{i}, t)u^{i},

and

T r_{\underline{Tor}_{·}(M,k)}(ϕ_{−1}, t, u) = ∑_{i≥0}(−1)^{i}T r_{\underline{Tor}_{i}(M,k)}(\underline{ϕ}_{i}, t)u^{i}.

Since we have assumed that M has an FMFR, both are actually polynomials in u and Laurent polynomials in t. We may write P^{i} as V^{i} ⊗ A and hence \underline{Hom}(P^{i}, k) = (V^{i})^{∗} and P^{i} ⊗_{A} k = V^{i}. By the k-linear duality between (V^{i})^{∗} and V^{i}, we obtain that T r_{\underline{Ext}^{i}(M,k)}(\underline{ϕ}_{i}, t^{−1}) = T r_{\underline{Tor}_{i}(M,k)}(\overline{ϕ}_{i}, t). Hence

T r_{\underline{Ext}^{·}(M,k)}(ϕ_{−1}, t^{−1}, u) = T r_{\underline{Tor}_{·}(M,k)}(ϕ_{−1}, t, u).  (4-1)

By the proof of Theorem 2.3.1, (2-5) and (4-1), we have

T r_{M}(ϕ_{−1}, t) = T r_{A}(σ, t){∑_{i≥0}(−1)^{i}T r_{V_{i}}(ϕ′_{i}, t)}

= T r_{A}(σ, t)\{∑_{i≥0}(−1)^{i}T r_{\underline{Tor}_{i}(M,k)}(\underline{ϕ}_{i}, t)\}

= T r_{A}(σ, t)T r_{\underline{Tor}_{·}(M,k)}(ϕ_{−1}, t, 1) = T r_{A}(σ, t)T r_{\underline{Ext}^{·}(M,k)}(ϕ_{−1}, t^{−1}, 1).

A special case is when M = k and ϕ_{−1} = id_{k}, written as σ_{−1}, as a σ-linear automorphism of k for any σ ∈ Aut_{gr}(A). Hence T r_{k}(σ_{−1}, t) = 1 implies that

T r_{A}(σ, t) = T r_{\underline{Ext}^{·}(k,k)}(σ_{−1}, t^{−1}, 1)^{−1}.  (4-2)

Note that we do assume k has an FMFR (and hence A has finite global dimension) in (4-2). So we can not apply (4-2) to graded algebras with infinite global dimension.

Both ⨁_{i≥0} \underline{Ext}^{i}(k, k) and ⨁_{i≥0} \underline{Tor}_{i}(k, k) are ℤ^{2}-graded vector spaces because \underline{Ext}^{i}(k, k) and \underline{Tor}_{i}(k, k) have ℤ-grading. By definition, for every i > 0, nonzero homogeneous elements in \underline{Ext}^{i}(k, k) (and in \underline{Tor}_{i}(k, k) respectively) have negative degree (positive degree respectively). For every σ ∈ Aut_{gr}(A) the σ-linear map σ_{−1} of k_{A} induces a ℤ^{2}-graded automorphisms \overline{σ} := \{\overline{σ}_{i}\} on ⨁_{i≥0} \underline{Ext}^{i}(k, k) and \underline{σ} := \{\underline{σ}_{i}\} on ⨁_{i≥0} \underline{Tor}_{i}(k, k).

Lemma 4.1. Let A be a graded algebra and σ, τ ∈ Aut_{gr}(A). Suppose that k_{A} has an FMFR. Let \overline{σ} and \overline{τ} be defined as above.

1. \overline{(στ)} = \overline{τ}\overline{σ} on ⨁_{i≥0} \underline{Ext}^{i}(k, k). Namely, Aut_{gr}(A)^{op} acts naturally on ℤ^{2}-graded vector space ⨁_{i≥0} \underline{Ext}^{i}(k, k).

2. \underline{(στ)} = \underline{σ}\underline{τ} on ⨁_{i≥0} \underline{Tor}_{i}(k, k). Namely, Aut_{gr}(A) acts naturally on ℤ^{2}-graded vector space ⨁_{i≥0} \underline{Tor}_{i}(k, k).

Proof. 1. By Lemma 3.2.1, there are automorphisms σ^{·} = {σ_{i}} and τ^{·} = {τ_{i}} of the minimal free resolution P^{·} of k_{A}, where σ_{−1} = id_{k_{A}}, σ_{0} = σ and τ_{−1} = id_{k_{A}}, τ_{0} = τ. The composition σ^{·}τ^{·} is a στ-linear automorphism of P^{·}. Hence σ^{·}τ^{·} = (στ)^{·}. Applying \underline{Hom}(−, k), we obtain that

\underline{Hom}((στ)^{·}, k) = \underline{Hom}((τ)^{·}, k)\underline{Hom}((σ)^{·}, k).

Thus \overline{(στ)}_{i} = \overline{τ}_{i}\overline{σ}_{i} for all i ≥ 0 and \overline{(στ)} = \overline{τ}\overline{σ}. Part 2 can be proved similarly. □

By using the Yoneda product, ⨁_{i≥0} \underline{Ext}^{i}(k, k) is a ℤ^{2}-graded algebra, which is denoted by A^{#}. Next lemma shows that \overline{σ} is also an algebra automorphism.

Lemma 4.2. Let A be a graded algebra and let k_{A} have an FMFR. Then every element in the group Aut_{gr}(A)^{op} acts naturally as a ℤ^{2}-graded automorphism of A^{#}.

Remark: To define A^{#} and \overline{σ} it is sufficient to assume that every free module in the minimal free resolution of k_{A} is finitely generated. Hence Lemma 4.2 holds even for such algebras with infinite global dimension.

Proof. By Lemma 4.1, it remains to show that \overline{σ} := \{\overline{σ}_{i}\} is an algebra homomorphism for all σ ∈ Aut_{gr}(A).

Let P^{·} be a minimal free resolution and σ^{·} = {σ_{i}} be a σ-linear automorphism of P^{·} with σ_{−1} = id_{k} and σ_{0} = σ. By definition, \overline{σ}_{i} sends f to fσ_{i} for all f ∈ \underline{Hom}(P^{i}, k). For every f ∈ \underline{Hom}(P^{i}, k) and every g ∈ \underline{Hom}(P^{j}, k), we lift g to a complex morphism \{g_{s}\} where g_{s} is an A-homomorphism from P^{j+s} to P^{s} for all s as in [Sm, (2-2)]. By the discussion of [Sm, Sec 2], the Yoneda product of f and g is f · g = fg_{i} ∈ \underline{Hom}(P^{i+j}, k). We can lift σ(g) = gσ_{j} to {σ_{s}^{−1}g_{s}σ_{s+j}} by checking that {σ_{s}^{−1}g_{s}σ_{s+j}} are A-homomorphisms commuting with the boundary maps. Hence

\overline{σ}(f) · \overline{σ}(g) = fσ_{i} · σ_{i}^{−1}g_{i}σ_{i+j} = fgσ_{i+j} = \overline{σ}(fg).

Thus \overline{σ} is an algebra homomorphism as required. □

By definition A^{#} is ℤ^{2}-graded. The trace of any ℤ^{2}-graded endomorphism ϕ is defined by

T r_{A^{#}}(ϕ, t, u) = ∑_{(n,m)∈ℤ^{2}} tr(ϕ|_{A_{n,m}})t^{n}u^{m}.

By definition, we have

T r_{A^{#}}(\overline{σ}, t, u) = T r_{\underline{Ext}^{·}(k,k)}(σ_{−1}, t, −u),

and by (4-2), we have

T r_{A}(σ, t) = T r_{\underline{Ext}^{·}(k,k)}(σ_{−1}, t^{−1}, 1)^{−1} = T r_{A^{#}}(\overline{σ}, t^{−1}, −1)^{−1}.  (4-3)

If A is regular, then by [SZ] A^{#} is Frobenius. Similar to Theorem 3.4, we can prove that, for every ℤ^{2}-graded automorphism ϕ of A^{#},

T r_{A^{#}}(ϕ, t, u) = at^{c}u^{d}T r_{A^{#}}(ϕ^{−1}, t^{−1}, u^{−1})

for some a ∈ k^{×} and c, d ∈ ℕ (the details left to the interesting reader). In particular, we have

T r_{A^{#}}(\overline{σ}, t^{−1}, −1) = (−1)^{d}at^{−c}T r_{A^{#}}(\overline{σ^{−1}}, t, −1)

which recovers Proposition 3.1 by (4-3).

It is not clear if Aut_{gr}(A)^{op} ≅ Aut_{gr}(A^{#}) holds in general. However if A is Koszul, then Aut_{gr}(A)^{op} ≅ Aut_{gr}(A^{#}). Let us first review some basic definitions and properties related to Koszul algebras and then prove (4-3) for Koszul algebras in a different way.

A graded algebra A is called quadratic if all defining relations of A generate a subspace R_{A} in A_{1} ⊗ A_{1}. Its dual algebra is then defined as A^{!} := T (A_{1}^{∗})/⟨R_{A}^{⊥}⟩, where R_{A}^{⊥} is the orthogonal subspace of R_{A} in A_{1}^{∗} ⊗ A_{1}^{∗}. If A is quadratic, then the graded subalgebra on the diagonal ⊕_{n}A_{−n,n}^{#} is isomorphic to A^{!}. Fix a basis {x_{i}} for A_{1} and the dual basis {x^{i}} for A_{1}^{∗} and set

e_{A} := ∑(x^{i})^{∗} ⊗ x_{i}.

Then e_{A}^{2} = 0, which defines the the Koszul complex of right A-modules

⋯ → (A_{n}^{!})^{∗} ⊗ A → ⋯ → (A_{1}^{!})^{∗} ⊗ A → k ⊗ A → k_{A} → 0  (4-4)

with left multiplication by e_{A} as the boundary maps (or called differentials). We say that A is Koszul if the Koszul complex (4-4) is exact. A Koszul algebra may not have finite global dimension. But every free module in the minimal free resolution of the trivial module is finitely generated, whence A^{#} is defined. It is easy to show that a graded algebra A is Koszul if and only if A^{#} = A^{!} (in this case we set deg(x) = n for all x ∈ A_{−n,n}^{#}). For more information on Koszul algebras see [Sm, Sec. 5].

Let σ be a graded endomorphism of a quadratic algebra A. Then σ|_{A_{1}} is an n × n-matrix. The transpose of σ|_{A_{1}} gives arise the dual graded endomorphism σ^{τ} for the dual algebra A^{!}. In the matrix format if σ(x_{j}) = ∑_{i} c_{ij}x_{i} for c_{ij} ∈ k, then σ^{τ} (x^{j}) = ∑_{i} c_{ji}x^{i}. It is routine to check that σ is a graded endomorphism of A, namely R_{A} is σ-invariant subspace of A_{1} ⊗ A_{1}, if and only if σ^{τ} is a graded endomorphism of A^{!}, namely, the orthogonal subspace R_{A}^{⊥} ⊂ A_{1}^{∗} ⊗ A_{1}^{∗} is a σ^{τ} -invariant subspace of (A_{1} ⊗ A_{1})^{∗} ≃ A_{1}^{∗} ⊗ A_{1}^{∗}. This construction shows that there is one-to-one corresponding between graded endomorphisms of A and those of A^{!}. In particular, Aut_{gr}(A)^{op} ≅ Aut_{gr}(A^{!}). If A is Koszul, then Aut_{gr}(A)^{op} ≅ Aut_{gr}(A^{#}) and it is easy to check that \overline{σ} = σ^{τ} (which is implicit in the proof of Theorem 4.3 next).

We now re-prove (4-3) for Koszul algebras, which is also a generalization of corresponding one for Hilbert series.

Theorem 4.3. Let A be a Koszul algebra. Suppose σ is a graded algebra endomorphism of A and σ^{τ} is the dual graded algebra endomorphism of A^{!}. Then

T r_{A}(σ, t) · T r_{A^{!}}(σ^{τ} , −t) = 1.

Proof. Let B = A^{!}. Consider the Koszul complex, which is also a minimal free resolution of k_{A}, P^{·} = {B_{n}^{∗} ⊗A} with the left multiplication e = ∑_{i}(x^{i})^{∗} ⊗x_{i} as the boundary map. We claim that the k-linear maps (σ^{τ} |_{B_{n}})^{∗} ⊗ σ : B_{n}^{∗} ⊗ A → B_{n−1}^{∗} ⊗ A make the following diagram commutative:

⋯ ⟶^{e} B_{n}^{∗} ⊗ A ⟶^{e} B_{n−1}^{∗} ⊗ A ⟶^{e} ⋯ ⟶^{e} k_{A} ⟶ 0
(σ^{τ})^{∗}⊗σ↓  ^{(σ^{τ})^{∗}⊗σ}↓  ↓
⋯ ⟶^{e} B_{n}^{∗} ⊗ A ⟶^{e} B_{n−1}^{∗} ⊗ A ⟶^{e} ⋯ ⟶^{e} k_{A} ⟶ 0

Let σ_{i}^{τ} denote σ^{τ} |_{B_{i}} for all i. For every a ∈ A and y^{∗} ∈ B_{n}^{∗},

((σ_{n−1}^{τ})^{∗} ⊗ σ)e(y^{∗} ⊗ a) = ∑_{i}(σ_{n−1}^{τ})^{∗}(x_{i})^{∗}y^{∗} ⊗ σ(x_{i}a)

= ∑_{i j} c_{ji}(σ_{n−1}^{τ})^{∗}(x_{i})^{∗}y^{∗} ⊗ x_{j}σ(a)

For every b ∈ B_{n−1}, we have

∑_{i} c_{ji}(σ_{n−1}^{τ})^{∗}(x_{i})^{∗}y^{∗}(b) = ∑_{i} c_{ji}(x^{i})^{∗}y^{∗}(σ_{n−1}^{τ}(b))

= ∑_{i} c_{ji}y^{∗}(x^{i}σ_{n−1}^{τ}(b)) = y^{∗}(σ_{1}^{τ} (x^{i})σ_{n−1}^{τ}(b))

= (σ_{n}^{τ})^{∗}y^{∗}(x^{i}b) = (x^{i})^{∗}(σ_{n})^{τ})^{∗}y^{∗}(b).

Hence

((σ_{n−1}^{τ})^{∗} ⊗ σ)e(y^{∗} ⊗ a) = ∑_{j}(x^{i})^{∗}(σ_{n})^{τ})^{∗}y^{∗} ⊗ x_{j}σ(a) = e((σ_{n}^{τ})^{∗} ⊗ σ)(y^{∗} ⊗ a),

which is exactly ((σ_{n−1}^{τ})^{∗} ⊗ σ)e = e((σ_{n}^{τ})^{∗} ⊗ σ). By Lemmas 2.1 and 2.2 we obtain

0 =1 − ∑_{n=0}^{∞}(−1)^{n}T r((σ_{n}^{τ})^{∗} ⊗ σ, t)
=1 − ∑_{n=0}^{∞}(−1)^{n}tr(σ_{n}^{τ})t^{n}T r_{A}(σ, t)
=1 − T r_{A^{!}}(σ^{τ} , −t)T r_{A}(σ, t).

Thus T r_{A}(σ, t)T r_{A^{!}}(σ^{τ} , −t) = 1. □

As an application of Theorem 4.3 we calculate the trace for graded automorphisms of the Sklyanin algebra. For simplicity we assume the base field k is the complex number field ℂ in this portion. For detail information about the Sklyanin algebra see [SSf] and [SSn]. Let α_{1}, α_{2}, α_{3} ∈ ℂ satisfying

α_{1} + α_{2} + α_{3} + α_{1}α_{2}α_{3} = 0

and (α_{1}, α_{2}, α_{3}) ≠ (−1, 1, α_{3}), (α_{1}, −1, 1), (1, α_{2}, −1). The Sklyanin algebra S is the connected graded ℂ-algebra with four generators x_{0}, x_{1}, x_{2}, x_{3} and six relations

x_{0}x_{i} − x_{i}x_{0} = α_{i}(x_{j}x_{k} + x_{k}x_{j}), x_{0}x_{i} + x_{i}x_{0} = x_{j}x_{k} − x_{k}x_{j}

where (i, j, k) is a cyclic permutation of (1, 2, 3). The Sklyanin algebra S is a regular domain of global dimension 4 [SSf, 0.3] and a Koszul algebra with its Koszul dual being Frobenius [SSf, 4.7 and 4.9].

The scalars α_{i} can be parametrized by points on an elliptic curve. Let θ_{00}, θ_{01}, θ_{10}, θ_{11} be Jacobi’s four theta functions [Mu, p.17] associated to the lattice Λ = ℤ ⊕ ℤη, where η ∈ ℂ with Im(η) > 0. Then

α_{1} = [\frac{θ_{11}(τ)θ_{00}(τ)}{θ_{01}(τ)θ_{10}(τ)}]^{2}, α_{2} = −[\frac{θ_{11}(τ)θ_{01}(τ)}{θ_{00}(τ)θ_{10}(τ)}]^{2}, α_{3} = [\frac{θ_{11}(τ)θ_{10}(τ)}{θ_{01}(τ)θ_{00}(τ)}]^{2}

for some τ ∈ ℂ, not of order 4 in E = ℂ/Λ [SSf]. Here we also restrict ourselves to the case when τ has infinite order (or equivalently the Sklyanin algebra S is not PI).

The graded automorphism group of the Sklyanin algebra S was studied in [SSn]. There is an exact sequence

1 ⟶ ℂ^{×} ⟶ Aut_{gr}(S) ⟶ E_{4} ⟶ 1

where the automorphisms corresponding to ℂ^{×} and E_{4} are described next. If r ∈ ℂ^{×}, the graded automorphism g_{r} is defined by g_{r} : x ⟶ r^{n}x for all x ∈ S_{n} − {0}. It is easy to see that ℂ^{×} is in the center of Aut_{gr}(S). For every m, E_{m} denotes the points in E with m-torsion. Then E_{4} = {a + bη | a, b ∈ {0, 1/4, 1/2, 3/4}}. The corresponding graded automorphisms are Φ(ξ) for all ξ ∈ E_{4}. In [SSn, Sec 2], seven automorphisms are listed. For example, the automorphism Φ(\frac{1}{4}) is determined by

Φ(\frac{1}{4})(x_{0}, x_{1}, x_{2}, x_{3}) = (i\frac{θ_{11}(τ)}{θ_{10}(τ)}x_{3}, i\frac{θ_{00}(τ)}{θ_{01}(τ)}x_{2}, −i\frac{θ_{01}(τ)}{θ_{00}(τ)}x_{1}, i\frac{θ_{10}(τ)}{θ_{11}(τ)}x_{0}).

Every graded automorphism σ of S has a form g_{r}Φ(ξ)(= Φ(ξ)g_{r}) for r ∈ ℂ^{×} and ξ ∈ E_{4}. Hence

T r(σ, t) = T r(Φ(ξ), rt).

For each ξ ∈ E_{4}, T r(Φ(ξ), t) can be computed explicitly. By Theorem 4.3, we only need to compute T r_{S^{!}}(Φ(ξ)^{τ} , t) where Φ(ξ)^{τ} is the dual graded algebra automorphism of the dual algebra S^{!}. Let us first recall the Koszul dual algebra of S [SSf, Sec. 4]. The dual algebra S^{!} = k ⊕ S_{1}^{!} ⊕ S_{2}^{!} ⊕ S_{3}^{!} ⊕ S_{4}^{!} is spanned by the following elements:

S_{0}^{!} : 1
S_{1}^{!} : x^{0}, x^{1}, x^{2}, x^{3}
S_{2}^{!} : x^{0}x^{1}, x^{0}x^{2}, x^{0}x^{3}, x^{1}x^{0}, x^{2}x^{0}, x^{3}x^{0}
S_{3}^{!} : x^{0}x^{1}x^{0}, x^{0}x^{2}x^{0}, x^{0}x^{3}x^{0}, x^{1}x^{0}x^{1}
S_{4}^{!} : x^{0}x^{1}x^{0}x^{1}

with the following relations:

(x^{0})^{2} = (x^{1})^{2} = (x^{2})^{2} = (x^{3})^{2} = 0
(α_{i} + 1)x^{0}x^{i} − (α_{i} − 1)x^{i}x^{0} + 2x^{j}x^{k} = 0
(α_{i} − 1)x^{0}x^{i} − (α_{i} + 1)x^{i}x^{0} + 2x^{k}x^{j} = 0

where (i, j, k) is a cyclic permutation of (1, 2, 3). For later use we also list the consequent relations:

x^{0}x^{i}x^{0}x^{j} = −δ_{ij}x^{i}x^{0}x^{j}x^{0}, ∀ i, j
x^{0}x^{3}x^{0}x^{3} = \frac{1 + α_{1}}{1 − α_{3}} x^{0}x^{1}x^{0}x^{1},
x^{0}x^{2}x^{0}x^{2} = \frac{(1 + α_{1})(1 + α_{3})}{(1 − α_{3})(1 − α_{2})}x^{0}x^{1}x^{0}x^{1}.

It is easy to check that tr(Φ(ξ)^{τ} |_{S_{1}^{!}}) = 0 for all ξ ≠ 0. Since Φ : E_{4} ⟶ Aut_{gr}(S)/ℂ^{×} is a group homomorphism, Φ : E_{4} ⟶ Aut_{gr}(S^{!})/ℂ^{×} is a group homomorphism. Hence by Corollary 3.5, tr(Φ(ξ)^{τ} |_{S_{3}^{!}}) = tr(rΦ(−ξ)^{τ} |_{S_{1}^{!}}) = 0 where (Φ(ξ)^{τ})^{−1} = g_{r}Φ(−ξ)^{τ}. Therefore T r(Φ(ξ)^{τ} , t) = 1 + a_{2}t^{2} + a_{4}t^{4} for all ξ ≠ 0.

Proposition 4.4. Let σ be a graded automorphism of the Sklyanin algebra S. Then

T r(σ, t) = \{ \begin{matrix}(1 − rt)^{−4} & if & σ = g_{r} \\ (1 + a_{2}t^{2} + a_{4}t^{4})^{−1} & otherwise. \end{matrix}

□

Next we give the details for T r(Φ(\frac{1}{4}), t). By definition (and using the transpose of the matrix), we have

(Φ(\frac{1}{4}))^{τ} (x^{0}, x^{1}, x^{2}, x^{3}) = (i\frac{θ_{10}}{θ_{11}} x^{3}, −i\frac{θ_{01}}{θ_{00}} x^{2}, i\frac{θ_{00}}{θ_{01}} x^{1}, i\frac{θ_{11}}{θ_{10}} x_{0})

18 NAIHUAN JING AND JAMES J. ZHANG

where θ_{ij} = θ_{ij}(τ). Then

(Φ(\frac{1}{4}))^{τ} (x^{0}x^{1}, x^{0}x^{2}, x^{0}x^{3}, x^{1}x^{0}, x^{2}x^{0}, x^{3}x^{0})
= (\frac{θ_{10}θ_{01}}{θ_{11}θ_{00}} (\frac{α_{1} + 1}{2}x^{1}x^{0} − \frac{α_{1} − 1}{2}x^{0}x^{1}), −\frac{θ_{10}θ_{00}}{θ_{11}θ_{01}} (\frac{α_{2} − 1}{2}x^{2}x^{0} − \frac{α_{2} + 1}{2}x^{0}x^{2}), −x^{3}x^{0},

(\frac{θ_{01}θ_{10}}{θ_{00}θ_{11}} (\frac{α_{1} − 1}{2}x^{1}x^{0} − \frac{α_{1} + 1}{2}x^{0}x^{1}), −\frac{θ_{00}θ_{10}}{θ_{01}θ_{11}} (\frac{α_{2} + 1}{2}x^{2}x^{0} − \frac{α_{2} − 1}{2}x^{0}x^{2}), −x^{0}x^{3}),

and

tr(Φ(\frac{1}{4})^{τ} |_{S_{2}^{!}}) = −\frac{θ_{10}θ_{01}}{θ_{11}θ_{00}} \frac{α_{1} − 1}{2} + \frac{θ_{10}θ_{00}}{θ_{11}θ_{01}} \frac{α_{2} + 1}{2} + \frac{θ_{10}θ_{01}}{θ_{11}θ_{00}} \frac{α_{1} − 1}{2} − \frac{θ_{10}θ_{00}}{θ_{11}θ_{01}} \frac{α_{2} + 1}{2} = 0.

Similarly,

(Φ(\frac{1}{4}))^{τ} (x^{0}x^{1}x^{0}x^{1}) = (\frac{θ_{10}θ_{01}}{θ_{11}θ_{00}} )^{2}x^{3}x^{2}x^{3}x^{2} = \frac{1}{4α_{1}} ((α_{1} − 1)^{2} − (α_{1} + 1)^{2})x^{0}x^{1}x^{0}x^{1}

and

tr(Φ(\frac{1}{4})^{τ} |_{S_{4}^{!}}) = \frac{1}{4α_{1}} ((α_{1} − 1)^{2} − (α_{1} + 1)^{2}) = −1.

Therefore

T r(Φ(\frac{1}{4}), t) = \frac{1}{T r(Φ(\frac{1}{4})^{τ} , −t)} = \frac{1}{1 − t^{4}} .

We list the traces for the seven automorphisms given in [SSn, Sec. 2].

T r(Φ(0), t) = \frac{1}{(1 − t)^{4}} ,
T r(Φ(\frac{1}{2}), t) = T r(Φ(\frac{η}{2} ), t) = T r(Φ(\frac{1}{2} + \frac{η}{2} ), t) = \frac{1}{(1 − t^{2})^{2}} ,
T r(Φ(\frac{1}{4}), t) = T r(Φ(\frac{η}{4} ), t) = \frac{1}{1 − t^{4}} ,
T r(Φ(\frac{1}{4} + \frac{η}{4} ), t) = \frac{1}{1 + t^{4}} .

For other ξ ∈ E_{4}, T r(Φ(ξ), t) also has a simple form. It is routine to check that

T r(Φ(ξ_{1} + ξ_{2}), t) := T r(Φ(ξ_{1})Φ(ξ_{2}), t) = T r(Φ(ξ_{1}), t)

for all ξ_{1} ∈ \{\frac{1}{4}, \frac{η}{4} , \frac{1}{4} + \frac{η}{4} \} and ξ_{2} ∈ \{\frac{1}{2}, \frac{η}{2} , \frac{1}{2} + \frac{η}{2} \}.

5. Subring of invariants

Let A be a (graded) ring and G a group. A group action of G on A is a group homomorphism η : G ⟶ Aut(A) (or η : G ⟶ Aut_{gr}(A) in graded case). For simplicity, we still use g for the image η(g). For example, we will use T r(g, t) instead of T r(η(g), t). The (sub)ring of invariants is

A^{G} := {a ∈ A | g(a) = a, ∀ g ∈ G}.

Subrings of invariants was studied in many details, for commutative case see the book [Be], and for noncommutative case see the book [Mo]. In this section we will discuss some basic properties of the subring of invariants of noncommutative graded algebras.

From now on we assume G is a finite subgroup of Aut(A). As noticed more than one hundred years ago, there is a projection of A associated to G

π_{G} : a ⟶ \frac{1}{|G|} ∑_{g∈G} g(a).  (5-1)

The part 1 of the following lemma is [Mo, 2.1.1] and part 2 is in [Mo, 1.12.1].

Lemma 5.1. Let A be an algebra, G a finite subgroup of Aut(A), and let π_{G} be defined as (5-1).

1. π_{G} is an A^{G}-bimodule projection from A to A^{G}, namely, A = A^{G} ⊕ C, where C = ker(π_{G}), as A^{G}-bimodule and π_{G}|_{A^{G}} = id|_{A^{G}}.

2.If A is right noetherian, then so is A^{G}. □

If A is finitely generated but not noetherian, A^{G} may not be finitely generated. Let R be the free algebra k⟨x_{1}, x_{2}⟩ and G = {id, g} where g : x_{1} → −x_{1}, x_{2} → x_{2}. By [Co, 6.10.3 and 6.10.4], R^{G} is an infinitely generated free algebra. In this case it is easy to see that R^{G} is the free subalgebra generated by {x_{2}, x_{1}x_{2}^{n}x_{1} | n ≥ 0} and that

R = R^{G} ⨁ ⊕_{n≥0}x_{2}^{n}x_{1}R^{G} = R^{G} ⨁ ⊕_{n≥0}R^{G}x_{1}x_{2}^{n}.

Of course the idea of T. Molien (see [Be, Sec. 2.5]) works well for noncommutative algebras. Suppose now G is a subgroup of Aut_{gr}(A) (although in most cases, it is not necessary to assume that G acts on A faithfully). Then A^{G} is also graded. By Lemma 5.1, π_{G} is a projection on A with image A^{G}. Hence

H_{A^{G}}(t) = T r_{A}(π_{G}, t) = T r_{A}(\frac{1}{|G|} ∑_{g∈G} g, t) = \frac{1}{|G|} ∑_{g∈G} T r_{A}(g, t).  (5-2)

More generally, let H be the subgroup of Aut_{gr}(A) defined by

H := {h ∈ Aut_{gr}(A) | h(A^{G}) = A^{G}}.

Then G is a normal subgroup of H. The fact group H/G is a subgroup of Aut_{gr}(A^{G}).

Lemma 5.2. Let A, G and H be as above. For every h ∈ H, as a graded automorphism of A^{G},

T r_{A^{G}}(h, t) = \frac{1}{|G|} ∑_{g∈G} T r_{A}(hg, t).

Proof. Define π_{h} to be hπ_{G} = \frac{1}{|G|} ∑_{g∈G} hg. By Lemma 5.1, π_{h}|_{C} = 0 and π_{h}|_{A^{G}} = h|_{A^{G}}. Hence

T r_{A^{G}}(h, t) = T r_{A}(π_{h}, t) = \frac{1}{|G|} ∑_{g∈G} T r_{A}(hg, t).

□

A noncommutative version of Molien’s theorem [Be, 2.5.2] follows easily.

Theorem 5.3. Let A be a graded algebra such that k_{A} has an FMFR. Suppose G is a finite subgroup of Aut_{gr}(A) and H = {h ∈ Aut_{gr}(A) | h(A^{G}) = A^{G}}. Then, for every h ∈ H viewed as in Aut_{gr}(A^{G}), T r_{A^{G}}(h, t) is a rational function.

Proof. By Theorem 2.3.3, T r_{A}(g, t) is a rational function for all g ∈ Aut_{gr}(A). By Lemma 5.2, T r_{A^{G}}(h, t) is a rational function. □

Using Lemma 5.2, we can derive a formula for the Hilbert series of Veronese subrings.

Example 5.4. Veronese subrings of a graded algebra.

Let A be a graded algebra and n a positive integer. The n-th Veronese subring A^{(n)} is ⊕_{i}A_{in}. There are two obvious gradings, one is deg(x) = i and the other is deg(x) = in for all x ∈ A_{in} − {0}. In this section we consider A^{(n)} as a graded subring of A. The Veronese subrings can be viewed as subrings of invariants as follows. Let r ∈ k^{×} := k − {0}, define g_{r} : x ⟼ r^{n}x for all x ∈ A_{n} − {0}. Then g_{r} is a graded automorphism of A. In fact, r ⟼ g_{r} is a group homomorphism from k^{×} to Aut_{gr}(A) with image in the center of Aut_{gr}(A). If ζ_{n} ∈ k is a n-th primitive root of 1, then the subring of invariants of A under the group action G = {g_{ζ_{n}}^{i} | i = 0, ⋯ n − 1} is the Veronese subring A^{(n)}. Therefore

H_{A^{(n)}}(t) = \frac{1}{n} ∑_{i=0}^{n−1} T r_{A}(g_{ζ_{n}}^{i}, t) = \frac{1}{n} ∑_{i=0}^{n−1} H_{A}(ζ_{n}^{i}t).  (5-3)

Note that the above formula (the left end equals the right end) can be proved easily for any field k, without using group action and the trace function, where ζ_{n} is taken to be a n-th root of 1 in ℂ. Let h be any graded automorphism of A. Then h is also a graded automorphism of A^{(n)}. By Lemma 5.2, we have

T r_{A^{(n)}}(h, t) = \frac{1}{n} ∑_{i=0}^{n−1} T r_{A}(hg_{ζ_{n}}^{i}, t) = \frac{1}{n} ∑_{i=0}^{n−1} T r_{A}(h, ζ_{n}^{i}t).

□

In the rest of this section we consider Frobenius algebras.

Proposition 5.5. Let A be a graded finite dimensional Frobenius algebra with highest degree n and G a finite subgroup of Aut_{gr}(A). If g|_{A_{n}} = id_{A_{n}} for all g ∈ G, then A^{G} is Frobenius.

Proof. Let A^{G} = ⊕_{i=0}^{n}B_{i}. By assumption, B_{n} = A_{n} ≅ k. For every nonzero element x ∈ B_{i}, xA_{n−i} = A_{n−i}x = A_{n} by the Frobenius property of A. Applying π_{G}, we have xB_{n−i} = B_{n−i}x = B_{n}. Hence A^{G} is Frobenius. □

Without the condition g|_{A_{n}} = id_{A_{n}}, A^{G} may not be Frobenius. First such counterexample was given in [PV]. Here is an easy way to product counterexamples. Let A be a finite dimensional Frobenius algebra with highest degree n. Let m be a integer such that (i) n = qm + i with 0 < i < m and dim(A_{qm}) > 1. Then by [Sm, 3.2], A^{(m)} is not Frobenius. If A is not isomorphic to k[x]/(x^{s}), then such m always exists. Also note that if m > n, then A^{(m)} = k is trivially Frobenius.

6. Gorenstein property

A variation of Watanabe’s theorem [Be, 5.3.2] states that if k[V] is the polynomial algebra over a finite dimensional vector space V and Λ(V^{∗}) is the exterior algebra, and G is a finite subgroup of GL(V), then the subring of invariants (k[V] ⊗ Λ(V^{∗}))^{G} is Gorenstein. Note that k[V] is a noetherian Koszul regular algebra and Λ(V^{∗}) is the Koszul dual of k[V]. Let A be a regular Koszul algebra and σ a graded automorphism of A. Then σ ⊗ σ^{τ} is a graded algebra automorphism of A ⊗ A^{!}. In this way we consider Aut_{gr}(A) as a subgroup of Aut_{gr}(A ⊗ A^{!}). Naturally we make the following conjecture.

Conjecture. Let A be a (non-commutative) noetherian Koszul regular algebra and G a finite subgroup of Aut_{gr}(A). Is then (A ⊗ A^{!})^{G} Gorenstein?

In this section we will study graded rings A⊗A^{!} and (A⊗A^{!})^{G} and prove the above conjecture for some special groups G.

Let A be a regular Koszul algebra of global dimension d. Then k has a minimal free resolution of length d such that the i-th free module in the free resolution is A(−i)^{⊕m_{i}} where m_{0} = m_{d} = 1 and m_{i} = m_{d−i} for all i. In particular, \underline{Ext}^{d}(k, A) ≅ \underline{Ext}^{d}(k, k) ≅ k(d). The Koszul dual of A is Frobenius [Sm, 5.10] with highest degree d.

Proposition 6.1. Let A be a regular Koszul algebra with its Koszul dual A^{!} and σ a graded automorphism of A.

1. T r_{A}(σ^{−1}, t^{−1}) = at^{l}T r_{A}(σ, t) for some a ∈ k^{×}, where −l is the degree of \underline{Ext}^{d}(k_{A}, A) (equal to the highest degree of A^{!}).

2. T r_{A⊗A^{!}}(σ ⊗ σ^{τ} , t) = T r_{A^{!}}(σ^{τ} , t)/T r_{A^{!}}(σ^{τ} , −t).

3. T r_{A⊗A^{!}}((σ ⊗ σ^{τ})^{−1}, t^{−1}) = (−1)^{l}T r_{A⊗A^{!}}(σ ⊗ σ^{τ} , t).

4. H_{A⊗A^{!}}(t^{−1}) = (−1)^{l}H_{A⊗A^{!}}(t).

5. Let G be a finite subgroup of Aut_{gr}(A) acting naturally on A ⊗ A^{!}, and (A ⊗ A^{!})^{G} the subring of invariants. Then H_{(A⊗A^{!})^{G}}(t^{−1}) = (−1)^{l}H_{(A⊗A^{!})^{G}}(t).

As in the commutative case, part 5 is a supporting evidence for the conjecture.

Proof. 1. By Theorem 4.3, it is equivalent to show that T r_{A^{!}}((σ^{τ})^{−1}, t^{−1}) = bt^{−l}T r_{A^{!}}(σ^{τ} , t), which follows from Theorem 3.4, because A^{!} is Frobenius [Sm, 5.10].

Part 2 follows from (2-4) and Theorem 4.3, part 3 follows from part 2 and Theorem 3.4, and part 4 is a special case of part 3.

5. By (5-2) and part 3, we have

H_{(A⊗A^{!})^{G}}(t^{−1}) = \frac{1}{|G|} ∑_{σ∈G} T r_{A⊗A^{!}}(σ ⊗ σ^{τ} , t^{−1}) = \frac{1}{|G|} ∑_{σ∈G} T r_{A⊗A^{!}}((σ ⊗ σ^{τ})^{−1}, t^{−1})

= \frac{(−1)^{l}}{|G|} ∑_{σ∈G} T r_{A⊗A^{!}}(σ ⊗ σ^{τ} , t) = (−1)^{l}H_{(A⊗A^{!})^{G}}(t).

□

It is relatively easy to show that A ⊗ A^{!} is Gorenstein. We have the following general fact.

Lemma 6.2. Let A be a graded algebra and B a graded Frobenius algebra.

1. The (right) injective dimensions of A ⊗ B and A are the same.

2. A ⊗ B is Gorenstein if and only if A is.

Proof. 1. Since k has global dimension zero and B is a finite dimensional algebra of injective dimension zero, by the Künneth formula for double complexes we have

Ext_{A⊗B}^{i}(M ⊗ N, A ⊗ B) ≅ Ext_{A}^{i}(M, A) ⊗ Hom_{B}(N, B)  (6-1)

for every right (or left) A-module M and every finite dimensional right (or left) B-module N. Since Hom_{B}(N, B) ≠ 0, (6-1) implies that injdim(A) ≤ injdim(A ⊗ B). It remians to show that

Ext_{A⊗B}^{i}(L, A ⊗ B) = 0  (6-2)

for all A ⊗ B-module L, and for all i > injdim(A). If L = M ⊗ N, it follows from (6-1). In general we consider a chain of submodules,

L ⊃ LJ ⊃ ⋯ ⊃ LJ^{l} ⊃ LJ^{l+1} = 0

where J = A ⊗ B_{≥1} and l is the largest degree of B. Hence it suffices to show (6-2) for all L^{s} := LJ^{s}/LJ^{s+1}. By definition, L^{s}J = 0, and it implies that L_{A⊗B}^{s} ≅ L_{A}^{s} ⊗ k_{B}. Hence (6-2) follows from (6-1). Thus injdim(A) = injdim(A ⊗ B).

2. A graded version of (6-1) holds and let M = k_{A} and N = k_{B}, we have

\underline{Ext}_{A⊗B}^{i}(k_{A⊗B}, A ⊗ B) ≅ \underline{Ext}_{A}^{i}(k_{A}, A) ⊗ \underline{Hom}_{B}(k_{B}, B) ≅ \underline{Ext}_{A}^{i}(k_{A}, A) ⊗ k(−l)  (6-3)

where l is the largest degree of B. Hence A is Gorenstein if and only if A ⊗ B is. □

If A is a regular Koszul algebra of global dimension d. Then \underline{Ext}_{A}^{d}(k_{A}, A) = k(d) and \underline{Hom}_{A^{!}}(k_{A^{!}}, A^{!}) = k(−d). Hence by Lemma 6.2 and (6-3), we obtain the following.

Corollary 6.3. Let A be a regular Koszul algebra of global dimension d.

1. A ⊗ A^{!} is a Gorenstein algebra of injective dimension d.

2. \underline{Ext}_{A⊗A^{!}}^{d}(k_{A⊗A^{!}}, A ⊗ A^{!}) = k. □

Next we consider the Veronese subrings A^{(m)} of a Gorenstein ring A. We give a necessary and sufficient condition for A^{(m)} being Gorenstein [Proposition 6.5].

Lemma 6.4. Let A be a graded noetherian algebra generated in degree 1 and m a positive integer.

1. A^{(m)} is a graded noetherian algebra generated in degree 1 and GKdim(A^{(m)}) = GKdim(A).

2. If I is a graded injective module of A contains no finite dimensional submodules, then I^{(m)} is a graded injective module over A^{(m)}.

Proof. 1. By [AZ, 5.10], A^{(m)} is a noetherian algebra generated in degree 1 and A is a finitely generated A^{(m)}-module. By [MR, 8.2.9.(ii)], GKdim(A) = GKdim(A^{(m)}).

2. A graded module M is called torsion if it is a union of finite dimensional submodules and M is called torsion-free if it contains no non-trivial finite dimensional submodules. We denote by Tors the subcategory of Gr– A of all torsion modules. Let QGr– A be the quotient category Gr– A/ Tors. Let π : Gr– A → QGr– A be the canonical functor let ω : QGr– A → Gr– A be a right adjoint to π (see [AZ, (2.2.2)]). Similarly we have an adjoint pair π^{(m)} and ω^{(m)} between Gr– A^{(m)} an QGr– A^{(m)}. Consider the functor (−)^{(m)} : Gr– A → Gr– A^{(m)} defined by taking the m-th Veronese of graded modules, it induces a functor, also denoted by (−)^{(m)} : QGr– A → QGr– A^{(m)} by sending π(M) to π^{(m)}(M^{(m)}). Hence we have an commutative diagram

Gr– A ⟶^{(−)^{(m)}} Gr– A^{(m)}
π↓  ↓_{π^{(m)}}
QGr– A ⟶^{(−)^{(m)}} QGr– A^{(m)}

By [AZ, 5.10(3)] the bottom functor π(M) → π^{(m)}(M^{(m)}) is a category equivalence, whence it sends injectives to injectives. By definition, it sends the shift π(M(im)) to π^{(m)}(M^{(m)}(i)).

Now let I be a graded injective A-module containing no finite dimensional submodules. Hence I is torsion-free and by [AZ, 7.1(3)], I = ω(π(I)). By [Po, 4.5.4] π(I) is injective and hence π^{(m)}(I^{(m)}) is injective in QGr– A^{(m)}. Thus, by [Po, 4.5.3(2)] and [AZ, 7.1.(3)] ω^{(m)}π^{(m)}(I^{(m)}) is an injective A^{(m)}-module. By [AZ, (2.2.2)], we have

ω^{(m)}π^{(m)}(I^{(m)}) = \underline{Hom}(A^{(m)}, ω^{(m)}π^{(m)}(I^{(m)}))
= ⊕_{i∈ℤ} Hom(A^{(m)}, ω^{(m)}π^{(m)}(I^{(m)})(i))
= ⊕_{i∈ℤ} Hom(π^{(m)}(A^{(m)}), π^{(m)}(I^{(m)})(i)) [AZ, (2.2.2)]
= ⊕_{i∈ℤ} Hom(π(A), π(I)(im))
= (⊕_{i∈ℤ} Hom(A, I(i)))^{(m)} [AZ, (2.2.2)]
= I^{(m)}

Therefore I^{(m)} is an injective A^{(m)}-module. □

Proposition 6.5. Let A be a graded noetherian Gorenstein algebra, generated in degree 1, of injective dimension d. Suppose \underline{Ext}^{d}(k_{A}, A) = k(l) and let m be a positive integer.

1. If m|l, then the Veronese subring A^{(m)} is a Gorenstein algebra of injective dimension d.

2. Suppose m |̸l. If either GKdim(A) > 1, or dim(A_{n}) > 1 for all n > 0, then A^{(m)} has infinite injective dimension.

Proof. 1. Consider the minimal graded injective resolution of A_{A}:

0 → A_{A} → I^{0} → I^{1} → ⋯ → I^{d−1} → I^{d} → 0.

Since \underline{Ext}^{i}(k, A) = 0 for all i < d, I^{i} (for i < d) contains no finite dimensional submodules. By [Zh, 1.2.3] I^{d} ≅ A^{∗}(l). Take the m-th Veronese, we have

0 → A^{(m)} → (I^{0})^{(m)} → (I^{1})^{(m)} → ⋯ → (I^{d−1})^{(m)} → (A^{∗}(l))^{(m)} → 0.  (6-4)

By Lemma 6.4.2, every (I^{i})^{(m)} (for i < d) is injective A^{(m)}-module and obviously contains no finite dimensional submodules. In particular, (6-4) is a part of an injective resolution and hence \underline{Ext}_{A^{(m)}}^{i}(k, A^{(m)}) = 0 for all i < d.

1. If m|l, (A^{∗}(l))^{(m)} = (A^{(m)})^{∗}(l/m), which is an injective A^{(m)}-module. Hence (6-4) is an injective resolution of A^{(m)} and the injective dimension of A^{(m)} is no more than d. Applying \underline{Hom}_{A^{(m)}}(k_{A^{(m)}}, −) to (6-4), we obtain that \underline{Ext}_{A^{(m)}}^{i}(k, A^{(m)}) = 0 for all i ≠ d and \underline{Ext}_{A^{(m)}}^{d}(k, A^{(m)}) = k(l/m). Therefore A^{(m)} is a Gorenstein algebra of injective dimension d.

2. Suppose injdim(A^{(m)}) < ∞ and m does not divide l. We will prove that GKdim(A) ≤ 1 and dim(A_{s}) = 1 for some s > 0. By (6-4), \underline{Ext}_{A^{(m)}}^{i}(k, A^{(m)}) = 0 for all i < d and \underline{Ext}_{A^{(m)}}^{i}(k, A^{(m)}) = \underline{Ext}_{A^{(m)}}^{i−d}(k, (A^{∗}(l))^{(m)}) for all i ≥ d. Since (A(l))^{(m)} is locally finite and left bounded, by the k-linear graded duality, \underline{Ext}_{A^{(m)}}^{i−d}(k, (A^{∗}(l))^{(m)}) = \underline{Ext}_{A^{(m)}}^{i−d}((A(−l))^{(m)}, k) for all i ≥ d. Hence \underline{Ext}_{A^{(m)}}^{i}(k, (A^{∗}(l))^{(m)}) is finite dimensional for all i. By [Zh, 1.2.1], A^{(m)} is Gorenstein. We may assume that A is infinite dimensional, then A^{(m)} is not zero and \underline{Ext}_{A^{(m)}}^{d}(k, (A^{∗}(l))^{(m)}) = \underline{Hom}_{A^{(m)}}((A(l))^{(m)}, k) ≠ 0. Hence the injective dimension of A^{(m)} is d and (A^{∗}(l))^{(m)}) = (A^{(m)})^{∗}(l′). Since m does not divide l, l ≡ s (mod m) for some m > s > 0 and hence A_{s} ≅ k. By Lemma 6.4.1, A^{(s)} is generated in degree 1, whence GKdim(A^{(s)}) ≤ 1. By Lemma 6.4.1, GKdim(A) = GKdim(A^{(s)}) ≤ 1. □

Theorem 6.6. Let A be a noetherian regular Koszul algebra and G a finite subgroup of k^{×} ⊂ Aut_{gr}(A). Then (A ⊗ A^{!})^{G} is Gorenstein.

Proof. If G is a finite subgroup of k^{×}, then G is cyclic and generated by a m-th primitive root ζ_{m} of 1 and hence (A ⊗ A^{!})^{G} = (A ⊗ A^{!})^{(m)}. By Corollary 6.3, A ⊗ A^{!} is Gorenstein and \underline{Ext}_{A⊗A^{!}}^{d}(k_{A⊗A^{!}}, A ⊗ A^{!}) = k(0). Of course m|0 for all m, hence the statement follows from Proposition 6.5.1. □

Acknowledgment

We would like to thank S. P. Smith and R. Irving for useful conversation on the subject.

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26 NAIHUAN JING AND JAMES J. ZHANG

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(N. Jing) Department of Mathematics, North Carolina State University, Raleigh, NC 27695 E-mail: jing@eos.ncsu.edu (J. J. Zhang) Department of Mathematics, Box 354350, University of Washington, Seattle, WA 98195 E-mail: zhang@math.washington.edu