GK dimension; some basic facts (1)

Posted: April 26, 2011 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
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So far we have only defined the GK dimension for finitely generated algebras. We now extend the definition to all algebras.

Definition. Let A be a k-algebra. We define \displaystyle {\rm{GKdim}}(A)=\sup_B {\rm{GKdim}}(B), where \sup runs over all finitely generated k-subalgebras B of A.

Fact 1. Let A be a k-algebra, B a k-subalgebra of A and let I be a two-sided ideal of A. Then {\rm{GKdim}}(A) \geq \max \{{\rm{GKdim}}(B),{\rm{GKdim}}(A/I)\}.

Proof. Let A_0 and B_0 be any finitely generated subalgebras of A and B, respectively. Let V and W be frame of A_0 and B_0, respectively, which both contain 1. ThenV+W is also a frame of $A_0$ and so we may assume that W \subseteq V. Thus W^n \subseteq V^n, for all n, and so {\rm{GKdim}}(A_0) \geq {\rm{GKdim}}(B_0). Fixing A_0 and taking supremum over all finitely generated subalgebras B_0 of B gives us {\rm{GKdim}}(A_0) \geq {\rm{GKdim}}(B). Now taking supremum over all finitely generated subalgebras A_0 of A gives us {\rm{GKdim}}(A) \geq {\rm{GKdim}}(B). For the second inequality, let A_1 be a finitely generated subalgebra of A/I. Let \pi: A \longrightarrow A/I be the natural homomorphism. Then A_0=\pi^{-1}(A_1) is a finitely generated subalgebra of A. Let W be a frame of A_1. Then V = \pi^{-1}(W) is a frame of A_0 and clearly \dim_k V^n \geq \dim_k W^n, for all n.  Thus

{\rm{GKdim}}(A) \geq {\rm{GKdim}}(A_0) \geq {\rm{GKdim}}(A_1).

Taking supremum over all finitely generated subalgebras A_1 gives us {\rm{GKdim}}(A) \geq {\rm{GKdim}}(A/I). \Box

Fact 2. Let A be a k-algebra. Then {\rm{GKdim}}(A)=0 if and only if A is locally finite, i.e. every finitely generated subalgebra of A is finite dimensional, as a vector space, over k.

Proof. Suppose first that A is locally finite and let B be a finitely generated subalgebra of A. Then B is finite dimensional over k and so V=A is a frame of B. Clearly V^n=B and thus

\displaystyle {\rm{GKdim}}(B)=\limsup_{n\to\infty} \log_n (\dim B)=0,

because \dim B does not depend on n. Conversely, suppose that {\rm{GKdim}}(A)=0 and let B be a finitely generated subalgebra of A. Let V be a frame of B and suppose for now that V^n \subset V^{n+1} for all n. Then k \subset V \subset V^2 \subset \ldots and thus 1 < \dim V < \dim V^2 < \ldots. Hence \dim V > 1, \ \dim V^2 > 2 and in general \dim V^n > n. Therefore

0={\rm{GKdim}}(B)= \limsup_{n\to\infty} \log_n(\dim V^n) \geq \lim_{n\to\infty} \log_n n = 1,

which is absurd. So our assumption that V^n \subset V^{n+1}, for all n, is false. Hence V^n=V^{n+1}, for some integer n \geq 0 and so  B=\bigcup_{i=0}^{\infty} V^i=V^n. Thus B is finite dimensional and so A is locally finite. \Box

Fact 3. Let A be a k-algebra. If {\rm{GKdim}}(A) \neq 0, then {\rm{GKdim}}(A) \geq 1.

Proof. So there exists a finitely generated k-subalgebra B of A such that {\rm{GKdim}}(B) \neq 0.  Let V be a frame of B. If V^n = V^{n+1}, for some integer n \geq 0, then

B=\bigcup_{i=0}^{\infty}V^i=V^n

and so B is finite dimensional. But then {\rm{GKdim}}(B) = 0, by Fact 2, which is false. Thus k \subset V \subset V^2 \subset \ldots and hence \dim V^n > n, for all n \geq 0. Therefore

\displaystyle {\rm{GKdim}}(B)= \limsup_{n\to\infty} \log_n(\dim V^n) \geq 1

and so \displaystyle {\rm{GKdim}}(A)\geq {\rm{GKdim}}(B) \geq 1. \Box

So if 0 < \alpha < 1, then there is no algebra A with {\rm{GKdim}}(A)=\alpha. We will see later that for every integer m \geq 1 there exists an algebra A such that {\rm{GKdim}}(A)=m.

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