## 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. Then$V+W$ 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 ${\rm{GKdim}}(A)=m.$