## GK dimension; definition and basic remarks

Posted: April 26, 2011 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
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Throughout $k$ is a field. Let $A$ be a $k$-algebra. Let $V$ be a $k$-vector subspace of $A$ spanned by the set $\{a_1, \ldots , a_m \}.$ For any integer $n \geq 1$ we will denote by $V^n$ the the $k$-subspace of $A$ spanned by all monomials of length $n$ in $a_1, \ldots , a_m.$ We will also define $V^0=k.$ If $A=k[a_1, \ldots , a_m],$ then $V$ is called a generating subspace of $A.$ If a generating subspace $V$ of $A$ contains 1, then $V$ is called a frame of $A.$ We will denote by $V_n$ the subspace $\sum_{i=0}^n V^i$ of $A.$ Clearly if $V$ is a generating subspace of $A,$ then $A=\bigcup_{n=0}^{\infty}V_n.$

Remark 1. Let $A$ be a finitely generated $k$-algebra and suppose that $V$ and $W$ are two generating subspaces of $A.$ Then $\displaystyle \limsup_{n\to\infty} \log_n (\dim V_n) = \limsup_{n\to\infty} \log_n (\dim W_n).$

Proof. We have $A=\bigcup_{n=0}^{\infty}V_n=\bigcup_{n=0}^{\infty}W_n.$ Since both $V$ and $W$ are finite dimensional, there exist integers $r \geq 1$ and $s \geq 1$ such that $V \subseteq W_r$ and $W \subseteq V_s.$ Thus $V_n \subseteq W_{rn}$ and $W_n \subseteq V_{sn}$ for all integers $n \geq 0.$ Now, $\dim V_n \leq \dim W_{rn}$ implies that

$\log_n (\dim V_n) \leq \log_n(\dim W_{rn})=(1+\log_n r) \log_{rn}(\dim W_{rn}).$

Taking limsup of both sides of the above inequality will give us

$\displaystyle \limsup_{n\to\infty} \log_n(\dim V_n) \leq \limsup_{n\to\infty} \log_n(\dim W_n),$

because $\displaystyle \lim_{n\to\infty} (1+ \log_n r)=1$ and

$\displaystyle \limsup_{n\to\infty} \log_{rn}(\dim W_{rn}) \leq \limsup_{n\to\infty} \log_n (\dim W_n).$

Similarly $\dim W_n \leq \dim V_{sn}$ will imply

$\displaystyle \limsup_{n\to\infty} \log_n(\dim W_n) \leq \limsup_{n\to\infty} \log_n(\dim V_n),$

which completes the proof. $\Box$

So, by the above remark, the value $\displaystyle \limsup_{n\to\infty} \log_n (\dim V_n)$ does not depend on the generaing subspace $V$ and thus the following definition makes sense.

Definition. Let $A$ be a finitely generated $k$-algebra and let $V$ be a generating subspace of $A.$ The Gelfand-Kirillov dimension, or GK dimension, of $A,$ which we will  denote it by ${\rm{GKdim}}(A),$ is defined by

$\displaystyle {\rm{GKdim}}(A)=\limsup_{n\to\infty} \log_n (\dim V_n).$

The GK dimension of a algebra somehow measures how far an algebra is from being finite dimensional. We will see later that if $A$ is commutative, then the GK dimension of $A$ is nothing but the transcendence degree of $A.$

Remark 2. In the above definition if $V$ is a frame of $A,$ then $V_n=\sum_{i=0}^n V^i = V^n,$ because

$k=V^0 \subseteq V \subseteq \ldots \subseteq V^n.$

Thus we get this simple expression $\displaystyle {\rm{GKdim}}(A) = \limsup_{n\to\infty} \log_n (\dim V^n).$

1. Alex Youcis says:

Hey Yaghoub. Maybe it’s because I haven’t read enough of the after-this GK dimension posts, but what precisely is the point of it? I mean, I get roughly what it measures, but is there a typical place that GK dimension would come up?

• Yaghoub says:

Yes, GK dimension is a useful and important invariant of algebras, although it’s a relatively new subject in ring theory.
For example, if k is an algebraically closed filed and A is a finitely generated k-algebra as well as a domain, then GKdim(A) = 1 implies that A is commutative. This is an important result.
For commutative algebras GK dimension is not very interesting because it coincides with Krul dimension, which is a much more known invariant.
.

• Alex Youcis says:

Thank you! Do you do work with GK dimension?

• Yaghoub says:

Not directly. I’m working on the ring structure of the centralizer of an element in an algebra and GK dimension is a useful tool in my research.