## GK dimension of Weyl algebras

Posted: April 10, 2012 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
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We defined the $n$-th Weyl algebra $A_n(R)$ over a ring $R$ in here.  In this post we will find the GK dimension of $A_n(R)$ in terms of the GK dimension of $R.$ The result is similar to what we have already seen in commutative polynomial rings (see corollary 1 in here). We will assume that $k$ is a field and $R$ is a $k$-algebra.

Theorem. ${\rm{GKdim}}(A_1(R))=2 + {\rm{GKdim}}(R).$

Proof. Suppose first that $R$ is finitely generated and let $V$ be a frame of $R.$ Let $U=k+kx+ky.$ Since $yx = xy +1,$ we have

$\dim_k U^n = \frac{(n+1)(n+2)}{2}. \ \ \ \ \ \ \ \ \ (*)$

Let $W=U+V.$ Clearly $W$ is a frame of $A_1(R)$ and

$W^n = \sum_{i+j=n} U^i V^j,$

for all $n,$ because every element of $V$ commutes with every element of $U.$ Therefore, since $V^j \subseteq V^n$ and $U^i \subseteq U^n$ for all $i,j \leq n,$ we have $W^n \subseteq U^nV^n$ and $W^{2n} \supseteq U^nV^n.$ Thus $W^n \subseteq U^nV^n \subseteq W^{2n}$ and hence

$\log_n \dim_k W^n \leq \log_n \dim_k U^n + \log_n \dim_k V^n \leq \log_n \dim_k W^{2n}.$

Therefore ${\rm{GKdim}}(A_1(R)) \leq 2 + {\rm{GKdim}}(R) \leq {\rm{GKdim}}(A_1(R)),$ by $(*),$ and we are done.

For the general case, let $R_0$ be any finitely generated $k$– subalgebra of $R.$ Then, by what we just proved,

$2 + {\rm{GKdim}}(R_0)={\rm{GKdim}}(A_1(R_0)) \leq {\rm{GKdim}}(A_1(R))$

and hence $2+{\rm{GKdim}}(R) \leq {\rm{GKdim}}(A_1(R)).$ Now, let $A_0$ be a $k$-subalgebra of $A_1(R)$ generated by a finite set $\{f_1, \ldots , f_m\}.$ Let $R_0$ be the $k$-subalgebra of $R$ generated by all the coefficients of $f_1, \ldots , f_m.$ Then $A_0 \subseteq A_1(R_0)$ and so

${\rm{GKdim}}(A_0) \leq {\rm{GKdim}}(A_1(R_0))=2 + {\rm{GKdim}}(R_0) \leq 2 + {\rm{GKdim}}(R).$

Thus

${\rm{GKdim}}(A_1(R)) \leq 2 + {\rm{GKdim}}(R)$

and the proof is complete. $\Box$

Corollary. ${\rm{GKdim}}(A_n(R))=2n + {\rm{GKdim}}(R)$ for all $n.$ In particular, ${\rm{GKdim}}(A_n(k))=2n.$

Proof. It follows from the theorem and the fact that $A_n(R)=A_1(A_{n-1}(R)). \Box$

## A theorem of Borho and Kraft

Posted: April 2, 2012 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
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As usual, I’ll assume that $k$ is a field. Recall that if a $k$-algebra $A$ is an Ore domain, then we can localize $A$ at $S:=A \setminus \{0\}$ and get the division algebra $Q(A):=S^{-1}A.$ The algebra $Q(A)$ is called the quotient division algebra of $A.$

Theorem (Borho and Kraft, 1976) Let $A$ be a finitely generated $k$-algebra which is a domain of finite GK dimension. Let $B$ be a $k$-subalgebra of $A$ and suppose that ${\rm{GKdim}}(A) < {\rm{GKdim}}(B) + 1.$ Let $S:=B \setminus \{0\}.$ Then $S$ is an Ore subset of $A$ and $S^{-1}A=Q(A).$ Also, $Q(A)$ is finite dimensional as a (left or right) vector space over $Q(B).$

Proof. First note that, by the corollary in this post, $A$ is an Ore domain and hence both $Q(A)$ and $Q(B)$ exist and they are division algebras. Now, suppose, to the contrary, that $S$ is not (left) Ore. Then there exist $x \in S$ and $y \in A$ such that $Sy \cap Ax = \emptyset.$ This implies that the sum $By + Byx + \ldots + Byx^m$ is direct for any integer $m.$ Let $W$ be a frame of a finitely generated subalgebra $B'$ of $B.$ Let $V=W+kx+ky$ and suppose that $A'$ is the subalgebra of $A$ which is generated by $V.$ For any positive integer $n$ we have

$V^{2n} \supseteq W^n(kx+ky)^n \supseteq W^ny + W^nyx + \ldots + W^nyx^{n-1}$

and thus $\dim_k V^{2n} \geq n \dim_k W^n$ because the sum is direct. So $\log_n \dim_k V^{2n} \geq 1 + \log_n \dim_k W^n$ and hence ${\rm{GKdim}}(A) \geq {\rm{GKdim}}(A') \geq 1 + {\rm{GKdim}}(B').$ Taking supremum of both sides over all finitely generated subalgebras $B'$ of $B$ will give us the contradiction ${\rm{GKdim}}(A) \geq 1 + {\rm{GKdim}}(B).$ A similar argument shows that $S$ is right Ore. So we have proved that $S$ is an Ore subset of $A.$ Before we show that $S^{-1}A=Q(A),$ we will prove that $Q(B)A=S^{-1}A$ is finite dimensional as a (left) vector space over $Q(B).$ So let $V$ be a frame of $A.$ For any positive ineteger $n,$ let $r(n) = \dim_{Q(B)} Q(B)V^n.$ Clearly $Q(B)V^n \subseteq Q(B)V^{n+1}$ for all $n$ and

$\bigcup_{n=0}^{\infty}Q(B)V^n =Q(B)A$

because $\bigcup_{n=0}^{\infty}V^n=A.$ So we have two possibilities: either $Q(B)V^n=Q(B)A$ for some $n$ or the sequence $\{r(n)\}$ is strictly increasing. If $Q(B)V^n = Q(B)A,$ then we are done because $V^n$ is finite dimensional over $k$ and hence $Q(B)V^n$ is finite dimensional over $Q(B).$ Now suppose that the sequence $\{r(n)\}$ is strictly increasing. Then $r(n) > n$ because $r(0)=\dim_{Q(B)}Q(B)=1.$ Fix an integer $n$ and let $e_1, \ldots , e_{r(n)}$ be a $Q(B)$-basis for $Q(B)V^n.$ Clearly we may assume that $e_i \in V^n$ for all $i.$ Let $W$ be a frame of a finitely generated subalgebra of $B.$ Then

$(V+W)^{2n} \supseteq W^nV^n \supseteq W^ne_1 + \ldots + W^ne_{r(n)},$

which gives us

$\dim_k(V+W)^{2n} \geq r(n) \dim_k W^n > n \dim_k W^n,$

because the sum $W^ne_1 + \ldots + W^ne_{r(n)}$ is direct. Therefore ${\rm{GKdim}}(A) \geq 1 + {\rm{GKdim}}(B),$ which is a contradiction. So we have proved that the second possibility is in fact impossible and hence $Q(B)A$ is finite dimensional over $Q(B).$ Finally, since, as we just proved, $\dim_{Q(B)}Q(B)A < \infty,$ the domain $Q(B)A$ is algebraic over $Q(B)$ and thus it is a division algebra. Hence $Q(B)A=Q(A)$ because $A \subseteq Q(B)A \subseteq Q(A)$ and $Q(A)$ is the smallest division algebra containing $A. \Box$

## A theorem of Smith and Zhang

Posted: May 11, 2011 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
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For the definition of a PI-algebra see this post.

Lemma. Let $k$ be a field and let $A$ be a finitely generated $k$-algebra which is a domain. If $A$ is not PI, then ${\rm{GKdim}}(A) \geq 2.$

Proof. If ${\rm{GKdim}}(A)=0,$ then $A$ is finite dimensional over $k,$ and so over its center, and hence it is PI. If ${\rm{GKdim}}(A) = 1,$ then $A$ is again PI (see here). Also, there is no algebra whose GK dimension is strictly between 1 and 2, by the Bergman’s gap theorem. Thus ${\rm{GKdim}}(A) \geq 2. \ \Box$

We are now going to refine the result given in the above lemma. We will write ${\rm{GKdim}}_F$ for the GK dimension of an algebra viewed as an $F$-algebra if $F$ is not $k.$

Theorem. (Smith and Zhang, 1996) Let $k$ be a field and let $A$ be a finitely generated $k$-algebra which is a domain. If $A$ is not PI, then ${\rm{GKdim}}(A) \geq 2 + {\rm{GKdim}}(Z(A)),$ where $Z(A)$ is the center of $A.$

Proof. By the above lemma, we may assume that $2 \leq {\rm{GKdim}}(A) < \infty$ and ${\rm{GKdim}}(Z(A)) \geq 1.$ Let $Q_Z(A)$ be the central localization of $A$ and let $F$ be the center of $Q_Z(A).$ Clearly $F$ is just the quotient field of $Z(A).$ Recall, from the theorem in here, that

${\rm{GKdim}}(Q_Z(A))={\rm{GKdim}}(A) \geq 2$  and ${\rm{GKdim}}(F)={\rm{GKdim}}(Z(A)) \geq 1.$

Let $0 \leq d < {\rm{GKdim}}(F)$ and $0 \leq e < {\rm{GKdim}}_F(Q_Z(A)).$ Then there exist a finite dimensional $k$-vector subspace $V$ of $F$ which contains 1 and $\dim_k V^n \geq n^d$ for all large enough integers $n.$ Also, there exists a finite dimensional $F$-vector subspace $W \supseteq V$ of $Q_Z(A)$ which contains 1 and $\dim_F W^n \geq n^e$ for all large enough integers $n.$ Hence, for large enough integers $n$ we have

$\dim_k W^{2n} \geq \dim_k (W^nV^n) \geq (\dim_F W^n)(\dim_k V^n) \geq n^{e+d}.$

Thus ${\rm{GKdim}}(Q_Z(A)) \geq e+d.$ Since the above inequality holds for all real number $0 \leq d < {\rm{GKdim}}(F)$ and $0 \leq e < {\rm{GKdim}}_F(Q_Z(A)),$ we have

${\rm{GKdim}}(Q_Z(A)) \geq {\rm{GKdim}}_F(Q_Z(A)) + {\rm{GKdim}}(F). \ \ \ \ \ \ \ \ (*)$

Now, since $A$ is finitely generated as a $k$-algebra, $Q_Z(A)$ is a finitely generated $F$-algebra and clearly $Q_Z(A)$ is not PI because $A$ is not PI. Therefore ${\rm{GKdim}}_F(Q_Z(A)) \geq 2$ by the above lemma. We also have ${\rm{GKdim}}(F)={\rm{GKdim}}(Z(A))$ and now the result follows from $(*). \ \Box$

## GK dimension and localization

Posted: April 26, 2011 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
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Theorem. Let $A$ be a $k$-algebra and suppose that $S$ is a regular submonoid of $A$ contained in the center of $A.$ Then ${\rm{GKdim}}(S^{-1}A) = {\rm{GKdim}}(A).$

Proof. Let $T$ be a finitely generated $k$-subalgebra of $S^{-1}A$ and suppose that $W=\{w_1=1, \ldots , w_m\}$ is a frame of $T.$ Choose $s \in S$ and $a_1, \ldots, a_m \in A$ such that $w_i=s^{-1}a_i$ for all $i.$ Let $B$ be the $k$-subalgebra of $A$ generated by $a_i$ and let $V$ be the $k$-subspace generated by $1$ and $a_i.$ Now, since $S$ is in the center of $A,$ we have $s^nW^n \subseteq V^n.$ Thus $\dim W^n = \dim s^nW^n \leq \dim V^n.$ Therefore

${\rm{GKdim}}(T) \leq {\rm{GKdim}}(B) \leq {\rm{GKdim}}(A),$

for every finitely generated $k$-subalgebra of $T$ of $S^{-1}A,$ and so ${\rm{GKdim}}(S^{-1}A) \leq {\rm{GKdim}}(A).$ On the other hand, $A \subseteq S^{-1}A,$ because $S$ is regular, and thus ${\rm{GKdim}}(A) \leq {\rm{GKdim}}(S^{-1}A). \Box$

Using the above result we can now evaluate the GK dimension of a Laurent polynomial ring.

Corollary. Let $A$ be a $k$-algebra and let $x$ be a variable over $A.$ Then ${\rm{GKdim}}(A[x,x^{-1}])=1+ {\rm{GKdim}}(A).$

Proof. Since $A[x,x^{-1}]$ is the localization of $A[x]$ at the central regular submonoid $S=\{1,x,x^2, \ldots \},$ we have ${\rm{GKdim}}(A[x,x^{-1}])={\rm{GKdim}}(A[x]).$ The result now follows from Theorem 1. $\Box$

Let $A$ be a $k$-algebra. We showed that ${\rm{GKdim}}(A)=0$ if and only if $A$ is locally finite. We also saw that there is no algebra of GK dimension strictly between 0 and one. Now, what can we say about the case ${\rm{GKdim}}(A)=1$? There is a partial answer to this question and we will see it in the next post.

## GK dimension of finite extensions

Posted: April 26, 2011 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
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The goal is to prove the following fundamental result: if an algebra $A$ is a finite module over some subalgebra $B,$ then ${\rm{GKdim}}(A)={\rm{GKdim}}(B).$

Lemma. Let $B$ be a subalgebra of a $k$-algebra $A.$ Suppose that, as a (left) module, $A$ is finitely generated over $B.$ Then  ${\rm{GKdim}}({\rm{End}}_B(A)) \leq {\rm{GKdim}}(B).$

Proof. So $A=\sum_{i=1}^n Ba_i$ for some $a_i \in A.$ Define $\varphi: B^n \longrightarrow A$ by $\varphi(b_1, \ldots , b_n)=\sum_{i=1}^n b_ia_i$ and let $I = \ker \varphi.$ Let

$C=\{f \in {\rm{End}}_B(B^n): \ f(I) \subseteq I \}.$

Clearly $C$ is a subalgebra of ${\rm{End}}_B(B^n) \cong M_n(B).$ Now, given $f \in C$ define $\overline{f}: A \longrightarrow A$ by  $\overline{f}(a)=\varphi f (u),$ where $u$ is any element of $B^n$ with $\varphi(u)=a.$ Note that $\overline{f}$ is well-defined because if $\varphi(v)=a$ for some other $v \in B^n,$ then $u-v \in I$ and so $f(u-v) \in I.$ Hence $0=\varphi f(u-v)=\varphi f(u) - \varphi f(v)$ and so $\varphi f(u) = \varphi f(v).$ It is easy to see that $\overline{f} \in {\rm{End}}_B(A)).$ Finally, define $\psi: C \longrightarrow {\rm{End}}_B(A))$ by $\psi(f)=\overline{f}.$ Then $\psi$ is an $k$-algebra onto homomorphism and hence

${\rm{GKdim}}({\rm{End}}_B(A)) \leq {\rm{GKdim}}(C) \leq {\rm{GKdim}}(M_n(B)) ={\rm{GKdim}}(B),$

by Fact 1 and Fact 5. $\Box$

Theorem. Let $B$ be a subalgebra of a $k$-algebra $A$ and suppose that, as a (left) module, $A$ is finitely generated over $B.$ Then ${\rm{GKdim}}(A)={\rm{GKdim}}(B).$

Proof. The algebra $A$ has a natural embedding into ${\rm{End}}_B(A)$ and so ${\rm{GKdim}}(A) \leq {\rm{GKdim}}({\rm{End}}_B(A)).$ Thus ${\rm{GKdim}}(A) \leq {\rm{GKdim}}(B),$ by the lemma. $\Box$

An important consequence of the above theorem is the following result. It shows that for finitely generated commutative algebras, GK dimension is nothing but the transcendence degree of the algebra over the base field.

Corollary. If $A$ is a finitely generated commutative $k$-algebra, then ${\rm{GKdim}}(A)={\rm{tr.deg}}(A/k).$

Proof. Let $m={\rm{tr.deg}}(A/k).$ Then $A$ contains a polynomial $k$-algebra $B=k[x_1, \ldots, x_m]$ such that $A$ is a finitely generated $B$-module, by the  Noether normalization theorem. Thus, by Corollary 2 and this theorem, ${\rm{GKdim}}(A) = {\rm{GKdim}}(B)=m. \Box$

## GK dimension of polynomial rings

Posted: April 26, 2011 in Gelfand-Kirillov Dimension, Noncommutative Ring Theory Notes
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Theorem 1. Let $A$ be a $k$-algebra. If $x$ is a variable over $A,$ then ${\rm{GKdim}}(A[x])=1+{\rm{GKdim}}(A).$

Proof.  Let $B_0$ be a finitely generated subalgebra of $A[x]$ generated by $f_1, \ldots , f_m \in A[x].$ Let $B$ be the subalgebra of $A$ generated by the coefficients of $f_i, \ i =1, \ldots , m.$ Then clearly $B$ is a finitely generated subalgebra of $A$ and $B_0 \subseteq B[x].$ Now, let $W$ be a frame of $B.$ Let $V=W+kx.$ Then $V$ is a generating subspace of $B[x]$ and clearly $V^n =(W+kx)^n \subseteq \bigoplus_{i=0}^nW^nx^i,$ for all integers $n \geq 0.$ Hence $\dim V^n \leq (n+1) \dim W^n$ and so

$\displaystyle {\rm{GKdim}}(B_0) \leq {\rm{GKdim}}(B[x]) \leq \lim_{n\to\infty} \log_n(n+1) + {\rm{GKdim}}(B)$

$= 1 + {\rm{GKdim}}(B) \leq 1 + {\rm{GKdim}}(A).$

Therefore ${\rm{GKdim}}(A[x]) \leq 1 + {\rm{GKdim}}(A).$ It is also clear that $V^{2n}=(W+kx)^{2n} \supseteq \bigoplus_{i=0}^n W^{2n}x^i,$ for all integers $n \geq 0.$ Thus $\dim V^{2n} \geq (n+1) \dim W^{2n}$ and so

$\displaystyle {\rm{GKdim}}(A[x]) \geq {\rm{GKdim}}(B[x]) \geq \lim_{n\to\infty} \log_n(n+1) + {\rm{GKdim}}(B) = 1 +$ ${\rm{GKdim}}(B).$

Therefore ${\rm{GKdim}}(A[x]) \geq 1 + {\rm{GKdim}}(A)$ and the result follows. $\Box$

Corollary 1. Let $A$ be a $k$-algebra. Then ${\rm{GKdim}}(A[x_1, \ldots , x_m])=m+{\rm{GKdim}}(A).$

Corollary 2. ${\rm{GKdim}}(k[x_1, \ldots , x_m])=m.$

An immediate result of corollary 2 is that if $X$ is an infinite set of commuting variables, then ${\rm{GKdim}}(k[X])=\infty.$

So, by corollary 2, for any integer $m \geq 1$ there exists an algebra $A$ such that ${\rm{GKdim}}(A)=m.$ In fact, for any real number $\alpha \geq 2$ there exists a $k$-algebra $A$ such that ${\rm{GKdim}}(A)=\alpha.$ It is also known that if $1 < \alpha < 2,$ then there is no algebra $A$ with ${\rm{GKdim}}(A)=\alpha.$ This result is called the Bergman’s gap theorem. We now look at the GK dimension of noncommutative polynomial algebras.

Theorem 2. If $X$ is a set of noncommuting variables with $|X| \geq 2,$ then ${\rm{GKdim}}(k \langle X \rangle)=\infty.$

Proof. Let $x, y \in X$ and consider the $k$-subalgebra $B=k \langle x, y \rangle$ of $A.$ Choose the generating subspace $V=k+kx + ky.$ Then it follows easily that $\dim V^n = 1 + 2 + \cdots + 2^n \geq 2^n$ and thus

$\displaystyle {\rm{GKdim}}(A) = \limsup_{n\to\infty} \log_n (\dim V^n) \geq \lim_{n\to\infty} n \log_n 2 = \infty.$

Therefore ${\rm{GKdim}}(A)=\infty.$ $\Box$

Corollary. Let $A$ be a $k$-algebra which is a domain. If ${\rm{GKdim}}(A) < \infty,$ then $A$ is Ore.

Proof. If $A$ contains a copy of $k \langle x, y \rangle,$ where $x$ and $y$ are noncommuting variables, then ${\rm{GKdim}}(A)=\infty,$ by Theorem 2. Thus $A$ does not contain such a subalgebra and hence $A$ is an Ore domain by the lemma in this post. $\Box$

## 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 ${\rm{GKdim}}(A)=m.$