Proof. Observe that $R_n = \left (\bigoplus _{i+j\leq n} R_y^x(n;i,j)\right )\oplus kx^n$ . Fix a k-linear basis $\mathcal {B}=\mathcal {B}_1\cup \mathcal {B}_2\cup \cdots $ of A, consisting of monomials in some homogeneous generating set of A (say, a basis of $A_1$ ). For each $i\geq 1$ , fix a basis element $w[i]\in \mathcal {B}_i=\mathcal {B}\cap A_i$ and let

$$ \begin{align*}W={\operatorname{Span}}_k\{w[i]\ |\ i=1,2,\dots\},\ W^{\perp}={\operatorname{Span}}_k \left( \mathcal{B} \setminus \{w[i]\ |\ i=1,2,\dots\}\right),\end{align*} $$

so $A=W\oplus W^{\perp }$ . Consider the (non-unital) free product

$$\begin{align*}A *_k \left(k[y]/\left<y^2\right>\right) = A \oplus A^1yA^1 \oplus A^1 y A y A^1 \oplus \cdots \end{align*}$$

with basis $\{b_0,b_0y,yb_0,b_0yb_1,b_0yb_1y,yb_0yb_1,b_0yb_1yb_2,\dots \ |\ b_0,b_1,b_2,\dots \in \mathcal {B}\}$ . For each $y^2$ -free monomial divisible by y (namely, a nonzero monomial in $\left <y\right>\triangleleft k\left <x,y\right>/\left <y^2\right>$ ),

$$\begin{align*}\xi = x^{i_0}yx^{i_1}\cdots x^{i_{s-1}}yx^{i_s}, \end{align*}$$

and for every $w_0\in \mathcal {B}_{i_0},w_1\in \mathcal {B}_{i_{s}}$ , consider the following element in the above free product:

$$\begin{align*}\pi_{w_0,w_1}(\xi) = w_0yw[i_1]\cdots w[i_{s-1}]yw_1. \end{align*}$$

In case that $\xi =x^{i_0}$ , we let $\pi _{w_0,w_1}(\xi )=w_0$ , and if $i_0=0$ or $i_s=0$ , we let $\pi _{w_0,w_1}(\xi )=yw[i_1]\cdots w[i_{s-1}]yw_1,w_0yw[i_1]\cdots w[i_{s-1}]y$ , respectively. We formally set $\pi _{w_0,w_1}(0)=0$ .

Consider the ideals

$$\begin{align*}J = \left<yW^{\perp}y\right>= \left< yuy\ |\ u\in \mathcal{B}\setminus \{w[1],w[2],\dots\} \right> \triangleleft A *_k \left(k[y]/\left<y^2\right>\right) \end{align*}$$

and

$$\begin{align*}\widehat{I} = \left< \pi_{w_0,w_1}(\xi)\ |\ \xi\in {\operatorname{Mon}}(I),\ w_0\in \mathcal{B}_{{\operatorname{pre}}_x(\xi)},\ w_1\in \mathcal{B}_{{\operatorname{suf}}_x(\xi)} \right> \triangleleft A *_k \left(k[y]/\left<y^2\right>\right). \end{align*}$$

Notice that $I\subseteq \left <y\right>\triangleleft k\left <x,y\right>$ , for otherwise, since I is a monomial ideal of $k\left <x,y\right>$ , it would contain some $x^n$ , making $\left <y\right>$ finite-codimensional in $R=k\left <x,y\right>/I$ ; since by assumption, y generates a locally nilpotent ideal in R, it follows that R is finite-dimensional, a contradiction to the assumption.

We also note that every element in $\widehat {I}$ is a linear combination of elements of the form $\pi _{w_0,w_1}(\xi ),\ \xi \in I$ and elements from J. Indeed, fix $\xi =x^{i_0}y\cdots yx^{i_s}\in {\operatorname {Mon}}(I)$ and $w_0\in \mathcal {B}_{{\operatorname {pre}}_x(\xi )=i_0},w_1\in \mathcal {B}_{{\operatorname {suf}}_x(\xi )=i_s}$ :

• • Given any $u\in \mathcal {B}_d$ , we can write $uw_0=\sum _{i=1}^n c_i w^{\prime }_i$ for some $w^{\prime }_i\in \mathcal {B}_{d+i_0},c_i\in k$ , since $u,w_0$ are homogeneous; we compute that $u\pi _{w_0,w_1}(\xi )=uw_0yw[i_1]\cdots w[i_{s-1}]yw_1=\sum _{i=1}^n c_i w^{\prime }_i yw[i_1]\cdots w[i_{s-1}]yw_1 = \sum _{i=1}^n c_i \pi _{w^{\prime }_i,w_1}(x^d \xi )$ . An analogous outcome holds for $\pi _{w_0,w_1}(\xi )u$ .

• • We compute $y\pi _{w_0,w_1}(\xi )=yw_0yw[i_1]\cdots w[i_{s-1}]yw_1$ , which is either in J if $w_0\neq w[i_0]$ , or otherwise, $w_0=w[i_0]$ and $y\pi _{w_0,w_1}(\xi )=\pi _{w_0,w_1}(y\xi )$ . An analogous outcome holds for $\pi _{w_0,w_1}(\xi )y$ .

It follows by induction that every element in $\widehat {I}$ takes the desired form, and moreover, for every monomials $\xi ,\xi '$ and $w_0,w_1,w^{\prime }_0,w^{\prime }_1\in \mathcal {B}$ of compatible degrees, we have that

(3.1) $$ \begin{align} \pi_{w_0,w_1}(\xi)\pi_{w^{\prime}_0,w^{\prime}_1}(\xi') \equiv \sum_{i=1}^n c_i \pi_{u^i_0,u^i_1}(\xi \xi')\quad \mod{J} \end{align} $$

for suitable $u^i_0,u^i_1\in \mathcal {B}$ and $c_i\in k$ .

Finally, let

$$\begin{align*}\widehat{A} = \frac{ A *_k \left( k[y]/\left<y^2\right>\right) }{ J + \widehat{I}}.\end{align*}$$

Note that $\widehat {A}$ is naturally graded, extending the grading from A by letting $\deg (y)=1$ ; indeed, this defines a grading on $A*_k \left ( k[y]/\left <y^2\right>\right ) $ , and both $J,\widehat {I}$ are homogeneous with respect to this grading, which thus induces a grading on the quotient ring. Furthermore, $\deg _{\widehat {A}}(\pi _{w_0,w_1}(\xi ))=\deg _R(\xi )$ .

Consider the following homogeneous set:

(3.2) $$ \begin{align} \widehat{\mathcal{B}} = \{ \pi_{w_0,w_1}(\xi)\ |\ \xi\in {\operatorname{Mon}}(R)\setminus\{1\},\ w_0\in \mathcal{B}_{{\operatorname{pre}}_x(\xi)},\ w_1\in \mathcal{B}_{{\operatorname{suf}}_x(\xi)} \} \subseteq \widehat{A}. \end{align} $$

Notice that $\mathcal {B}\subseteq \widehat {\mathcal {B}}$ as seen by considering $\xi =x^n$ , $n\in \mathbb {N}$ . We claim that $\widehat {\mathcal {B}}$ spans $\widehat {A}$ . Indeed, $A *_k \left (k[y]/\left <y^2\right>\right )/J$ is spanned by

$$\begin{align*}\mathcal{B} \cup \Big\{ \pi_{w_0,w_1}(\xi)\ \Big|\ \xi\in {\operatorname{Mon}}\left(k\left<x,y\right>/\left<y^2\right>\right)\cap \left<y\right>,\ w_0\in \mathcal{B}_{{\operatorname{pre}}_x(\xi)},\ w_1\in \mathcal{B}_{{\operatorname{suf}}_x(\xi)}\Big\}, \end{align*}$$

and modulo $\widehat {I}$ , we may further assume that $\xi \notin I$ – namely, $\xi \in {\operatorname {Mon}}(R)$ .

We now claim that $\widehat {\mathcal {B}}$ is linearly independent. We first observe that $\mathcal {\widehat {B}}$ is a linearly independent subset of $A *_k \left (k[y]/\left <y^2\right>\right )$ . Next, if some nontrivial linear combination $\sum _{i=1}^{n} \alpha _i \pi _{w_0^i,w_1^i}(\xi _i)$ with $\xi _i\in {\operatorname {Mon}}(R)$ belongs to $\widehat {I}+J$ (as an element of $A *_k \left (k[y]/\left <y^2\right>\right )$ ), then we claim that it in fact belongs to $\widehat {I}$ . Recall that every element of $\widehat {I}$ is a linear combination of elements of the form $\pi _{w_0,w_1}(\xi )$ , where $\xi \in I$ (for possibly different $w_0,w_1,\xi $ ) and elements from J. Write

$$\begin{align*}\sum_{i=1}^{n} \alpha_i \pi_{w_0^i,w_1^i}(\underbrace{\xi_i}_{\in {\operatorname{Mon}}(R)}) = \underbrace{f}_{\in J} + \sum_{j=1}^m \beta_j \pi_{u_0^j,u_1^j}(\underbrace{\xi^{\prime}_j}_{\in {\operatorname{Mon}}(I)}) \end{align*}$$

so

$$\begin{align*}\sum_{i=1}^{n} \alpha_i \pi_{w_0^i,w_1^i}(\xi_i) - \sum_{j=1}^m \beta_j \pi_{u_0^j,u_1^j}(\xi^{\prime}_j) = f \in J. \end{align*}$$

However, the left-hand side belongs to the vector space

$$ \begin{align*}A\oplus A^1yA^1 \oplus A^1yWyA^1\oplus A^1yWyWyA^1\oplus \cdots\end{align*} $$

while

$$ \begin{align*}J\subseteq \bigoplus_{i,j\geq 1} A^1\underbrace{yA\cdots Ay}_{i\ \text{times}\ y}W^{\perp}\underbrace{yA\cdots Ay}_{j\ \text{times}\ y}A^1,\end{align*} $$

so f must vanish and $\sum _{i=1}^{n} \alpha _i \pi _{w_0^i,w_1^i}(\xi _i) = \sum _{j=1}^m \beta _j \pi _{u_0^j,u_1^j}(\xi ^{\prime }_j)$ . Since ${\operatorname {Mon}}(I),{\operatorname {Mon}}(R)$ are disjoint sets of monomials in $k\left <x,y\right>$ , we have by Remark 3.2 that $\sum _{i=1}^{n} \alpha _i \pi _{w_0^i,w_1^i}(\xi _i) = 0$ in the free product $A *_k \left (k[y]/\left <y^2\right>\right )$ , a contradiction.

Let us now provide a formula for products of basis elements from $\widehat {\mathcal {B}}$ . Let $\psi _n\colon A\rightarrow k$ be the linear functional projecting an element to the (possibly zero) coefficient of $w[n]$ of it, when written uniquely as a linear combination of $\mathcal {B}$ . If $w_1,w_2\in A$ multiply as $w_1w_2=\sum _{b\in \mathcal {B}} \alpha _b b$ , then

$$\begin{align*}w_0yw[i_1]\cdots w[i_{s-1}]yw_1 * w_2 = \sum_{b\in \mathcal{B}} \alpha_b w_0yw[i_1]\cdots w[i_{s-1}]yb \end{align*}$$

$$\begin{align*}w_1*w_2yw[i_1]\cdots w[i_{s-1}]yw_3 = \sum_{b\in \mathcal{B}} \alpha_b byw[i_1]\cdots w[i_{s-1}]yw_3 \end{align*}$$

and

$$ \begin{align*} & w_0yw[i_1]\cdots w[i_{s-1}]yw_1 * w^{\prime}_0yw[j_1]\cdots w[j_{t-1}]yw^{\prime}_1 \\ & \quad =\psi_p(w_1w^{\prime}_0)\cdot w_0yw[i_1]\cdots yw[p]y\cdots w[j_{t-1}]yw^{\prime}_1, \end{align*} $$

where $p=\deg (w_1)+\deg (w^{\prime }_0)$ .

Let $\sum _{i=1}^l c_i \pi _{w^i_0,w^i_1}(\xi _i)\in \left <y\right>\triangleleft \widehat {A}$ . Then $\xi _1,\dots ,\xi _l\in \left <y\right>\triangleleft R$ . By assumption, $\left <y\right>$ is a locally nilpotent ideal of R, so there exists some d such that

$$\begin{align*}\forall 1\leq i_1,\dots,i_d\leq l,\ \xi_{i_1}\cdots \xi_{i_d}=0, \end{align*}$$

and it follows that

$$\begin{align*}\left(\sum_{i=1}^l c_i \pi_{w^i_0,w^i_1}(\xi_i)\right)^d \in {\operatorname{Span}}_k \{\pi_{*,*}(\xi_{i_1}\cdots\xi_{i_d})\ |\ 1\leq i_1,\dots,i_d\leq l\} = 0 \end{align*}$$

as seen by (3.1). Consequently, $\left <y\right>\triangleleft \widehat {A}$ is a nil ideal. Notice that $\widehat {A}/\left <y\right>\cong A$ , which is nil. Therefore,

$$\begin{align*}0\rightarrow \left<y\right>\rightarrow \widehat{A} \rightarrow A \rightarrow 0 \end{align*}$$

is an extension of two nil algebras; hence, $\widehat {A}$ is nil.

As seen by (3.2),

$$\begin{align*}\widehat{A}_n \cong A_n \oplus \bigoplus_{i+j\leq n} A_i \otimes_k R_y^x(n;i,j) \otimes_k A_j \end{align*}$$

and

$$\begin{align*}\widehat{A}_{\leq n} \cong A_{\leq n} \oplus \bigoplus_{i+j\leq n} A_i \otimes_k R_y^x({\leqslant} n;i,j) \otimes_k A_j \end{align*}$$

isomorphisms of vector spaces (formally taking $A_0=k$ ); hence,

$$\begin{align*}\dim_k \widehat{A}_{\leq n} = \dim_k A_{\leq n} + \sum_{i+j\leq n} \dim_k R_y^x({\leqslant} n;i,j)\cdot \dim_k A_i \cdot \dim_k A_j, \end{align*}$$

as claimed.

Proof of Theorem 1.2.

Let $f\colon \mathbb {N}\rightarrow \mathbb {N}$ be the growth function of any finitely generated infinite-dimensional algebra, not of linear growth. We may assume that f is subexponential (nil algebras of exponential growth exist by the Golod-Shafarevich theorem, and all of the exponential growth functions are equivalent to each other). By [Reference Bell and Zelmanov9, Theorem 1.1], f satisfies $f'(n)\geq n+1$ and $f'(n)\leq f'(m)^2$ for every $m\leq n\leq 2m$ . Moreover, by the same theorem, f is then equivalent to the growth function of an algebra R explicitly constructed therein. Let us recall now the general structure of that algebra. This is a monomial algebra $R=k\left <x,y\right>/I$ in which the set of nonzero monomials is

$$ \begin{align*}\bigcup_{n\geq 1} T(d_n,n) \cup \bigcup_{n\geq 1} x^{e_n}T(d_n-1,n-e_n),\end{align*} $$

where

$$ \begin{align*}T(d,n) = \{x^n\} \cup \{x^iyx^{a_1}yx^{a_2}\cdots x^{a_s}yx^j\ \text{of length } n \text{ with}\ a_1,\dots,a_s\geq d \}\end{align*} $$

and for some sequences $\{d_n\}_{n=1}^{\infty },\{e_n\}_{n=1}^{\infty }$ of positive integers, explicitly constructed in the proof of [Reference Bell and Zelmanov9, Theorem 1.1]. Furthermore, the sequence $\{d_n\}_{n=1}^{\infty }$ tends to infinity since the realized growth function f is subexponential (see [Reference Bell and Zelmanov9, Page 691]). Notice also that $y^2=0$ . Since every monomial w in the ideal $\left <y\right>^2=RyRyR$ contains an occurrence of $yx^iy$ for some $i>0$ , it follows that for some n (in fact, for every n such that $d_n>i+1$ ), $wR_n=R_nw=0$ , and therefore, $\left <y\right>$ is locally nilpotent. Furthermore,

(3.3) $$ \begin{align} \dim_k \left<y\right>_{\leq n} = \dim_k R_{\leq n} - \dim_k \left(k+kx+\cdots+kx^n\right) = \dim_k R_{\leq n}-(n+1). \end{align} $$

Let us now conclude the proof, using Theorem 3.1.

Case I: Countable base fields. Assume that k is countable. We use R from the above discussion as a base ring for the construction given in Theorem 3.1, along with the nil graded algebra A from [Reference Lenagan, Smoktunowicz and Young25], constructed over an arbitrary countable field. As proven in [Reference Lenagan, Smoktunowicz and Young25, Theorem 7.5], $\dim _k A_n\leq Cn^2\log ^6 n$ for all n and for some constant $C>0$ . Consider the finitely generated nil graded algebra $\widehat {A}$ resulting by applying Theorem 3.1 to this setting. Then, for every $\varepsilon>0$ ,

$$ \begin{align*} \dim_k \widehat{A}_{\leq n} & = \dim_k A_{\leq n} + \sum_{i+j\leq n} \dim_k R_y^x({\leqslant} n;i,j)\cdot \dim_k A_i \cdot \dim_k A_j \\ & \leq Cn^3\log^6 n + \sum_{i+j\leq n} \dim_k R_y^x({\leqslant} n;i,j) \cdot Ci^2\log^6 i \cdot Cj^2\log^6 j \\ & \leq 2C^2n^4\log^{12} n\dim_k R_{\leq n}\preceq n^{4+\varepsilon}f(n). \end{align*} $$

However, by Theorem 3.1,

$$ \begin{align*} \dim_k \widehat{A}_{\leq n} \geq \dim_k A_{\leq n} + \sum_{i+j\leq n} \dim_k R_y^x({\leqslant} n;i,j) \geq n + \left<y\right>_{\leq n} = \dim_k R_{\leq n} - 1 \sim f(n). \end{align*} $$

Case II: Arbitrary base fields. Assume that k is of arbitrary cardinality. We use R from the above discussion as a base ring for the construction given in Theorem 3.1, along with a nil graded algebra A from [Reference Bell and Young8], constructed over an arbitrary base field. As proven in [Reference Lenagan, Smoktunowicz and Young25], for any super-polynomial function, one can construct a nil graded algebra A over k whose growth is bounded from above by that function. Given a function $\omega (n)\xrightarrow {n\rightarrow \infty } \infty $ , let A be a nil graded algebra over k whose growth satisfies $\gamma _A(n)\preceq n^{\frac {1}{2}\omega (n)}$ . Consider the finitely generated nil graded algebra $\widehat {A}$ resulted by applying Theorem 3.1 to this setting. Then

$$ \begin{align*} \dim_k \widehat{A}_{\leq n} & = \dim_k A_{\leq n} + \sum_{i+j\leq n} \dim_k R_y^x({\leqslant} n;i,j)\cdot \dim_k A_i \cdot \dim_k A_j \\ & \leq \dim_k A_{\leq n} + \sum_{i+j\leq n} \dim_k R_y^x({\leqslant} n;i,j) \cdot \left(n^{\frac{1}{2}\omega(n)}\right)^2 \\ & \leq n^{\frac{1}{2}\omega(n)} + n^{\omega(n)}\dim_k R_{\leq n}\sim n^{\omega(n)}f(n). \end{align*} $$

However, by Theorem 3.1,

$$ \begin{align*} \dim_k \widehat{A}_{\leq n} \geq \dim_k A_{\leq n} + \sum_{i+j\leq n} \dim_k R_y^x({\leqslant} n;i,j) \geq n + \left<y\right>_{\leq n} = \dim_k R_{\leq n} - 1 \sim f(n). \end{align*} $$

The proof is completed.

Proof. Let A be a nil associative algebra. For every $a\in A$ , there is some d such that $a^d=0$ . Since

$$\begin{align*}{\operatorname{ad}}_a^m(x) = \sum_{i=0}^m (-1)^{m-i}{m \choose i} a^i x a^{m-i} \end{align*}$$

for every $m\geq 0$ , we obtain that ${\operatorname {ad}}_a^{2d+1}(x)$ is a linear combination of monomials of the form $a^ixa^j$ with $i+j=2d+1$ , so ${\operatorname {ad}}_a^{2n+1}(x)\in a^d A+A a^d=0$ . It follows that L is a nil Lie algebra. If $A=\bigoplus _{n=0}^{\infty } A_n$ is a graded-nil associative algebra, then the above argument applies for every homogeneous element a (and arbitrary x), proving that L is a graded-nil Lie algebra.

Proof of Corollary 1.3.

Fix an arbitrary $\alpha \in (0,1)$ . There exists a finitely generated algebra B whose growth function is $\gamma _B(n)\sim \exp (n^\alpha )$ (see, for example, [Reference Smoktunowicz and Bartholdi35, Corollary D]). By Theorem 1.2, it follows that, for every increasing, super-polynomial function $\omega \colon \mathbb {N}\rightarrow \mathbb {N}$ , there exists a finitely generated nil graded algebra over an arbitrary base field whose growth is

$$\begin{align*}\gamma_B(n)\preceq \gamma_A(n)\preceq \omega(n)\gamma_B(n). \end{align*}$$

Taking $\omega (n)=n^{\log n}$ , we get, for $n\gg _\alpha 1$ , that

$$ \begin{align*} \exp(n^\alpha)\preceq \gamma_A(n) & \preceq n^{\log n}\exp(n^\alpha) \\ & \leq \exp((2n)^\alpha)\sim \exp(n^\alpha), \end{align*} $$

as claimed.

Now let $L=[A^{(-)},A^{(-)}]$ . By [Reference Alahmadi and Alharthi1, Theorem 2], L is a finitely generated graded Lie algebra and $\gamma _L(n)\sim \gamma _A(n)\sim \exp (n^\alpha )$ , and by Lemma 4.1, L is a nil Lie algebra.

Proof of Corollary 1.4.

Fix an arbitrary $\alpha \geq 6$ . Let A be a finitely generated nil graded algebra of ${\operatorname {GKdim}}(A)\leq 3$ (see [Reference Lenagan, Smoktunowicz and Young25]); denote $\rho :={\operatorname {GKdim}}(A)$ . Let $d\in \mathbb {Z}_{\geq 0}$ and $0<\beta \leq 1$ be such that $\alpha =2\rho +d\beta $ (in particular, if $\alpha>6$ , then $d=\lceil \alpha -2\rho \rceil $ , and $\beta =\frac {\alpha -2\rho }{d}$ and if $\alpha =6$ , take $d=0,\beta =1$ ). Let $S=\{\lceil n^{1/\beta } \rceil \ |\ n\in \mathbb {N}\}$ and observe that $\# \left (S\cap [1,n]\right ) = \lfloor n^\beta \rfloor $ .

Consider the monomial algebra

$$\begin{align*}R = \frac{k\left<x,y\right>}{\left<y^2\right>+\left<y\right>^{d+2}+\left<yx^i y\ |\ i\notin S \right>}. \end{align*}$$

Notice that the ideal $\left <y\right>\triangleleft R$ is nilpotent. Now for any $i,j$ ,

(4.1) $$ \begin{align} \dim_k R_y^x({\leqslant} n;i,j) & \leq \sum_{c=1}^{d+1} \#\Big\{ \vec{t}\in S^{c-1}\ \Big|\ \sum_{i=1}^{c-1} t_i\leq n \Big\} \\ & \leq \sum_{c=1}^{d+1} \# \left( S\cap [1,n] \right)^{c-1} \nonumber \leq (d+1) \cdot \# \left( S\cap [1,n] \right)^d \leq (d+1)n^{d\beta}. \nonumber \end{align} $$

Moreover, if $i+j\leq 2n$ , then, given $\vec {t}\in \left (S\cap [1,n]\right )^d$ , we have a nonzero monomial $\xi _{\vec {t}}\in R_y^x( {\leqslant } Kn;i,j)$

$$\begin{align*}x^iyx^{t_1}y\cdots yx^{t_d}yx^j \end{align*}$$

for some constant K (in fact, one can take any K such that $Kn\geq dn+(d+1)+i+j$ , so we can take $K=5d$ ), and the assignment $\vec {t}\mapsto \xi _{\vec {t}}$ is injective. It follows that

(4.2) $$ \begin{align} \dim_k R_y^x({\leqslant} Kn;i,j) & \geq \# \left(S\cap [1,n]\right)^d \\ & \geq \left(\lfloor n^\beta \rfloor\right)^d\geq 2^{-d} n^{d\beta} \nonumber \end{align} $$

(the last inequality follows since $\lfloor x \rfloor \geq \frac {1}{2}x$ for every $x\geq 1$ .) Thus, by Theorem 3.1, there exists a finitely generated nil graded algebra $\widehat {A}$ such that

(4.3) $$ \begin{align} \dim_k \widehat{A}_{\leq n} = \dim_k A_{\leq n} + \sum_{i+j\leq n} \dim_k R_y^x({\leqslant} n;i,j)\cdot \dim_k A_i \cdot \dim_k A_j. \end{align} $$

Hence, by (4.1),

(4.4) $$ \begin{align} \dim_k \widehat{A}_{\leq n} & \leq \dim_k A_{\leq n} + \sum_{i+j\leq n} (d+1)n^{d\beta} \cdot \dim_k A_i \cdot \dim_k A_j \\ & \leq \dim_k A_{\leq n} + (d+1)n^{d\beta} \left( \dim_k A_{\leq n} + 1 \right)^2 \nonumber \\ & \leq (d+2)n^{d\beta} \left( \dim_k A_{\leq n} + 1\right)^2 \nonumber, \end{align} $$

and by (4.2),

(4.5) $$ \begin{align} \dim_k \widehat{A}_{\leq K n} & \geq \sum_{i+j\leq 2 n} \dim_k R_y^x({\leqslant} K n;i,j) \cdot \dim_k A_i \cdot \dim_k A_j \\ & \geq 2^{-d} n^{d\beta} \sum_{i+j\leq 2 n} \dim_k A_i \cdot \dim_k A_j \nonumber \\ & \geq 2^{-d} n^{d\beta} \left(\dim_k A_{\leq n}\right)^2. \nonumber \end{align} $$

Combining (4.4) and (4.5), it follows that

$$\begin{align*}{\operatorname{GKdim}}(\widehat{A}) = \limsup_{n\rightarrow \infty} \frac{\log \dim_k \widehat{A}_{\leq n}}{\log n} = 2\rho+d\beta = \alpha, \end{align*}$$

as required.

Now let $L=[A^{(-)},A^{(-)}]$ . By [Reference Alahmadi and Alharthi1, Theorem 2], L is a finitely generated graded Lie algebra and ${\operatorname {GKdim}}(L) = {\operatorname {GKdim}}(A) = \alpha $ , and by Lemma 4.1, L is a nil Lie algebra.