Hall–Mal’cev Coordinates and Bass–Guivarc’h Growth: Examples

Addition in ${\mathbb{Z}}^d$ is commutative, so ${\gamma }_2=1$. Its only nonzero lower-central rank is $r_1=d$, hence $D=d$. The standard basis is a finite generating set, and the group is nilpotent of class one when d>0, so F1 applies.

For an integer vector a, each generator letter changes one coordinate by 1 in absolute value. Thus any representing word has at least ${\sum }_j\mathrm{\mid }{a}_j\mathrm{\mid }$ letters; writing each coordinate power attains that number. Consequently $[-\lfloor n\mathrm{/}d\rfloor ,\lfloor n\mathrm{/}d\rfloor {]}^d\cap {\mathbb{Z}}^d\subseteq B(n)\subseteq [-n,n{]}^d\cap {\mathbb{Z}}^d$ for d>0. For n>=d, the left cube has at least $(n\mathrm{/}d{)}^d$ points, and the right has at most $(3n{)}^d$. For $1\le n<d$ the ball contains 1 and $(n\mathrm{/}d{)}^d<=1$, so these bounds persist.

For d=0 there is one empty vector, the empty word has length zero, and every ball has size 1. The dimension is the empty sum 0 and $n^0=1$. Thus the degree and both estimates include the zero-dimensional case.

Matrix multiplication gives $(a,b,c)(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }})=(a+a^{\mathrm{\prime }},b+b^{\mathrm{\prime }},c+c^{\mathrm{\prime }}+a{b}^{\mathrm{\prime }})$. The identity is (0,0,0), and substitution on both sides gives $(a,b,c{)}^{-1}=(-a,-b,-c+ab)$. Put $x=(1,0,0)$, $y=(0,1,0)$, $z=(0,0,1)$. Their integer powers satisfy $x^a{y}^b{z}^k=(a,b,ab+k)$, so every matrix is uniquely $x^a{y}^b{z}^{c-ab}$.

Using the product and inverse formulas gives $[(a,b,c),(a^{\mathrm{\prime }},b^{\mathrm{\prime }},c^{\mathrm{\prime }})]=(0,0,a{b}^{\mathrm{\prime }}-a^{\mathrm{\prime }}b)$. In particular $[x,y]=z$. Every commutator is a power of z and z itself is a commutator, so $[H,H]=⟨z⟩$. Elements (0,0,c) commute with every triple by the product rule, hence ${\gamma }_3=1$. Conversely, if $(a,b,c)$ commutes with $x$ and $y$, its commutators have central entries $-b$ and $a$, so $a=b=0$. Thus the center is exactly $⟨z⟩$. Since z has infinite order, H has class exactly two. The elements x,y generate because z=[x,y] and step 1.1 gives every matrix.

The homomorphism $(a,b,c)\mapsto (a,b)$ is onto ${\mathbb{Z}}^2$ with kernel $⟨z⟩$, while $c\mapsto (0,0,c)$ identifies the kernel with $\mathbb{Z}$. Thus $r_1=2$ and $r_2=1$, giving $D=1\cdot 2+2\cdot 1=4$. F1 applies because x,y are finite generators and H has class two. The identity has a=b=k=0; the matrix entry c equals ab+k and is not generally the normal coordinate k.

For integers a,b the commutator formula gives $[x^a,y^b]=z^{ab}$. For $m\ge 1$ put $q=\lceil \sqrt{m}\rceil$ and write $m=aq+r$ with $0\le r<q$ and $0\le a\le q$. Then $[x^a,y^q][x^r,y]=z^{aq+r}=z^m$. Its length is at most $2a+2q+2r+2\le 6q$. Since $q\le \sqrt{m}+1\le 2\sqrt{m}$ for $m\ge 1$, this is at most $12\sqrt{m}$. In particular $m=1$ gives the word $[x,y]$ of length four. Thus the stated upper bound holds.

For a horizontal word of length n let (a_j,b_j,c_j) be the prefix value after j letters, starting at zero. Multiplication by x or its inverse changes only a by 1; multiplication by y or its inverse changes b by 1 and changes c by plus or minus the preceding a. Therefore $\mathrm{\mid }{a}_j\mathrm{\mid }\le j$ and $\mathrm{\mid }{c}_n\mathrm{\mid }\le {\sum }_{j=0}^{n-1}j=n(n-1)\mathrm{/}2\le n^2$. If the word represents z^m, its final c is m, so $\mathrm{\mid }m\mathrm{\mid }\le n^2$ and $n\ge \sqrt{\mathrm{\mid }m\mathrm{\mid }}$.

Invert the word of step 1.1 for negative m; it represents z^m with unchanged length. At m=0 use the empty word. Intrinsically $\mathrm{\mid }{z}^m{\mathrm{\mid }}_z=\mathrm{\mid }m\mathrm{\mid }$, since the exponent sum of an intrinsic n-letter word has absolute value at most n and m identical signed letters attain |m|. Thus $\mathrm{\Delta }(n)\le n^2$. For $n\ge 18$ choose $m=\lfloor (n\mathrm{/}12{)}^2\rfloor$; then $m\ge n^2\mathrm{/}288$ and the upper length bound puts z^m in B(n). Hence $n^2\mathrm{/}288\le \mathrm{\Delta }(n)\le n^2$ for $n\ge 18$. This also verifies the hypotheses and conclusion of F2, since the center is infinite and equals gamma_2.

The matrix-unit rule is $E_{ij}{E}_{kl}={\delta }_{jk}{E}_{il}$. Hence $T_{ij}(a{)}^{-1}=T_{ij}(-a)$, $T_{ij}(a)T_{ij}(b)=T_{ij}(a+b)$, and for $i<j<k$ direct expansion yields $[T_{ij}(a),T_{jk}(b)]=T_{ik}(ab)$. In that expansion the only surviving cross product is $ab{E}_{ik}$, since $E_{ij}^2=E_{jk}^2=E_{jk}{E}_{ij}=0$.

The ordered product $T_{12}(a)T_{23}(b)T_{34}(c)T_{13}(d)T_{24}(e)T_{14}(f)$ has entries (12,23,34,13,24,14) equal to $(a,b,c,ab+d,bc+e,abc+ae+f)$. Given entries (A,B,C,D,E,F), the unique exponents are $a=A,b=B,c=C,d=D-AB,e=E-BC,f=F-AE$. Thus this is a six-integer normal form. The three adjacent transvections with parameter 1 generate: step 1.1 gives T_13(1), T_24(1), and then T_14(1), and all integer powers give every factor.

Let J_i be the additive group of strictly upper triangular matrices supported at distances at least i, with J_4=0. Matrix-unit multiplication gives $J_i{J}_j\subseteq J_{i+j}$. The groups $F_i=1+J_i$ are closed under multiplication and inversion, since $(1+u{)}^{-1}=1-u+u^2-u^3$. In the quotient ring by $J_{i+j}$, the images of u in J_i and v in J_j have both uv=vu=0, so 1+u and 1+v commute. It follows that $[F_i,F_j]\subseteq F_{i+j}$. Since G=F_1, induction gives ${\gamma }_i(G)\subseteq F_i$.

Conversely F_2 is generated by T_13(1), T_24(1), T_14(1), with arbitrary integer powers: its first superdiagonal is zero and the product has precisely those three independent remaining entries. The first two are adjacent commutators by step 1.1. The third is $[T_{13}(1),T_{34}(1)]$; since T_13(1) is in gamma_2, F2 places this commutator in gamma_3, hence also gamma_2. Therefore $F_2\le {\gamma }_2$ and $F_3=⟨T_{14}(1)⟩\le {\gamma }_3$. Along with step 2.2, this gives ${\gamma }_2=F_2$, ${\gamma }_3=F_3$, ${\gamma }_4=1$.

Taking the entries on superdiagonal i is an onto homomorphism $F_i\to {\mathbb{Z}}^{4-i}$ with kernel $F_{i+1}$: cross products have strictly greater distance. Hence the ranks are 3,2,1; gamma_3 is nontrivial, so the class is three. Thus $h=3+2+1=6$ and $D=1\cdot 3+2\cdot 2+3\cdot 1=10$. Apply F1 using the three explicit finite generators in step 2.1. All six zero exponents give I.

For H, the factor ranks (2,1) give $h=2+1=3$ and $D=1\cdot 2+2\cdot 1=4$. Their difference is $(2-1)\cdot 1=1$, contributed by the central second-layer free generator.

For ${\mathrm{UT}}_4(\mathbb{Z})$ the ranks (3,2,1) give $h=3+2+1=6$ and $D=1\cdot 3+2\cdot 2+3\cdot 1=10$. The difference is $(2-1)\cdot 2+(3-1)\cdot 1=4$. Thus h counts each free coordinate once, while D records its lower-central layer; these two explicit nilpotent groups have different values of the two invariants.

For every integer $N\ge 1$ the explicit word $[x^N,y^N]=x^N{y}^N{x}^{-N}{y}^{-N}=z^{N^2}$ has length at most 4N. Its normal coordinate tuple is (0,0,N^2), since the ordered form is unique. Hence the central coordinate in B(4N) can equal N^2.

If a fixed linear bound $\mathrm{\mid }k\mathrm{\mid }\le Cn$ held in every ball B(n), step 1.1 would imply $N^2\le 4CN$ for every positive integer N. Taking any integer N>4C contradicts this. Thus the claimed uniform linear bound fails. Moreover H has three independent coordinates but ball growth degree four by F1, so the predicted degree three from weight-one counting is false; positive multiples of n^4 cannot be bounded above by a fixed multiple of n^3 as n grows.

The product of (a,pb,c) and (a prime,pb prime,c prime) is $(a+a^{\mathrm{\prime }},p(b+b^{\mathrm{\prime }}),c+c^{\mathrm{\prime }}+pa{b}^{\mathrm{\prime }})$, and the inverse is $(-a,-pb,-c+pab)$. Thus G is a subgroup. Set x=(1,0,0), y=(0,p,0), z=(0,0,1). Then $x^a{y}^b{z}^k=(a,pb,pab+k)$, giving a unique normal form and a finite generating list x,y,z.

Commutators are $(0,0,p(a{b}^{\mathrm{\prime }}-a^{\mathrm{\prime }}b))$, and $[x,y]=z^p$. Thus $[G,G]=⟨z^p⟩$: every commutator lies there and z^p is a commutator. This subgroup is central and nontrivial, so G is nilpotent of class two. The map $\psi (a,pb,c)=(a,b,c<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>p)$ is an onto homomorphism to ${\mathbb{Z}}^2\oplus \mathbb{Z}\mathrm{/}p\mathbb{Z}$, since the extra product term pab prime vanishes modulo p. Its kernel is exactly $⟨z^p⟩$. The induced quotient map is therefore a bijective homomorphism, with injectivity given by this kernel calculation.

For a positive integer m, repeated multiplication gives $(a,pb,c{)}^m=(ma,mpb,mc+pabm(m-1)\mathrm{/}2)$; induction follows by adding c+p(ma)b at the next multiplication. If this power is the identity, ma=mpb=0 forces a=b=0, and then mc=0 forces c=0. Hence every nonidentity element has infinite order. Nevertheless z[G,G] has order exactly p: z^j belongs to <z^p> exactly when p divides j. Since $p\ge 2$, this is nontrivial torsion in the abelianization.

For the mixed lower-central form choose first-layer lifts x,y,z, with the z exponent reduced to a residue r in {0,...,p-1}, and second-layer generator z^p. For a normal exponent k divide k=pq+r; then $x^a{y}^b{z}^k=x^a{y}^b{z}^r(z^p{)}^q$. The identity is the zero tuple. Thus the p-th power of the finite factor lift is a nontrivial carry into layer two, exactly as retained by F2. Setting p=1 would remove the torsion and is explicitly excluded.