Free abelian groups have degree equal to rank

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.