Growth of algebras

Remarks

The growth does not depend on $V$. If $W$ is another such subspace, then $W \subseteq V^c$ for some $c$, because $W$ is finite-dimensional and $\bigcup_n V^n = A$; so $d_W(n) \le d_V(cn)$, and symmetrically.

The algebra $A$ has polynomial growth if $d_V(n) \le C n^r$ for some constants $C, r$. It has exponential growth if $d_V(n) \ge a^n$ for some $a > 1$ and all $n$. Otherwise it has intermediate growth.

Example: for $A = k[x, y]$ and $V = k + kx + ky$, the space $V^n$ has the basis of monomials $x^i y^j$ with $i + j \le n$. So $d_V(n) = (n+1)(n+2)/2$, and $A$ has polynomial growth.

Non-example of the setting: the field $k(x)$ of rational functions is not a finitely generated $k$-algebra, so no growth function is defined for it in this way.

For a finitely generated group $G$ with a finite symmetric generating set $S$, take $A = k[G]$, the group algebra, and $V$ the span of $\set{1} \cup S$. Then $V^n$ has the basis of group elements of word length at most $n$, so the growth of $k[G]$ is the growth of $G$.