The Heisenberg center is quadratically distorted

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.