Let $G$ be a finite group, $S\subset G$ a generating set, $|g| = |g|_S = $ word length with respect to $S$. Define the "defect" of $g,h$ to be

$$\psi(g,h) = |g|+|h|-|gh|$$

Then $\psi:G\times G \rightarrow \mathbb{Z}$ is a $2$-cocycle and satisfies the normalizing condition $\psi(g,1) = \psi(1,g) = 0$.

Construct a group $\mathbb{Z} \times G$ by defining:

$$(a,g)\oplus(b,h) := (a+b+\psi(g,h), gh)$$

This is associative, since $\psi$ is a 2-cocycle. Furthermore, define a norm on $\mathbb{Z}\times G$ by setting: $$|(a,g)| := |a|+|g|$$ Then this norm is subadditive: $$|x \oplus y| \le |x| + |y|$$ If we restrict this norm on $\mathbb{N}_0 \times G$ then it is additive. Using this norm one can define a metric on $\mathbb{Z}\times G$ by setting: $$d(x,y):= \max\{ |x^{-1} \oplus y|,|y^{-1} \oplus x|\}$$ Furthermore, the group $G$ operates on $\mathbb{Z} \times G$ by setting: $$g \cdot (a,h) := (a+\psi(g,h),gh) = (0,g)\oplus(a,h)$$ This group acts isometrically on $\mathbb{Z}\times G$ as we have: $$d(g\cdot x, g\cdot y) = d(x,y)$$

I came across this construction by trying to mimic "elementary addition and the carrying process", but I do not know how this construction is called and what further properties it has. Any hint would be very nice!

Thanks for your help!

Edit 23.05.2019: Is there a description of finite index subgroups of $\mathbb{Z}\times G$ in terms of subgroups of $G$ and $\mathbb{Z}$?

By observation of Ashot Minasyan, this group is isomorphic as groups to $\mathbb{Z} \times G=$ the direct product, via:

$$\beta : \mathbb{Z} \times G \rightarrow \mathbb{Z} \times_S G: (a,g) \mapsto (a-|g|,g)$$

It is to me unclear however, if this group homomorphism is an isometry of metric spaces?

  • 2
    $\begingroup$ If you divide this quantity by 2, it becomes Gromov product, denoted $(g,h)_e$, which is defined for general metric spaces, not just for Cayley graphs. $\endgroup$ – Misha May 19 at 18:12
  • $\begingroup$ @Misha: Thanks for your comment. Which quantity do you mean? $\endgroup$ – orgesleka May 19 at 18:19
  • 1
    $\begingroup$ Sorry, it is $\psi(g,h^{-1})$, and assuming symmetric generating set so that $|h|=|h^{-1}|$. $\endgroup$ – Misha May 20 at 13:57
  • 2
    $\begingroup$ Since your $2$-cocycle is defined as the coboundary of the $1$-cochain $g \mapsto |g|$, the group that you get is the genuine direct product $\mathbb{Z} \times G$. In fact, the map $g \mapsto (|g|,g)$ defines an embedding of $G$ into your construction, which commutes with the natural copy of $\mathbb{Z}$. $\endgroup$ – Ashot Minasyan May 23 at 12:45
  • 1
    $\begingroup$ @orgesleka: your map $\beta$ should go in the opposite direction, from the genuine direct product to your construction. The formula is the same, and it is a group isomorphism, as you can check directly. $\endgroup$ – Ashot Minasyan May 23 at 13:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.