Let $G$ be a metric group, and let $h$ be the associated Hausdorff dimension function on subsets of $G$. (See for instance Barnea and Shalev, Hausdorff dimension, pro-p groups and Kac-Moody algebras, Trans. AMS 1997.) When do we have $h(AB) = h(A) + h(B) - h(A \cap B)$ for normal subgroups of $G$? If this property fails, is there still a general way to use $h$ (or some similar 'dimension' function) to construct a pseudometric on the lattice of normal subgroups of $G$? What I am looking for here are some informative examples of bad behaviour, particularly for profinite groups.



Hausdorff dimension of products of normal subgroups
In the first conference I ever went to Slava Grigorchuk asked me a similar question and I didn’t have an answer. But when I have got back to Jerusalem I have talked with Elon Lindenstrauss about it and he suggested the following easy counterexample. Take $G=\mathbb{F}\_p[[t]]$. Pick $S$ to be a subset of the integers with density one and with infinite complement $T$. Say $S=\left\{ n\_i \right\}$ and $T=\left\{ m\_i \right\}$. Take $A=\overline{< t^{n\_i}>}$ and take $B=\overline{\left< t^{n\_i} +t^{m\_i} \right>}$. Cleary, $AB=G$, $h(A)=h(B)=1$, but $A \cap B=\emptyset$.


Now, $G$ is not finitely generated, if you would like to have a counterexample which is finitely generated, then you can take $G=SL\_d(F\_p[[t]])$ and construct in a similar way to the above $A$ which is made from upper triangular matrices and $B$ which is made from lower triangular matrices. However, $A$ and $B$ will not be normal any more.


I am not familiar with a counterexample in which $A$ and $B$ are normal and $G$ is finitely generated. I am also not familiar with a counterexample in which $A$ and $B$ are finitely generated. But as you can deduce from my story above this does not mean much.


