The completion of the space of finite groups

Question

Edit: I revise the question based on the comment conversations

Let $\mathcal{F}$ be the set of all equivalence classes of finite groups under the "Isomorphism" equivalence relation. We define a pseudo metric $d$ on $\mathcal{F}$ as follows:

$$d(G,H)= \inf \{Hd(\tilde {G}_{n},\tilde{H}_{n})\} $$

where $\inf$ is taken over all arbitrary isomorphic copies $\tilde{G}_{n}$ and $\tilde{H}_{n}$ of $G$ and $H$ in $Gl(n,\mathbb{R})$, respectively, while $Hd$ is the Hausdorff distance in $GL(n,\mathbb{R})$ induced by its standard left invariant metric.

The definition of this metric is motivated by the Hausdorff Gromov metric on the space of compact Riemannian manifolds.

Is $d$ a metric on $\mathcal{F}$? If the answer is yes, we denote by $\bar{\mathcal{F}}$ the completion of $\mathcal{F}$. What can be said about an object $Z$ in $\bar{\mathcal{F}}$?

Can one consider the unit circle, in some reasonable sense, as an object in this completion? Is there a natural group structure on every element $Z\in \bar{\mathcal{F}}$? Is there a natural topology on $Z$?

Is $\bar{\mathcal{F}}$ a compact space?

Answer 1

i don't think it's a metric. Take a large prime $p$. By embedding $\mathbb Z/p\mathbb Z$ and $\mathbb Z/(p^2+p)\mathbb Z$ in the circle $S^1 \subseteq GL(2,\mathbb R)$, one sees that the distance between them is at most $O(1/p)$. By embedding $\mathbb Z/p \mathbb Z \times \mathbb Z/p \mathbb Z$ and $\mathbb Z/p \mathbb Z \times \mathbb Z/(p+1)\mathbb Z$ in the torus $S^1 \times S^1 \subseteq GL(4,\mathbb R)$, one again sees that the distance between them is at most $O(1/p)$.

But of course $\mathbb Z/(p^2+p)$ and $\mathbb Z/p \mathbb Z \times \mathbb Z/(p+1)\mathbb Z$ are isomorphic, so one would be forced to conclude by the triangle inequality that the distance between $\mathbb Z/p\mathbb Z$ and $\mathbb Z/p \mathbb Z \times \mathbb Z/p\mathbb Z$ is $O(1/p)$.

But this is false. If they had that distance in some $GL(n,\mathbb R)$, then by pidgeonhole, $p$ different elements of $\mathbb Z/p \mathbb Z \times \mathbb Z/p\mathbb Z$ would have to be within $O(1/p)$ of some element of $\mathbb Z/p\mathbb Z$ and hence within $O(1/p)$ of each other. By left invariance, $p$ different elements of $\mathbb Z/p \mathbb Z \times \mathbb Z/p\mathbb Z$ would have to be within $O(1)$ of the identity.

But in any representation of $\mathbb Z/p\mathbb Z \times \mathbb Z/p\mathbb Z$, only $o(p)$ elements have eigenvalues within $o(1/\sqrt{p})$ of the identity, since we can write the representation as a sum of characters, the eigenvalues on each character must be $p$th roots of unity, and each element is determined by its eigenvalues on two independent characters.

So we just need to check that every element within $O(1/p)$ of the identity has eigenvalues within $o(1/\sqrt{p})$ of the identity.

In fact we can show more is true, and an element within $d$ of the identity matrix can't move any vector of length one by a distance of greater than $e^{d}-1$. Since $e^{O(1/p)}-1 = O(1/p) = o(1/\sqrt{p})$, we obtain the desired conclusion. To check this, differentiate $Mv$ with respect to $M$ and observe that its operator norm with respect to your metric is the operator norm of $M$, so if $f(x)$ is the maximum total distance moved a vector of length one by a matrix within $x$ of the identity, $df/dx \leq 1+f$ so $f(x) \leq e^x-1$.