Covering a Lie group with balls of radius $\epsilon$.

Question

Suppose $G$ is a compact lie group with a bi-invariant metric (a metric under which multiplications from right and left are isometries). How can one show that there is a uniform $C>0$ so that for any $\epsilon > 0$, $G$ can be covered with at most $\left\lceil C\dfrac{\text{Vol}(G)}{\epsilon^d}\right\rceil$ balls?

Answer 1

For Lie groups, we can use the $3r$-covering Lemma in real analysis. Since the metric is bi-invariant, all balls of the same radius have the same volume (this is the use of the fact that we have a lie group). Now, the $3r$-covering lemma works for every metric space. If we use it here on a finite sub-covering of balls $\left\{B_{\epsilon/3}(g)\right\}_{g \in G}$ like ${B_\epsilon/3(g_1), B_\epsilon/3(g_2), ..., B_\epsilon/3(g_n)}$, then there are disjoint balls $B_{\epsilon/3}(g_{i_1}), B_{\epsilon/3}(g_{i_2}), ..., B_{\epsilon/3}(g_{i_k})$ where $1 \le i_1, ..., i_k \le n$, so that after making those balls 3 times fater, $G$ is covered. Now the idea of having disjoint balls is that, when we don't allow intersection, we have a bound like $\left\lceil\dfrac{Vol(G)}{Vol(B_{\epsilon/3}(e))}\right\rceil$ on their cardinality. This now gives the assymptotic bound.

(The core idea of the solution is by my friend Hesam.)