Topology of the double of the double of the double …

In Sub-riemannian geometry and Lie groups , Part I, section 4.2, I introduced the strange notion of “uniform group”. Instead of saying that an uniform group is just a topological group (which has an unique uniformity associated to it), I proposed the following construction.

1. Double of a group.  To any group G we associate its double group G^{(2)} = G \times G with the group operation

(x,u) \, (y,v) = (xy, y^{-1}u y v)

I introduce also the following three functions:

op: G^{(2)} \rightarrow G , op(x,u) = xu,

i': G \rightarrow G^{(2)} , i'(x) = (x,e)   (where e is the neutral element of the group G),

i": G \rightarrow G^{(2)} , i"(x) = (x, x^{-1}).

Remark that all these functions are group morphisms. Notice especially the morphism op, which is nothing but the group operation of G, seen as a group morphism from G^{(2)} to G.

2. Uniform group. For my purposes I introduced the notion of an uniform group: an uniform group is a group G, together with two uniformities, one on G, the other on G^{(2)}, such that the three morphisms from the point 1. are uniformly continuous.

So, instead of one uniformity, now I use two. Why? Well, we may eliminate one of the uniformities, the one on G. Indeed, suppose that  G^{(2)}  is an uniform group. Then take on G the smallest uniformity which makes the three morphisms uniformly continuous. Now we have two uniformities, one on the double of G, the other on the double of the double of G. We may repeat the procedure indefinitely, pushing to infinity the pair of uniformities.

Wait, what?

(TO BE CONTINUED)

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