# Malcev completion of free groups

Let $$K$$ be a field with $$\operatorname{char} K=0$$, $$\hat{L}_n$$ the complete free Lie algebra of $$n$$ variables $$x_1,\dotsc,x_n$$ and $$\exp(\hat{L}_n)$$ its associated group with the product given by BCH formula (which is the Malcev completion of the free group). Is $$\mathrm{exp}(\hat{L}_n)$$ generated group-theoretically by the set $$\{\lambda x_i\colon \lambda\in K, 1\leq i\leq n\}$$?

I guess the statement is false; Although all elements of $$\exp(\hat{L}_n)$$ can be written using commutator expansion, i.e. infinite product of the form $$g_1 g_2\cdots$$ with $$g_i$$ being a product of $$i$$-fold group commutators of $$\lambda x_i$$'s (hence the convergence), some elements like $$[x_1,x_2]$$ seems to be impossible to realize by applying finitely many products.

• Very nice question. There are "Lazard formulas" (inversion of the Hausdorff formula) to define the addition and Lie bracket in terms of the elements (and their rational multiples) and using only the group law, and these are indeed infinite formulas. This makes it indeed very reasonable to conjecture that the Lie algebra bracket $[x_1,x_2]$ and the sum $x_1+x_2$ cannot be obtained by a finite group product. But this not a proof; I'll think about it.
– YCor
May 11 at 15:50

Here is a proof that the statement is, as you expected, false (for $$n\ge 2$$), in case the field $$K$$ is $$\mathbf{R}$$ or $$\mathbf{C}$$. Unfortunately it does not give anything explicit.

Let $$B$$ be its closed unit ball. Then $$\hat{L}_n$$ is a Polish topological group.

If by contradiction, $$\hat{L}_n$$ is group-wise generated by $$\bigcup_{i=1}^n Kx_i$$, then it is also generated by the compact symmetric subset $$S=\bigcup_{i=1}^n Bx_i$$. So $$\hat{L}_n=\bigcup_m S^m$$. By Baire's theorem, there exists $$m$$ such that the compact subset $$S^m$$ has nonempty interior. So the topological group $$\hat{L}_n$$ is locally compact, contradiction. (In general, a Polish group generated by a compact subset is locally compact: this is well-known.)

Added: this is a contradiction simply because $$\hat{L}_n$$ is not locally compact (for $$n\ge 2$$) and this is something which can be seen directly. Indeed, if $$V$$ is a neighborhood of 0, then it contains the kernel $$U_m=(\hat{L}_n)_{\ge m}$$ for some $$m$$ (which is the $$m$$-term in the lower central series). Then choosing nonzero $$x$$ in $$U_m$$ and considering its scalar multiples, we see that $$U_m$$ does not have compact closure ($$U_m$$ is closed, so I'm just saying $$U_m$$ is not compact). Hence $$\hat{L}_n$$ is not locally compact.

The argument also works if $$K$$ can be made a Polish ring for which it is a countable union of compact subsets (say $$B_m$$ with $$0\in B_m\subset B_{m+1}$$). In this case the same argument works using the exhaustion by the $$(\bigcup_{i=1}^nB_mx_i)^m$$.

This applies to $$K$$ countable (well, obvious since the $$\bigcup Kx_i$$ is countable), and also to $$K$$ $$p$$-adic field.

• Thank you for the answer. I don't see any contradiction when you said that $\hat{L}_n$ is locally compact: it does not prevent the group from being a Polish group generated by a compact subset, as you stated. May 12 at 14:35
• @Qwert This is a contradiction simply because $\hat{L}_n$ is not locally compact (for $n\ge 2$) and this is something which can be seen directly. I added the argument.
– YCor
May 12 at 14:47
• Ah, now I get it. Thank you so much! May 12 at 14:48