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A group $G$ is divisible if for all $g \in G$ and $k \in \mathbb{Z}_+$ there is an element $h \in G$ such that $h^k = g$; we call such an $h$ a $k$th root of $g$. In an answer to a recent question asking for nonabelian examples, Micah pointed out than any if the exponential map of a Lie group $G$ is surjective, then the group is divisible, as any element $g \in G$ can written as $g = \exp X$ for some $X \in T_0 G$, and for any such $X$, $\exp(\frac{1}{k} X)$ is a $k$th root of $g$ for all $k$.

On the other hand, there are examples of connected Lie groups that are not divisible---for example, $\left(\begin{array}{c}-1&0\\0&-2\end{array}\right)$ has no square root in $GL_+(n, \mathbb{R}) := \{A \in GL(2, \mathbb{R}) : \det A > 0\}$.

Are there examples of connected, divisible Lie groups whose exponential map is not surjective? In other words, is surjectivity here (not) necessary?

Community
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Travis Willse
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1 Answers1

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The answer seems to be yes, according to a theorem of Hoffman and Lawson at http://jlms.oxfordjournals.org/content/s2-27/3/427.extract

orangeskid
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    Perfect, thanks @orangeskid. – Travis Willse Sep 10 '14 at 03:15
  • In case someone isn't able to access the above pdf, the relevant result is Theorem A in the linked article, which says that any closed semigroup $S$ in a connected Lie group (and in particular the Lie group itself) is divisible iff the exponential map on $S$ is onto $L(S)$, where $S \mapsto L(S)$ generalizes the construction of a Lie algebra from a Lie group in a suitable way. – Travis Willse Sep 10 '14 at 03:19
  • no worries @Travis. what if you only assume square or cubic or $p$ roots? ( + connected ) – orangeskid Sep 10 '14 at 04:19
  • I suspect that if we assume only existence of $p$th roots we still get equivalence: If $G$ divisible, for each $g \in G$ we get an algebraic homomorphism $f: \mathbb{Q} \to G$ such that $f(1) = g$, and the proof of Theorem A just uses the divisibility hypothesis to guarantee the existence of such copies of $\mathbb{Q}$ and in particular exploits its density in $\mathbb{R}$. – Travis Willse Sep 10 '14 at 04:55
  • Now, existence of $p$th roots means for each $g$ we have an algebra homomorphism $f$ from $\mathbb{Q}_p$ (the ring of rationals of the form $\frac{m}{p^n}$ for $m, p \in \mathbb{Z}$) to $G$ such that $f(1) = g$. But (for $p > 1$) $\mathbb{Q}_p$ is dense in $\mathbb{Q}$, so if density in $\mathbb{R}$ is really all that is needed, the proof should apply mutatis mutandis. (In short, in the Lie group setting, one can approximate $n$th roots by products of $(p^n)$th roots as closely as one would like.) This is of course just a heuristic argument and would require careful checking. – Travis Willse Sep 10 '14 at 05:00
  • In fact, we don't need to rewrite the proof, using their Theorem $A$ it's enough to show that if a connected Lie group admits $p$th roots for all elements $g$, then it is divisible, and again one can approximate any rational $\frac{1}{n}$ arbitrarily closely by elements $t_n$ of $\mathbb{Q}_p$. This seems to reduce the problem to showing that $f(t_n)$ converges in $G$ (since if it does converge, it must converge to an $n$th root of $g$). – Travis Willse Sep 10 '14 at 05:09
  • Yes, that seems plausible....What if a single given element has lots of divisors? Is it then an $\exp$? On another note, I recall seeing in a book by Hochschild that for complex connected Lie groups the $\exp$ is surjective. – orangeskid Sep 10 '14 at 06:18
  • The proof in the article works on a per-divisor basis, so if the above heuristic works, it works for all elements separately, i.e., it would be the case that an element of the connected component of the identity of a Lie group is in the image of the exponential map iff it has a divisor. I don't believe that $\exp$ is onto for $SL(2, \mathbb{C})$ (unless I've missed something, it misses the Jordan block $J_2(-1)$ for example), but I believe that its exponential is onto for $GL(n, \mathbb{C})$, which may be the result you have in mind. – Travis Willse Sep 12 '14 at 02:18
  • Correct, no $\log$ for $J_2(-1)$ and no square root either. But we have roots of any odd order in $SL(2, \mathbb{R})$ or $SL(2, \mathbb{C})$. Now I have to look up several things.. – orangeskid Sep 12 '14 at 08:32
  • I see---when we try to approximate, e.g., $\frac{1}{2}$ in $\mathbb{Q}_p$, with the sequence $\frac{1}{3}, \frac{1}{3} + \frac{1}{9}, \ldots$ to find a square root of $J_2(-1) \in SL(2, \mathbb{R})$, the diagonal elements of the $f(t_n)$ alternate between $1$'s and $-1$'s, and so the $f(t_n)$ do not converge as desired. Do post another comment if you find out anything interesting. – Travis Willse Sep 12 '14 at 10:25