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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?

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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

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    $\begingroup$ Perfect, thanks @orangeskid. $\endgroup$ – Travis Willse Sep 10 '14 at 3:15
  • $\begingroup$ 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. $\endgroup$ – Travis Willse Sep 10 '14 at 3:19
  • $\begingroup$ no worries @Travis. what if you only assume square or cubic or $p$ roots? ( + connected ) $\endgroup$ – orangeskid Sep 10 '14 at 4:19
  • $\begingroup$ 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 $\mathbb{R}$. $\endgroup$ – Travis Willse Sep 10 '14 at 4:55
  • $\begingroup$ 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. $\endgroup$ – Travis Willse Sep 10 '14 at 5:00

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