Suppose $G$ is a simple (linear) algebraic group over an algebraically closed field of characteristic zero, that $n$ is a positive natural number, and that $g\in G$. Can we always find an $h\in G$ such that $h^n=g$?

(It appears to be possible to check this for the classical algebraic groups by direct computations in each case, but covering the exceptional Lie Algebras this way seems like it might be tricky, and anyhow, I'm inclined to think that a case analysis is probably not the optimal way of approaching this problem!)

Note added: Kovalev made a comment showing that the answer is `no' in general. The counterexamples appear to revolve around non-semisimple elements. I wonder whether the answer becomes positive if one restricts oneself to $g$ of finite order?

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    $\begingroup$ The counterexample in the paper seems to give a counterexample to the question here too; it gives an example of an element of Sp(4,\Qbar) which, not being in the image of the exponential map when considered as an element of Sp(4,\C), cannot have all nth roots. $\endgroup$ – blt Feb 9 '10 at 20:56

A semisimple element lies in a maximal torus, so you can extract any root from it inside this torus.


As Pavel says, any semisimple element lies in a maximal torus, where you can take any root. On the other hand, if $g$ is unipotent, then it is in the image of the exponential map, so you can make sense of $g^\lambda$ for any $\lambda \in \mathbb C$, so counterexamples must have nontrivial semisimple and unipotent parts.

Suppose $g=s.u=u.s$ is the Jordan decomposition of an element. Then $s^n$ and $u^n$ are semisimple and unipotent respectively, so they are the Jordan decomposition of $g^n$. Thus the existence of roots is compatible with Jordan decomposition.

Now take a semisimple element $s$ such that $Z_G(s)^0$ does not contain a central torus (so its center is finite) -- if $G$ is simply connected then in fact $Z_G(s)$ is connected, so I'll assume that. Now pick a regular unipotent element $u$ in $Z_G(s)$ and consider $g =s.u$. I want to claim $g$ is a counterexample. Indeed suppose for each $n$ we have $h_n$ an $n$-th root, and $h=s_nu_n$ is its Jordan decomposition. Then $s_n^n =s$ and $u_n^n=u$, and both $s_n$ and $u_n$ lie in $Z_G(s)$. Then we see that $s_n$ centralizes $u$ for all $n$, but since $u$ is regular in $Z_G(s)$ and $s_n$ semisimple it follows that $s_n$ must lie in the centre of $Z_G(s)$. But then taking, say, $n$ equal to the order of that centre (which is finite) we get a contradiction, as $s_n^n$ must then be $1$.

Semisimple elements $s$ such that $Z_G(s)$ does not contain a central torus exist, but there are only finitely many conjugacy classes of them, as was essentially shown in the paper of Borel and de Siebenthal. In fact I think that paper establishes that there are $r+1$ such classes where $r$ is the rank of $G$, so this would give a negative answer for the exceptional groups also. I suspect that these might somehow be the only counterexamples?

  • $\begingroup$ I may be being stupid, but this appears to contradict P. Etingof's answer above. Is there something I'm missing here? $\endgroup$ – blt Feb 10 '10 at 0:41
  • $\begingroup$ I also did not understand this argument. The group $Z$ could be smaller than $Z_G(x)$, so it could contain a central torus, even if $Z_G(x)$ does not. E.g. take $x$ to be central in $G$ (say, $-1$ in $SL(2)$). $\endgroup$ – Pavel Etingof Feb 10 '10 at 1:59
  • $\begingroup$ My post yesterday was pretty incoherent, but hopefully the edit is better! $\endgroup$ – Kevin McGerty Feb 10 '10 at 15:23
  • $\begingroup$ Now I understand. So, the simplest example is the non-semisimple element of $SL(2)$ with eigenvalue -1. Thanks! $\endgroup$ – Pavel Etingof Feb 10 '10 at 17:41
  • $\begingroup$ Oops yes thanks, I meant to spell that out in my answer -- it's a simpler example than the one in the paper mentioned in Kovalev's comment (which uses the $SL_2\times SL_2$ subgroup in $Sp_4$). $\endgroup$ – Kevin McGerty Feb 10 '10 at 18:52

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