Taking roots in simple linear algebraic groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T07:53:56Z http://mathoverflow.net/feeds/question/14801 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14801/taking-roots-in-simple-linear-algebraic-groups Taking roots in simple linear algebraic groups blt 2010-02-09T19:43:17Z 2010-02-10T15:20:35Z <p>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 $h^n=g$?</p> <p>(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!)</p> <p>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?</p> http://mathoverflow.net/questions/14801/taking-roots-in-simple-linear-algebraic-groups/14814#14814 Answer by Kevin McGerty for Taking roots in simple linear algebraic groups Kevin McGerty 2010-02-09T21:41:15Z 2010-02-10T15:20:35Z <p>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. </p> <p>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. </p> <p>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 $1$.</p> <p>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?</p> http://mathoverflow.net/questions/14801/taking-roots-in-simple-linear-algebraic-groups/14816#14816 Answer by Pavel Etingof for Taking roots in simple linear algebraic groups Pavel Etingof 2010-02-09T22:03:08Z 2010-02-09T22:03:08Z <p>A semisimple element lies in a maximal torus, so you can extract any root from it inside this torus. </p>