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Let U be a subgroup of unipotents matrices(A - E is nilpotent). U is affine algebraic group. How to describe finite($\phi^n = id$) algebraic automorphisms $\phi$ of U such that $U^{\phi}$ is finite?

I solved this in case dim U = 1. In this case U = $\mathbb C$. Any automorphism is described by $a \in \mathbb C^*$. $\phi_a(x) = ax$.

$U^{\phi_a} = {0}$ therefore is finite. $\phi_a^n = \phi_{a^n} = 1$ so $a^n = 1$. So all finite automorphisms with finite $U^{\phi}$ are described by $a \in \mathbb C^*$, such that $a^n = 1$.

What to do if dim U > 1. Any help is appreciated.

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I will give some hints.

Case (i) $U$ is commutative: then you can find vector space automorphisms of finite order. It is a matrix $A$ such that $A^n=I$, you can easily find reflection matrices.

Case (ii) Non-commutative. See if you can embed $U$ as a normal subgroup of some linear algebraic group. Typically, in some parabolic $P$ of a reductive group. Then any inner automorphism of finite order in $P$ will do. Take a Levi decomposition and an element of finite order in the Levi subgroup.

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  • $\begingroup$ Thanks. Case(i) is the same as dim U = 1 as if U is commutitative then its $\mathbb C^n$. So my solutions works as well. I missunderstood case(ii), will it give me ALL automorphisms? Why any automorphism of U is a reducion of automorphism of P? $\endgroup$ – SaveMyLife Apr 25 '17 at 13:28
  • $\begingroup$ I am not claiming we will get all automorphisms this way. $\endgroup$ – P Vanchinathan Apr 25 '17 at 16:27
  • $\begingroup$ Okay, this was not clear for me. $\endgroup$ – SaveMyLife Apr 25 '17 at 16:53

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