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Let $k$ be an arbitrary field (in my case $k = \mathbb Q_p$), and $G\subset \mathrm{GL}(n)_{/k}$ a reductive group. Let $G^0$ be its identity connected component.

Suppose that $G^0(k)$ contains an element with pairwise distinct eigenvalues (in the natural representation $G(k)\subset \mathrm{GL}(n, k)$), and hence a Zariski-dense subset of such elements.

Is it the case that every connected component of $G(k)$ contains an element with pairwise distinct eigenvalues?

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  • $\begingroup$ "distinct" means "not all equal" or "pairwise distinct"? $\endgroup$
    – YCor
    Jul 3, 2017 at 17:26
  • $\begingroup$ For that matter, just to be clear, by "eigenvalues" do you mean "… in the natural representation of $\mathrm{GL}(n)$", "… in the adjoint representation of $G$", or something else? $\endgroup$
    – LSpice
    Jul 3, 2017 at 18:26
  • $\begingroup$ @YCor distinct means pairwise distinct. Specifically, I want to show that if $V$ is the closed subvariety of elements with a repeated eigenvalue, then $V$ has dimension strictly smaller than the dimension of $V$. I therefore need to rule out the case that $V$ contains a connected component of $G$. $\endgroup$ Jul 3, 2017 at 18:40
  • $\begingroup$ @LSpice I mean in the natural representation $G(k)\subset GL(n, k)$ $\endgroup$ Jul 3, 2017 at 18:49

1 Answer 1

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No. Consider the group $H$ of $2\times 2$ matrices of the form $d_t=\begin{pmatrix}t & 0 \\ 0 & t^{-1}\end{pmatrix}$ or $c_t=\begin{pmatrix} 0 & u^{-1} \\ u & 0\end{pmatrix}$ with $t\in k^*$. Let $G$ be the group of $4\times 4$ matrices of the form either $D_{t,u}=\begin{pmatrix}d_t & 0 \\ 0 & d_u\end{pmatrix}$ or $C_{t,u}=\begin{pmatrix}c_t & 0 \\ 0 & c_u\end{pmatrix}$, $t,u\in k^*$.

Its identity component contains a matrix with no double eigenvalue (any $D_{t,u}$ with $t,u,t^{-1},u^{-1}$ pairwise distinct. On the other hand, the other component consists of the $C_{t,u}$, which have the spectrum $(1,-1,1,-1)$.

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  • $\begingroup$ Thanks for the answer! Do you know if there are any conditions that could be imposed to make this true? For example, in the case that I'm trying to apply this to, I have some information about the Lie algebra of $G$ -- in particular, I know that it's non-trivial. $\endgroup$ Jul 4, 2017 at 10:58
  • $\begingroup$ "non-trivial" is not clearly defined: you probably mean non-abelian. Maybe you can tensor this rep to the standard rep of $SL_2$ and get something similar with non-abelian Lie algebra. $\endgroup$
    – YCor
    Jul 4, 2017 at 11:29

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