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Perhaps there is a simple answer, but I'm very puzzled by the following question:

Question: Does there exist a (smooth, connected) algebraic group $G$ such that the general centralizer (i.e. the centralizer in a Zariski open set) is finite?

I'm pretty sure the answer is no (maybe some extra assumptions such as reductive, algebraically closed field... must be added), but I have no clue about how to prove it.

Of course, typical groups such as $GL_n, SL_n, O_n...$ are not counterexamples.

Thank you!

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    $\begingroup$ In a connected algebraic group of positive dimension, there is no element at all with finite centralizer. Indeed such an element would have an open conjugacy class, and this is not possible. $\endgroup$
    – YCor
    Mar 29 at 17:17
  • $\begingroup$ #YCor could you please give some references for these statements? $\endgroup$
    – a_g
    Mar 29 at 17:40
  • $\begingroup$ I mean, I obviously understand why such element has an open conjugacy class, but I don't see why is that a problem. $\endgroup$
    – a_g
    Mar 29 at 20:26
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    $\begingroup$ Passing to a quotient, we can suppose $G$ simple or abelian, and the abelian case is trivial to discard, so we can suppose $G$ simple. Then the condition would also mean that a generic element has finite order. Say, over the complex numbers, this implies that there exists $n$ such that $g^n=1$ for all $g$. But a simple algebraic group contains up to isogeny a copy of $\mathrm{SL}_2$. $\endgroup$
    – YCor
    Mar 29 at 21:27
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    $\begingroup$ @YCor Your argument works over fields of arbitrary characteristic. One can finish over the complex numbers by noting that $g^n=1$ is a contradiction for $g$ in a small neighborhood of $1$, but this doesn't work in characteristic $p$. However, your argument with splitting into cases and using $SL_2$ does. $\endgroup$
    – Will Sawin
    Mar 29 at 22:29

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Every element $g\in G$ is contained in a Borel subgroup $B\subseteq G$. The quotient $B^{ab}:=B/(B,B)$ has positive dimension since $B$ is solvable. Moreover, the $B$-conjugacy class of $g$ maps to a point in $B^{ab}$. Hence $\dim B/C_B(g)\le\dim (B,B)$ and therefore $\dim C_G(g)\ge\dim C_B(g)\ge \dim B^{ab}>0$.

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