General centralizer of algebraic group

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!

• 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.
– YCor
Mar 29 at 17:17
• #YCor could you please give some references for these statements?
– a_g
Mar 29 at 17:40
• I mean, I obviously understand why such element has an open conjugacy class, but I don't see why is that a problem.
– a_g
Mar 29 at 20:26
• 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 $\mathrm{SL}_2$.
– YCor
Mar 29 at 21:27
• @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. Mar 29 at 22:29

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 $$\dim C_G(g)\ge\dim C_B(g)\ge \dim B^{ab}>0$$.