Let $G$ be a connected algebraic group over an algebraically closed field $k$. I'd like to see that the following two definitions are equivalent.
- An element $g\in G$ is regular if $\dim(C_G(g))$ is minimal.
- An element $g\in G$ is regular if $\dim(C_G(g))=\mathrm{rank}(G)$.
To show that these definitions are equivalent, one must show that $\dim(C_G(g))\ge\mathrm{rank}(G)$ for all $g\in G$ and that there is some element yielding equality. How can I show these two facts? Recall that the rank of $G$ is the dimension of a maximal torus $T\subset G$.
An element need not be in a maximal torus, but in a connect algebraic group over an algebraically closed field, every element $g$ does lie in some Borel subgroup, $B$. Let $U$ be the unipotent radical of $B$. Then $B/U$ is abelian so $(xU)^b = xU$ for all $b \in B$, and so $x^b \in xU$ for all $b \in B$. That means the dimension of the $B$-conjugacy class of $x$ is at most $\dim(U)$, so: $$\dim( C_G(x) ) \geq \dim( C_B(x) ) = \dim(B) - \dim( x^B ) \geq \dim(B) - \dim(U) = \dim(T) = \operatorname{rank}(G).$$
If $x$ is semisimple, then you can show there is a maximal torus $T$ with $x \in T \leq C_G(x)$, but if $x$ is unipotent then the centralizer is a combination of unipotent and semi-simple elements (and at least to me they look pretty weird; take a regular unipotent element of GL, the centralizer is a triangular Toeplitz matrix).