I'm studying on this Kostant article, section 3.1/3.2. I'm dealing with regular elements in a Lie groups and I'm a bit confused about some Lie groups theory background that I don't have. I got two questions.
So let $K$ be a compact connected Lie group and $T\subset K$ a maximal torus. Then we consider the root system associated with the pair $(\mathfrak{g},\mathfrak{h})$, where $\mathfrak{g}$ and $\mathfrak{h}$ are the complexifications of the Lie algebras of $K$ and $T$.
We say that $a \in K$ is regular if it's contained in exactly one maximal torus (T.Brocker,T. tom Dieck, "Representation of Compact Lie Groups"). Now, in section 3.1 of Kostant article he says that, in case $K$ is also simply connected,checking regularity of an element $a$ reduces to checking if the centralizer $K^a$ is a torus.
Why? I see that $a$ is regular iff $K^a$ is a maximal torus. How can I prove that maximality isn't necessary? I think is a matter of connectedness, but why is $K^a$ connected? I think there's some homotopy exact sequence here, but I'd need some references or some alternative proofs.
Next in the section he states, without references, that, given $a \in T$, $a$ is regular if and only if
$$ e^\beta(a) \neq 1, \ \text{ for all roots } \beta $$
where $e^\beta(a)=e^{\beta(x)}$ with $x \in \mathfrak{t}$ such that
$exp(x)=a.$
I know that a similar results stands for semisimple Lie algebras... but how does it work in Lie groups?
Thanks to everyone for the attention,
Bye
To put my comments together into an answer. Firstly, the natural basic definition of a regular element (in the Lie group or in the Lie algebra) is that it has centraliser of lowest possible dimension. We observe separately that the centraliser of a regular element turns out to be a maximal torus (or toral subalgebra in the Lie algebra version). Combining those two facts, if we have an element $g$ whose centraliser $C(g)$ is a torus then $C(g)$ must be contained in a maximal torus $T$ but there are no elements with centralisers of lower dimension than a regular element so $C(g)$ must be a torus of the same dimension as $T$ (all maximal tori have the same dimension). Then we can quickly see $C(g)$ must be a maximal torus itself, $C(g) = T$.
Next we note that an equivalent definition of regular can be framed as follows. Take a maximal torus $T\leq G$ or corresponding toral subalgebra/Cartan subalgebra $\mathfrak{t}\leq \mathfrak{g}$. Note we have just seen every regular element is contained in one of these given by its centraliser. Not every element in the torus is regular but we can identify them by the conditons:
Lie algebra version: $x \in \mathfrak{t}$ is regular if $\beta(x) \neq 0$ for every root in the root system corresponding to $(\mathfrak{g}, \mathfrak{t})$.
Lie group version: $a \in T$ is regular if $e^{\beta(x)} \neq 1$ for every root in the root system corresponding to $(\mathfrak{g}, \mathfrak{t})$ where $a = \exp(x)$.
Note this is not saying $a = \exp(x)$ is regular if $x$ is. Indeed there is a definite problem with that idea. If $\beta(x)$ is imaginary for some $x$ then for some $\lambda \in \mathbb{R}\setminus \{0\}$, we will have $e^{\beta(\lambda x)} = 1$ but $\lambda x$ should still be regular if $x$ is, by the Lie algebra definition. For example, if $\beta(x) = i$ then $\beta(2\pi x) = 2\pi i$ so that $e^{\beta(2\pi x)} = 1$.
This is to be expected of course, $\exp$ is not injective so we may well have another $y \in \mathfrak{t}$ such that $a = \exp(y)$ as well and this $y$ might not be regular. I suspect but don't know a proof off the top of my head that the converse is true i.e. if $a\in T$ is regular, then any $x\in \mathfrak{t}$ with $a=\exp(x)$ will be regular.