Here is a definition which I invented and which I would like to understand better.
Let $ A $ be a complex affine algebraic group. Let $ X \in \mathfrak g $ be an element in its Lie algebra. We say that $ X $ generates $ A $ if there is no proper subgroup of $ A $ whose Lie algebra contains $ X $.
Is this a standard notion? Are the following two facts true?
(i) If $ A $ is generated by $ X $, then $ A $ is abelian.
(ii) If $ V $ is a representation of $ A $ and $ X $ acts by $ 0 $ on $ V $, then $ A $ acts by the identity on $ V$.
The main example that I have in mind is as follows. Let $ G $ be a reductive group and let $ X $ be a regular element. Let $ A $ be the centralizer of $ X $ in $ G$. Then $ A $ is generated by $ X $.
Even more, I would like to know if this works in families. Take $ A $ to be the group scheme of regular centralizers, viewed as a group scheme over $ \mathfrak t / W $. Then I would like to know that $ A $ is generated by the universal section $X $ and I would like to know that the conclusion of (ii) holds in this case.
So to ask some specific questions:
Does this notion of "generation" exist in the literature?
Is (ii) above always true?
Can I apply this to the group of regular centralizers?
Update: In light of Jim's answer below, I realize that the centralizer of a principal nilpotent is not generated by the principal nilpotent. (I was just being stupid.) To make this question a little more interesting, I offer up one more claim:
- A complex algebraic group $ A $ is generated by an element of its Lie algebra if and only if $ A = T \rtimes V $ where $ T $ is a torus, $ V $ is a vector space (viewed as a group under addition) such that $ T $ acts linearly on $ V $ with one-dimensional weight spaces.
For example, this means that any torus is generated by an element of its Lie algebra (the case $ V = 0 $) and that the only unipotent group generated by an element of its Lie algebra is $ \mathbb G_a $ (the case $ T = \{1 \} $).