# Semisimple algebraic group vs semisimple Lie algebra

Let $G$ be a connected affine algebraic group over $\mathbb{C}$ of positive dimension.

By definition:

1. $G$ is semisimple as an algebraic group if it has no non-trivial connected, normal, abelian algebraic subgroups.
2. $G$ is semisimple as a complex Lie group if it has no non-trivial connected, normal, abelian Lie subgroups.
3. $G$ has semisimple Lie algebra if its Lie algebra has no nonzero abelian ideals.

My question is: Are 1. 2. 3. all equivalent?

I think 1. and 2. are equivalent because any connected, normal, abelian subgroup is automatically both an algebraic subgroup and a Lie subgroup. Is that right? It seems that 2. and 3. are also equivalent, yes?

The fact that 1. 2. 3. are all equivalent would be consistent with the Encyclopedia of Mathematics which says here that "If the ground field is the field $\mathbb{C}$ of complex numbers, a semi-simple algebraic group is nothing but a semi-simple Lie group over $\mathbb{C}$". But the reason I am being wary is that they say the same thing with reductive groups, however I have been told that an affine algebraic group over $\mathbb{C}$ is reductive if and only if its Lie algebra is reductive and its center is of multiplicative type. I want to make sure that such a precaution is not necessary for the notion of semisimplicity. Could we not imagine that $G$ has a finite center which is not of multiplicative type? Would such a group still be considered semisimple?

• Yes, it sounds good. These issues have been discussed here; in particular see the answer of Qiaochu Yuan. – Dietrich Burde Oct 24 '16 at 8:36

It is pretty clear that 3. $\implies$ 2. $\implies$ 1. (A connected Lie subgroup is detected by its Lie algebra, and a connected algebraic subgroup is necessarily also a connected Lie subgroup.)
So one approach to getting the equivalence is to show that 1. $\implies$ 3. This can be done by sufficiently developing the theory of linear algebraic groups.
Alternatively, one could show that 2. and 3. are equivalent, and then show 1. $\implies$ 2. by showing that a Lie subgroup is necessarily an algebraic subgroup. This is not a general fact about algebraic groups; you would have to use something more (that $G$ is linear, or perhaps even 1.) to help show it.
[To see why, consider the complex torus $\mathbb C^2/\mathbb Z[i]^2$, which is a product of two elliptic curves, and so is naturally an algebraic group. This has lots of Lie subgroups, obtained by looking at the image of the $\mathbb CP^1$ worth of lines through the origin in $\mathbb C^2$. But the lines of irrational slope (i.e. whose slope is not in $\mathbb Q(i)$) will not have algebraic image. This shows that Lie subgroups of complex algebraic groups aren't automatically algebraic subgroups.]
The key difference with reductive groups is that the trivial Lie algebra $\mathbb C$ has two algebraic "models": the additive group $\mathbb G_a,$ and the multiplicative group $\mathbb G_m$. And one doesn't want to include the former under the umbrella of the word reductive.