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For an affine algebraic group over $\mathbb{C}$ which is simple, in the sense of no normal subgroups closed in the Zariski topology except for finite central subgroups and the whole thing, I'm trying to refresh my memory of how to show that in that case your Lie algebra over C is going to be simple in the sense of no ideals except for zero and the whole thing.

We have a correspondence between subalgebras and subgroups, and any subalgebra of your Lie algebra will give rise to a Lie group and closed immersion of that (relative to the strong topology) into your original Lie group, but I would presume that there will be no guarantee the range will be Zariski closed. So I'm just trying to remind myself how to fill in the rest of the argument. Is this result stated in my first paragraph actually correct for $\mathbb{C}$?

My main reason for asking this is that I'm wondering whether it can be generalised to non-archimedean local fields, but I thought it might be convenient to start by thinking about the case of the complex numbers.

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    $\begingroup$ For a Lie group, "Zariski closed" makes no sense in general. It turns out to make sense in a unique way in a simple complex Lie group, but this takes some energy to prove. Maybe you just want to start with $G(\mathbf{C})$ for some complex algebraic group. $\endgroup$
    – YCor
    Commented Mar 27, 2020 at 9:56
  • $\begingroup$ Thank you for reminding me of that point, so I made a few modifications to my question which I hope now make it a more sensible question. $\endgroup$
    – Rupert
    Commented Mar 27, 2020 at 10:00
  • $\begingroup$ The sketch would be: if $\mathfrak{g}$ has a nontrivial abelian ideal, then $G(\mathbf{C})$ has a nontrivial abelian normal subgroup (taking the exponential), hence its Zariski closure yields a contradiction. So $\mathfrak{g}$ is semisimple (and nontrivial: I assume the group has positive dimension since otherwise the claim is false). If $\mathfrak{g}$ were not simple, it would have a simple factor: then the centralizer of this factor is not trivial, and hence the centralizer of this factor in the group is a nontrivial proper Zariski-closed subgroup. $\endgroup$
    – YCor
    Commented Mar 27, 2020 at 10:02
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    $\begingroup$ Yes: if $G$ is a (finite-dimensional) Lie group over a complete normed field of characteristic zero (a) if its Lie algebra is not semisimple, then $G$ has a nontrivial abelian normal subgroup (b) if its Lie algebra is semisimple and not simple, then $G$ has a positive-dimensional proper normal subgroup, obtained as centralizer of some simple factor. $\endgroup$
    – YCor
    Commented Mar 27, 2020 at 10:12
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    $\begingroup$ There are some very good books of Tony Springer, of J. Humphreys, and of A. Borel, all with the title Linear Algebraic Groups, which cover this material in detail. $\endgroup$
    – Ben McKay
    Commented Mar 27, 2020 at 12:57

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