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Question part 1: Considering reductive algebraic groups over $\Bbb C$. Let $G$ be such a group, then that means that the largest connected, normal, solvable subgroup $H\subset G$ is actually the trivial group.

What conditions are necessary for such a group to be contained in $\text{GL}(n,\Bbb C)$ for some $n$?

I know that if it is a linear algebraic group, then this follows. But I want to try to understand all representations of reductive algebraic groups over $\Bbb C$, which I know is more tangible than over $\text{char}\ne 0$.

[Perhaps to give some insight into what I am thinking here: it would be a dream for $G$ reductive to imply $G$ is a linear algebraic group, but I doubt that is true. So what more do I need?]

Question part 2: What are all of the representations of $\text{GL}(n,\Bbb C)$? Can I get all of these if I know all of the $\text{SL}(n,\Bbb C)$ representations?

Representation theory
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  • Every reductive algebraic group is linear algebraic: https://math.stackexchange.com/questions/217993/what-is-reductive-group-intuitively – Dustan Levenstein Aug 31 '17 at 13:16
  • @DustanLevenstein The dream is real?! – Representation theory Aug 31 '17 at 13:32
  • @DustanLevenstein Do you perhaps know where I can find a proof of that? – Representation theory Aug 31 '17 at 13:40
  • Unfortunately no. I just found that by googling. – Dustan Levenstein Aug 31 '17 at 17:06
  • The proof of the fact that all algebraic groups (not just reductive ones) are linear can be found in (I think) all the standard references on the topic, such as the books by Humphreys, Springer or Serre. – Tobias Kildetoft Aug 31 '17 at 18:28
  • As for understanding the representations, this is almost as easy for reductive groups as for semisimple groups, at least as long as you only consider rational representations. Basically, reductive implies linearly reductive, and the simple modules have characters given by Weyl's character formula. – Tobias Kildetoft Aug 31 '17 at 18:32
  • @TobiasKildetoft Isn't any abelian variety a counterexample to the claim that all reductive algebraic groups are affine? (I say it like this to ensure there is no confusion on my part between linear algebraic group and linearly reductive etc)

    I guess I am not sure what a unipotent element would be if it is not an element of a ring, or of a matrix group.

     – Representation theory Sep 01 '17 at 03:52
  • It is customary (but potentially unfortunate) to include affine in the definition of algebraic group, at least in the context of reductive groups. I should have been more clear about including it here to avoid confusion. – Tobias Kildetoft Sep 01 '17 at 04:57

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What you've defined (no non-trivial connected normal solvable subgroup) is the definition of a semisimple group, not of a reductive group. Semisimple implies reductive, but not vice-versa. For example, under your definition $GL_n$ would not be reductive, since its centre (the subgroup of diagonal matrices with all entries the same, which is canonically isomorphic to $GL_1$) is a connected normal solvable subgroup. A reductive group is by definition a linear algebraic group with non non-trivial connected normal unipotent subgroup (i.e. consisting of unipotent elements).

So, in particular, reductive groups are linear by definition. So, as you mention, they automatically admit a closed immersion into $GL_n$ for some $n$.

P...
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  • This is not the standard (or at least the only) definition of reductive. The usual definition does not require the group to be linear, but rather that the maximal closed, connected, solvable subgroup is a torus. As it happens, any algebraic group is in fact linear, though this is a non-trivial fact. – Tobias Kildetoft Aug 31 '17 at 18:27