algebraic complex group and reductive algebraic group

Question

A complex algebraic simply-connected group is always reductive? I saw that you con defined a langlands dual group for a reductive algebraic group in general but then I've found out a text that use the langlands dual group of a simple, simplyconnected, complex algebraic, group and I don't know why. Maybe is a trivial question but I am at the begin of these theory so I'm confused

Answer 1

I think you have been confused by a collision of terminology. It is not true that if $G$ is an algebraic group over $\mathbb{C}$, that is, a complex algebraic variety with compatible group structure, that is moreover simply connected as a real manifold, then $G$ is a reductive algebraic group. For example, the additive group $(\mathbb{C}, +)=\mathbb{G}_a(\mathbb{C})$ is abelian and unipotent.

Suppose however that $G$ is a reductive algebraic group over $\mathbb{C}$. Then it has a coroot and coweight lattice. The coroot lattice sits inside the coweight lattice. Then $G$ is said to be simply-connected if they are equal. It is a fact that the quotient group agrees with the fundamental group of $G$ viewed as a real manifold. When $G$ is simply-connected, the Langlands dual will be of adjoint type (the root lattice will equal the weight lattice) essentially by definition. In general, Langlands duality will exchange the fundamental group and the centre.