Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am studying Geometric invariant theory and wonder how I should understand linearly reductive algebraic group. We say that an affine algebraic group $G$ is linearly reductive if all finite dimensional $G$-modules are semi-simple.

I am not sure if linearly reductive groups are the same as reductive groups, which are defined as algebraic groups $G$ over algebraically closed field such that the unipotent radical of $G$ is trivial. But this definition is still beyond my intuition.

Are there any good way to understand (linearly) reductive groups? It would especially be nice if reductive (Lie) groups can be characterized in geometry.

share|improve this question

1 Answer 1

up vote 6 down vote accepted

"Linearly reductive" and "reductive" are equivalent when the base field is of characteristic zero, but for prime characteristic they are different -- in fact, in prime characteristic, the only connected linearly reductive groups are algebraic tori, though for example $\operatorname{GL}_n$ is reductive.

There is a nice characterization of reductive groups over $\mathbb{C}$: they are precisely the complexifications of compact connected Lie groups. More precisely: every compact connected Lie group is actually a real algebraic group, and every smooth homomorphism of compact Lie groups is algebraic. We can then look at the points of this group over $\mathbb{C}$, and thereby obtain a complex algebraic group. For example, the complexification of $U(n)$ is $\operatorname{GL}_n(\mathbb{C})$ -- this can be seen at the level of Lie algebras from $\mathfrak{gl}_n(\mathbb{C}) = \mathfrak{u}(n) \oplus i\mathfrak{u}(n)$. It turns out that in fact every complex reductive group arises in this way from a maximal compact subgroup, and this gives an equivalence between the category of compact connected Lie groups (with smooth homomorphisms) and the category of complex reductive groups (with algebraic homomorphisms).

share|improve this answer
    
Your answer makes my understanding much clearer. Thanks! –  M. K. Oct 21 '12 at 16:12
    
@Brad I can't seem to find a source for the statement that every complex reductive algebraic group is the complexification of a compact Lie group. Do you happen to know one? –  Victor Jun 17 '13 at 19:27
1  
@Victor See, for example, "The Structure of Complex Lie Groups" by Dong Hoon Lee: tinyurl.com/lkqlfoy –  Brad Jun 19 '13 at 4:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.