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 want to understand the regular representation of an affine algebraic group. An affine algebraic group as I know it, is a functor from the category of $k $ -algebras to groups that is representable when considered as a functor from $k$ algebras to sets. The coordinate ring $ \mathcal{O} (G) = Nat (G,\mathbb{A}^1)$ where $ \mathbb{A} ^1 $ the functor from $k$-algebras to sets.

It is written in Milne's book on Affine Group Schemes that the regular representation may be defined such that for $g \in G(R)$, $f \in \mathcal{O} (G)$ and $x \in G(R)$, Then $(gf)_R (x) = f_R (xg)$.

I don't understand why this gives us another element in the coordinate ring (i.e. an element in $Nat(G,\mathbb{A}^1)$ ). For instance if we pick another algebra $Z$ what what is then $(gf)_Z (z)$ where $g\in G(R)$ and $z \in G(Z)$? Why is this data enough to give a linear representation?

share|improve this question
2  
I don't have time to write up the details right now, but you might be able to find more details in Jantzen's Representations of Algebraic Groups (at least I recall it having the details of this). –  Tobias Kildetoft Jun 3 '13 at 0:05

1 Answer 1

up vote 2 down vote accepted

The definition that Milne uses for a representation on $V$ is that it is a natural transformation of functors $$G \to \operatorname{End}(V)$$ such that the components $$G(R) \to \operatorname{End}_R(V \otimes_k R)$$ are homomorphisms. If $V = k[G]$ is the coordinate ring of a scheme then $V \otimes_k R = k[G] \otimes_k R = R[G]$ is the coordinate ring of the scheme $G_R$ obtained by restricting $G$ to $R$-algebras.

For you this means that the function $gf$ will lie in $R[G] = \operatorname{Nat}(G_R, \mathbb A^1_R)$, so in particular you only have to define it's value on elements $x \in G(A)$ where $A$ is an $R$-algebra.

To do this note that being an $R$-algebra means there is a ring homomorphism $R \to A$. Then $G$ being a functor we get a homomorphism $G(R) \to G(A)$ so we can push $g$ into $G(A)$ in order to multiply it with $x$. Then apply $f_A$ to the result.

share|improve this answer
    
Thank you Jim, it is clear to me now! –  Anette Jun 3 '13 at 14:49

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.