4
$\begingroup$

I've been reading Waterhouse's book "Introduction to affine group schemes", in part to help prepare myself for an (oral) advanced topic exam in algebraic geometry. There is one exercise in chapter 1 that has been giving me trouble. Let $G$ be an affine group scheme with associated Hopf algebra $A$. The exercise says that I should prove the following Hopf-algebraic fact about $A$ by translating it to a basic fact about the group theory of $G$ : "The map $A \otimes A \rightarrow A \otimes A$ sending $a \otimes b$ to $(a \otimes 1)(\Delta(b))$ is an algebra isomorphism".

The other parts of the exercise give Hopf-algebraic facts corresponding to really basic group theory facts, like $(x^{-1})^{-1} = x$ and $(xy)^{-1} = y^{-1} x^{-1}$ and $1^{-1} = 1$. However, I can't figure out which group-theoretic fact the above corresponds to. It almost seems like it is saying that there is some automorphism of the group corresponding to the above Hopf-algebra isomorphism; however, the only group automorphisms I know that exist in general are the inner ones, and those don't seem to do the job.

$\endgroup$
0

2 Answers 2

10
$\begingroup$

You are confusing algebra isomorphisms with Hopf algebra isomorphisms. The map $A\otimes A \to A\otimes A$ given by $a\otimes b\mapsto \left(a\otimes 1\right)\left(\Delta\left(b\right)\right)$ is an algebra isomorphism but not a Hopf algebra isomorphism in general. So it corresponds not to a group automorphism of $G\times G$, but to an automorphism of the affine scheme $G\times G$. This automorphism is the one that sends $\left(x,y\right)$ to $\left(x,xy\right)$ in terms of points.

$\endgroup$
0
3
$\begingroup$

Consider the map of groups $G\times G \to G\times G$, $(g_1,g_2) \mapsto (g_1,g_1\cdot g_2)$. Clearly, it is an isomorphism, hence induces an isomorphism of the corresponding Hopf algebras, which is given by precisely the map you are asking about.

$\endgroup$
3
  • $\begingroup$ It's not a map of groups, though, unless you assume $G$ to be abelian. $\endgroup$ Commented Dec 11, 2012 at 5:46
  • $\begingroup$ @darij grinberg : Yes, I figured that out an instant ago, so I switched the answer to yours. $\endgroup$
    – Catherine
    Commented Dec 11, 2012 at 5:47
  • $\begingroup$ Of course it is not an isomorphism of groups, but still it gives an isomorphism of algebras. $\endgroup$
    – Sasha
    Commented Dec 11, 2012 at 5:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .