**Background:**

Let $G$ be a linear algebraic group over an algebraically closed field $k$ and let $I \subseteq k[G]$ be the ideal of the identity element. The hyperalgebra $U(G)$ of $G$ is defined to be the subspace of the linear dual of $k[G]$ consisting of all $f$ such that $f(I^n) = 0$ for some $n > 0$; then there is a natural Hopf algebra structure on $U(G)$.

Given any Hopf algebra $A$ over a field $k$, one can also define the Hopf dual $A^*$ of as follows: Let $A^*$ be the subspace of the full linear dual of $A$ consisting of elements that vanish on some two-sided ideal of $A$ of finite codimension. Then $A^*$ has a natural Hopf algebra structure.

EDIT: As pointed out in comments, I am using what might be nonstandard terminology -- perhaps I should write $A^\circ$ instead of $A^*$.

**Questions**:

I now have two related questions.

1) I assume that in general the hyperalgebra of $G$ is not the same as the Hopf algebra dual of $k[G]$. What *is* the Hopf dual of $k[G]$?

2) Although one obtains $U(G)$ from $k[G]$ by a dual construction, you can't in general obtain $k[G]$ from $U(G)$ -- for example, if $G$ is reductive, you can't get $k[G]$ from $U(G)$ by some sort of duality since the hyperalgebra doesn't see isogeny. Is there something one can say in the reductive case about the structure of the Hopf dual of $U(G)$? More generally, for which algebraic groups $G$ can we obtain $k[G]$ from $U(G)$ by some sort of duality? And when we can do so, is it as simple as just taking the Hopf dual?

`$A^*$`

not as the full dual, but as only those linear maps $A \to k$ that factor through some finite-dimensional quotient algebra. This "Hopf dual" I think is a Hopf algebra with the algebraic tensor product, isn't it? I've usually seen it written not as`$A^*$`

, which should mean the full dual, but as something like $A^\circ$. $\endgroup$ – Theo Johnson-Freyd Jul 9 '10 at 19:54