Let $G$ be an algebraic group over an algebraically closed field $k$ of arbitrary characteristic and let $U$ be the hyperalgebra of $G$. Recall that $U$ is defined as the subspace of the full linear dual of $k[G]$ consisting of elements that vanish on some power of the ideal $I \subseteq k[G]$ defining the identity in $G$. Then one has the standard degree filtration $U_{\leq n}$ on $U$ given by setting $$ U_{\leq n} := ( k[G]/I^{n+1} )^* . $$
Set $\mathfrak g := $ Lie$(G)$. In characteristic 0, $U$ is just the enveloping algebra of $\mathfrak g$, and the PBW theorem tells us that the ring gr $ U $ is isomorphic to $S(\mathfrak g)$. Moreover, there is a $G$-equivariant symmetrization isomorphism $S(\mathfrak g) \to U$ of vector spaces over $k$, where we take the conjugation action of $G$ on $U$ (note that this is NOT an algebra isomorphism, however). Remark that the symmetrization isomorphism is given explicitly by $$ x_1 \cdots x_n \mapsto \frac 1 {n!} \sum_{\sigma \in S_n} x_{\sigma(1)} \cdots x_{\sigma(n)} . $$
Now assume that char $k = p > 0$. In this case, gr $U$ is a commutative algebra -- let's call it $S'(\mathfrak g)$ -- that is not a polynomial ring. (One way of seeing this is that the ring $S'(\mathfrak g)$ is infinitely generated as a $k$-algebra). Note that $S'(\mathfrak g)$ carries a natural $G$-module structure coming from the conjugation $G$-action on $U$. I'm wondering if one still can construct a $G$-equivariant "symmetrization" isomorphism $S'(\mathfrak g) \to U$ of $k$-vector spaces in this case as well. (Clearly the formula for the symmetrization morphism in characteristic 0 doesn't work, since $n!$ isn't invertible in $k$ for $n \geq p$).
