Distributions of a group scheme as differential operators. I'm trying to understand the distributions on an affine group scheme act as differential operators. Let $X$ be a scheme and $G$ a group scheme both affine over some commutative ring $k$. Suppose $\alpha:G\times X\to X$ is an action on $X$. Then there is a morphism between the coordinate algebras $\Delta:k[X] \to k[G]\otimes k[X]$, which makes $k[X]$ a comodule.
Let $I$ be the kernel of the augmentation map $\varepsilon:k[G]\to k$. A distribution of order $\leq n$ on $G$ is a $k$-linear map $\mu:\frac{k[G]}{I^{n+1}} \to k$. Each distribution induces a $k$-linear endomorphism on $k[X]$ via
$$D_{\mu}: k[X] \to_\Delta k[G]\otimes k[X] \to _{\mu\otimes \text{id}} k\otimes k[X] \to_\cong k[X] $$
Jantzen claims in chapter 7 of Representation of Algebraic Groups, that this endomorphism $D_{\mu}$ is a differential operator on $k[X]$. He cites a book by Demazure and Gabriel, which I've also looked at, but they don't provide a proof either. I've been trying to prove it for quite a while now, but haven't made any progress. Does anybody know a proof of this or a resource to find one?
 A: Question: "I've been trying to prove it for quite a while now, but haven't made any progress. Does anybody know a proof of this or a resource to find one?"
Answer: Let $k$ be a fixed commutative unital ring and let $G:=Spec(R)$ be a $k$-group scheme
($R$ is a finitely generated $k$-algebra).
Let $\epsilon: R \rightarrow k$ be any $k$-valued point $\epsilon \in G(k)$ with $I:=ker(\epsilon)$  and let
$$\mu\in Hom_k(R/I^{n+1}, R/I):=Dist^n(R,I):=Dist^n(G,I).$$
It seems to me that there is an isomorphism
$$Dist^n(R,I) \cong Diff^n_k(R/I^{n+1}, R/I)$$
where $Diff^n_k(R/I^{n+1}, R/I)$ is the module of $n$'th order differential operators from $R/I^{n+1}$ to $R/I$.
You may see this as follows: There is a splitting $s:R/I\rightarrow R/I^2$ and let $a:=s \circ p$ where $p: R/I^2 \rightarrow R/I$ is the canonical map. Any element $u\in I^2$ is a sum of elements on the form
$$u:=(x-a(x))(y-a(y))z$$
with $x,y,z \in R$. Since $\mu$ is well defined we get
$$A1.\text{  }0=\mu((x-a(x))(y-a(y))z)=$$
$$\mu(xyz)-x\mu(yz)-y\mu(xz)+xy\mu(z)=[[\mu, \phi_x] \phi_y](z)=0$$
where $\phi_x(w):=xw$ is the multiplication map. Hence $\mu$ is a differential operator of order one. An induction proves the result for all $n\geq 1$.
Definition: If $k\rightarrow R$ is a map of commutative rings and $E,F$ are $R$-modules you define $Diff^n_k(E,F)$ inductively as the set of maps $D\in Hom_k(E,F)$
with
$$[\cdots [D, \phi_{a_0}]\cdots ]\phi_{a_{n}}]=0$$
for all elements $a_0,..,a_n\in R$. Here $[D,\phi_a](u):=D(au)-aD(u)$.
Hence the map $\phi_a$ is multiplication with $a$ and the product $[,]$ is the Lie product. By definition
$$Diff_k(E,F):= \cup_n Diff_k^n(E,F).$$
You must verify that your map $D_{\mu}$ satisfies the conditions in the definition.
You get a canonical map
$$ \phi:R\otimes_k R \rightarrow R/I^{n+1}\otimes_k R \rightarrow^{\mu \otimes 1}R/I\otimes_k R \cong R$$
defined by
$$ \phi(a\otimes b):= \mu(\overline{a})b \in R$$
which is $k$-linear.  Let $J\subseteq R\otimes_k R$ be the ideal of the diagonal and let $J^{n+1}$ be its $n+1$'th power. It seems to me that $\phi(J^{n+1})=0$ in $R$ hence you get a canonical $k$-linear map
$$\phi_{\mu}: R\otimes_k R/J^{n+1} \rightarrow R.$$
This is proved with an argument similar to the one in A1 above - the calculation is similar.
If you change the definition of the map $\phi$ to the map
$$\psi: R\otimes_k R \rightarrow R$$
with $\psi(a\otimes b):=a\mu(\overline{b})$ you get a well defined left $R$-linear map
$$\psi_{\mu}: R\otimes_k R/I^{n+1} \rightarrow R.$$
For the composed map
$$D_{\mu}:=\phi \circ \Delta: R \rightarrow R$$
to be a differential operator, you must prove it factors via an $R$-linear map
$$\phi_1: R\otimes_k R/J^{n+1} \rightarrow R$$
with $D_{\mu}=\phi_1 \circ d^n$
where $d^n$ is the "canonical differential operator" $d^n: R \rightarrow R\otimes_k R/J^{n+1}.$
The map $E_{\mu}:= \psi_{\mu} \circ d^n$ is by the above argument a differential operator of order $\leq n$.
Jantzen defines $Dist(G,I):=\cup_u Dist^n(G,I)$ by taking the union.
Maybe the fact that $\mu$ is a differential operator helps in the case when $G$ acts on itself.
In Jantzens book he defines the  distribition $Dist(G)$ using
$Dist(G,I)$ where $I$ corresponds to the unit element $e\in G$. If $k$ is a field of characteristic zero and $Lie(G)$ is the Lie algebra of $G$ there is an isomorphism
$$Dist(G) \cong U(Lie(G))$$
where $U(Lie(G))$ is the universal enveloping algebra of $Lie(G)$. There is a canonical filtration $U^n(LieG))\subseteq U(Lie(G))$ and an induced isomophism
$$Dist^n(G) \cong U^n(Lie(G)).$$
Hence the algebra of distributions generalize the universal enveloping algebra of the Lie algebra of $G$ to group schemes over Dedekind domains or more general base schemes $S$.
Another approach: If $k$ is a field and if $P^n_{G/k}:=R\otimes_k R/J^{n+1}$ it follows there is an isomorphism
$$R/I^{n+1} \cong \mathcal{O}_{G,x}/\mathfrak{m}_x^{n+1}$$
where $x\in G$ is the point corresponding to $I \subseteq R$. There is an exact sequence
$$0 \rightarrow I^{n+1}\rightarrow R \rightarrow R/I^{n+1} \rightarrow 0$$
and localizing (let $S:=R-I$) you get since $R/I^{n+1}$ is a local ring
$$S^{-1}(R/I^{n+1}) \cong R/I^{n+1} \cong S^{-1}R/S^{-1}I^{n+1}\cong \mathcal{O}_{G,x}/\mathfrak{m}_x^{n+1}.$$
There is moreover
an isomorphism
$$P^n_{G/k} \otimes \kappa(x) \cong \mathcal{O}_{G,x}/\mathfrak{m}_x^{n+1}.$$
By definition
$$Hom_R(P^n_{G/k}, R) \cong Diff^n_k(R,R)$$
which is the module of n'th order differential operators on $G$. There is an equality
$$Hom_k(R/I^{n+1}, R/I) \cong Hom_{\kappa(x)}(P^n_{G/k}\otimes \kappa(x), \kappa(x)), $$
hence the space of distributions $Dist^n(G,I)$ is the dual of the fiber $P^n_{G/k}\otimes \kappa(x)$. There is an isomorphism $P^1_{G/k}\cong R \oplus \Omega^1_{G/k}$ and hence
$$P^1_{G/k}\otimes \kappa(x) \cong \kappa(x)\oplus \Omega^1_{G/k}\otimes \kappa(x).$$
The dual $T_{G/k}:=(\Omega^1_{G/k})^*$ has
$$ T_{G/k}\otimes \kappa(x) \cong Lie(G)$$
is the Lie algebra $Lie(G)$ of $G$.
Addendum: Let me add that in the Waterhouse book he defines for any affine algebraic group $G$ with coordinate ring $k[G]$, the $k$-Lie algebra $Lie(G)$ as a Lie sub algebra (of left invariant vector fields) of $Der_k(k[G])$. When $G$ is smooth it follows $Diff_k(k[G])$ is the ring of differential operators on $G$ and there is a canonical map
$$U(Lie(G)) \rightarrow Diff_k(k[G]),$$
hence when the characteristic is zero you get a canonical map
$$Dist(G) \rightarrow Diff_k(k[G]).$$
A: I've been meaning to post an update for a while now. I managed to use one of the exercises in Waterhouse's book ``Introduction to affine group schemes" to find a proof that more closely follows Jantzen's use of distributions.
The  following argument is my own though similar ones may exist somewhere. Huge thanks to @hm2020 for the fantastic book recommendation and helpful comments.
Following the notation of my original post, let $\alpha : G\times X \to X$ be an action of an affine group scheme $G$ on an affine $X$ scheme, both over a commutative unital ring $k$. Let $m:k\otimes k[X] \to k[X]$ be the multiplication isomorphism, $\varepsilon:k[G]\to k$ be the counit, and $\Delta :k[G]\otimes k[X] \to k[X]$ be the comodule map. Any linear map $\psi:k[G]\to k$ acts on an element $x\in k[X]$ by $\psi(x):= m\circ(id \otimes \psi)\circ \Delta$.
The following result is described in Chapter 12, Excercise 10 of Waterhouse.
(*) Let $n\geq 0$ be an integer. A $k$-linear map $\mu:k[G]\to k$ is a distribution of order $\leq n$ if and only if for all $g\in k[G]$, the linear map $\varepsilon\circ ad_{g} \mu$ is a distribution of order $\leq n-1$.
The forward implication follows from Jantzen's definition of a distribution (chapter 7), the converse follows by induction on the order $n\geq 0$.
We are now in a position to prove the claim:
(**) Each distribution $\mu:k[G]\to k$ acts as a differential operator of order $\leq n$ via the composite endomorphsim $m\circ (\mu \otimes id_{k[X]})\circ\Delta$.
We proceed by induction the order $n\geq 0$ of distributions.
The base case ($n=0$) follows from the fact that every distribution of order $\leq 0$ has the form $a\varepsilon$, for some $a\in k$.
Induction: Suppose $n> 1$ and for all $n> m \geq 0$, the map $m\circ (\mu \otimes id_{k[X]})\circ\Delta$ is a differential operator of order $\leq m$, whenever $\mu$ is a distribution of order $\leq m$.
Let $x,f \in k[X]$ and $\mu:k[G] \to k$ be a distribution of order $\leq n$.
We use the following properties from chapter 2 of Jantzen.

*

*for some $x_i,f_i \in k[X]$ and $g_i,h_i\in k[G]$ write $\Delta(x) =
   \sum_i x_i \otimes g_i$ and $\Delta(f) = \sum_j f_j \otimes h_j$.


*One of the axioms of a comodule $M$: $id_M =m\circ (id_{k[G]} \otimes
   \varepsilon)\circ \Delta_M$.


*Since $\alpha$ is a morphism of schemes the map $\Delta$ is an
algebra morphism so $\Delta(fx) = \Delta(f)\Delta(x)$.
Some computation gives us the following
\begin{align*}
(ad_f\mu)(x) =f[\mu(x)] - \mu(fx) 
&= [m\circ (id \otimes  \varepsilon)\circ \Delta_M (f)][\mu(x)] - \mu(fx) \\
& =\sum_j\sum_i x_i f_j\bigg(\varepsilon(h_j)\mu(g_i) -\mu( h_j g_i)\bigg) \\
& = \sum_j f_j ~\sum_{i} x_i (\varepsilon\circ ad_{h_j} \mu)(g_i) \\
& = m\circ \sum_j (f_j \otimes (\varepsilon\circ ad_{h_j} \mu)\sum_i x_i \otimes g_i \\
& = \left(m\circ \sum_j(f_j \otimes (\varepsilon\circ ad_h \mu)\circ \Delta\right)(x) ~.
\end{align*}
Now $\mu$ is a distribution of order $\leq n$ therefore by (*) $(\varepsilon\circ ad_h\mu)$ is a distribution of order $\leq n-1$, for all $h\in k[G]$. Then by the induction hypothesis $ad_f\mu =m\circ \left(\sum_j(f_j \otimes (\varepsilon\circ ad_h \mu)\right)\circ \Delta$ is a differential operator of order $\leq n-1$.
Therefore $ \mu$ is a differential operator of order $n$.
