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Setup:

Let $k$ be a field and $G \subseteq \mathrm{GL}_n(k)$ an algebraic group, reductive if that makes a difference. Let $\mathfrak g \subseteq \mathfrak{gl}_n(k)$ be the Lie algebra of $G$ with enveloping algebra $U(\mathfrak g)$ and let $M$ be a $G$-module, hence also a $U(\mathfrak g)$-module.

Let $X \in \mathfrak g$ be given by a nilpotent matrix ($X^{n + 1} = 0$) so that $\exp(X) = \sum_{t = 0}^n\frac{1}{t!}X^t$ is a well defined element of $\mathrm{GL}_n(k)$ (if the characteristic of $k$ is positive then assume $n$ is less than the characteristic). Note that that expression also gives a well defined element $\exp(X) \in U(\mathfrak g)$.

Question:

Assume $\exp(X) \in \mathrm{GL}_n(k)$ lies in $G$. Does the action of $\exp(X)$ on $M$ when considered as an element of $G$ agree with the action of $\exp(X)$ on $M$ when considered as an element of $U(\mathfrak g)$?

You can directly compute that if $M = k^n$ is the standard representation of $\mathrm{GL}_n(k)$ then this is true. There is a paper by Carlson, Lin, and Nakano, "Support Varieties for modules over Chevalley groups and classical Lie algebras", where they claim (unless I've read it wrong) that this is true when $\exp$ is the exponential map of the unipotent radical of a parabolic. Their reference for this fact is Seitz's paper "Unipotent elements, tilting modules, and saturation" but I can't find this claim in there. I thought a fact like this about the exponential map should be easy enough to be in some elementary exposition but after looking I've come up dry. Do any of you know anything about this?

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So as it turns out the answer is no. Here's an example:

Let $G$ be the set of matrices of the form $$g = \begin{bmatrix} 1 & a & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ the map $\Phi\colon G \to \mathrm{GL}_3(k)$ given by $$\begin{bmatrix} 1 & a & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \to \begin{bmatrix} 1 & a & a^p \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}$$ is a homomorphism and hence defines a $3$-dimensional module $M$. The Lie algebra of $G$ is $1$-dimensional and the natural inclusion $G \subseteq \mathrm{GL}_3(k)$ gives $\mathfrak g$ as the set of matrices of the form $$X = \begin{bmatrix} 0 & a & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$$ They are all nilpotent and $\exp(X)$ is exactly the matrix $g$ above. The derivative of the representation $\Phi$ is simply this inclusion $\mathfrak g \to \mathfrak{gl}_3(k)$. Thus $\exp(X) \in G$ acts on $M$ via the matrix $$\begin{bmatrix} 1 & a & a^p \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix},$$ but $\exp(X) \in U(\mathfrak g)$ acts on $M$ via the matrix $$\begin{bmatrix} 1 & a & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.$$

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