Let $k$ be a field (of possibly positive characteristic), let $U\_n$ denote the space of all $n \times n$ unipotent upper triangular matrices over $k$, and let $G$ be an algebraic subgroup of $U\_n$ (hence a unipotent algebraic group itself). Then each $X \in \text{Lie}(G)$ (thought of as a member of $\text{Lie}(U\_n)$, i.e. a strictly upper triangular $n \times n$ matrix) is nilpotent, so it makes sense to define


$\text{exp}(X) = 1 + X + X^2/2! + \dots + X^{n-1}/(n-1)!$


(This definition makes sense even in characteristic $p > 0$ so long as $p \geq n$, i.e. so that $p$ never divides $1!, 2!, \dots, (n-1)!$). We can also define, for $g \in G$,
since $g-1$ is nilpotent,


$\log(g) = (g-1) - (g-1)^2/2 + (g-1)^3/3 - \dots \pm (g-1)^{n-1}/(n-1) $


Obviously $\text{exp}$ and $\log$ define maps from $\text{Lie}(U\_n)$ to $U\_n$ and back to $\text{Lie}(U\_n)$, and are bijective, being inverses of one another.


**My Question**: If $g \in G$, is $\log(g) \in \text{Lie}(G)$? Or, equivalently, for $X \in \text{Lie}(G)$, is $\text{exp}(X) \in G$?


I feel like there should be an obvious proof of this, but I don't see it. If $G$ were a Lie group, the Lie algebra of $G$ would often just be defined to
be all $X$ such that $e^{tX} \in G$ for all $t \in \mathbb{R}$, and so for Lie groups $\text{exp}$ maps from $\text{Lie}(G)$ to $G$ simply by definition. In the algebraic group context this definition no longer makes sense generally, and even when it does, is not used in the literature (so far as I've seen),
so I tried using each of the following equivalent definitions of $\text{Lie}(G)$, with no success:


* $\text{Lie}(G) = \text{Dist}\_1^+(G)$ (distributions of order no greater than $1$ without constant term)
* $\text{Lie}(G) = $ the subspace of $\text{Lie}(U\_n) = \text{Dist}\_1^+(U\_n)$ which kills $I = (\text{defining polynomials of $G$})$
* $\text{Lie}(G) = \{M \in \text{Lie}(U\_n): 1 + \tau M \in G(k[\tau]) \}$ where $\tau^2 = 0$
* $\text{Lie}(G) = \{M \in \text{Lie}(U\_n) : 1 + \tau M \text{ satisfies the defining polynomials of } G \}$, again where $\tau^2 = 0$
* $\text{Lie}(G) = $ left invariant derivations on the Hopf algebra of $G$


It is certainly believable on it's face; we have that $\text{Lie}(G) \stackrel{\text{exp}}{\longrightarrow} U\_n \stackrel{\log}{\longrightarrow} \text{Lie}(G)$
composes to the identity, similarly for $G \stackrel{\log}{\longrightarrow} \text{Lie}(U\_n) \stackrel{\text{exp}}{\longrightarrow} G$, but I don't see why in the
meantime that $\log(G) \subset \text{Lie}(G)$ or that $\text{exp}(\text{Lie}(G)) \subset G$.


If it makes a difference, I'm actually only interested in the case where the defining polynomials of $G$ have integer (perhaps mod $p$) coefficients.


Thanks in advance for any help.


**EDIT**: Here's a more basic question, one which might help answer the above.


Suppose $k = \mathbb{R}$. Then $G$ is also a Lie group, and it is customary to define


$\text{Lie}(G) = \{ X \in \text{Lie}(U\_n): e^{tX} \in G \text{ for all } t \in \mathbb{R} \}$


Can someone explain, or point me to a reference explaining, why this definition is equivalent to any of the above definitions for $\text{Lie}(G)$ as an algebraic group?



exponential/logarithm for unipotent algebraic groups
This is false in characteristic $p$, no matter how large $p$ is. The counterexample is the group parameterized by


$\begin{pmatrix} 1 & t & t^p \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$


Its Lie algebra is generated by the matrix


$\begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}$


whose exponential does not lie in the group.


