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?

  • 2
    $\begingroup$ "unipotent group, hence a subgroup of $U_n$" is not very well-said: you're making a choice of $n$ and embedding into $U_n$. In other words, it's fine to start with a closed subgroup $G$ of $U_n$ and proceed; what follows does not only depend on the isomorphism class of $G$ (notably the fact that $n$ exceeds the characteristic ought to play a role). $\endgroup$
    – YCor
    Mar 31, 2019 at 9:02
  • 1
    $\begingroup$ You're using intrinsic definitions of $Lie(G)$, but $G$ being embedded into an affine space (the space of matrices), you could use, as a definition, the definition of its Zariski tangent subspace of the subvariety $G$ at $I_n$. $\endgroup$
    – YCor
    Mar 31, 2019 at 10:12
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    $\begingroup$ en.wikipedia.org/wiki/Baker–Campbell–Hausdorff_formula ensures that exp(Lie algebra) is some group and vice versa log(group) = Lie algebra, so in sense that is one of the ways to define Lie group<->algebra correspondence, it seems you wanna try to prove that "some" group would be exactly the group you started with , is it correct ? $\endgroup$ Mar 31, 2019 at 21:00
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    $\begingroup$ (cont.) Namely, we may assume that $G$ and $X$ are defined over a field $K$ that is finitely generated over $\mathbb Q$, and we embed $K$ into $\mathbb C$. $\endgroup$ Mar 31, 2019 at 22:38
  • 1
    $\begingroup$ @LSpice Yes, but unpublished as of yet. The finished manuscript can be found in the below link (it's sitting in my Github for now). Mikhail's contribution, the subject under discussion, can be found in the appendix. raw.githubusercontent.com/mikecrumley/Publications/master/… $\endgroup$ Aug 16, 2020 at 1:54

4 Answers 4


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.

  • 2
    $\begingroup$ Assume that over group $G$ over a field $k$ of a characteristic $p$ is the reduction module $p$ of a smooth group scheme $\mathfrak G$ over the Witt ring $O=W(k)$, such that ${\rm Lie}(G)$ is the reduction modulo $p$ of ${\rm Lie}(\mathfrak G)$. Assume that our nilpotent matrix $X$ in ${\rm Lie}(G)$ lifts to a nilpotent matrix $X_O$ in ${\rm Lie}(\mathfrak G)$. Is it possible to prove that $\exp_{<p}(X)\in G(k)$ because it is the reduction of $\exp(X_O)\in \mathfrak G(O)$ ? $\endgroup$ Apr 9, 2019 at 15:02
  • 1
    $\begingroup$ @WillSawin: Even though I write "a smooth group scheme $\mathfrak G$ over the Witt ring", I do not understand what a "smooth group scheme" means. Where can I read about this? $\endgroup$ Apr 9, 2019 at 20:25
  • 1
    $\begingroup$ @MikhaiBorovoi Just a couple thoughts. Will's counterexample is in fact a counterexample to several other "nice" properties of unipotent groups in positive characteristic that I had conjectured (relating to their distribution algebras). The problem obviously is the the presence of a "Frobenius twist", i.e. the $t^p$ term. This seems to me to be irreparable and illustrates why postive characteristic unipotent groups are so nasty in general. $\endgroup$ Apr 11, 2019 at 4:07
  • 1
    $\begingroup$ (cont.) But, if we could place conditions on a unipotent group so that counterexamples such as Will's are ruled out, I think the rest of the results will go through (for whatever it turns out to be worth). $\endgroup$ Apr 11, 2019 at 4:07
  • 1
    $\begingroup$ (cont.) For example, the conjecture is obviously true for the group $\begin{pmatrix} 1 & x & y \\ 0 & 1 & x \\ 0 & 0 & 1 \end{pmatrix}$ in characteristic $p > 3$, because you don't get that Frobenius garbage. If we were able to state in formal terms how to rule out examples such as Will's, and if I can verify that my other conjectures are true for such groups, some of this might be worth something. $\endgroup$ Apr 11, 2019 at 4:07

EDIT: I roll back to the previous proof, in characteristic 0 only. My last proof including characteristic $p$ was false (thanks to Will Sawin for noticing this).

Let $G={\rm GL}_{n,k}$\,, where $k$ is a field of characteristic 0. Consider the formal power series $$\exp(x)=1 + x+\frac{1}{2!}x^2+\dots$$ over ${\mathbb{Q}}$ and the polynomial $$\exp_{<n}(x)=1+x+\dots+\frac{1}{(n-1)!}x^{n-1},$$ which is defined also in characteristic $p\ge n$.

Theorem 1. Let $k$ be a field of characteristic 0. Let $H\subset G={\rm GL}_{n,k}$ be an algebraic subgroup defined over $k$. Let $X\in{\rm Lie}(H)\subset{\mathfrak{gl}}_{n,k} = M_n(k)$ be a nilpotent matrix. Then $\exp_{<n}(X)\in H(k)$.

Note that since $X\in M_n(k)$ is nilpotent, we have $X^n=0$.

The version of Theorem 1 in positive characteristic is false, see the answer of Will Sawin.

Set $V=k^n$. Write $$T^{ij}(V)=V\otimes\dots\otimes V\otimes V^*\otimes\dots\otimes V^*$$ ($i$ times $V$, $j$ times $V^*$), where $V^*$ is the dual space to $V$, and set $$W=W^{\le N}=\bigoplus_{0\le i,j\le N} T^{ij}(V).$$ Let $\theta=\theta^{\le N}$ denote the natural representation of $G={\rm GL}(V)$ in $W$, and let $d\theta$ denote the corresponding representation of ${\rm Lie}(G)={\mathfrak{gl}}(V)$ in $W$. Since $X$ is nilpotent and $\theta\colon {\rm GL}(V)\to {\rm GL}(W)$ is a homomorphism of linear algebraic groups, the linear operator $(d\theta)(X)\in{\rm End}(W)$ is nilpotent; see Springer, Linear Algebraic Groups (2nd ed.), Theorem 4.4.20. Note that in general it is not true that $((d\theta)(X))^n=0$, but one can show that $((d\theta)(X))^{2nN}=0$.

Theorem 2. Assume that ${\rm char}(k)=0$. Then for any nilpotent matrix $X\in {\rm End}(V)={\mathfrak{gl}}(V)$ we have $$\exp((d\theta)(X))=\theta(\exp(X)).$$

Note that $\theta(\exp(X))$ and $\exp((d\theta)(X))$ are defined because both $X\in {\rm End}(V)$ and $(d\theta)(X)\in{\rm End}(W)$ are nilpotent operators and because ${\rm char}(k)=0$.

We deduce Theorem 1 from Theorem 2. There exists a natural number $N=N_H$ and a tensor $t=t_H\in W=W^{\le N}$ such that $H$ is the stabilizer in $G$ of the line $k\cdot t\subset W$ with respect to $\theta$ and such that ${\rm Lie}(H)$ is the stabilizer in ${\rm Lie} (G)$ of this line with respect to $d\theta$; see Springer's book, Lemmas 5.5.1 and 5.5.2. Since $X\in {\rm Lie}(H)$, we have $(d\theta)(X)\cdot t=\lambda t$ for some $\lambda\in k$, and we have $\lambda=0$ since $(d\theta)(X)$ is nilpotent. Now it follows from Theorem 2 that $$\theta(\exp(X))\cdot t=\exp((d\theta)(X))\cdot t=t,$$ and therefore, $\exp(X)\in H(k)$, which proves Theorem 1.

Proof of Theorem 2. Let $A$ be a Lie group over $k=\mathbb{R}$ or $k={\mathbb{C}}$. As usual, for $X\in{\rm Lie}(A)$ we define the exponential map $Z(s)=\exp(sX)\in A$ as the solution of the differential equation $\frac{d}{ds} Z(s)=X\cdot Z(s)$ with initial condition $Z(0)=1_A$. Then for $A={\rm GL}(V)$ the exponential map is defined by the convergent series above. If $\phi\colon A\to B$ is a homomorphism of Lie groups, then the following diagram commutes: $$ \require{AMScd} \begin{CD} {\rm Lie}(A) @>{d\phi}>> {\rm Lie}(B);\\ @V{\exp_A}VV @VV{\exp_B}V \\ A @>{\phi}>> B; \end{CD} $$ Indeed, both composite maps are solutions of the same differential equation with the same initial condition.

Note that the algebraic groups ${\rm GL}(V)$ and ${\rm GL}(W^{\le N})$ and the homomorphism $\theta=\theta^{\le N}$ are all defined over ${\mathbb{Q}}$. We use the idea of the Lefschetz principle. We consider the finitely generated field $l={\mathbb{Q}}(x_{ij})$, where $x_{ij}\in k$ for $1\le i,j\le n$ are the matrix elements of $X$. We embed $l$ into ${\mathbb{C}}$. We obtain that $$\exp((d\theta)(X))=\theta(\exp(X))$$ over ${\mathbb{C}}$. Since $X$ and $(d\theta)(X)$ are nilpotent, the expressions in the formula above are actually polynomials of $(d\theta)(X)$ and $X$, respectively. This completes the proof of Theorem 2.

  • $\begingroup$ Thanks so much Mikhail. Just a couple points of clarification: 1) The representation $\theta:H \rightarrow \text{Aut}(T(V^*))$. What you mean is, for $h \in H$, $f \in V^*$, and $v \in V$, $ \theta(h)(f)(v) = f(hv)$ and then extend to $T(V^*)$ in the obvious manner. Correct? 2) Is the identity $\theta(\text{exp}(X)) = \text{exp}((d \theta)(X))$ as obvious as it looks? I don't immediately see it (probably because I'm not used to thinking about differentials). Thanks again. $\endgroup$ Apr 1, 2019 at 21:52
  • $\begingroup$ Mikhail, with your permission I'd like to use this proof in a paper I'm about to submit (with attribution of course). Do you see any problem with this working in positive characteristic? Again, we can assume that both the series $\text{exp}(X)$ and $\text{log}(g)$ are defined, since we are assuming that $X$ and $g-1$ are both nilpotent of order $\leq n \leq p$. $\endgroup$ Apr 5, 2019 at 7:37
  • $\begingroup$ Dear Mike, yes, you may use this proof in your paper. Do use the last version, it is correct and not that long. Unfortunately, I cannot deal with the case of positive characteristic. The reason is that we may not assume that $((d\theta)(X))^n=0$, and I cannot control the big powers of $(d\theta)(X)$. I will keep thinking. $\endgroup$ Apr 5, 2019 at 17:21
  • $\begingroup$ @MikhailBorovoi, since you can increase $n$ to a $p$-power, is it really illegal to assume that $((\mathrm d\theta)(X))^n = 0$? $\endgroup$
    – LSpice
    Apr 5, 2019 at 18:18
  • $\begingroup$ @LSpice: I use $\exp((d\theta)(X))$, and in characteristic $p$ the expression $\frac{1}{p!}((d\theta)(X))^p$ is not defined. $\endgroup$ Apr 5, 2019 at 21:12

In characteristic $p>0$ you can expect a property of this kind in nice situations with certain precautions.

First precaution is that a Lie subalgebra ${\mathfrak g} \leq Lie (GL_n)$ may be tangent to several distinct connected algebraic subgroups. This is brilliantly explored by Will Sawin in his answer.

In practice, your ${\mathfrak g}$ and $G$ are often God given, i.e., they are stabilizers of a $k$-tensor $t$ on $V$. Then you will have your desired exponential and logarithm as soon as you have a second precaution: you need to require $X^{p/k}=0$ or $(1-g)^{p/k}=0$.


It seems to me that a reasonable way to prove this result in characteristic 0 is to use the standard Lie group/Lie algebra argument, but to use the algebraic exponential function as defined in Demazure-Gabriel Groupes algebriques to supplement the explicit version used in the question.


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