Given a (finite dimensional) Lie group $G$ (real $k=\mathbb{R}$ or complex $k=\mathbb{C}$) and its Lie algebra $\mathfrak{g}$, one can prove (a basis $B=(b_i)_{1\leq i\leq n}$ of $\mathfrak{g}$ being given) that there exists a neighbourhood $W$ of $1_G$ (in $G$) and $n$ local coordinate analytic functions $$ W\rightarrow k,\ (t_i)_{1\leq i\leq n} $$ such that, for all $g\in W$ $$ (*)\quad g=\prod_{1\leq i\leq n}^{\rightarrow} e^{t_i(g)b_i}=e^{t_1(g)b_1}e^{t_2(g)b_2}\dots e^{t_n(g)b_n} $$ to see this, just remark that $$ (t_1,t_2,\cdots t_n)\rightarrow exp(t_1b_1)exp(t_2b_2)\cdots exp(t_nb_n) $$ is a local diffeomorphism from $k^n$ to $G$ in a neighbourhood of $0$ and take the inverse.

My questions are the following

Let us loosely call infinite dimensional a Lie group whose Lie algebra is not finite dimensional (this includes the example below and infinite dimensional Banach-Lie groups for instance).

Q1) Can you provide examples of infinite dimensional Lie groups where the exponential map $L(G)\rightarrow G$ is locally surjective ? (locally surjective means here an open map $L(G)\to G$)

Q2) If, in these examples, $L(G)$ admits a topological basis in some sense, does it provide a formula like (*) ?

The case of Schützenberger's factorization.-

Below an example of an infinite dimensional Lie group and such a system of coordinates.

All definitions of : algebra (resp. large algebra) of a monoid, Lie algebra, enveloping algebra, used here are standard and can be taken e.g. from [1] and [2] (I can detail interactively on request).

Let $X$ be a set (of variables, or indeterminates, or an alphabet), $k$ a $\mathbb{Q}$-algebra and let $$ k\langle X\rangle, k\langle \langle X\rangle\rangle, \mathcal{L}_k\langle X\rangle, \mathcal{L}_k\langle\langle X\rangle\rangle $$ be respectively the free algebra (i.e. the algebra of noncommutative polynomials or the algebra of the free monoid $X^*$), the algebra of noncommutative formal power series (i.e. the large algebra of the free monoid $X^*$) [1], the free Lie algebra and the Lie algebra of Lie series [2]. We will use the natural pairing between $k\langle \langle X\rangle\rangle=k^{X^*}$ and $k\langle X\rangle=k^{(X^*)}$ given by the following sum on the words $$ \langle S|P\rangle=\sum_{w} coeff(S,w)coeff(P,w) $$ It is well known that $$ k\langle X\rangle=\mathcal{U}(\mathcal{L}_k\langle X\rangle)\ . $$ As such, it admits a structure of Hopf algebra $$ (k\langle X\rangle, conc, 1_{X^*}, \Delta_{shuffle},\epsilon,S) $$ $conc$ being the concatenation, $\Delta_{shuffle}$ being the dual law of the shuffle product, $\epsilon(P)=\langle P|1_{X^*}\rangle$ (constant term) and $S(a)=-a$ for all $a\in X$;

Every basis ($B=(b_i)_{i\in I};\ I$ totally ordered) of $\mathcal{L}_k\langle X\rangle$ (which is free, for all rings $k$) can be extended to a Poincaré-Birkhoff-Witt basis of $k\langle X\rangle$, parametrized by the multiindices of $\mathbb{N}^{(I)}$. The multi-index product is defined as follows. For every $\alpha\in \mathbb{N}^{(I)}$, we set
$$ B^\alpha=b_{i_1}^{\alpha_1}b_{i_2}^{\alpha_2}\cdots b_{i_m}^{\alpha_m} $$ with $supp(\alpha)=\{i_1<i_2<\cdots i_m\}$. Now, if $B$ is multi-homogeneous (w.r.t. the $\mathbb{N}^{(X)}$-grading), so is $(B^\alpha)_{\alpha\in \mathbb{N}^{(I)}}$ and there is a unique family of polynomials $B_\alpha$ such that $$ \langle B_\alpha|B^\beta \rangle=\delta_{\alpha,\beta}\qquad\qquad \mathrm{(Dual-Basis)} $$
Now within the algebra of double series (whose support is $k^{X^*\otimes X^*}$ endowed with the law $shuffle\hat{\otimes} conc$, Schützenberger (see [3,4]) gave the beautiful formula $$ (**)\ \sum_{w\in X^*}w\hat{\otimes} w=\prod_{i\in I}^{\rightarrow} e^{B_{e_i}\hat{\otimes} b_i} $$ where $e_i$ are the irreducibles of the monoid $\mathbb{N}^{(I)}$ defined by $e_i(j)=\delta_{i,j}$ (in particular $B^{e_i}=b_i$).

This can be used to provide a system of local coordinates on the Hausdorff group [2] $$ Haus_k(X)=\{e^L\}_{L\in \mathcal{L}_k\langle \langle X\rangle\rangle}=\{S\in k\langle \langle X\rangle\rangle|\epsilon(S)=1,\, \Delta_{shuffle}(S)=S\hat{\otimes}S\} $$ because, in this case $S\otimes Id$ is compatible with the law of the double algebra and then, applying this operator to $(**)$, we get $$ S=(S\hat{\otimes} Id)(\sum_{w\in X^*}w\hat{\otimes} w)=\prod_{i\in I}^{\rightarrow} e^{\langle S|B_{e_i} \rangle\,b_i} $$ which is a system of local coordinates of the type $(*)$ for the group $Haus_k(X)$.

Application to Riemann zeta functions When one multiplies several zeta values $$ \zeta(s)=\sum_{n\geq 1} \frac{1}{n^s} $$ multi-zeta values do appear, they are defined by $$ (***)\ \zeta(s_1,s_2,\cdots s_k)=\sum_{n_1>n_2>\cdots n_k\geq 1} \frac{1}{n_1^{s_1}n_2^{s_2}\cdots n_k^{s_k}}\ . $$ When $s_1,s_2,\cdots s_k$ are integers, the link with the shuffle product is that the quantity $(***)$ converges when $s_1>1$ and, coding $(s_1,s_2,\cdots s_k)$ by the word (here $X=\{x_0,x_1\}$) $w=(x_0^{s_1-1}x_1x_0^{s_1-1}x_1\cdots x_0^{s_k-1}x_1)$ and recoding $(***)$ by $\tilde{\zeta}(w)=\zeta(s_1,s_2,\cdots s_k)$ one can prove that $\tilde{\zeta}$ can be extended uniquely as a shuffle character of $\mathbb{Q}\langle X\rangle$ satisfying $\tilde{\zeta}(x_0)=\tilde{\zeta}(x_1)=0$ so that, applying (**) we get $$ \tilde{\zeta}=(\tilde{\zeta}\hat{\otimes} Id)(\sum_{w\in X^*}w\hat{\otimes} w)=\prod_{i\in I}^{\rightarrow} e^{\tilde{\zeta}(B_{e_i})\,b_i} $$ for every multihomogeneous basis $B$ of the free Lie algebra $\mathcal{L}_{\mathbb{Q}}\langle X\rangle$.

Coda: I guess (but have not set it technically yet) that, given a Lie $\mathfrak{g}$ algebra (finite or infinite dimensional), which is free as a $k$-module ($k$ is, as above, a $\mathbb{Q}$-algebra), given any ordered basis $B=(b_i)_{i\in I}$ of $\mathfrak{g}$, you can consider the convolution ($\ast$) algebra $$ \mathcal{A}=span_k\{(B_\alpha)|\alpha\in \mathbb{N}^{(I)}\}\subset \mathcal{U}^*(\mathfrak{g}) $$ which is closed by $\ast$ and every character $\chi$ of $\mathcal{A}$ could be factorized in the same way $$ \chi=\prod_{i\in I}^{\rightarrow} e^{\chi(B_{e_i})\,b_i} $$ for the topology of pointwise convergence on $\mathcal{A}$ ($k$ being discrete and the notation of $B_{\alpha}$ being those of $\mathrm{(Dual-Basis)}$).


[1] Bourbaki, Algebra, Ch III par. 2, Springer

[2] Bourbaki, Lie groups and Lie algebras, Ch II, Springer

[3] Schützenberger, LITP Report

[4] Christophe Reutenauer, Free Lie Algebras, Oxford University Press.

[5] Georges Racinet, Séries génératrices non-commutatives de polyzetas et associateurs de Drinfeld, Thèse (2 Nov 2006)

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    $\begingroup$ Will it be correct to reformulate your question as follows: when is the exponential map $\exp:L(G)\to G$ (locally) surjective? $\endgroup$ Apr 24, 2015 at 8:28
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    $\begingroup$ Gérard, some infinite dimensional locally compact groups have exponential maps that are surjective. For example, the infinite power ${\mathbb T}^{\mathfrak m}$ (where ${\mathfrak m}$ is any infinite cardinal) of the circle $\mathbb T$ has this property. $\endgroup$ May 4, 2015 at 7:24
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    $\begingroup$ Gérard, the Lie algebra for ${\mathbb T}^{\mathfrak m}$ is ${\mathbb R}^{\mathfrak m}$, the ${\mathfrak m}$-th power of $\mathbb R$. $\endgroup$ May 4, 2015 at 17:34
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    $\begingroup$ It would have been useful to define "locally surjective". Does it mean that the exponential of any neigh. of 0 is a neigh. of 1? Does it mean that the exponential of some neigh. of 0 is a neigh. of 1? Does it mean something not only at 0? $\endgroup$
    – YCor
    May 4, 2015 at 18:08
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    $\begingroup$ @YCor, I thought this means that there is a neighbourhood $U$ of 1 in $G$, and a neighbourhood $V$ of 0 in $L(G)$ such that $\exp(V)=U$. $\endgroup$ May 4, 2015 at 19:30

2 Answers 2


I can only provide partial answers to your questions. Namely, to Q1:

(Infinite-dimensional) Lie groups whose Lie group exponential map is a local diffeomorphism are in the literature called "locally exponential Lie groups". Information on locally exponential Lie groups along with examples and further references should be contained in Neeb's survey

where also some structure theory on these groups is summarised. Examples include Banach Lie groups, unit groups of continuous inverse algebras and character groups of (graded and connected) Hopf algebras.

However, diffeomorphism groups are in general the first example of a Lie group where the Lie group exponential map is not locally surjective. It is even not locally surjective in a very strong sense. I think the strongest results in this direction were proved by Grabowski: If I recall it correctly he proved that there are curves converging towards the identity which are not contained in the image of the exponential map. Picking up Sergei Akbarovs answer: the diffeomorphism group on the circle is NOT locally exponential. I think Peter Michor explained the details of this example some time ago in a post on MO. However, I am also quite sure that the details are also contained in his book "The convenient setting of global analysis" (see Example 42.3).

  • $\begingroup$ @Schmeding Thank you for your answer and references (that's what I was looking for) +1 $\endgroup$ May 4, 2015 at 20:24
  • $\begingroup$ .@AlexanderSchmeding I downloaded (2) thank you ! $\endgroup$ May 4, 2015 at 20:35
  • $\begingroup$ Downloaded (1) and began to read it, thanks again (cannot put more than one point, but I re-express my gratitude). $\endgroup$ Oct 21, 2019 at 9:16

The group of diffeomorphisms $G=\text{Diff}({\mathbb T})$ of the circle ${\mathbb T}$ has the Lie algebra consisting of all smooth vector fields on ${\mathbb T}$ $$ L(G)=\text{Vect}({\mathbb T}). $$ Each smooth vector field $X\in\text{Vect}({\mathbb T})$ is defined by the coefficient $\varphi\in {\mathcal C}^\infty({\mathbb T})$ in the decomposition $$ X=\varphi\cdot\frac{d}{d t} $$ where $\frac{d}{d t}$ is the usual operator of taking derivative on $\mathbb T$. So the space $\text{Vect}({\mathbb T})$ of vector fields is naturally isomorphic to the space ${\mathcal C}^\infty({\mathbb T})$ of smooth functions on ${\mathbb T}$: $$ L(G)=\text{Vect}({\mathbb T})={\mathcal C}^\infty({\mathbb T}). $$ But ${\mathcal C}^\infty({\mathbb T})$, as a Fréchet space, has a natural basis consisting of trigonometric functions $$ 1,\quad \cos nx,\quad \sin nx,\quad n\in{\mathbb N}. $$ I believe, your formula (*) must be true in this case (since everything is too natural here), but it is a supplementary work to understand in which sense.

EDIT. I am sorry, I got some doubts that the exponential map is locally surjective here.

And another discordant detail: $\text{Diff}({\mathbb T})$ is not a Lie group in the sense of Bourbaki, since its Lie algebra $\text{Vect}({\mathbb T})={\mathcal C}^\infty({\mathbb T})$ is not a Banach space.

  • $\begingroup$ @Segei Thank you for the example (I need to constitute a list of such examples/references) . On the other hand with the ${\mathbb T}^{\mathfrak m}$, the Lie algebra is ${\mathbb R}^{\mathfrak m}$ and $exp$ is not locally invertible (contrariwise to $\text{Diff}({\mathbb T})$). I agree that $\text{Diff}({\mathbb T})$ is not a Banach but a Frechet Lie group (but my combinatorial example is not a Banach lie group either) $\endgroup$ May 4, 2015 at 20:17
  • $\begingroup$ Gérard, as far as I understand, the exponential map of $\text{Diff}({\mathbb T})$ is also not locally invertible. I think, you should accept Alexander Schmeding's answer, it sounds much more useful. $\endgroup$ May 4, 2015 at 20:27
  • $\begingroup$ You wrote [I think, you should accept Alexander Schmeding's answer]. You are right, I just done it because it corresponds to my request. $\endgroup$ Jul 28, 2023 at 3:04

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