Consider a Lie algebra $\mathfrak{g}$ with elements $\{g_1, g_2,\ldots,g_N\}$, with a Lie group defined by the exponential map $\exp(g)$ for $g\in\mathfrak{g}$. Given an arbitrary general element $g=\sum_{i}^{N}\alpha_{i}g_{i}$, what facts about the Lie algebra makes it true that we can express $\exp(g)$ in the form $$ \exp(g)=\exp(\beta_{1}g_1)\exp(\beta_{2}g_2)\ldots\exp(\beta_{N}g_N)? $$ I'm coming from a physics background so please excuse my attempts to get the mathematical language correct (feel free to correct it).

An example from physics is the Lie algebra spanned by $\{K_0,K_+,K_-\}$ with the following commutation relations: $$ [K_+,K_-]=-2K_{0};\quad [K_0,K_{\pm}]=\pm K_{\pm}. $$ In this case we can write, for example, $$ \exp(\alpha K_{+}+\beta K_{-})=\exp(\gamma K_{+})\exp(\eta K_{-})\exp(\xi K_{0}). $$ So this is a case where the nested commutators in the Zassenhaus formula do not terminate but can be summed so that only a finite number of exponential factors are required.

I also have the very related question of under what circumstances we can write $$ \exp(\alpha_{1}g_2)\exp(\alpha_{2}g_1)=\exp(\beta_{1}g_1)\exp(\beta_{2}g_2)\ldots\exp(\beta_{N}g_N). $$ That is, a reordering of group elements written in terms of a product of elements in the group (with each element only appearing at most once).

I will clarify if necessary. Thank you.

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    $\begingroup$ In your example, is one of the $K_+$ supposed to be a $K_-$? $\endgroup$
    – Ted
    May 24, 2020 at 3:09
  • $\begingroup$ thanks, I have edited. $\endgroup$ May 24, 2020 at 4:22

2 Answers 2


Here is an answer to the first question.

Your Lie algebra is $\mathfrak{sl(2, \mathbb{R})}$. (Ok, you didn't specify that you are working over $\mathbb{R}$ and in computations that stay strictly within the Lie algebra you might probably even work over $\mathbb{C}$, which is advantageous in many situations, but when we are talking about the exponential map this makes only sense for Lie algebras over $\mathbb{R}$.)

A very concrete definition of the corresponding group $SL(2, \mathbb{R})$ is that of all 2-by-2 matrices over $\mathbb{R}$. The following (providing a 'YES' answer) is true for all groups of matrices of any finite dimension over either $\mathbb{R}$ or $\mathbb{C}$. So that included pretty much every group you ever need: $SL(n, \mathbb{R})$, $SO(p, q)$, $SU(p, q)$ the (three dimensional) Heisenberg group etc. A more abstract version of the argument also applies to wilder groups like $E_8$ and the universal cover of $SL(2, \mathbb{R})$ but I won't discuss these here.

So your group $G$ is sitting inside $GL(n, \mathbb{C})$ for some $n$. Let $A$ be the set of all diagonal matrices in $G$, $N$ be the set of all upper triangular matrices in $G$ with 1's on the diagonal and $\overline{N}$ be the set of all lower triangular matrices with $1$'s on the diagonal. We note that all three of $A$, $N$, $\overline{N}$ are subgroups of $G$.

We write $\mathfrak{a}, \overline{\mathfrak{n}}, \mathfrak{n}$ for their Lie algebras, viewed as subalgebras of $\mathfrak{g}$. Since $\mathfrak{g}$ is a subalgebra of $\mathfrak{gl}(n, \mathbb{C})$ which consists of all $n$-by-$n$-matrices, we find that $\mathfrak{a}$ to consist of the diagonal matrices in $\mathfrak{g}$, $\mathfrak{n}$ of upper triangular matrices with $0$'s on the diagonal and $\overline{\mathfrak{n}}$ of lower triangular matrices with $0$'s on the diagonal.

In your example $K_0$ spans $\mathfrak{a}$, $K_+$ spans $\mathfrak{n}$ and $K_-$ spans $\overline{\mathfrak{n}}$

Now there are three facts that are relevant here:

  1. Every $G$ can be written as a product $bac$ with $b \in \overline{N}$, $a \in A$, $c \in N$. (In numerical mathematics this is called the LDU decomposition)

  2. The exponential map is surjective when viewed as a map from $\mathfrak{a}$ to $A$, also when viewed as a map from $\mathfrak{n}$ to $N$ and also when viewed as map from $\overline{\mathfrak{n}}$ to $\overline{N}$.

This (statement 2) is quite special because the exponential map is in general not surjective when viewed as map form $\mathfrak{g}$ to $G$.

Combining 1) and 2) we get that

Every $g \in G$ can be written as a product $\exp(X_1)\exp(X_2)\exp(X_3)$ with $X_1 \in \overline{\mathfrak{n}}$, $X_2 \in \mathfrak{a}$ and $X_3 \in \mathfrak{n}$.

In the special case that $\overline{\mathfrak{n}}$, $\mathfrak{a}$ and $\mathfrak{n}$ are one-dimensional (as in your example) this means we are done. We get an expression of the form you ask about, not only for elements of the form $\exp(X)$ with $X \in \mathfrak{g}$ but for all elements in the group.

In the higher dimensional case it seems you want something more: you want express the group element as a product of exponentials of scalar multiples of fixed basis elements. In view of the result I highlighted we can get that if we show that:

  1. Your conjecture is true for the special cases of Lie groups $A$, $N$ and $\overline{N}$.

Now for $A$ this is really really easy because $A$ is commutative ($ab = ba$ for all $a, b \in A$, and subsequently $[X, Y] = 0$ for all $X, Y \in \mathfrak{a}$).

$\mathfrak{n}$ and $\overline{\mathfrak{n}}$ are in general not commutative but they are something which is close enough for our purposes: they are nilpotent matrices. Concretely: the Lie bracket of two upper triangular matrices will have its non-zero entries on a 'higher' diagonal than the ones you started with. Since there are only finitely many diagonals all nested commutators of sufficient length will be zero and hence there are only finitely many terms in the Baker Cambell Hausdorff formula. This can then be used to prove both statements 3) and 2) for $\mathfrak{n}$ and $N$ and a mirror image of the argument works for the lower diagonal matrices of $\overline{\mathfrak{n}}$ and $\overline{N}$.

So what remains is to verify 1) but this is essentially Gaussian elimation and understand why 2) holds for $\mathfrak{a}$ and $A$ but this last thing is competely trivial: the exponential of a diagonal matrix with entries $a_1, \ldots, a_n$ is simply the diagonal matrix with entries $\exp(a_1), \ldots, \exp(a_n)$.

I leave quite some details to you, but let me know if you have any questions!

  • $\begingroup$ I'm still absorbing the answer but does the disentangling work if there are more than 3 generators? $\endgroup$ Aug 18, 2020 at 21:33
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    $\begingroup$ Yes, the point is that you choose your generators in a special way. For instance for $G = GL(n, \mathbb{R})$ and hence $\mathfrak{g} = \mathfrak{gl}(n, \mathbb{R})$ we have that the first $n(n-1)/2$ generators (let's call them $Y_1, \ldots, Y_{n(n-1)/2}$ span $\overline{\mathfrak{n}}$, the next $n$ generators (let's say $H_1, \ldots, H_n$) span $\mathfrak{a}$ and the last $n(n-1)/2$ generators ($X_1, \ldots, X_n$). Now if you have decomposed a group element $g$ as $g = yax$ as in the corollary to 1) en 2) we only nee three decompositions each involving only one of the subgroups: (ctd below) $\endgroup$
    – Vincent
    Aug 18, 2020 at 21:41
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    $\begingroup$ $$y = \exp(\alpha_1 Y_1)\cdots \exp(\alpha_{n(n-1)/2} Y_{n(n-1)/2}) $$ $$a = \exp(\beta_1 H_1)\cdots \exp(\beta_{n} Y_{n})$$ $$x = \exp(\alpha_1 X_1)\cdots \exp(\alpha_{n(n-1)/2} X_{n(n-1)/2}) $$ $\endgroup$
    – Vincent
    Aug 18, 2020 at 21:43
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    $\begingroup$ So if your conjecture holds for each of the three subgroups separately (which I claim it does in statement 3) then 1) and 2) together guarantee that it holds for all of $G$ by just multipying the three expressions above into a long expression $\endgroup$
    – Vincent
    Aug 18, 2020 at 21:45
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    $\begingroup$ Yes, but it is even stronger than that. Not every element of $G$ is of the form $\exp(X)$ for $X \in \mathbb{g}$ but still all of them can be composed in this way $\endgroup$
    – Vincent
    Aug 18, 2020 at 21:53

Vincent explained why your first question has positive answer. I will explain why the answer is negative. More precisely, even for ${\mathfrak g}= sl(2, {\mathbb R})\cong o(2,1)$ and $G=PSL(2, {\mathbb R})\cong SO(2,1)_0$, when the exponential map is surjective, there are plenty of elements $\xi_1, \xi_2, \xi_3$ forming a basis of ${\mathfrak g}$, such that not every element of $G$ can be written as a product of the form $$ \exp(t_1 \xi_1) \exp(t_2 \xi_2) \exp(t_3 \xi_3). $$ (I do not like using the notation $g$ for elements of a Lie algebra, I will use the letter $g$ to denote elements of a Lie group.)

  1. Assume that $\xi_1, \xi_2, \xi_3$ are compact elements of the Lie algebra, meaning that each subgroup $G_k=\exp({\mathbb R}\xi_k)$ is compact, isomorphic to $S^1=U(1)$. I will assume that $\xi_1, \xi_2, \xi_3$ are chosen to form a basis of the Lie algebra ${\mathfrak g}$: This is true if you pick compact elements of the Lie algebra generically.

The product (as a topological space!) $$ M=G_1\times G_2\times G_3 $$ is also compact (it is the 3-dimensional torus). Therefore, the image of this product under the Lie-group product map $$ (g_1,g_2,g_3)\in M \mapsto g_1 g_2 g_3\in G $$ has compact image $C$. Since the group $G$ is very much noncompact, there are many elements of $G$ which cannot be written as products $$ g_1 g_2 g_3= \exp(t_1 \xi_1) \exp(t_2 \xi_2) \exp(t_3 \xi_3) $$ no matter what the real numbers $t_1, t_2, t_3$ are. At the same time, since $\xi_1, \xi_2, \xi_3$ span the Lie algebra, every element $g\in G$ can be written as $$ \exp(t_1 \xi_1 + t_2\xi_2 + t_3\xi_3) $$ for suitable choices of real numbers $t_1, t_2, t_3$.

  1. This phenomenon is by no means limited to triples of compact elements of the Lie algebra. I will need a bit of geometry. The group $G= PSL(2, {\mathbb R})$ acts via linear-fractional transformations on the upper half-plane $$ U=\{(x, y): y> 0\}. $$ Take three nested round disks $D_1, D_2, D_3$ centered on the x-axis: $$ D_1\subset D_2\subset D_3 $$ (I do not want to assume that these disks have the same center!) I will denote the diameters of these disks on the x-axis $p_1q_1, p_2q_2, p_3q_3$ (the points $p_i, q_i$ lie on the boundary of $D_i$). Now, the subgroup $G_i$ of $G$ preserving $D_i$ is a 1-parameter noncompact subgroup isomorphic to ${\mathbb R}$, let $\xi_i$ denote generating vectors of the Lie algebras of $G_i$, $i=1, 2, 3$. Then the elements $\xi_1, \xi_2, \xi_3$ again span the Lie algebra ${\mathfrak g}$ provided that the disks are chosen generically. However, not every element $g\in G$ can be written as a product $$ g_1 g_2 g_3, $$ where $g_i\in G_i$, $i=1, 2, 3$. The reason is that $G$ acts on $U$ transitively: every point can be moved to any other point by an element of $G$. However, none of the product elements as above can move a point outside of $D_3$ to a point inside of $D_1$. (I leave this as an exercise.)

Edit. All nontrivial 1-parameter subgroups $\exp({\mathbb R}\xi)$ of $G=PSL(2, {\mathbb R})$ fall into three classes: Elliptic (compact, equivalently, $\xi$ is conjugate to an anti-symmetric matrix), parabolic (equivalently, $\det(\xi)=0$), hyperbolic ($\xi$ is conjugate to a symmetric matrix). The parabolic case is non-generic, I will describe below which triples $G_1, G_2, G_3$ of elliptic/hyperbolic 1-parameter subgroups of $G$ satisfy $G=G_1 G_2 G_3$.

  1. EHE case ($G_1, G_3$ are elliptic, $G_2$ is hyperbolic). Then $G=G_1 G_2 G_3$ if and only if an element of $G_2$ conjugates $G_1$ to $G_3$. (The Cartan decomposition of $G$, $G=KAK$, is the standard example.)

  2. HEH case ($G_1, G_3$ are hyperbolic, $G_2$ is elliptic). Then $G=G_1 G_2 G_3$ if and only if the subgroups $G_1, G_3$ "cross" in the following sense: Each hyperbolic 1-parameter subgroup $H$ has a unique invariant hyperbolic geodesic $\alpha_H$ in the upper half-plane $U$ (each hyperbolic geodesic is either a vertical line in $U$ or the semi-circle with the center on the x-axis). Then $G_1$ and $G_3$ "cross" means that their axes $\alpha_{G_1}, \alpha_{G_3}$ cross at a single point in $U$. This HEH decomposition came as a bit of a surprise to me, I do not think such decompositions have a name (but special cases are surely known since they correspond to non-Riemannian symmetric spaces).

In all other generic cases $G\ne G_1 G_2 G_3$. A proof of this is a bit tedious case-by-case analysis involving fixed points which I omit. (Given vastness of the literature on Lie groups, most likely, this result is known.) The example of $K_0, K_\pm$ in your question is non-generic: The subalgebras $K_\pm$ correspond to parabolic 1-parameter subgroups.

Given this, it is clear to me that the question about the equality $$ G= \prod_{i=1}^n \exp({\mathbb R} \xi_i) $$ for general Lie groups (with surjective exponential map) and even generic tuples of elements of the Lie algebra, does not have a nice answer.

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    $\begingroup$ +1 The first two sentences are pretty funny. The short summary of the two answers combined is that 'yes you can find a basis with the desired property, but no not any basis will do'. $\endgroup$
    – Vincent
    Aug 19, 2020 at 14:31
  • $\begingroup$ BTW I remembered that $\exp$ is not surjective onto $PSL(2, \mathbb{R})$, although now I find that it's not easy to give a counterexample (i.e. unreachable element). Why do you write that it is? $\endgroup$
    – Vincent
    Aug 19, 2020 at 14:35
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    $\begingroup$ @Vincent: It is surjective in this case (but not for $SL(2,R)$). See link.springer.com/article/10.1007/BF01765631#page-1 for a comprehensive treatment of surjectivity. $\endgroup$ Aug 19, 2020 at 14:51
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    $\begingroup$ @SaurabhU.Shringarpure No, different. $\endgroup$ Aug 20, 2020 at 19:31
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    $\begingroup$ @SaurabhU.Shringarpure This would require some thought, since the question does not fit any standard decomposition. Most likely, the answer is negative since it is similar to an HHE decomposition that I can rule out in the case of real coefficients. $\endgroup$ Aug 20, 2020 at 19:40

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