In this question, the answer by twistor59 says

by using the exponential map on linear combinations of [the lie algebra basis vectors], you generate (at least locally) a copy of the Lie group.

I'm interested in the details of "at least locally": Given a fixed basis $A_1, A_2, ... A_k$ of the lie algebra, under what conditions can every element of the lie group be obtained as:

$$ \exp(A_1 a_1 + A_2 a_2 + ... + A_k a_k)\tag{1}$$

and in what cases can it be represented as

$$ \exp(A_1 a_1) \cdot \exp(A_2 a_2) \cdot ... \cdot \exp(A_k a_k) ~?\tag{2}$$

If the $A_i$ commute, $[A_i, A_j] = 0 \;\; \forall i , j $, then the two forms are of course equivalent, and by adding up infinitesimal steps, either of the two forms should be valid, so this question is about the remaining cases.

The last answer here by Arnold Neumaier seems to have an answer, but since it has only one upvote: Can anyone confirm it and / or give more details on how he concludes the following:

If $G$ is a simply connected Lie group with associated Lie algebra $g $, any basis of $g$ is referred to as a (minimal) set of generators of both $G$ and $g$. Indeed, the elements of $g$ are generated by taking linear combinations of generators, while the elements of $G$ are generated by taking products of exponentials $e^{\alpha A_i}$ with real or complex $α$ and a generator $A_i$ . In the compact case, the elements $G$ are also generated by taking all exponentials $e^{∑_i α_i A_i}$.

I have limited background in Lie groups/algebras, so I'm for example not entirely sure about 'simply connected' (for me this means every element is connected to every other element by a path through the lie group, and there are no holes), or 'compact' (to me that means limit points are also contained in the Lie group).


3 Answers 3


The main points are that given a Lie group $G$,

  • the Lie algebra $\mathfrak{g}=T_eG$ is the tangent space of the identity element $e\in G$.

  • generators are a basis for the Lie algebra $\mathfrak{g}$.

  • there exists an exponential map $\exp:\mathfrak{g}\to G$ into the Lie group.

  • The image ${\rm Im}(\exp)\subseteq G $ of the expential map is a neighborhood of $e$, but the exponential map is not necessarily surjective for a generic group $G$. See e.g. this math.SE post and this Phys.SE post for counterexamples. It is however surjective for a group $G$ that is connected and compact.

  • the relation between OP's eqs. (1) and (2) is described by (various versions of) the BCH formula.


Maybe as a simple example, consider $O(2)$ (real $2\times 2$ matrices $M$ with $M^T M=\mathbb{1}$). You can see that $\det M=\pm1$, so the group is disconnected (you cannot have a continuous path from a matrix $M$ with $\det M=1$ to a matrix $N$ with $\det N=-1$, becaude the determinant would hav to jump somewhere in between). Explicitly, you can write group elements as $$M=\begin{pmatrix} \cos \phi & -\sin \phi\\ \sin\phi & \cos\phi\end{pmatrix}$$ or $$N=\begin{pmatrix} -\cos \psi & \sin \psi\\ \sin\psi & \cos\psi\end{pmatrix}$$ with some angles $\phi$ and $\psi$. Note that matrices of the form $M$ form the subgroup $SO(2)$; the $N$ matrices form the other connected component which is not a subgroup.

You can generate all matrices of type $M$ as the exponential of the generator $$m=\begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix}\,,$$ but you will never reach $N$.


Let us consider simply connected Lie groups first. Because all of them are groups, the group property allows every element of the group to be generated by the exponential map on the Lie algebra space of the identity element, so $$e^{\sum_ia_iA_i}$$ definitely works. You can get from identity element to any element in a simply connected Lie group like that. A general Lie group might have discrete symmetries like reflections, and those will separate into multiple copies of simply connected Lie groups, so you can add those discrete symmetries separately.

I am not familiar with whether $$\prod_ie^{a_iA_i}$$ works, and I do not think it matters that much. It probably should, but I am happy with the earlier form. Do you know that there is a https://en.wikipedia.org/wiki/Baker%E2%80%93Campbell%E2%80%93Hausdorff_formula that helps you with this?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.