Let $G$ be a Lie group of matrix. I can define two Lie algebras from there :

  • $G'$: the set of matrices obtained by computing componentwise the derivative $\gamma'(0)$ of every paths $\gamma$ in $G$ such that $\gamma(0)=e$. The elements of $G'$ are matrices and I can show that this is a Lie algebra for the usual commutator of matrices $[X,Y]=XY-YX$ mainely following the steps here.

  • The tangent space $T_eG$ is also a Lie algebra with the commutator given by the commutator of vector fields on the manifold $G$. That Lie algebra can be identified with the left-invariant vectors fields on $G$; I have no specific problems with that.

Are these two Lie algebras isomorphic? I need a proof.

This can be a duplicate of Matrix Lie algebras but the accepted answer there does not answers my question here (in particuar, it does not provide a proof).

The answer of Computing the Lie bracket on the Lie group $GL(n, \mathbb{R})$ does not satisfies me either because it assumes that the exponential from $G'$ has its values in $G$, which remains unclear to me. (In fact, that question adresses the special case $G=GL(n)$)

EDIT: due the comment of Charlie Frohman (which I understood reading the preamble by Mike Miller -- my bad), I precise the set $G'$.

EDIT: A precise statement:

I believe that the following map is a Lie algebra isomorphism in the case of $G=GL(n)$. $\phi: G'\to T_eG$ given by $\phi(X)=\frac{d}{dt}\big( e^{tX} \big)$.


  • $X$ is a matrix (I can proce that G' is the set of all matrices in the case of $G=GL(n)$)
  • on the right hand side, $e^{tX}$ is the matrix exponential
  • on the right hand side, $\frac{d}{dt}\big( e^{tX} \big)$ is the differential operator whose action is $\frac{d}{dt}\big( e^{tX} \big)f=\frac{d}{dt}f\big( e^{tX} \big)$.

I need to prove that for every matrices $X$ and $Y$ we have $\phi[X,Y]=[\phi(X),\phi(Y)]$ (equality of differential operators) where

  • $[X,Y]$ is the usual matrix commutator
  • $[\phi(X),\phi(Y)]$ is the commutator of tangent vectors as differential operators which is defined trough the commutator of the left-invariant vector field.

EDIT: I wrote a complete proof in the case $G=GL(n)$ with a precise statement (as far as In understand something) here: https://laurent.claessens-donadello.eu/pdf/giulietta.pdf

(search for the section intutilated "Matrix lie group and its algebra " around page 2251)

  • $\begingroup$ You are asking what the linear span of $AX$ Is where $A\in G$ and $X\in T_eG$ . It depends on the embedding of $G$ in matrices. The answer will be messy unless you add some more hypotheses. $\endgroup$ Dec 28, 2018 at 13:29
  • $\begingroup$ Do you know more about your matrix group? Is it all matrices that preserve some bilinear form? That might have a more sensible answer. Think about embedding $\mathbb{R}^n\in \mathbb{R}^N$ by appending zeroes and let your matrix group be $GL(\mathbb{R}^n$ embedded in $GL(\mathbb{R}^N)$ by adding a block matrix of the identity. $\endgroup$ Dec 28, 2018 at 13:33
  • $\begingroup$ Here is a sort of lame answer. If $G\leq M_n(\mathbb{R})$ where $G$ is the group and $M_n(\mathbb{R})$ is $n\times n$ matrices with real coefficients, then the Lie algebra spanned by $G$ will just be the linear span of $G$ in $M_n(\mathbb{R})$. $\endgroup$ Dec 28, 2018 at 13:40
  • $\begingroup$ @charlieFrohman Do you know a counter-example if I do not add an hypothesis ? If yes, I am mainely interseted in the groups involved in quantum field theory: SO(n), SU(n), Lorentz. $\endgroup$ Dec 28, 2018 at 14:41
  • $\begingroup$ Yes. Consider $Gl(5,\mathbb{R})\leq Gl(7,\mathbb{R})$ by adjoining a $2\times2$ identity matrix. In the cases you are interested in it looks like everything. $\endgroup$ Dec 28, 2018 at 17:46

1 Answer 1


I assume you are trying to show that the two natural brackets on $\mathfrak g$ - one Lie-theoretic and one via commutators of matrices - agree. As Charlie Frohman points out in the comments above, the "set of derivatives of all paths on $G$" is not the right object to think about; this is the collection of all $AX$, where $A \in G$ and $X \in \mathfrak g$. This is much larger and not a linear space unless you pass to the closure under span, and that has no natural/interesting Lie bracket. What you want is the set of derivatives $\gamma'(0)$ of curves with $\gamma(0) = e$.

Starting from your second bullet point, recall that a homomorphism $f: G \to H$ of Lie groups induces a homomorphism $df_e: \mathfrak g \to \mathfrak h$ of Lie algebras. In particular, take $\rho: G \to GL_n$ to be your defining faithful representation (that is, $\rho$ is injective); then $d\rho_e: \mathfrak g \to \mathfrak{gl}_n$ gives the isomorphism of $\mathfrak g$ with its image inside the space of matrices; its image is what you call $G'$.

In particular, because $d\rho_e$ is injective, it gives an isomorphism of Lie algebras between $\mathfrak g$ and your $G'$, the latter equipped with the bracket on $\mathfrak gl_n$. But you already know that the bracket on $\mathfrak gl_n$ is the usual commutator of matrices, so this gives an isomorphism between $\mathfrak g$ with the left invariant vector field bracket and $G'$ equipped with the matrix-commutator bracket. This is what you wanted.

(Because the Lie-theoretic exponential map is also natural under homomorphisms, one also sees immediately that $\text{exp}(G') \subset \rho(G)$ from this, where because the Lie-theoretic exponential map on $\mathfrak{gl}_n$ is $X \mapsto \sum_{n \geq 0} \frac{X^n}{n!}$, the same is true for the subspace $G'$.)

For self-containedness, here is a proof that the Lie bracket on $\mathfrak{gl}(n)$ is the matrix commutator.

Given any Lie group $G$, there is a map $G \to GL(T_e G)$, given by taking the derivative at $e$ of the conjugation action of $G$ on itself. Taking the derivative of this map we obtain a Lie algebra map $\mathfrak g \mapsto \mathfrak{gl}(\mathfrak g)$; this is the map $X \mapsto [X, -]$, the Lie algebra commutator.

For $G = GL_n$, let $x(t) = I + tX$ for some vector $X$, and similarly $y(s) = I + sY$. These live in $GL_n$ for small $t$, and $x^{-1}(t) = I - tX + O(t^2)$. Then $x(t) \cdot y(s) \cdot x^{-1}(t) = I + sY + tsXY - tsYX + O(s^2, t^2)$. Thus taking the derivative of the map $G \to \text{Aut}(G)$ given by conjugation obtains one the map $G \to \text{Aut}(\mathfrak g)$ given by sending $I+tX$ to the operator $(I+tX)(Y) = Y + t[X, Y] + O(t^2)$, where the commutator is the Lie group commutator. Now taking the derivative as $t \to 0$ we obtain that the map $\mathfrak gl_n \to \text{End}(\mathfrak gl_n)$ is given by $X \mapsto [X, -]$.

  • $\begingroup$ "But you already know that the bracket on 𝔤l(n) is the usual commutator of matrices, ..." My very point is that I do not know that. The Lie algebra of GL(n) are left-invariant vector fields, not matrices. But yes, your answer shows that if I have an isomorphism in the case of the full GL(n), then I can deduce the desired isomorphism for (Lie) subgroups. $\endgroup$ Dec 28, 2018 at 18:04
  • $\begingroup$ @Laurent That's what your second linked answer shows. It seemed to me your only objection was that it covered a special case. $\endgroup$
    – user98602
    Dec 28, 2018 at 18:04
  • $\begingroup$ You are right. I'll perform all the verifications to be 100% sure (math is full of surpises). Then I'll mark your answer as the correct one. $\endgroup$ Dec 28, 2018 at 18:16
  • $\begingroup$ @Laurent I added some more words to make this self-contained. The additional argument already essentially works for any matrix group, and the formalism I added at the start was mainly to avoid reproducing this proof. For instance, naturality shows that the Lie group exponential is natural under Lie group homomorphisms, and so $GL_n$'s matrix exponential restricts to the usual Lie-theoretic exponential on any subgroup. $\endgroup$
    – user98602
    Dec 28, 2018 at 18:33
  • $\begingroup$ I just noticed Charlie's comments above. I added a preamble in response. $\endgroup$
    – user98602
    Dec 28, 2018 at 18:37

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