# Prove the equivalence of the definitions of matrix Lie algebra via 1) differentiation of curves and 2) tangent space to the identity

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 $$\phi(X)=\frac{d}{dt}\big( e^{tX} \big)$$.

Where:

• $$X$$ 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)

• 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. – Charlie Frohman Dec 28 '18 at 13:29
• 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. – Charlie Frohman Dec 28 '18 at 13:33
• 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 $M_n(\mathbb{R})$. – Charlie Frohman Dec 28 '18 at 13:40
• @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. – Laurent Claessens Dec 28 '18 at 14:41
• 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. – Charlie Frohman Dec 28 '18 at 17:46

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 $$X \mapsto [X, -]$$.

• "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. – Laurent Claessens Dec 28 '18 at 18:04
• @Laurent That's what your second linked answer shows. It seemed to me your only objection was that it covered a special case. – user98602 Dec 28 '18 at 18:04
• 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. – Laurent Claessens Dec 28 '18 at 18:16
• @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. – user98602 Dec 28 '18 at 18:33
• I just noticed Charlie's comments above. I added a preamble in response. – user98602 Dec 28 '18 at 18:37