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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)

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  • 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
  • Thinking about it, I think your statement is getting more plausible, and my counterexample isn't a counterexample. – Charlie Frohman Dec 29 '18 at 02:45

1 Answers1

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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 l(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. –  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. –  Dec 28 '18 at 18:33
  • I just noticed Charlie's comments above. I added a preamble in response. –  Dec 28 '18 at 18:37
  • I wrote a precise statement of what I thing must be true. But I cannot make the link between the map $\phi$ I defined and the text you write. For example, when you write "$x(t)=I+tX$ for some vector $X$", you mean that $X$ is a matrix, or a tangent vector as differential operator ? I think that there is somewhere an identification of vector space to be made more explicit. – Laurent Claessens Dec 29 '18 at 07:14
  • @Laurent In this answer, $X$ is a matrix, thought of as a tangent vector to $GL_n$, which is open in all matrices. The Lie bracket may also be phrased in terms of the second derivative at the identity of $G \times G \to G$, the conjugation map. The map you call $\phi$ is the identity map (by definition of the exponential!) I initially wrote the computation in terms of exponentials before realizing there was no reason to. –  Dec 29 '18 at 12:48
  • While the notation can be confusing, the map $\phi$ is not the identity. Indeed, $X$ is a matrix while $\phi(X)$ is the differential operator defined by $\phi(X)f=\frac{d}{dt}f(e^X)$. This is the 'true' signification of the derivative of a path in differential geometry. It turns out that the Lie algebra of these differential operators, in the case of $G=GL(n)$ is isomorphic to the Lie algebra of matrices. I believe that $\phi$ is the isomorphism and the sense of my question is : is $\phi$ an isomorphisme ? Can one prove it ? – Laurent Claessens Dec 29 '18 at 20:51
  • @Laurent Your notation is not confusing. On $GL_n$, one may identify a neighborhood of the identity with a neighborhood of the identity in $\mathfrak{gl}_n$, the set of $n \times n$ matrices, by the simple map from the latter to the former: $X \mapsto I + X$. The first theorem one proves about tangent spaces is that every linear operator $D$ on germs of smooth functions near $I$ satisfying the Leibniz rule is necessarily of the form $Df = \frac{d}{dt} f(\gamma(t))$, for a smooth curve $\gamma$ with $\gamma(0) = I$. This functional is completely determined by $\gamma'(0)$... –  Dec 29 '18 at 21:02
  • ...(thus giving the identification of two notions of tangent space). If $X \in \mathfrak{gl}_n$, call the associated operator as above $D_X$. You give the operator $L_X f = \frac{d}{dt}(e^{tX})$. The curve $\gamma(t) = e^{tX}$ is a smooth curve with $\gamma(0) = I$ and $\gamma'(0) = X$. Therefore, $L_X = D_X$. Your map just sends a tangent vector to the associated functional; this is the canonical identification of these two notions of tangent space. –  Dec 29 '18 at 21:02
  • It is reasonable to say that this "is the identity", because when one speaks of tangent spaces as tangent vectors to curves and as sets of derivations in the same breath, one is implicitly passing through this canonical identification. –  Dec 29 '18 at 21:03
  • Ok. This time I got it. I had some difficulties to relate all that with the "abstract" definition of commutator of vector fileds (which is not the "second derivative" of the adjoint action). I'll typeset a detailed proof for the record and add a link in the next days. Thanks for your help. – Laurent Claessens Dec 30 '18 at 18:34