Computing the Lie bracket on the Lie group $GL(n, \mathbb{R})$

Question

Consider the Lie group $GL(n, \mathbb{R})$. Since $GL(n, \mathbb{R})$ is an open subset of the space $M_{n,n}(\mathbb{R})$ of $n \times n$ matrices, we can identify the tangent space (Lie algebra) $T_{e}GL(n, \mathbb{R})$ with $M_{n,n}(\mathbb{R})$, where $e$ is the identity matrix. I'm trying to understand the following:

For all $x, y \in M_{n,n}(\mathbb{R}) = T_{Id}GL(n, \mathbb{R})$ we have $[x, y] = xy -yx$ where $[\cdot, \cdot]$ is the Lie algebra bracket.

According to the question Is the Lie bracket of a Lie Algebra of a Matrix Lie Group always the commutator?, this seems to be the case for matrix Lie groups in general.

So far, I've worked out the following: Let $x \in M_{n,n}(\mathbb{R}) = T_{Id}GL(n, \mathbb{R})$ and let $X \in \Gamma^{inv}(TG)$ be the left invariant vector field corresponding to $x$, i.e. with $X(e) = x$. By openness, we can choose $\epsilon > 0$ such that the curve $(-\epsilon, \epsilon) \ni t \mapsto e + tx$ through the identity is contained in $GL(n, \mathbb{R})$. For $g \in GL(n, \mathbb{R})$ let $L_g$ be the left multiplication by $g$ in $GL(n, \mathbb{R})$. Using left invariance of $X$ calculate: $$X(g) = ((L_g)_{*}X)(g) = D_eL_g(X(e)) = D_eL_g(x) = \frac{d}{dt}|_{t=0}g(e+tx) = gx$$ Now, the above implies for $f \in C^{\infty}(GL(n, \mathbb{R}))$ and $g \in GL(n, \mathbb{R})$: $$X(g)(f) = (D_gf)(gx)$$ where we identify the tangent vector $X(g)$ with the corresponding derivation at $g$.

Now how do I calculate $[x,y] \stackrel{def.}{=} [X, Y](e) \stackrel{\text{WHY?}}{=}xy-yx$ from this?

Answer 1

I've now been able to do the calculation myself with a bit of inspiration from freeRmodule's answer:

Using the calculations presented in the question, we calculate the adjoint operator for the Lie group $G := GL(n, \mathbb{R})$. For $g \in G$ and $x \in M_{n,n}(\mathbb{R})=T_{Id}G$ with corresponding left invariant vector field $X$ we have: $$Ad(g)(x) = ((R_{g^{-1}})_{*}X)(Id) = D_gR_{g^{-1}}(\overbrace{X(g)}^{=gx}) = gxg^{-1}$$

By a statement which holds for Lie groups in general we then calculate the Lie bracket of $x$ and $y$ (noting that the Lie group exponential $exp_G$ on $G$ is simply the matrix exponential $Exp$): \begin{align*} [x,y] &= \frac{d}{dt}|_{t=0} Ad(exp_G(tx))(y) = \frac{d}{dt}|_{t=0}Exp(tx)yExp(-tx) \\ &= \left( \frac{d}{dt}|_{t=0}Exp(tx) \right)\cdot y \cdot Id + Id \cdot y \cdot \left( \frac{d}{dt}|_{t=0}Exp(-tx) \right) \\ &= xy -yx \end{align*}

Answer 2

Perhaps this won't satisfy you, but we know that $ad:\mathfrak{g}\to\mathfrak{gl}(\mathfrak{g})$ and $Ad:G\to GL(\mathfrak{g})$ satisfy $d(Ad)_e=ad$, for any Lie group. Here, since we are working with matrices in $\mathbb{R}^{n^2}$, this implies:

If $X,Y\in\mathfrak{gl}_n$, and $g(t)$ is a path of matrices in $GL_n$ such that $g(0)=I_n$ and $g'(0)=X$, then: $$ ad(X)(Y)=\frac{d}{dt}Ad(g(t))(Y)\bigg|_{t=0}=\frac{d}{dt}g(t)Yg(t)^{-1}\bigg|_{t=0}=XY-YX $$ with the last line being by the product rule.

EDIT: Here's a more hands-on approach, following some notes by Wolfgang Ziller (I always get confused during these calculations! I hope this helps someone.)

Let $X,Y\in T_eG$ be tangent vectors at the identity ($G=GL_n(\mathbb{R}))$. Then since $G$ is an open submanifold of $Mat_n(\mathbb{R})$, we can make a canonical identification of all its tangent spaces with $Mat_n(\mathbb{R})$. Once we do this, if $g\in GL_n(\mathbb{R})$ then $(dL_g)_e(X)=gX$, (since matrix multiplication is linear) so the left invariant vector field associated to $X$ is $gX$ at the point $g\in G$. Similarly the one associated to $Y$ is $gY$ at $g$.

Next, notice that if $y_{ij}$ is the coordinate function on matrices picking out the $(i,j)$th entry in the tangent space $T_{h}G$, then $(gX)(y_{ij})=dy_{ij}(gX)=(gX)_{ij}$ because $y_{ij}$ is a linear function. Note that $(gX)_{ij}$ should be thought of as the function defined in a small neighborhood of $h$ given by $g\mapsto (gX)_{ij}$.

Now we can compute the value of the bracket of $X$ and $Y$ at the identity $e$:

$$ [X,Y](y_{ij})=X(Y(y_{ij}))-Y(X_{ij}))=X(g\mapsto(gY)_{ij})-Y(g\mapsto(gX)_{ij}) $$ Now $d(g\mapsto (gY)_{ij})=\sum\limits_{k=1}^ny_{kj}dy_{ik}$, so $$ X(g\mapsto(gY)_{ij})=\sum\limits_{k=1}^ny_{kj}dy_{ik}(X)=\sum\limits_{k}x_{ik}y_{kj}=(XY)_{ij} $$ Similary, $Y(g\mapsto(gX)_{ij})=(YX)_{ij}$, so $[X,Y](y_{ij})=(XY-YX)_{ij}$. This says that the entries of $[X,Y]$ are the same as the entries of $(XY-YX)$, so we're done.

Answer 3

I know how to do this for linear algebraic groups over an algebraically closed field $k$. I assume it works the same for Lie groups? I don't know Lie groups. Anyway, I hope this is helpful to you.

Let $G = \textrm{GL}_n$, let $R = k[T_{ij} : 1 \leq i, j \leq n]$, and let $A$ be the localization of $R$ at the function $\textrm{Det}$. Then $A = k[G]$ is the coordinate ring of $G$.

By definition, the tangent space of $G$ at $1_G$ is the $k$-vector space of derivations $\textrm{Der}_k(A,k)$ of $A$ into $k$, i.e. those $k$-linear maps $\delta: A \rightarrow k$ for which $\delta(fg) = f(1_G)\delta(g) + g(1_G) \delta(f)$ for all $f, g \in A$.

Every $\delta \in \textrm{Der}_k(R,k)$ extends uniquely, using the quotient rule, to an element of $\textrm{Der}_k(A,k)$. And $\textrm{Der}_k(R,k)$ is isomorphic as $k$-vector spaces to $M_n(k)$, the space of $n$ by $n$ matrices: every $\delta \in \textrm{Der}_k(R,k)$ is completely determined by its values on the coordinate functions $T_{ij}$, so associate $\delta$ to the matrix $(\delta(T_{ij})) \in M_n(k)$.

Now, let $L(G)$ be the subspace of $\textrm{Der}_k(A,A)$ (derivations from $A$ to itself) consisting of those $D$ which are left invariant: this means that $\lambda_x \circ D = D \circ \lambda_x$ for all $x \in G$. Here if $x \in G$, $\lambda_x$ is the $k$-algebra isomorphism $A \rightarrow A$ given by $\lambda_x(f)(y) = f(x^{-1}y)$ for all $y \in G$.

One can show that $\textrm{Der}_k(A,k)$ is isomorphic as $k$-vector spaces of $L(G)$: if $D \in L(G)$, we associate the element $\delta \in \textrm{Der}_k(A,k)$ given by $\delta(f) = D(f)(1_G)$. The inverse map is more complicated to describe: if $\delta \in \textrm{Der}_k(A,k)$, then we associate the element $D \in L(G)$ which sends an $f \in A$ to the element of $A$ given by the formula $x \mapsto \delta(\lambda_{x^{-1}}(f))$.

Now, $L(G)$ is naturally a Lie algebra with the bracket operation $[D,D'] = D \circ D' - D' \circ D$. Using the $k$-vector space isomorphism $L(G) \rightarrow \textrm{Der}_k(A,k)$, we transfer the bracket operation to $\textrm{Der}_k(A,k)$ to give $\textrm{Der}_k(A,k)$ the structure of a Lie algebra. The vector space $\textrm{Der}_k(A,k)$, considered as a Lie algebra, is called the Lie algebra of $G$ and is often denoted $\mathfrak g$.

Now you identify the Lie algebra of $G$ with $M_n(k)$, and you want to describe the bracket operation on $M_n(k)$ explicitly. So you need to transfer the bracket operation on $L(G)$ using the isomorphisms

$$M_n(k) \rightarrow \textrm{Der}_k(R,k) \rightarrow \textrm{Der}_k(A,k) \rightarrow L(G)$$

If you go through these isomorphisms, you will see that an element $D \in L(G)$, identified as an element of $M_n(k)$, is the $n$ by $n$ matrix whose $ij$th entry is $D(T_{ij})(1_G)$. Conversely, if you begin with an $n$ by $n$ matrix $A = (a_{ij})$, the corresponding element $D$ in $L(G)$ is given by the formula

$$D(T_{ij}) = \sum\limits_l T_{il}a_{lj}$$

Now, let $A, B \in M_n(k)$, and let $D,D'$ be the corresponding elements of $L(G)$. To show that the Lie bracket transfers over the way you want, it is enough to show that $D \circ D'$ is the element of $L(G)$ which corresponds to the product $AB$.

Let $C$ be the matrix corresponding to $D \circ D'$. The $ij$th entry of $C$ is $(D \circ D')(T_{ij})(1_G)$.

Now $D'(T_{ij}) = \sum\limits_l T_{il}b_{lj}$, and $D(T_{il}) = \sum\limits_s T_{is}a_{sl}$. Hence

$$D \circ D'(T_{ij}) = \sum\limits_{l,s} T_{is}a_{sl}b_{lj}$$ and so $c_{ij}$ is the evaluation of this polynomial at the identity matrix $1_G$. Now $T_{is}(1_G) = \delta_{is}$, and so

$$c_{ij} = \sum\limits_l a_{il}b_{lj}$$ done.