The lie bracket appears in manifolds and matrix groups. For manifolds a tangent vector $X$ is $$X(p)=\sum_{i=1}^n a_i(p) \frac{\partial}{\partial x_i}$$ where there is the parametrization $\mathbf{x}:U\subset\mathbb{R}^n\to M$ and $\frac{\partial}{\partial x_i}$ is the basis associated with $\mathbf{x}$. Then the change in a function $f$ in the direction of $X$ is $$(Xf)(p) = \sum_i a_i(p) \frac{\partial f}{\partial x_i}(p)$$ The lie bracket of $[X,Y]$ is $(XY-YX)f$ and that expression amounts to applying the above formula for $(Xf)(p)$ multiple times.

Consider a matrix lie group $G$ with lie algebra $T_1(G)$, for example orthogonal matrices and skew-symmetric matrices. The lie bracket is also $XY-YX$ but this is matrix multiplication. This expression can come from considering $C_s(t)=A(s)B(t)A(s)^{-1}$, $C_s'(0)=A(s)YA(s)^{-1}$, $D(s)=A(s)YA(s)^{-1}$,$D'(0)=XY-YX$. But is there a way this can be understood in the previous manifold description of $XY-YX$ which involves tangent vectors where $XYf$ is applying the dot product of a tangent vector with the gradient of a function and doing this twice whereas in the matrix context this was matrix multiplication. Can this all be reconciled?


Remark that $Gl(n,\mathbb{R})$ is an open subset of the vector space $M(n,\mathbb{R})$ so for every $g\in Gl(n,\mathbb{R})$, $T_gGl(n,\mathbb{R})=M(n,\mathbb{R})$. Let $X$ be an element of $M(n,\mathbb{R})$ it defines a vector field on $Gl(n,\mathbb{R})$ such that $X(g)=gX\in T_gGl(n,\mathbb{R})$ where we consider the multiplication of matrices.

For every function $f$ defined on $Gl(n,\mathbb{R})$, $(Xf)(g)=df_g(X(g)=df_ggX$, we deduce that $(Y(Xf)(g)=d^2f.X(g).Y(g)+YX(g)$ since the differential of $l_X:g\rightarrow gX$ is $l_X$ since $l_X$ is linear.

This implies that $[X,Y](f)=[XY-YX]g$ since $d^2f.X(g).Y(g)=d^2f.Y(g).X(g).$

  • $\begingroup$ what is $df_g$ given that $f:Gl(n,\mathbf{R})\to \mathbf{R}$ $\endgroup$ – user782220 Jan 21 '19 at 0:22
  • $\begingroup$ It is the differential of $f$, you can consider $Gl(n,R)$ as an open subset of the vector space $M(n,R)$. $\endgroup$ – Tsemo Aristide Jan 21 '19 at 0:28
  • $\begingroup$ But what does that look like given $M(n,R)$. Normally the differential is in the context of something like $R^n \to R$ $\endgroup$ – user782220 Jan 21 '19 at 1:11
  • $\begingroup$ The differential of a function $f$ at $x$ is a linear map $df_x$ defined on the tangent space of $\mathbb{R}^n$ at $x$. $\endgroup$ – Tsemo Aristide Jan 21 '19 at 1:30

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