# Lie Bracket and Matrix Groups

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 $$d^2f.X(g).Y(g)=d^2f.Y(g).X(g).$$
• what is $df_g$ given that $f:Gl(n,\mathbf{R})\to \mathbf{R}$ – user782220 Jan 21 '19 at 0:22
• It is the differential of $f$, you can consider $Gl(n,R)$ as an open subset of the vector space $M(n,R)$. – Tsemo Aristide Jan 21 '19 at 0:28
• But what does that look like given $M(n,R)$. Normally the differential is in the context of something like $R^n \to R$ – user782220 Jan 21 '19 at 1:11
• The differential of a function $f$ at $x$ is a linear map $df_x$ defined on the tangent space of $\mathbb{R}^n$ at $x$. – Tsemo Aristide Jan 21 '19 at 1:30