Differentiable Manifolds/Lie algebras and the vector field Lie bracket

The vector field Lie bracket

Theorem 6.4: If $𝐕,𝐖$ are vector fields of class ${𝒞}^n$ on $M$, then $[𝐕,𝐖]$ is a vector field of class ${𝒞}^n$ on $M$ (i. e. $[\cdot ,\cdot ]$ really maps to $𝔛(M)$)

Proof:

1. We show that for each $p\in M$, $[𝐕,𝐖](p)\in T(p)M$. Let $\phi ,\vartheta \in {𝒞}^{\mathrm{\infty }}(M)$ and $c\in ℝ$.

1.1 We prove linearity:

$\begin{array}{cc}[𝐕,𝐖](p)(\phi +c\vartheta )& =𝐕(p)(𝐖(\phi +c\vartheta ))-𝐖(p)(𝐕(\phi +c\vartheta ))\\ & =𝐕(p)(𝐖\phi +c𝐖\vartheta )-𝐖(p)(𝐕\phi +c𝐕\vartheta )\\ & =𝐕(p)(𝐖\phi )-𝐖(p)(𝐕\phi )+c(𝐕(p)(𝐖\vartheta )-𝐖(p)(𝐕\vartheta ))\\ & =[𝐕,𝐖](p)(\phi )+c[𝐕,𝐖](p)(\vartheta )\end{array}$

1.2 We prove the product rule:

$\begin{array}{cc}[𝐕,𝐖](p)(\phi \vartheta )& =𝐕(p)(𝐖(\phi \vartheta ))-𝐖(p)(𝐕(\phi \vartheta ))\\ & =𝐕(p)(\phi 𝐖\vartheta +\vartheta 𝐖\phi )-𝐖(p)(\phi 𝐕\vartheta +\vartheta 𝐕\phi )\\ & =𝐕(p)(\phi 𝐖\vartheta )+𝐕(p)(\vartheta 𝐖\phi )-𝐖(p)(\phi 𝐕\vartheta )-𝐖(p)(\vartheta 𝐕\phi )\\ & =\phi (p)𝐕(p)(𝐖\vartheta )+\stackrel{=Y(p)(\vartheta )}{\overbrace{\mathbf{(}Y\vartheta )(p)}}𝐕(p)(\phi )+\vartheta (p)𝐕(p)(𝐖\phi )+\stackrel{=Y(p)(\phi )}{\overbrace{\mathbf{(}Y\phi )(p)}}𝐕(p)(\vartheta )\\ & -\phi (p)𝐖(p)(𝐕\vartheta )-\stackrel{=X(p)(\vartheta )}{\overbrace{\mathbf{(}X\vartheta )(p)}}𝐖(p)(\phi )-\vartheta (p)𝐖(p)(𝐕\phi )-\stackrel{=X(p)(\phi )}{\overbrace{\mathbf{(}X\phi )(p)}}𝐖(p)(\vartheta )\\ & =\phi (p)𝐕(p)(𝐖\vartheta )-\phi (p)𝐖(p)(𝐕\vartheta )+\vartheta (p)𝐕(p)(𝐖\phi )-\vartheta (p)𝐖(p)(𝐕\phi )\\ & =\phi (p)[𝐕,𝐖](p)(\vartheta )+\vartheta (p)[𝐕,𝐖](p)(\phi )\end{array}$

2. We show that $[𝐕,𝐖]$ is differentiable of class ${𝒞}^n$.

Let $\phi \in {𝒞}^n(M)$ be arbitrary. As $𝐕,𝐖$ are vector fields of class ${𝒞}^n$, $𝐕\phi$ and $𝐖\phi$ are contained in ${𝒞}^n(M)$. But since $𝐕,𝐖$ are vector fields of class ${𝒞}^n$, $𝐕(𝐖\phi )$ and $𝐖(𝐕\phi )$ are contained in ${𝒞}^n(M)$. But the sum of two differentiable functions is again differentiable (this is what theorem 2.? says), and thus $[𝐕,𝐖]\phi$ is in ${𝒞}^n(M)$, and since $\phi$ was arbitrary, $[𝐕,𝐖]$ is differentiable of class ${𝒞}^n$.$◻$

Theorem 6.5:

If $M$ is a manifold, and $[\cdot ,\cdot ]$ is the vector field Lie bracket, then $𝔛(M)$ and $[\cdot ,\cdot ]$ form a Lie algebra together.

Proof:

1. First we note that $𝔛(M)$ as defined in definition 5.? is a vector space (this was covered by exercise 5.?).

2. Second, we prove that for the vector Lie bracket, the three calculation rules of definition 6.1 are satisfied. Let $𝐕,𝐖,𝐔\in 𝔛(M)$ and $c\in ℝ$.

2.1 We prove bilinearity. For all $p\in M$ and $\phi \in {𝒞}^n(M)$, we have

$\begin{array}{cc}[𝐕,𝐖+c𝐔](p)(\phi )& =𝐕(p)((𝐖+c𝐔)\phi )-(𝐖+c𝐔)(p)(𝐕\phi )\\ & =𝐕(p)(𝐖\phi +c𝐔\phi )-𝐖(p)(𝐕\phi )-c𝐔(p)(𝐕\phi )\\ & =𝐕(p)(𝐖\phi )-𝐖(p)(𝐕\phi )+c𝐕(p)(𝐔\phi )-c𝐔(p)(𝐕\phi )\\ & =[𝐕,𝐖](p)(\phi )+c[𝐕,𝐔](p)(\phi )\end{array}$

and hence, since $p\in M$ and $\phi \in {𝒞}^n(M)$ were arbitrary,

$[𝐕,𝐖+c𝐔]=[𝐕,𝐖]+c[𝐕,𝐔]$

Analogously (see exercise 1), it can be proven that

$[𝐕+c𝐖,𝐔]=[𝐕,𝐔]+c[𝐖,𝐔]$

2.2 We prove skew-symmetry. We have for all $p\in M$ and $\phi \in {𝒞}^n(M)$:

$[𝐕,𝐖](p)(\phi )=𝐕(p)(𝐖\phi )-𝐖(p)(𝐕\phi )=-(𝐖(p)(𝐕\phi )-𝐕(p)(𝐖\phi ))=-[𝐖,𝐕](p)(\phi )$

2.3 We prove Jacobi's identity. We have for all $p\in M$ and $\phi \in {𝒞}^n(M)$:

$\begin{array}{cc}[𝐕,[𝐖,𝐔]](p)(\phi )+[𝐔,[𝐕,𝐖]](p)(\phi )+[𝐖,[𝐔,𝐕]](p)(\phi )& =𝐕(p)([𝐖,𝐔]\phi )-[𝐖,𝐔](p)(𝐕\phi )\\ & +𝐔(p)([𝐕,𝐖]\phi )-[𝐕,𝐖](p)(𝐔\phi )\\ & +𝐖(p)([𝐔,𝐕]\phi )-[𝐔,𝐕](p)(𝐖\phi )\\ & =𝐕(p)(𝐖(𝐔\phi )-𝐔(𝐖\phi ))-𝐖(p)(𝐔(𝐕\phi ))+𝐔(p)(𝐖(𝐕\phi ))\\ & +𝐔(p)(𝐕(𝐖\phi )-𝐖(𝐕\phi ))-𝐕(p)(𝐖(𝐔\phi ))+𝐖(p)(𝐕(𝐔\phi ))\\ & +𝐖(p)(𝐔(𝐕\phi )-𝐕(𝐔\phi ))-𝐔(p)(𝐕(𝐖\phi ))+𝐕(p)(𝐔(𝐖\phi ))\\ & =0\end{array}$

, where the last equality follows from the linearity of $𝐕(p),𝐖(p)$ and $𝐔(p)$.$◻$