space of all vector fields

$\mathfrak{X}(M)$ - the space of all vector fields

We can take the space $\mathfrak{X}(M)$ of vector fields on a smooth manifold $M$. This is a vector space, since elements can be added and scaled pointwise $(X+Y{)}_p=X_p+Y_p\forall X,Y\in \mathfrak{X}(M).$ In this way, it uses the vector space structures of the fibers.

As a Lie algebra

For a vector field $X\in \mathfrak{X}(G)$ and a smooth function $f\in {\Omega }^0(G)$, we can combine these in several ways. We can multiply the function and the vector field $(fX{)}_p=f(p)X_p.$ We can also evaluate $X$ on $f$ ($X$ is an operator on the smooth functions $G\to \mathbb{R}$.) If $f:U\to \mathbb{R}$ is defined on an open set $U\subset G$,

$$\begin{array}{rl}Xf:U& ⟶\mathbb{R}\\ p& ⟼X_p(f).\end{array}$$

Thus, for a function $f:G\to \mathbb{R}$, we can view $Xf:G\to \mathbb{R}$ as its own smooth function, and we can apply another vector field to it, $YXf=Y(Xf).$

The only problem is that this new object $YX$ is not guaranteed to be a derivation, so while it is an operator on smooth functions (i.e. $YX\in {\mathcal{O}}_{G,p}^{\ast }$) it may not be in the tangent space.

So, instead we can use the special commutator bracket $[\cdot ,\cdot ]:\mathfrak{X}(G)\to \mathfrak{X}(G)$ where $[X,Y](f)=X(Yf)-Y(Xf)\forall X,Y\in \mathfrak{X}(G).$

Proposition

For $X,Y\in \mathfrak{X}(G)$, $[X,Y]$ is a derivation, and thus $[X,Y]\in \mathfrak{X}(G)$.

Proof

Proof in here…todo

It really is a Lie algebra

$\mathfrak{X}(G)$ is a Lie algebra with Lie bracket $[\cdot ,\cdot ]$ as defined above.

Proof

todo Bilinearity:

Antisymmetry:

Jacobi Identity:

References

@lee2013