Vector Fields and Lie Algebra

Vector Fields on Manifolds

Given a differentiable manifold $M$, a vector field $X$ on $M$ is a section of the tangent bundle. Explicitly, $X:M\to T(M)$ is an assignment that sends each point $p\in M$ to a tangent vector $X(p)\in T_p(M)$. The space of all vector fields is denoted by $\mathfrak{X}(M)$. or $\mathcal{X}(M)$.

e.g. $\Gamma (M\times {\mathbb{R}}^k)\cong C^{\infty }(M{)}^{\oplus k}$.

Vector Fields are Derivations

One can see from the derivation definition of tangent vectors that a vector field $V$ can be identified as a derivation $C^{\infty }(M)\to C^{\infty }(M)$ that satisfies the equation: $(V(f))(x):=(V(x))(f).$

The space of vector fields has a rich structure:

Proposition

The space of vector fields $\mathfrak{X}(M)$ is a $\mathbb{R}$-vector space.

Proof It suffices to check that the scaler multiplication is valid. Indeed, $(cX)(p):=c\cdot X(p),\text{for all }c\in \mathbb{R},p\in M,$where the dot denotes the scalar multiplication in $T_p(M)$. It is clear that this satisfies distributivity and the associativity. $\square$

Proposition

The space of vector fields $\mathfrak{X}(M)$ is a module over the ring $C^{\infty }(M)$.

Proof We can define the the scaler multiplication of a vector field $X$ by a smooth function $f\in C^{\infty }(M)$ as follows: $(fX)(p):=f(p)\cdot X(p),\text{for all }p\in M,$where the dot denotes the scalar multiplication in $T_p(M)$. Clearly this gives a smooth vector field, and thus makes $\mathfrak{X}(M)$ a module over $C^{\infty }(M)$. $\square$

Pullback Vector Fields

Pullback Vector Field

Suppose $f:M\to N$ is a local diffeomorphism. The pullback of a vector field $X\in \mathfrak{X}(N)$ by $f$ is the pullback section of the tangent bundle. That is $(f^{\ast }X)(p):=({\text{d}}_{p}f{)}^{-1}(X(f(p)))$

Proof

Lie Bracket of Vector Fields

Lie Bracket of Vector Fields

Let $X,Y\in \mathfrak{X}(M)$. Then the Lie bracket $[X,Y]$ of $X$ and $Y$ is the vector field defined (as a derivation) by
$[X,Y]f:=X(Yf)-Y(Xf).$
The Lie bracket is thus the commutator of $X$ and $Y$ as operators on smooth functions.

Properties of Lie Bracket

The following hold for the Lie bracket of vector fields:

1. Bilinearity: $[X,aY+bZ]=a[X,Y]+b[X,Z]$.
2. Skew-symmetry: $[X,Y]=-[Y,X]$.
3. Jacobi Identity: $[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0$.

That is, $\mathfrak{X}(M)$ forms a real Lie algebra.

Proof Bilinearity and skew-symmetry follow from the definition directly. We now check Jacobi identity. For all $f\in C^{\infty }(M)$, there holds $\begin{array}{rl}& [X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]\\ =& X(Y(Zf))-X(Z(Yf))-Y(Z(Xf))+Z(Y(Xf))\\ & +Y(Z(Xf))-Y(X(Zf))-Z(X(Yf))+X(Z(Yf))\\ & +Z(X(Yf))-Z(Y(Xf))-X(Y(Zf))+Y(X(Zf))\\ =& 0\end{array}$ $\square$

Corollary

Let $M$ be a smooth manifold and let $S$ be an immersed submanifold with or without boundary in $M$. If $X$ and $Y$ are smooth vector fields on $M$ that are tangent to $S$, then $[X,Y]$ is also tangent to $S$.