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

Vector Fields on Lie Groups

Left-Invariant Vector Field

If $G$ is a Lie group and $v$ is a vector in $T_{e}G$, where $e$ is the identity element in $G$. Suppose ${\ell }_g:G\to G$ is the left multiplication map that sends $h$ to $gh$. Then we can use the maps ${\text{D}}_e{\ell }_g$ to define a natural vector field on $G$, for each $g\in G$. We set
$V(g):={\text{d}}_e{\ell }_g(v).$
The resulting vector field $V$ is called a left-invariant vector field.

e.g.

• Let Lie group $G=\text{C}\setminus \{0\}$, with the group operation of complex number multiplication. Fix some $v\in T_{1}G={\mathbb{R}}^2$, the left-invariant vector field $V$ is given by $V(z)={\text{d}}_1{\ell }_z(v)=zv$

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 $S$.

Lie Algebra of Lie Groups

Every Lie group $G$ has a canonical Lie algebra $\mathfrak{g}$, which has two natural representations, either as the tangent space at the identity element $e$ of $G$, or as the space of left-invariant vector fields on $G$. We will see that the two representations are isomorphic.

Proposition

Let $G$ be a Lie group, and suppose $X$ and $Y$ are smooth left-invariant vector fields on $G$. Then $[X,Y]$ is also left-invariant.

Proposition

The left-invariant vector fields on a Lie group form a Lie algebra.

Proof It suffices to check $\text{D}{\ell }_g{|}_e$

Theorem

The set of all left-invariant vector fields on a Lie group $G$ is isomorphic to the tangent space at the identity element $e$ of $G$, via evaluation at $e$. In other words, every tangent vector $v\in T_{e}G$ extends uniquely to a left-invariant vector field.

Proof

Lie Algebra of a Lie Group

The Lie algebra of a Lie group $G$, often denoted $\mathfrak{g}$, is up to isomorphic, the Lie algebra of left-invariant vector fields on $G$, or the tangent space at the identity element $e$ of $G$.

Remark

Elements of the Lie algebra represent “infinitesimal transformations” or “directions of motion” within the group starting from the identity.

Exponential Map

For any Lie group $G$ with Lie algebra $\text{g}$, the exponential map $\exp :\text{g}\to G$ is defined by $\exp (X)=\gamma (1),$where $\gamma$ is the unique integral curve of the left-invariant vector field $X$ such that $\gamma (0)=e$, or equivalently, $\gamma$ is the one-parameter subgroup generated by $X$.

e.g. The Lie algebra of $S^1$, denoted $\mathfrak{s}$, is isomorphic to $\mathbb{R}$. The exponential map of $S^1$ satisfies $\exp (tx)=e^{itx},\text{for all }x\in \text{s}\cong \mathbb{R},t\in \mathbb{R}.$

Proposition

Let $G$ be a Lie group with associated Lie algebra $\text{g}$. For any $X\in \text{g}$, $s\mapsto \exp (sX)$ is the one-parameter subgroup of $G$ generated by $X$. That is, $\exp (tX)=\gamma (t)$ for all $t\in \mathbb{R}$.

Proof

Fundamental Vector Field

Suppose $G$ is a Lie group with Lie algebra $\text{g}$, acting on a smooth manifold $M$ with the action $\ast :G\times M\to M$ that $(g,p)\mapsto g\ast p$. The fundamental vector field on $M$ generated by the infinitesimal action of $\xi \in \text{g}$, denoted by ${\xi }_M$ (or $X^{\xi }$), is defined at each point $p\in M$ as ${\xi }_M{|}_p:=\frac{\mathrm{d}}{\mathrm{d}t}{|}_{t=0}\left(\exp (t\xi )\ast p\right).$

e.g. Since the $n$-torus $T^n={\left(S^1\right)}^n$, its Lie algebra $\mathfrak{t}$ is isomorphic to ${\mathbb{R}}^n$. From the previous example of $S^1$, we can deduce that the exponential map of an $n$-torus takes the form $\exp (tX)=\left(e^{it{x}_1},e^{it{x}_2},\dots ,e^{it{x}_n}\right),\text{for all }X=(x_1,\dots ,x_n)\in \text{t}\cong {\mathbb{R}}^n.$Consider a torus $T^n$ acting on ${\text{C}}^n$ by $(e^{i{\theta }_1},\dots ,e^{i{\theta }_n})\ast (z_1,\dots ,z_n):=(e^{i{k}_1{\theta }_1}{z}_1,\dots ,e^{i{k}_n{\theta }_n}{z}_n),$for nonzero constants $k_1,\dots ,k_n\in \mathbb{Z}$. . Then for any $\xi =({\xi }_1,\dots ,{\xi }_n)\in \text{t}$ and $z=(z_1,\dots ,z_n)\in {\text{C}}^n$, there holds ${\xi }_M{|}_z:=\text{ddtz}\left(\exp (t\xi )\ast z\right)=(i{k}_1{\xi }_1{z}_1,\dots ,i{k}_n{\xi }_n{z}_n).$