8. Lie Theory

8.1Lie Derivative

Let $M$ be a smooth manifold, $f\in {\mathcal{C}}^{\infty }(M)={\mathcal{C}}^{\infty }(M,\mathbb{R})$ and $X\in \mathfrak{X}(M)$.

定义 8.1.1. $\begin{array}{rl}(L_{X}f)(p)& =\frac{\partial }{\partial t}{\phi }_t^{\ast }(f)(p){|}_{t=0},\text{where}\phi \text{is the flow of}X.\\ & =\frac{\partial }{\partial t}f({\phi }_t(p)){|}_{t=0}=\underset{t\to 0}{\lim }\frac{f({\phi }_t(p))-f({\phi }_0(p))}{t}\\ & =\underset{t\to 0}{\lim }\frac{f({\phi }_t(p))-f(p)}{t}=D_{p}f(X(p))=d_{p}f(X(p))\end{array}$

$$\begin{array}{rl}{D}_p& f:T_p\to T_{f(p)}\mathbb{R}=\mathbb{R}\\ \leftrightarrow d_p& f:T_p^{\ast }M\end{array}$$

The Lie derivative $L_X$ sends smooth functions to smooth functions$$L_X:{\mathcal{C}}^{\infty }(M)\to {\mathcal{C}}^{\infty }(M)$$

引理 8.1.2. $L_X(f\cdot g)=(L_{X}f)\cdot g+f\cdot (L_{X}g)$ for all $f,g\in {\mathcal{C}}^{\infty }$, where$$f\cdot g:M\to \mathbb{R}$$Like Leibniz rule in derivative $(fg{)}^{\prime }=f^{\prime }g+f{g}^{\prime }$, we have$$D_p(f\cdot g)=(D_{p}f)\cdot g+f\cdot (D_{p}g)$$We can see that$$\begin{array}{rl}L:\mathfrak{X}(M)& \to \mathrm{Der}({\mathcal{C}}^{\infty }(M))\\ X& \mapsto L_X\end{array}$$

定义 8.1.3. If $A$ is a $\mathbb{R}$-algebra, then$$\mathrm{Der}(A):=\{d:A\to A\mid d\text{is}\mathbb{R}\text{-linear and}d(a\cdot b)=d(a)\cdot b+a\cdot d(b)\}$$

If $\lambda \in \mathbb{R}$, then $L_{\lambda X}f=\lambda L_{X}f$, $\forall f\in {\mathcal{C}}^{\infty }(M)$. In fact, for all $g\in {\mathcal{C}}^{\infty }(M)$, $L_{gX}f=g{L}_{X}f$, $\forall f\in {\mathcal{C}}^{\infty }(M)$. Moreover, $L_{X+Y}(f)=L_{X}f+L_{Y}f$.

证明. $\forall p\in M$, we have$$\begin{array}{rl}{L}_X(f\cdot g)(p)& =\frac{\partial }{\partial t}(f\cdot g)({\phi }_t(p)){|}_{t=0}\\ & =\frac{\partial }{\partial t}(f({\phi }_t(p))\cdot g({\phi }_t(p))){|}_{t=0}\\ & =(L_{X}f)(p)\cdot g(p)+f(p)\cdot (L_{X}g)(p)\end{array}$$

$\square$

命题 8.1.4. The map $\begin{array}{rl}\mathfrak{X}(M)& \to \mathrm{Der}({\mathcal{C}}^{\infty }(M))\\ X& \mapsto L_X\end{array}$ is an isomorphism of vector spaces.

证明. (1) The map is linear.
(2) The map is injective: If $X\not\equiv 0$, then $\exists p\in M$, s.t. $X(p)\ne 0$. Consider $\phi :U_p\times (-\epsilon ,\epsilon )\to M$ be part of a local flow of $x$. $\exists f\in {\mathcal{C}}^{\infty }(U_p)$, s.t. $f({\phi }_t(p))\equiv t$. After multiplication with a suitable bump function and extension by $0$, we may arrange $f\in {\mathcal{C}}^{\infty }(M)$. $(L_{X}f)(p)=\left(\frac{\partial }{\partial t}t\right){|}_{t=0}=1$, so $L_X\not\equiv 0$.
(3) The map is surjective: Let $\Delta \in \mathrm{Der}({\mathcal{C}}^{\infty }(M))$.
Step 1: If $U\subset M$ is open, and $f\in {\mathcal{C}}^{\infty }(M)$ is such that $f{|}_U\equiv 0$, then $\Delta (f){|}_U\equiv 0$. For $x\in U$, take $\phi \in {\mathcal{C}}^{\infty }(M)$ with $\phi (x)=0$ and $\phi {|}_{M\setminus U}\equiv 1$.$$\begin{array}{rl}\Rightarrow \phi \cdot f=f& \Rightarrow \Delta (f)=\Delta (\phi )\cdot f+\Delta (f)\cdot \phi \\ & \Rightarrow (\Delta f)(x)=(\Delta \phi )(x)\cdot \underset{=0}{\underbrace{f(x)}}+(\Delta f)(x)\cdot \underset{=0}{\underbrace{\phi (x)}}=0\end{array}$$Step 2: If there is an open neighborhood $U$ of a point $x\in M$, such that $f{|}_U\equiv g{|}_U$, then $(\Delta f)(x)=(\Delta g)(x)$. (Apply Step 1 to $f-g$.)
Step 3: Let $G_x$ be the $\mathbb{R}$-vector space of germs of ${\mathcal{C}}^{\infty }$ functions at $x\in M$. We can define$$\begin{array}{rl}\Delta (x):G_x& \to \mathbb{R}\\ [f]& \mapsto (\Delta f)(x)\end{array}$$Step 2 says that this is well-defined. $\Delta (x)$ is a derivation on the algebra $G_x$. Using a chart, we may assume $M={\mathbb{R}}^n$, $x\in {\mathbb{R}}^n$, $\Delta (x)=\sum _{i=1}^n{\lambda }_i\frac{\partial }{\partial x_i}$. So $\Delta (x)$ is a tangent vector in $T_{x}M$, and it depends smoothly on $x$. Define $X\in \mathfrak{X}(M)$ by setting $X(x)=\Delta (x)$.

Thus, $\Delta =L_X$.

$\square$

引理 8.1.5. For $X$, $Y\in \mathfrak{X}(M)$, there is a unique $[X,Y]\in \mathfrak{X}(M)$, s.t. $L_X{L}_Y-L_Y{L}_X=L_{[X,Y]}$.

证明.  $$\begin{array}{rl}(L_X{L}_Y-L_Y{L}_X)(f\cdot g)& =L_X((L_{Y}f)\cdot g+f\cdot (L_{Y}g))-L_Y((L_{X}f)\cdot g+f\cdot (L_{X}g))\\ & =(L_X{L}_{Y}f)\cdot g+\cancel{(L_{Y}f)\cdot (L_{X}g)}+\cancel{(L_{X}f)\cdot (L_{Y}g)}+f\cdot (L_X{L}_{Y}g)\\ & -(L_Y{L}_{X}f)\cdot g-\cancel{(L_{X}f)\cdot (L_{Y}g)}-\cancel{(L_{Y}f)\cdot (L_{X}g)}-f\cdot (L_Y{L}_{X}g)\\ & =(L_X{L}_Y-L_Y{L}_X)(f)\cdot g+f\cdot (L_X{L}_Y-L_Y{L}_X)(g)\end{array}$$$\forall f,g\in {\mathcal{C}}^{\infty }(M)$, so $L_X{L}_Y-L_Y{L}_X$ is a derivation on ${\mathcal{C}}^{\infty }(M)$. By the surjectivity in the Proposition 8.1.4, $\exists [X,Y]\in \mathfrak{X}(M)$, s.t.$$L_{[X,Y]}=L_X{L}_Y-L_Y{L}_X$$By the injectivity in the Proposition 8.1.4, this vector field is unique.

$\square$

定义 8.1.6. $[X,Y]$ is the Lie bracket of $X$ and $Y$.
$[,]:\mathfrak{X}(M)\times \mathfrak{X}(M)\to \mathfrak{X}(M)$ is bilinear and skew-symmetric.

引理 8.1.7 (Jacobi Identity). $[[X,Y],Z]+[[Z,X],Y]+[[Y,Z],X]=0$, $\forall X,Y,Z\in \mathfrak{X}(M)$.

8.2Lie Algebra and Lie Group

定义 8.2.1. A Lie algebra $\mathfrak{g}$ is a $\mathbb{R}$-vector space, with a map $[,]:\mathfrak{g}\times \mathfrak{g}\to \mathfrak{g}$, which is bilinear, skew-symmetric, and satisfies the Jacobi identity.

$\mathfrak{X}(M)$ is a Lie algebra with the Lie bracket.

定义 8.2.2. A Lie group $G$ is a smooth manifold with a group structure$$\begin{array}{rl}m:G\times G& \to G\\ (g_1,g_2)& \mapsto g_1{g}_2=g_1\cdot g_2\end{array}$$s.t. $m$ and $\begin{array}{rl}i:G& \to G\\ g& \mapsto g^{-1}\end{array}$ are smooth maps.

例 8.2.3.  

(1)	$G=G{L}_k(\mathbb{R})\subset \mathrm{Mat}(k\times k,\mathbb{R})={\mathbb{R}}^{k^2}$.
(2)	Subgroups of $G{L}_k(\mathbb{R})$ which are also submanifolds, e.g. $G{L}_k^{+}(\mathbb{R})$, $O(k)$, $G{L}_k(\mathbb{C})\subset G{L}_{2k}(\mathbb{R})$.

If $G$ is a Lie group and $g\in G$, then$$\begin{array}{rl}\text{left multiplication}{l}_g:G& \to G\\ h& \mapsto g\cdot h\\ \text{right multiplication}{r}_g:G& \to G\\ h& \mapsto h\cdot g\end{array}$$are diffeomorphisms.$l_{g^{-1}}$ is also a smooth map $l_g\circ l_{g^{-1}}=l_{g^{-1}}\circ l_g={\mathrm{Id}}_G$.

For every $g\in G$,$$D_e{l}_g:T_{e}G\to T_{g}G$$is an isomorphism.
$\dim G=n$, $T_{e}G={\mathbb{R}}^n$$$\begin{array}{rl}G\times T_{e}G& \stackrel{t}{\to }TG\\ (g,v)& \mapsto (D_e{l}_g)(v)\end{array}$$

引理 8.2.4. This is an isomorphism of vector bundle, so $TG$ is trivial.

证明. $t$ is smooth. $D_e{l}_g$ is an isomorphism $T_{e}G\to T_{g}G$ for any $g\in G$.

$\square$

定义 8.2.5. $X\in \mathfrak{X}(G)$ is left-invariant if $X(g)=(D_e{l}_g)X(e)$.

引理 8.2.6. If $X$ is left-invariant, then $(D_g{l}_h(X(g))=X(h\cdot g)$.

证明. $((D_g{l}_h)(D_e{l}_g)(X(e)))=(D_e{l}_{h\cdot g})(X(e))=X(h\cdot g)$.

$\square$

定义 8.2.7. $\mathfrak{g}\subset \mathfrak{X}(G)$ is linear subspace of left-invariant vector field.

$[,]$ sends pairs of left-invariant vector fields to a left-invariant vector field. $\Rightarrow \mathfrak{g}\subset \mathfrak{X}(G)$ is a sub-Lie algebra.

定义 8.2.8. $\mathfrak{g}=L(G)$ is the Lie algebra of the Lie group $G$. $\dim \mathfrak{g}=\dim G$.

定义 8.2.9. $X,Y\in \mathfrak{X}(M)$. ${\phi }_t$ is the flow of $x$.$$\begin{array}{rl}{L}_{X}Y& =\text{“}\frac{\partial }{\partial t}Y({\phi }_t(p)){|}_{t=0}\text{”},Y({\phi }_t(p))\in T_{{\phi }_t(p)}M\\ & =\frac{\partial }{\partial t}{D}_{{\phi }_t(p)}{\phi }_{-t}(Y({\phi }_t(p))){|}_{t=0}\\ & =\underset{t\to 0}{\lim }\frac{D_{{\phi }_t(p)}{\phi }_{-t}(Y({\phi }_t(p)))-Y(p)}{t}\end{array}$$

Define $g(t,x)={\int }_0^1{f}^{\prime }(ts,x)\mathrm{d}s$, where $f^{\prime }(u,x)=\frac{\partial f}{\partial u}$ and $f(u,x)=f({\phi }_u(x))$, for any $f\in {\mathcal{C}}^{\infty }(M)$.$$\begin{array}{rl}tg(t,x)& ={\int }_0^1{f}^{\prime }(ts,x)\cdot t\cdot \mathrm{d}s={\int }_0^t{f}^{\prime }(u,x)\mathrm{d}u,\text{where}u=ts\\ & =f(t,x)-f(0,x)=f(t,x)-f(x)\end{array}$$$\Rightarrow f(t,x)=f(x)+tg(t,x)$, $f\circ {\phi }_{-t}=f(-t,x)$.

Claim 8.2.10. $g(0,x)=(L_{X}f)(x)$.

证明. $g(0,x)=\underset{t\to 0}{\lim }g(t,x)=\underset{t\to 0}{\lim }\frac{1}{t}(f(t,x)-f(x))=(L_{X}f)(x)$.

$\square$

定理 8.2.11. $L_{X}Y=[X,Y]$, $\forall X,Y\in \mathfrak{X}(M)$.

证明. Using the isomorphism of $\mathfrak{X}(M)$ and $\mathrm{Der}({\mathcal{C}}^{\infty }(M))$, we need to prove$$L_{L_{X}Y}f=L_{[X,Y]}f,f\in {\mathcal{C}}^{\infty }(M)$$Let ${\phi }_t$ be the flow of $X$ and $\begin{array}{rl}f(t,x)& =f({\phi }_t(x))\\ & =f(x)+tg(t,x)\end{array}$ with $g(0,x)=L_{X}f$.
$L_Y{L}_{X}f=L_{Y}g(0,-)=\underset{t\to 0}{\lim }{L}_{Y}g(t,-)$.
$Z_t=\frac{1}{t}(D_{{\phi }_t(p)}{\phi }_{-t}(Y({\phi }_t)(p))-Y(p))$, so that$$\begin{array}{rl}{L}_{X}Y& =\underset{t\to 0}{\lim }{Z}_t\\ L_{L_{X}Y}f& =\underset{t\to 0}{\lim }{L}_{Z_{t}f}=\underset{t\to 0}{\lim }\frac{1}{t}(L_{D_{{\phi }_t}{\phi }_{-t}(Y)}f-L_{Y}f)\\ & =\underset{t\to 0}{\lim }\frac{1}{t}(L_{Y({\phi }_{-t}(-))}(f\circ {\phi }_{-t})-L_{Y}f)\\ & =\underset{t\to 0}{\lim }\frac{1}{t}(L_{Y({\phi }_t(p))}(f-t{g}_{-t})-L_{Y(p)}f)\\ & =\underset{t\to 0}{\lim }\frac{1}{t}(L_{Y({\phi }_t(p))}(f-t{g}_{-t})-L_{Y(p)}f)\\ & =\underset{t\to 0}{\lim }\frac{1}{t}(L_{Y({\phi }_t(-))}f-L_{Y(-)}f)-\underset{t\to 0}{\lim }{L}_{Y({\phi }_t(-))}{g}_{-t}\\ & =L_X{L}_{Y}f-L_Y{L}_{X}f=L_{[X,Y]}f\end{array}$$

$\square$

定理 8.2.12. Let $X,Y\in \mathfrak{X}(M)$, ${\phi }_t$, ${\phi }_s$ flows for $x$ respectively $Y$. Then $[X,Y]\equiv 0$ $\iff$ ${\phi }_t\circ {\phi }_s={\phi }_s\circ {\phi }_t$, $\forall s,t$.

证明.  

“$\Leftarrow$”

${\phi }_t$, ${\phi }_s$ commuting means that ${\phi }_t$ maps flowlines of $Y$ to flowlines of $Y$. $\Rightarrow$ $D{\phi }_t(Y)=Y$.$$[X,Y]=L_{X}Y=\underset{t\to 0}{\lim }\frac{1}{t}(D{\phi }_t(Y)-Y)=0$$

“$\Rightarrow$”

For $p\in M$, consider $$\begin{array}{rl}v(t)& =D_{{\phi }_t(p)}{\phi }_{-t}Y({\phi }_t(p))\in T_{p}M\\ \dot{v}(t)& =\frac{\partial }{\partial t}v(t){|}_{t=0}=(L_{X}Y)(p)=[X,Y](p)=0\end{array}$$Take $p={\psi }_s(q)$, then $\frac{\partial p}{\partial s}=Y$. $$\frac{\partial }{\partial s}{\phi }_t(p)=(D{\phi }_t)\left(\frac{\partial p}{\partial s}\right)=D{\phi }_t(Y)=Y$$since $v(t)$ is independent of $t$. So ${\phi }_t({\psi }_s(q))$ is a flowline of $Y$ starting at $p={\phi }_t(q)$ at time $s=0$. By the uniqueness of the flowline of $Y$ through $p$, we have $${\phi }_t{\psi }_s(q)={\psi }_s(p)={\psi }_s({\phi }_t(q))$$${\phi }_t\circ {\psi }_s={\psi }_s\circ {\phi }_t$ whenever both sides are defined.

$\square$

定理 8.2.13. Let $X_1,\dots ,X_k\in \mathfrak{X}(M)$, s.t. $[X_i,X_j]\equiv 0$ for all $i$, $j$, and $X_1(p),\dots ,X_k(p)$ are linearly independent in $T_{p}M$ for all $p\in M$. Then around every point $p\in M$, there is a chart $(U,\phi )$, such that $D_q\phi (X_i(q))=\frac{\partial }{\partial X_i}$ for all $i$ and all $q\in U$.

证明. The problem is local, so we may assume $M$ is ${\mathbb{R}}^n$.

After a linear change of basis for ${\mathbb{R}}^n$, we may assume $X_i(0)=\frac{\partial }{\partial x_i}$ for $i\in \{1,\dots ,k\}$. So $X_1(0),\dots ,X_k(0),\frac{\partial }{\partial x_{k+1}},\dots ,\frac{\partial }{\partial x_n}$ is a basis for ${\mathbb{R}}^n=T_0{\mathbb{R}}^n$. $\exists$ open neighborhood $U$ of $0$ in ${\mathbb{R}}^n$ and an $\epsilon >0$, s.t. the local flows ${\phi }^i$ of $X_i$ are defined for all $(p,t)\in U\times (-\epsilon ,\epsilon )$. Define $f:U\to {\mathbb{R}}^n$ by$$f(x_1,\dots ,x_n)={\phi }_{X_1}^1\circ {\phi }_{X_2}^2\circ \cdots \circ {\phi }_{X_k}^k(0,\dots ,0,x_{k+1},\dots ,x_n)$$Without loss of generality, this is defined for all $(x_1,\dots ,x_n)\in U$. By the assumption $[X_i,X_j]\equiv 0$, the ${\phi }^i$ and ${\phi }^j$ commute.

$f$ is smooth and$$\begin{array}{rlrl}\frac{\partial f}{\partial x_i}(0)& =X_i(0)& \text{for}i\in \{1,\dots ,k\},& \text{We also have}\frac{\partial f}{\partial x_i}(x)=X_i(x)\\ \frac{\partial f}{\partial x_i}(0)& =\frac{\partial }{\partial x_i}& \text{for all}i& \\ f(0)& =0& & \end{array}$$$f(0)=0$.
$D_{0}f\left(\frac{\partial }{\partial x_i}\right)=\frac{\partial }{\partial x_i}$ for $i⩾k+1$.
For any $x\in U$, we have $D_{x}f\left(\frac{\partial }{{\partial }_i}\right)=X_i(f(x))$ for $i⩽k$.
If $U$ is small enough, then $f:U\to f(U)$ is a diffeomorphism. Define $\phi :=f^{-1}$, $D_p\phi (X_i)=\frac{\partial }{\partial x_i}$ for all $p\in f(U)$.

$\square$