Differentiable Manifolds/Product manifolds and Lie groups

The product manifold

Lemma 10.3:

Let $O\subseteq {ℝ}^{2d}$ be open. Then for each $(x,y)\in O$, there exists ${ϵ}_{(x,y)}>0$ such that

$B_{{ϵ}_{(x,y)}}(x)\times B_{{ϵ}_{(x,y)}}(y)\subset O$

Proof:

Let $(x,y)\in O$ be arbitrary. We choose $\delta >0$ such that $B_{\delta }((x,y))\subseteq O$. Then we define ${ϵ}_{(x,y)}:=\frac{\delta }{2}$. Then, by the triangle inequality we have for every $(z,w)\in B_{{ϵ}_{(x,y)}}(x)\times B_{{ϵ}_{(x,y)}}(y)$:

$\Vert (z,w)-(x,y)\Vert =\Vert (z,0)+(0,w)-(x,0)-(0,y)\Vert \le \Vert (z,0)-(x,0)\Vert +\Vert (0,w)-(0,y)\Vert =\Vert z-x\Vert +\Vert w-y\Vert <2{ϵ}_{(x,y)}=\delta$

Therefore, $(z,w)\in B_{\delta }((x,y))\subseteq O$, and since $(z,w)\in B_{{ϵ}_{(x,y)}}(x)\times B_{{ϵ}_{(x,y)}}(y)$ was arbitrary:

$B_{{ϵ}_{(x,y)}}(x)\times B_{{ϵ}_{(x,y)}}(y)\subseteq O$$◻$

Lemma 10.4:

Let $O,U\subseteq {ℝ}^d$ be open sets. Then $O\times U$ is open in ${ℝ}^{2d}$.

Proof:

Due to the openness of $O$ and $U$ for each point $(x,y)$ in $O\times U$, we find ${ϵ}_1,{ϵ}_2\in {ℝ}_{>0}$ such that $B_{{ϵ}_1}(x)\subseteq O$ and $B_{{ϵ}_2}(x)\subseteq U$. If we define $ϵ:=\min \{{ϵ}_1,{ϵ}_2\}$, we have for all $(z,w)\in B_{ϵ}((x,y))$:

$\Vert x-z{\Vert }^2=\Vert (x,y)-(z,w){\Vert }^2-\Vert y-w{\Vert }^2\le \Vert (x,y)-(z,w){\Vert }^2$

and

$\Vert y-w{\Vert }^2=\Vert (x,y)-(z,w){\Vert }^2-\Vert x-z{\Vert }^2\le \Vert (x,y)-(z,w){\Vert }^2$

and since the function $\sqrt{\cdot }$ is monotonely increasing on $[0,\mathrm{\infty })$, it follows

$\Vert x-z\Vert <ϵ$

and

$\Vert y-w\Vert <ϵ$

and thus:

$(z,w)\in O\times U$$◻$

Lemma 10.5: Let $X,Y,Z,W$ be sets and let $f:X\to Z$ and $g:Y\to W$ be functions. If $A\subseteq Z$ and $B\subseteq W$, then

$(f\times g{)}^{-1}(A\times B)=f^{-1}(A)\times g^{-1}(B)$

Proof: See exercise 2.

Theorem 10.6: The product manifold of a manifold $M$ of class ${𝒞}^n$, where $n\in {\mathbb{N}}_0\cup \{\mathrm{\infty }\}$ with atlas $\{(O_{\upsilon },{\varphi }_{\upsilon })|\upsilon \in \mathrm{{\rm Y}}\}$ really is a manifold; i. e. $\{(O_{\upsilon }\times O_o,{\varphi }_{\upsilon }\times {\varphi }_o)|\upsilon ,o\in \mathrm{{\rm Y}}\}$ really is an atlas of class ${𝒞}^n$.

Proof:

1. We show that for all $\upsilon ,o\in \mathrm{{\rm Y}}$ the set $({\varphi }_{\upsilon }\times {\varphi }_o)(O_{\upsilon }\times O_o)$ is open.

We have:

$\begin{array}{cc}({\varphi }_{\upsilon }\times {\varphi }_o)(O_{\upsilon }\times O_o)& =\{({\varphi }_{\upsilon }(p),{\varphi }_o(q))\in {ℝ}^{2d}|p\in O_{\upsilon },q\in O_o\}\\ & ={\varphi }_{\upsilon }(O_{\upsilon })\times {\varphi }_o(O_o)\\ \end{array}$

This set is open in ${ℝ}^{2d}$ due to lemma 10.4, since it is the cartesian product of two open sets.

2. We prove that for all $\upsilon ,o\in \mathrm{{\rm Y}}$, the function ${\varphi }_{\upsilon }\times {\varphi }_o$ is a homeomorphism.

2.1. For bijectivity, see exercise 1.

2.2. We prove continuity.

Let $O\subseteq ({\varphi }_{\upsilon }\times {\varphi }_o)(O_{\upsilon }\times O_o)$ be open. Due to the definition of the subspace topology,

Lemma 10.4 implies that we have

$O=\bigcup _{(x,y)\in O}{B}_{{ϵ}_{(x,y)}}(x)\times B_{{ϵ}_{(x,y)}}(y)$

(this equation can be proven by showing '$\subseteq$' and '$\supseteq$') and thus follows with lemma 10.5, that:

$\begin{array}{cc}({\varphi }_{\upsilon }\times {\varphi }_o{)}^{-1}(O)& =({\varphi }_{\upsilon }\times {\varphi }_o{)}^{-1}(\bigcup _{(x,y)\in O}{B}_{{ϵ}_{(x,y)}}(x)\times B_{{ϵ}_{(x,y)}}(y))\\ & =\bigcup _{(x,y)\in O}({\varphi }_{\upsilon }\times {\varphi }_o{)}^{-1}(B_{{ϵ}_{(x,y)}}(x)\times B_{{ϵ}_{(x,y)}}(y))\\ & =\bigcup _{(x,y)\in O}{\varphi }_{\upsilon }^{-1}(B_{{ϵ}_{(x,y)}}(x))\times ({\varphi }_o{)}^{-1}(B_{{ϵ}_{(x,y)}}(y))\end{array}$

, which is open as the union of open sets (since we had equipped $M\times M$ with the product topology).

2.3. We prove continuity of the inverse.

Let $O\subseteq M\times M$ be open.

3. We prove that

Lie groups

Theorem 10.11: Let $G$ be a $d$-dimensional Lie group of class ${𝒞}^n$. Then for each $g\in G$ the respective left multiplication function and the respective right multiplication are diffeomorphisms of class ${𝒞}^n$ from $G$ to itself.

Proof:

We only prove the claim for the left multiplication. The proof for the right multiplication goes the same way.

In this proof, the group operation of $G$ is denoted by $*$.

1. We show that $L_g$ is differentiable of class ${𝒞}^n$.

Let $g\in G$ be arbitrary. Since $G$ is a Lie group, the function

${\psi }_{*}:G\times G\to G,{\psi }_{*}(g,h)=g*h$

is differentiable of class ${𝒞}^n$, where $G\times G$ is equipped with the product manifold structure.

Let now $(O,\varphi )$ and $(U,\theta )$ be two arbitrary elements in the atlas of $G$. We choose $(V,\chi )$ in the atlas of $G$ such that $g\in V$. As ${\psi }_{*}$ is differentiable of class ${𝒞}^n$, the function

$\varphi {|}_{*({\psi }_{*}^{-1}(O)\cap V\times U)}\circ {\psi }_{*}{|}_{{\psi }_{*}^{-1}(O)\cap V\times U}\circ (\chi \times \theta ){|}_{{\psi }_{*}^{-1}(O)\cap V\times U}^{-1}$

is contained in ${𝒞}^n({ℝ}^{2d},{ℝ}^d)$. Therefore, also the function

$f:{ℝ}^d\to {ℝ}^d,f(x)=\varphi {|}_{{\psi }_{*}({\psi }_{*}^{-1}(O)\cap V\times U)}\circ {\psi }_{*}{|}_{{\psi }_{*}^{-1}(O)\cap V\times U}\circ (\chi \times \theta ){|}_{{\psi }_{*}^{-1}(O)\cap V\times U}^{-1}({\chi }^{-1}(g),x)$

is contained in ${𝒞}^n({ℝ}^{2d},{ℝ}^d)$; the partial derivatives exist and are equal to the partial derivatives of the last $d$ variables of the function

$\varphi {|}_{{\psi }_{*}({\psi }_{*}^{-1}(O)\cap V\times U)}\circ {\psi }_{*}{|}_{{\psi }_{*}^{-1}(O)\cap V\times U}\circ (\chi \times \theta ){|}_{{\psi }_{*}^{-1}(O)\cap V\times U}^{-1}$

. But we have for all $x\in \theta (L_g^{-1}(O)\cap U)$:

$f(x)=(\varphi {|}_{L_g({|}_{L_g^{-1}(O)\cap U})}\circ L_g{|}_{L_g^{-1}(O)\cap U}\circ \theta {|}_{L_g^{-1}(O)\cap U}^{-1})(x)$

and therefore the function

$\varphi {|}_{L_g({|}_{L_g^{-1}(O)\cap U})}\circ L_g{|}_{L_g^{-1}(O)\cap U}\circ \theta {|}_{L_g^{-1}(O)\cap U}^{-1}$

is contained in ${𝒞}^n({ℝ}^d,{ℝ}^d$, which means the definition of differentiability of class ${𝒞}^n$ is fulfilled.

2. We show that $L_g$ is bijective.

We do so by noticing that an inverse function of $L_g$ is given by $L_{g^{-1}}$: For arbitrary $h\in G$, we have:

$L_{g^{-1}}(L_g(h))=g*(g^{-1}*h)=(g*g^{-1})*h=e*h=h$

and

$L_g(L_{g^{-1}}(h))=g^{-1}*(g*h)=(g^{-1}*g)*h=e*h=h$

3. We note that the inverse function is differentiable of class ${𝒞}^n$:

We use 1. with $g^{-1}$; $g^{-1}$ also is an element of $G$, and 1. proved that the left multiplication function is differentiable for every element of $G$, including $g^{-1}$.$◻$

Left invariant vector fields

Let us repeat the definition of a Lie subalgebra:

Definition 6.2:

Let $L$ with $[\cdot ,\cdot ]$ be a Lie algebra. A subset of $L$ which is a Lie algebra with the restriction of $[\cdot ,\cdot ]$ on that subset is called a Lie subalgebra.

Proof:

Let $𝐕,𝐖\in 𝔤$. It suffices to show that $[𝐕,𝐖]\in 𝔤$, because then $𝔤$ is a Lie algebra with the restriction of the vector field Lie bracket.

Indeed, we have for all $g,h\in G$ and $\phi \in {𝒞}^n(G)$:

$\begin{array}{cc}(d{L}_g{)}_h([𝐕,𝐖](h))(\phi )& =[𝐕,𝐖](h)(L_g^{*}\phi )\\ & =𝐕(h)(𝐖(L_g^{*}\phi ))-𝐖(h)(𝐕(L_g^{*}\phi ))\\ & =𝐕(h)((d{L}_g{)}_{\cdot }(𝐖\phi ))-𝐖(h)((d{L}_g{)}_{\cdot }(𝐕\phi ))\\ & =𝐕(h)(𝐖(g\cdot )(\phi ))-𝐖(h)(𝐕(g\cdot )(\phi ))\\ & =𝐕(h)(𝐖\phi \circ L_g)-𝐖(h)(𝐕\phi \circ L_g)\\ & =𝐕(h)(L_g^{*}(𝐖\phi ))-𝐖(h)(L_g^{*}(𝐕\phi \circ L_g))\\ & =(d{L}_g{)}_h(𝐕(h))(𝐖\phi )-(d{L}_g{)}_h(𝐖(h))(𝐕\phi )\\ & =𝐕(gh)(𝐖\phi )-𝐖(gh)(𝐕\phi )\\ & =[𝐕,𝐖](gh)(\phi )\end{array}$

, where by $(d{L}_g{)}_{\cdot }(𝐖\phi )$ the function

$p\mapsto (d{L}_g{)}_p(𝐖\phi )$

is meant and by $𝐖(g\cdot )(\phi )$ the function

$p\mapsto 𝐖(gp)(\phi )$

is meant (as both are equal to $𝐖(L_g^{*}\phi )$, both are differentiable of class ${𝒞}^n$).$◻$

Proof:

We choose the function

$\mathrm{\Theta }:𝔤\to T_{e}G,\mathrm{\Theta }(𝐕):=𝐕(e)$

We will now show that this function is the desired isomorphism.

1. We prove linearity: Let $𝐕,𝐖\in 𝔤$ and $c\in ℝ$. We have:

$\begin{array}{cc}\mathrm{\Theta }(𝐕+c𝐖)& =(𝐕+c𝐖)(e)\\ & =𝐕(e)+c𝐖(e)\\ & =\mathrm{\Theta }(𝐕)+c\mathrm{\Theta }(𝐖)\end{array}$

2. We prove bijectivity.

2.1. We prove injectivity.

Let $\mathrm{\Theta }(𝐕)=\mathrm{\Theta }(𝐖)$ (i. e. $𝐕(e)=𝐖(e)$) for $𝐕,𝐖\in 𝔤$. Since $𝐕,𝐖$ are left invariant, it follows for all $g\in G$ and $\phi \in {𝒞}^n$, that

$\begin{array}{cc}𝐕(g)(\phi )& =𝐕(ge)(\phi )\\ & =(d{L}_g{)}_e(𝐕)(\phi )\\ & =𝐕(e)(L_g^{*}(\phi ))\\ & =𝐖(e)(L_g^{*}(\phi ))\\ & =(d{L}_g{)}_e(𝐖)(\phi )\\ & =𝐖(ge)(\phi )\\ & =𝐖(g)(\phi )\end{array}$

2.2. We prove surjectivity.

Let ${𝐕}_e\in T_{e}G$ be arbitrary. We define

${𝐔}_{{𝐕}_e}(g):=(d{L}_g{)}_e{𝐕}_e$

Due to theorem 2.19, this is a vector field. It is also left invariant because for all $g,h\in G$ and $\phi \in {𝒞}^n(M)$, we have:

$\begin{array}{cc}(d{L}_g{)}_h({𝐔}_{{𝐕}_e}(h))(\phi )& ={𝐔}_{{𝐕}_e}(h)(L_g^{*}(\phi ))\\ & =(d{L}_h{)}_e{𝐕}_e(L_g^{*}(\phi ))\\ & ={𝐕}_e(L_h^{*}(L_g^{*}(\phi )))\\ & ={𝐕}_e(\phi \circ L_g\circ L_h)\\ & ={𝐕}_e(\phi \circ L_{gh})\\ & =(d{L}_{gh}{)}_e{𝐕}_e(\phi )\\ & ={𝐔}_{{𝐕}_e}(gh)(\phi )\end{array}$

Further, we have for all $\phi \in {𝒞}^n(M)$:

$\begin{array}{c}\mathrm{\Theta }({𝐔}_{{𝐕}_e})(\phi )={𝐔}_{{𝐕}_e}(e)(\phi )=(d{L}_e{)}_e{𝐕}_e(\phi )={𝐕}_e(\phi )\end{array}$

3. We note that the inverse of $\mathrm{\Theta }$ is linear since the inverse of a linear bijective function is always linear.

$◻$

The next theorem shows that in a Lie group, all left invariant vector fields are complete.

Lemma 10.15:

Let $G$ be a Lie group, let $𝐕\in 𝔤$ and let ${\mathrm{\Phi }}_{𝐕}$ be the flow of $𝐕$. Then for all $g\in G$ and all $(x,h)$ in the domain of $\mathrm{\Phi }𝐕$, we have:

${\mathrm{\Phi }}_{𝐕}(x,g*h)=h*{\mathrm{\Phi }}_{𝐕}(x,g)$

Proof:

Let $h\in G$ be arbitrary, and let <mat>I_h</math> be the unique largest interval such that $0\in I_h$ and there exists a unique integral curve ${\gamma }_h:I_h\to M$ such that

${{\gamma }_h}^{\prime }$

Proof:

Let $g\in G$ be arbitrary and let ${\gamma }_g$ be an integral curve at $g$.

The adjoint function

For the next definition, we recall that the automorphism group of a group was given by the set of group isomorphisms from the group to itself with composition as the group operation. Indeed, this is a group (see exercise 3).

We further recall that for a group $G$, the automorphism group is denoted by $\text{Aut}(G)$.

Theorem 10.19:

Let $G$ be a manifold of class ${𝒞}^n$, where $n\in {\mathbb{N}}_0\cup \{\mathrm{\infty }\}$. For each $g\in G$, $\text{Ad}(g)$ is of class ${𝒞}^n$.

Proof:

We have:

$\text{Ad}(g)=L_g\circ R_{g^{-1}}$

Therefore, the claim follows from theorems 2.29 and 10.11.$◻$

Exercises

1. Let $X,Y,Z,W$ be sets $f:X\to Z$ and $g:Y\to W$ be two bijective functions. Prove that $f\times g$ is bijective.
2. Prove lemma 10.5.
3. Let $G$ be a group and $\text{Aut}(G)$ be the set of group isomorphisms from $G$ to $G$. Prove that $\text{Aut}(G)$ together with the composition as operation is a group.