# Differentiable Manifolds/Product manifolds and Lie groups

 Differentiable Manifolds ← Group actions and flows Product manifolds and Lie groups Vector (sub-)bundles, sections, foliations, distributions and Frobenius' theorem →

## The product manifold

Definition 10.1:

Let ${\displaystyle X,Y,Z,W}$  be sets and ${\displaystyle f:X\to Z}$  and ${\displaystyle g:Y\to W}$  be functions. Then the cartesian product of ${\displaystyle f}$  and ${\displaystyle g}$ , denoted by ${\displaystyle f\times g}$ , is defined to be the function

${\displaystyle f\times g:X\times Y\to Z\times W,(f\times g)(x,y)=(f(x),g(y))}$

Definition 10.2:

Let ${\displaystyle M}$  be a manifold with atlas ${\displaystyle \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}$ . The product manifold of ${\displaystyle M}$  is defined to be the product space ${\displaystyle M\times M}$  with the atlas

${\displaystyle \{(O_{\upsilon }\times O_{\mathrm {o} },(\phi _{\upsilon },\phi _{\mathrm {o} }))|\upsilon ,\mathrm {o} \in \Upsilon \}}$

Lemma 10.3:

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

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

Proof:

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

${\displaystyle \|(z,w)-(x,y)\|=\|(z,0)+(0,w)-(x,0)-(0,y)\|\leq \|(z,0)-(x,0)\|+\|(0,w)-(0,y)\|=\|z-x\|+\|w-y\|<2\epsilon _{(x,y)}=\delta }$

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

${\displaystyle B_{\epsilon _{(x,y)}}(x)\times B_{\epsilon _{(x,y)}}(y)\subseteq O}$ ${\displaystyle \Box }$

Lemma 10.4:

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

Proof:

Due to the openness of ${\displaystyle O}$  and ${\displaystyle U}$  for each point ${\displaystyle (x,y)}$  in ${\displaystyle O\times U}$ , we find ${\displaystyle \epsilon _{1},\epsilon _{2}\in \mathbb {R} _{>0}}$  such that ${\displaystyle B_{\epsilon _{1}}(x)\subseteq O}$  and ${\displaystyle B_{\epsilon _{2}}(x)\subseteq U}$ . If we define ${\displaystyle \epsilon :=\min\{\epsilon _{1},\epsilon _{2}\}}$ , we have for all ${\displaystyle (z,w)\in B_{\epsilon }((x,y))}$ :

${\displaystyle \|x-z\|^{2}=\|(x,y)-(z,w)\|^{2}-\|y-w\|^{2}\leq \|(x,y)-(z,w)\|^{2}}$

and

${\displaystyle \|y-w\|^{2}=\|(x,y)-(z,w)\|^{2}-\|x-z\|^{2}\leq \|(x,y)-(z,w)\|^{2}}$

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

${\displaystyle \|x-z\|<\epsilon }$

and

${\displaystyle \|y-w\|<\epsilon }$

and thus:

${\displaystyle (z,w)\in O\times U}$ ${\displaystyle \Box }$

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

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

Proof: See exercise 2.

Theorem 10.6: The product manifold of a manifold ${\displaystyle M}$  of class ${\displaystyle {\mathcal {C}}^{n}}$ , where ${\displaystyle n\in \mathbb {N} _{0}\cup \{\infty \}}$  with atlas ${\displaystyle \{(O_{\upsilon },\phi _{\upsilon })|\upsilon \in \Upsilon \}}$  really is a manifold; i. e. ${\displaystyle \{(O_{\upsilon }\times O_{\mathrm {o} },\phi _{\upsilon }\times \phi _{\mathrm {o} })|\upsilon ,\mathrm {o} \in \Upsilon \}}$  really is an atlas of class ${\displaystyle {\mathcal {C}}^{n}}$ .

Proof:

1. We show that for all ${\displaystyle \upsilon ,\mathrm {o} \in \Upsilon }$  the set ${\displaystyle (\phi _{\upsilon }\times \phi _{\mathrm {o} })(O_{\upsilon }\times O_{\mathrm {o} })}$  is open.

We have:

{\displaystyle {\begin{aligned}(\phi _{\upsilon }\times \phi _{\mathrm {o} })(O_{\upsilon }\times O_{\mathrm {o} })&=\{(\phi _{\upsilon }(p),\phi _{\mathrm {o} }(q))\in \mathbb {R} ^{2d}{\big |}p\in O_{\upsilon },q\in O_{\mathrm {o} }\}\\&=\phi _{\upsilon }(O_{\upsilon })\times \phi _{\mathrm {o} }(O_{\mathrm {o} })\\\end{aligned}}}

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

2. We prove that for all ${\displaystyle \upsilon ,\mathrm {o} \in \Upsilon }$ , the function ${\displaystyle \phi _{\upsilon }\times \phi _{\mathrm {o} }}$  is a homeomorphism.

2.1. For bijectivity, see exercise 1.

2.2. We prove continuity.

Let ${\displaystyle O\subseteq (\phi _{\upsilon }\times \phi _{\mathrm {o} })(O_{\upsilon }\times O_{\mathrm {o} })}$  be open. Due to the definition of the subspace topology,

Lemma 10.4 implies that we have

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

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

{\displaystyle {\begin{aligned}(\phi _{\upsilon }\times \phi _{\mathrm {o} })^{-1}(O)&=(\phi _{\upsilon }\times \phi _{\mathrm {o} })^{-1}\left(\bigcup _{(x,y)\in O}B_{\epsilon _{(x,y)}}(x)\times B_{\epsilon _{(x,y)}}(y)\right)\\&=\bigcup _{(x,y)\in O}(\phi _{\upsilon }\times \phi _{\mathrm {o} })^{-1}(B_{\epsilon _{(x,y)}}(x)\times B_{\epsilon _{(x,y)}}(y))\\&=\bigcup _{(x,y)\in O}\phi _{\upsilon }^{-1}(B_{\epsilon _{(x,y)}}(x))\times (\phi _{\mathrm {o} })^{-1}(B_{\epsilon _{(x,y)}}(y))\end{aligned}}}

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

2.3. We prove continuity of the inverse.

Let ${\displaystyle O\subseteq M\times M}$  be open.

3. We prove that

## Lie groups

Definition 10.7:

Let ${\displaystyle G}$  be a ${\displaystyle d}$ -dimensional manifold of class ${\displaystyle {\mathcal {C}}^{n}}$ , which is simultaneously a group, i. e. we have a group operation ${\displaystyle *:G\times G\to G}$  regarding to which there exist an identity, which we in the following shall denote by ${\displaystyle e}$ , and inverses in ${\displaystyle G}$ , and which is associative. We call ${\displaystyle G}$  a ${\displaystyle d}$ -dimensional Lie group of class ${\displaystyle {\mathcal {C}}^{n}}$  iff

• the function ${\displaystyle \psi _{*}:G\times G\to G,\psi _{*}((g,h)):=g*h}$  is differentiable of class ${\displaystyle {\mathcal {C}}^{n}}$ , where ${\displaystyle G\times G}$  has the product manifold atlas, AND
• the function ${\displaystyle G\to G,g\mapsto g^{-1}}$  is also of class ${\displaystyle {\mathcal {C}}^{n}}$ .

Definition 10.8:

Let ${\displaystyle M}$  be a manifold, and let ${\displaystyle N\subseteq M}$  be a subset of ${\displaystyle M}$  which is also a manifold. The inclusion of ${\displaystyle N}$  into ${\displaystyle M}$  is defined to be the function

${\displaystyle \iota :N\to M,\iota (q):=q}$

Definition 10.9:

Let ${\displaystyle G}$  be a Lie group of class ${\displaystyle {\mathcal {C}}^{n}}$ , and let ${\displaystyle H\subseteq G}$  be a subset of ${\displaystyle G}$ . We call ${\displaystyle H}$  a Lie subgroup of ${\displaystyle G}$  iff

• ${\displaystyle H}$  is a Lie group of class, but not necessarily with the subspace topology induced by ${\displaystyle G}$
• the inclusion of ${\displaystyle H}$  into ${\displaystyle G}$  is differentiable of class ${\displaystyle {\mathcal {C}}^{n}}$  and its differential is injective at every point.

Definitions 10.10:

Let ${\displaystyle G}$  be a Lie group with group operation ${\displaystyle *}$ , and let ${\displaystyle g\in G}$ . The left multiplication function with respect to ${\displaystyle g}$ , denoted by ${\displaystyle L_{g}}$ , is defined to be the function

${\displaystyle L_{g}:G\to G,L_{g}(h):=g*h}$

The right multiplication function with respect to ${\displaystyle g}$ , denoted by ${\displaystyle R_{g}}$ , is defined to be the function

${\displaystyle R_{g}:G\to G,R_{g}(h):=h*g}$

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

1. We show that ${\displaystyle L_{g}}$  is differentiable of class ${\displaystyle {\mathcal {C}}^{n}}$ .

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

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

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

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

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

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

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

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

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

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

${\displaystyle f(x)=(\phi |_{L_{g}(|_{L_{g}^{-1}(O)\cap U})}\circ L_{g}|_{L_{g}^{-1}(O)\cap U}\circ \theta |_{L_{g}^{-1}(O)\cap U}^{-1})(x)}$

and therefore the function

${\displaystyle \phi |_{L_{g}(|_{L_{g}^{-1}(O)\cap U})}\circ L_{g}|_{L_{g}^{-1}(O)\cap U}\circ \theta |_{L_{g}^{-1}(O)\cap U}^{-1}}$

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

2. We show that ${\displaystyle L_{g}}$  is bijective.

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

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

and

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

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

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

## Left invariant vector fields

Definition 10.12:

Let ${\displaystyle G}$  be a Lie group. We call a vector field ${\displaystyle \mathbf {V} \in {\mathfrak {X}}(M)}$  left invariant iff

${\displaystyle \forall g,h\in G:(dL_{g})_{h}(\mathbf {V} (h))=\mathbf {V} (gh)}$

The set of all left invariant vector fields of a lie group is denoted by the corresponding lower case fraktur letter; for example, the set of all left invariant vector fields of a Lie group ${\displaystyle G}$  one would denote by ${\displaystyle {\mathfrak {g}}}$ , and the set of all left invariant vector fields of a Lie group ${\displaystyle H}$  would be denoted by ${\displaystyle {\mathfrak {h}}}$ .

Let us repeat the definition of a Lie subalgebra:

Definition 6.2:

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

Theorem 10.13:

Let ${\displaystyle G}$  be a Lie group. Then ${\displaystyle {\mathfrak {g}}}$ , together with pointwise addition and the vector field Lie bracket, is a Lie subalgebra of ${\displaystyle {\mathfrak {X}}(M)}$  with pointwise addition and the vector field Lie bracket.

Proof:

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

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

{\displaystyle {\begin{aligned}(dL_{g})_{h}([\mathbf {V} ,\mathbf {W} ](h))(\varphi )&=[\mathbf {V} ,\mathbf {W} ](h)(L_{g}^{*}\varphi )\\&=\mathbf {V} (h)(\mathbf {W} (L_{g}^{*}\varphi ))-\mathbf {W} (h)(\mathbf {V} (L_{g}^{*}\varphi ))\\&=\mathbf {V} (h)((dL_{g})_{\cdot }(\mathbf {W} \varphi ))-\mathbf {W} (h)((dL_{g})_{\cdot }(\mathbf {V} \varphi ))\\&=\mathbf {V} (h)(\mathbf {W} (g\cdot )(\varphi ))-\mathbf {W} (h)(\mathbf {V} (g\cdot )(\varphi ))\\&=\mathbf {V} (h)(\mathbf {W} \varphi \circ L_{g})-\mathbf {W} (h)(\mathbf {V} \varphi \circ L_{g})\\&=\mathbf {V} (h)(L_{g}^{*}(\mathbf {W} \varphi ))-\mathbf {W} (h)(L_{g}^{*}(\mathbf {V} \varphi \circ L_{g}))\\&=(dL_{g})_{h}(\mathbf {V} (h))(\mathbf {W} \varphi )-(dL_{g})_{h}(\mathbf {W} (h))(\mathbf {V} \varphi )\\&=\mathbf {V} (gh)(\mathbf {W} \varphi )-\mathbf {W} (gh)(\mathbf {V} \varphi )\\&=[\mathbf {V} ,\mathbf {W} ](gh)(\varphi )\end{aligned}}}

, where by ${\displaystyle (dL_{g})_{\cdot }(\mathbf {W} \varphi )}$  the function

${\displaystyle p\mapsto (dL_{g})_{p}(\mathbf {W} \varphi )}$

is meant and by ${\displaystyle \mathbf {W} (g\cdot )(\varphi )}$  the function

${\displaystyle p\mapsto \mathbf {W} (gp)(\varphi )}$

is meant (as both are equal to ${\displaystyle \mathbf {W} (L_{g}^{*}\varphi )}$ , both are differentiable of class ${\displaystyle {\mathcal {C}}^{n}}$ ).${\displaystyle \Box }$

Theorem 10.14:

Let ${\displaystyle G}$  be a Lie group of class ${\displaystyle {\mathcal {C}}^{n}}$ . Then there exists a vector space isomorphism between ${\displaystyle {\mathfrak {g}}}$  and ${\displaystyle T_{e}G}$ .

Proof:

We choose the function

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

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

1. We prove linearity: Let ${\displaystyle \mathbf {V} ,\mathbf {W} \in {\mathfrak {g}}}$  and ${\displaystyle c\in \mathbb {R} }$ . We have:

{\displaystyle {\begin{aligned}\Theta (\mathbf {V} +c\mathbf {W} )&=(\mathbf {V} +c\mathbf {W} )(e)\\&=\mathbf {V} (e)+c\mathbf {W} (e)\\&=\Theta (\mathbf {V} )+c\Theta (\mathbf {W} )\end{aligned}}}

2. We prove bijectivity.

2.1. We prove injectivity.

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

{\displaystyle {\begin{aligned}\mathbf {V} (g)(\varphi )&=\mathbf {V} (ge)(\varphi )\\&=(dL_{g})_{e}(\mathbf {V} )(\varphi )\\&=\mathbf {V} (e)(L_{g}^{*}(\varphi ))\\&=\mathbf {W} (e)(L_{g}^{*}(\varphi ))\\&=(dL_{g})_{e}(\mathbf {W} )(\varphi )\\&=\mathbf {W} (ge)(\varphi )\\&=\mathbf {W} (g)(\varphi )\end{aligned}}}

2.2. We prove surjectivity.

Let ${\displaystyle \mathbf {V} _{e}\in T_{e}G}$  be arbitrary. We define

${\displaystyle \mathbf {U} _{\mathbf {V} _{e}}(g):=(dL_{g})_{e}\mathbf {V} _{e}}$

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

{\displaystyle {\begin{aligned}(dL_{g})_{h}(\mathbf {U} _{\mathbf {V} _{e}}(h))(\varphi )&=\mathbf {U} _{\mathbf {V} _{e}}(h)(L_{g}^{*}(\varphi ))\\&=(dL_{h})_{e}\mathbf {V} _{e}(L_{g}^{*}(\varphi ))\\&=\mathbf {V} _{e}(L_{h}^{*}(L_{g}^{*}(\varphi )))\\&=\mathbf {V} _{e}(\varphi \circ L_{g}\circ L_{h})\\&=\mathbf {V} _{e}(\varphi \circ L_{gh})\\&=(dL_{gh})_{e}\mathbf {V} _{e}(\varphi )\\&=\mathbf {U} _{\mathbf {V} _{e}}(gh)(\varphi )\end{aligned}}}

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

{\displaystyle {\begin{aligned}\Theta (\mathbf {U} _{\mathbf {V} _{e}})(\varphi )=\mathbf {U} _{\mathbf {V} _{e}}(e)(\varphi )=(dL_{e})_{e}\mathbf {V} _{e}(\varphi )=\mathbf {V} _{e}(\varphi )\end{aligned}}}

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

${\displaystyle \Box }$

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

Lemma 10.15:

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

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

Proof:

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

${\displaystyle \gamma _{h}'}$

Theorem 10.16:

Let ${\displaystyle G}$  be a Lie group and let ${\displaystyle \mathbf {V} \in {\mathfrak {g}}}$ . Then ${\displaystyle \mathbf {V} }$  is complete.

Proof:

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

## The exponential function

Definition 10.17:

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 ${\displaystyle G}$ , the automorphism group is denoted by ${\displaystyle {\text{Aut}}(G)}$ .

Definition 10.18:

Let ${\displaystyle G}$  be a Lie group, whose group operation we shall denote by ${\displaystyle *}$ . If for all ${\displaystyle g\in G}$  we define the function ${\displaystyle \psi _{g}:G\to G}$  by ${\displaystyle \psi _{g}(h):=g*h*g^{-1}}$ , the adjoint function of ${\displaystyle G}$ , denoted by ${\displaystyle {\text{Ad}}(g)}$  is given by

${\displaystyle {\text{Ad}}:G\to {\text{Aut}}(G),{\text{Ad}}(g)(h):=g*h*g^{-1}}$

Theorem 10.19:

Let ${\displaystyle G}$  be a manifold of class ${\displaystyle {\mathcal {C}}^{n}}$ , where ${\displaystyle n\in \mathbb {N} _{0}\cup \{\infty \}}$ . For each ${\displaystyle g\in G}$ , ${\displaystyle {\text{Ad}}(g)}$  is of class ${\displaystyle {\mathcal {C}}^{n}}$ .

Proof:

We have:

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

Therefore, the claim follows from theorems 2.29 and 10.11.${\displaystyle \Box }$

Theorem 10.20:

Let ${\displaystyle G}$  be a Lie group and ${\displaystyle \mathbf {V} \in {\mathfrak {X}}(G)}$ . Then

${\displaystyle }$

## Exercises

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

## Sources

 Differentiable Manifolds ← Group actions and flows Product manifolds and Lie groups Vector (sub-)bundles, sections, foliations, distributions and Frobenius' theorem →