Continuous homomorphisms of Lie groups are smooth

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I'm working on problem 20-11 from Lee's "Introduction to Smooth Manifolds", which asks us to prove:

1. Every continuous homomorphism $gamma : mathbb R to G$ is smooth ($G$ a Lie group).


2. Every continuous homomorphism $F : G to H$ of Lie groups is smooth.

The first part comes with a hint: let $V subseteq mathrmLie(G) = mathfrak g$ be a neighborhood of $0$ such that $exp: 2V to exp(2V)$ is a diffeomorphism (with $2V = 2X : X in V$). Choose $t_0$ small enough that $gamma(t) in exp(V)$ whenever $|t| leq t_0$, and let $X_0$ be the element of $V$ such that $gamma(t_0) = exp X_0$. Then one can show $gamma(qt_0) = exp(qX_0)$ whenever $q = m/2^n$ for some $m,n$.

I've been able to show all of this in the hint, but I'm not sure why that implies $gamma$ is smooth. Is it because it now depends smoothly on $X_0$, which is in one-to-one correspondence with $t_0$? But why should that be true? And why do we care about the dyadic rational $q$?

Part 2 also comes with a hint: show that there's a map $phi : mathfrak g to mathfrak h$ so that the following diagram commutes:
$requireAMScd$
beginCD
mathfrak g @>phi>> mathfrak h\
@V exp V V @VV exp V\
G @>>F> H
endCD
and then show $phi$ is linear. But without knowing whether we can talk about $dF_e$, how could we construct such a $phi$?

Any help with either of these problems would be greatly appreciated (or even a good resource on why continuous homomorphisms of Lie groups are automatically smooth).

differential-geometry lie-groups lie-algebras smooth-manifolds

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asked Jul 21 at 17:22

D Ford

417212

add a comment | 

up vote
3
down vote

favorite

1

I'm working on problem 20-11 from Lee's "Introduction to Smooth Manifolds", which asks us to prove:

1. Every continuous homomorphism $gamma : mathbb R to G$ is smooth ($G$ a Lie group).


2. Every continuous homomorphism $F : G to H$ of Lie groups is smooth.

The first part comes with a hint: let $V subseteq mathrmLie(G) = mathfrak g$ be a neighborhood of $0$ such that $exp: 2V to exp(2V)$ is a diffeomorphism (with $2V = 2X : X in V$). Choose $t_0$ small enough that $gamma(t) in exp(V)$ whenever $|t| leq t_0$, and let $X_0$ be the element of $V$ such that $gamma(t_0) = exp X_0$. Then one can show $gamma(qt_0) = exp(qX_0)$ whenever $q = m/2^n$ for some $m,n$.

I've been able to show all of this in the hint, but I'm not sure why that implies $gamma$ is smooth. Is it because it now depends smoothly on $X_0$, which is in one-to-one correspondence with $t_0$? But why should that be true? And why do we care about the dyadic rational $q$?

Part 2 also comes with a hint: show that there's a map $phi : mathfrak g to mathfrak h$ so that the following diagram commutes:
$requireAMScd$
beginCD
mathfrak g @>phi>> mathfrak h\
@V exp V V @VV exp V\
G @>>F> H
endCD
and then show $phi$ is linear. But without knowing whether we can talk about $dF_e$, how could we construct such a $phi$?

Any help with either of these problems would be greatly appreciated (or even a good resource on why continuous homomorphisms of Lie groups are automatically smooth).

differential-geometry lie-groups lie-algebras smooth-manifolds

share|cite|improve this question

asked Jul 21 at 17:22

D Ford

417212

add a comment | 

up vote
3
down vote

favorite

1

up vote
3
down vote

favorite

1

1

I'm working on problem 20-11 from Lee's "Introduction to Smooth Manifolds", which asks us to prove:

1. Every continuous homomorphism $gamma : mathbb R to G$ is smooth ($G$ a Lie group).


2. Every continuous homomorphism $F : G to H$ of Lie groups is smooth.

The first part comes with a hint: let $V subseteq mathrmLie(G) = mathfrak g$ be a neighborhood of $0$ such that $exp: 2V to exp(2V)$ is a diffeomorphism (with $2V = 2X : X in V$). Choose $t_0$ small enough that $gamma(t) in exp(V)$ whenever $|t| leq t_0$, and let $X_0$ be the element of $V$ such that $gamma(t_0) = exp X_0$. Then one can show $gamma(qt_0) = exp(qX_0)$ whenever $q = m/2^n$ for some $m,n$.

I've been able to show all of this in the hint, but I'm not sure why that implies $gamma$ is smooth. Is it because it now depends smoothly on $X_0$, which is in one-to-one correspondence with $t_0$? But why should that be true? And why do we care about the dyadic rational $q$?

Part 2 also comes with a hint: show that there's a map $phi : mathfrak g to mathfrak h$ so that the following diagram commutes:
$requireAMScd$
beginCD
mathfrak g @>phi>> mathfrak h\
@V exp V V @VV exp V\
G @>>F> H
endCD
and then show $phi$ is linear. But without knowing whether we can talk about $dF_e$, how could we construct such a $phi$?

Any help with either of these problems would be greatly appreciated (or even a good resource on why continuous homomorphisms of Lie groups are automatically smooth).

differential-geometry lie-groups lie-algebras smooth-manifolds

share|cite|improve this question

asked Jul 21 at 17:22

D Ford

417212

I'm working on problem 20-11 from Lee's "Introduction to Smooth Manifolds", which asks us to prove:

1. Every continuous homomorphism $gamma : mathbb R to G$ is smooth ($G$ a Lie group).


2. Every continuous homomorphism $F : G to H$ of Lie groups is smooth.

The first part comes with a hint: let $V subseteq mathrmLie(G) = mathfrak g$ be a neighborhood of $0$ such that $exp: 2V to exp(2V)$ is a diffeomorphism (with $2V = 2X : X in V$). Choose $t_0$ small enough that $gamma(t) in exp(V)$ whenever $|t| leq t_0$, and let $X_0$ be the element of $V$ such that $gamma(t_0) = exp X_0$. Then one can show $gamma(qt_0) = exp(qX_0)$ whenever $q = m/2^n$ for some $m,n$.

I've been able to show all of this in the hint, but I'm not sure why that implies $gamma$ is smooth. Is it because it now depends smoothly on $X_0$, which is in one-to-one correspondence with $t_0$? But why should that be true? And why do we care about the dyadic rational $q$?

Part 2 also comes with a hint: show that there's a map $phi : mathfrak g to mathfrak h$ so that the following diagram commutes:
$requireAMScd$
beginCD
mathfrak g @>phi>> mathfrak h\
@V exp V V @VV exp V\
G @>>F> H
endCD
and then show $phi$ is linear. But without knowing whether we can talk about $dF_e$, how could we construct such a $phi$?

Any help with either of these problems would be greatly appreciated (or even a good resource on why continuous homomorphisms of Lie groups are automatically smooth).

differential-geometry lie-groups lie-algebras smooth-manifolds

share|cite|improve this question

asked Jul 21 at 17:22

D Ford

417212

share|cite|improve this question

share|cite|improve this question

asked Jul 21 at 17:22

D Ford

417212

asked Jul 21 at 17:22

D Ford

417212

asked Jul 21 at 17:22

D Ford

417212

417212

add a comment | 

add a comment | 

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Those dyadic rationals are dense in $[-1,1]$. So, near $0$, we have, by the continuity of $gamma$, $gamma(ttimes t_0)=exp(tX_0)$. In other words, $gamma(t)=expleft(frac tt_0X_0right)$ and therefore $gamma$ is differentiable.

For the other question, let $(X_1,ldots,X_n)$ be a basis of $mathfrak g$ and consider the mapbeginarrayrcccalphacolon&mathbbR^n&longrightarrow&G\&(t_1,ldots,t_n)&mapsto&exp(t_1X_1)ldotsexp(t_nX_n).endarrayThen $alpha$ is smooth. If $(x_1,ldots,x_n)inmathbbR^n$, thenbeginalignleft.fracmathrm dmathrm dtalphabigl(t(x_1,ldots,x_n)bigr)right|_t=0&=left.fracmathrm dmathrm dtexp(tx_1X_1)ldotsexp(tx_nX_n)right|_t=0\&=x_1X_1+cdots x_nX_n.endalignand therefore $Dalpha_0$ is an isomorphism. So, $alpha$ induces a diffeomorphism from a neighborhood $U$ of $(0,0,ldots,0)$ onto a neighborhood $V$ of $e_G$. If $(t_1,ldots,t_n)in U$, thenbeginalignFbigl(alpha(t_1,ldots,t_n)bigr)&=Fbigl(exp(t_1X_1)ldotsexp(t_nX_n)bigr)\&=Fbigl(exp(t_1X_1)bigr)ldots Fbigl(exp(t_nX_n)bigr),endalignwhich is smooth. Since $F|_V$ is smooth, $F$ is smooth.

share|cite|improve this answer

answered Jul 21 at 17:57

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! I came across this same argument for part 2 shortly after writing this question. I like the explanation, though I would be curious about how to construct an explicit function $phi$ as hinted. It seems such a map could only be defined in a neighborhood of the identity of $mathfrak g$...
  – D Ford
  Jul 21 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  My guess is that $phi$ is defined as follows: first, you find, for each $kin1,ldots,n$, a $Y_kinmathfrak h$ such that $tmapstoexp(tY_k)$ is equal to $tmapsto Fbigl(exp(tX_k)bigr)$. Then you define $phi$ as$$exp(tX_1)ldotsexp(tX_n)mapstoexp(tY_1)ldotsexp(tY_n).$$But this only occurred to me only after answering your question.
  – José Carlos Santos
  Jul 21 at 18:40
  
  

  
  
  
  

add a comment | 

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Those dyadic rationals are dense in $[-1,1]$. So, near $0$, we have, by the continuity of $gamma$, $gamma(ttimes t_0)=exp(tX_0)$. In other words, $gamma(t)=expleft(frac tt_0X_0right)$ and therefore $gamma$ is differentiable.

For the other question, let $(X_1,ldots,X_n)$ be a basis of $mathfrak g$ and consider the mapbeginarrayrcccalphacolon&mathbbR^n&longrightarrow&G\&(t_1,ldots,t_n)&mapsto&exp(t_1X_1)ldotsexp(t_nX_n).endarrayThen $alpha$ is smooth. If $(x_1,ldots,x_n)inmathbbR^n$, thenbeginalignleft.fracmathrm dmathrm dtalphabigl(t(x_1,ldots,x_n)bigr)right|_t=0&=left.fracmathrm dmathrm dtexp(tx_1X_1)ldotsexp(tx_nX_n)right|_t=0\&=x_1X_1+cdots x_nX_n.endalignand therefore $Dalpha_0$ is an isomorphism. So, $alpha$ induces a diffeomorphism from a neighborhood $U$ of $(0,0,ldots,0)$ onto a neighborhood $V$ of $e_G$. If $(t_1,ldots,t_n)in U$, thenbeginalignFbigl(alpha(t_1,ldots,t_n)bigr)&=Fbigl(exp(t_1X_1)ldotsexp(t_nX_n)bigr)\&=Fbigl(exp(t_1X_1)bigr)ldots Fbigl(exp(t_nX_n)bigr),endalignwhich is smooth. Since $F|_V$ is smooth, $F$ is smooth.

share|cite|improve this answer

answered Jul 21 at 17:57

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! I came across this same argument for part 2 shortly after writing this question. I like the explanation, though I would be curious about how to construct an explicit function $phi$ as hinted. It seems such a map could only be defined in a neighborhood of the identity of $mathfrak g$...
  – D Ford
  Jul 21 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  My guess is that $phi$ is defined as follows: first, you find, for each $kin1,ldots,n$, a $Y_kinmathfrak h$ such that $tmapstoexp(tY_k)$ is equal to $tmapsto Fbigl(exp(tX_k)bigr)$. Then you define $phi$ as$$exp(tX_1)ldotsexp(tX_n)mapstoexp(tY_1)ldotsexp(tY_n).$$But this only occurred to me only after answering your question.
  – José Carlos Santos
  Jul 21 at 18:40
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

Those dyadic rationals are dense in $[-1,1]$. So, near $0$, we have, by the continuity of $gamma$, $gamma(ttimes t_0)=exp(tX_0)$. In other words, $gamma(t)=expleft(frac tt_0X_0right)$ and therefore $gamma$ is differentiable.

For the other question, let $(X_1,ldots,X_n)$ be a basis of $mathfrak g$ and consider the mapbeginarrayrcccalphacolon&mathbbR^n&longrightarrow&G\&(t_1,ldots,t_n)&mapsto&exp(t_1X_1)ldotsexp(t_nX_n).endarrayThen $alpha$ is smooth. If $(x_1,ldots,x_n)inmathbbR^n$, thenbeginalignleft.fracmathrm dmathrm dtalphabigl(t(x_1,ldots,x_n)bigr)right|_t=0&=left.fracmathrm dmathrm dtexp(tx_1X_1)ldotsexp(tx_nX_n)right|_t=0\&=x_1X_1+cdots x_nX_n.endalignand therefore $Dalpha_0$ is an isomorphism. So, $alpha$ induces a diffeomorphism from a neighborhood $U$ of $(0,0,ldots,0)$ onto a neighborhood $V$ of $e_G$. If $(t_1,ldots,t_n)in U$, thenbeginalignFbigl(alpha(t_1,ldots,t_n)bigr)&=Fbigl(exp(t_1X_1)ldotsexp(t_nX_n)bigr)\&=Fbigl(exp(t_1X_1)bigr)ldots Fbigl(exp(t_nX_n)bigr),endalignwhich is smooth. Since $F|_V$ is smooth, $F$ is smooth.

share|cite|improve this answer

answered Jul 21 at 17:57

José Carlos Santos

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! I came across this same argument for part 2 shortly after writing this question. I like the explanation, though I would be curious about how to construct an explicit function $phi$ as hinted. It seems such a map could only be defined in a neighborhood of the identity of $mathfrak g$...
  – D Ford
  Jul 21 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  My guess is that $phi$ is defined as follows: first, you find, for each $kin1,ldots,n$, a $Y_kinmathfrak h$ such that $tmapstoexp(tY_k)$ is equal to $tmapsto Fbigl(exp(tX_k)bigr)$. Then you define $phi$ as$$exp(tX_1)ldotsexp(tX_n)mapstoexp(tY_1)ldotsexp(tY_n).$$But this only occurred to me only after answering your question.
  – José Carlos Santos
  Jul 21 at 18:40
  
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Those dyadic rationals are dense in $[-1,1]$. So, near $0$, we have, by the continuity of $gamma$, $gamma(ttimes t_0)=exp(tX_0)$. In other words, $gamma(t)=expleft(frac tt_0X_0right)$ and therefore $gamma$ is differentiable.

For the other question, let $(X_1,ldots,X_n)$ be a basis of $mathfrak g$ and consider the mapbeginarrayrcccalphacolon&mathbbR^n&longrightarrow&G\&(t_1,ldots,t_n)&mapsto&exp(t_1X_1)ldotsexp(t_nX_n).endarrayThen $alpha$ is smooth. If $(x_1,ldots,x_n)inmathbbR^n$, thenbeginalignleft.fracmathrm dmathrm dtalphabigl(t(x_1,ldots,x_n)bigr)right|_t=0&=left.fracmathrm dmathrm dtexp(tx_1X_1)ldotsexp(tx_nX_n)right|_t=0\&=x_1X_1+cdots x_nX_n.endalignand therefore $Dalpha_0$ is an isomorphism. So, $alpha$ induces a diffeomorphism from a neighborhood $U$ of $(0,0,ldots,0)$ onto a neighborhood $V$ of $e_G$. If $(t_1,ldots,t_n)in U$, thenbeginalignFbigl(alpha(t_1,ldots,t_n)bigr)&=Fbigl(exp(t_1X_1)ldotsexp(t_nX_n)bigr)\&=Fbigl(exp(t_1X_1)bigr)ldots Fbigl(exp(t_nX_n)bigr),endalignwhich is smooth. Since $F|_V$ is smooth, $F$ is smooth.

share|cite|improve this answer

answered Jul 21 at 17:57

José Carlos Santos

114k1698177

Those dyadic rationals are dense in $[-1,1]$. So, near $0$, we have, by the continuity of $gamma$, $gamma(ttimes t_0)=exp(tX_0)$. In other words, $gamma(t)=expleft(frac tt_0X_0right)$ and therefore $gamma$ is differentiable.

For the other question, let $(X_1,ldots,X_n)$ be a basis of $mathfrak g$ and consider the mapbeginarrayrcccalphacolon&mathbbR^n&longrightarrow&G\&(t_1,ldots,t_n)&mapsto&exp(t_1X_1)ldotsexp(t_nX_n).endarrayThen $alpha$ is smooth. If $(x_1,ldots,x_n)inmathbbR^n$, thenbeginalignleft.fracmathrm dmathrm dtalphabigl(t(x_1,ldots,x_n)bigr)right|_t=0&=left.fracmathrm dmathrm dtexp(tx_1X_1)ldotsexp(tx_nX_n)right|_t=0\&=x_1X_1+cdots x_nX_n.endalignand therefore $Dalpha_0$ is an isomorphism. So, $alpha$ induces a diffeomorphism from a neighborhood $U$ of $(0,0,ldots,0)$ onto a neighborhood $V$ of $e_G$. If $(t_1,ldots,t_n)in U$, thenbeginalignFbigl(alpha(t_1,ldots,t_n)bigr)&=Fbigl(exp(t_1X_1)ldotsexp(t_nX_n)bigr)\&=Fbigl(exp(t_1X_1)bigr)ldots Fbigl(exp(t_nX_n)bigr),endalignwhich is smooth. Since $F|_V$ is smooth, $F$ is smooth.

share|cite|improve this answer

answered Jul 21 at 17:57

José Carlos Santos

114k1698177

share|cite|improve this answer

share|cite|improve this answer

answered Jul 21 at 17:57

José Carlos Santos

114k1698177

answered Jul 21 at 17:57

José Carlos Santos

114k1698177

answered Jul 21 at 17:57

José Carlos Santos

114k1698177

114k1698177

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! I came across this same argument for part 2 shortly after writing this question. I like the explanation, though I would be curious about how to construct an explicit function $phi$ as hinted. It seems such a map could only be defined in a neighborhood of the identity of $mathfrak g$...
  – D Ford
  Jul 21 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  My guess is that $phi$ is defined as follows: first, you find, for each $kin1,ldots,n$, a $Y_kinmathfrak h$ such that $tmapstoexp(tY_k)$ is equal to $tmapsto Fbigl(exp(tX_k)bigr)$. Then you define $phi$ as$$exp(tX_1)ldotsexp(tX_n)mapstoexp(tY_1)ldotsexp(tY_n).$$But this only occurred to me only after answering your question.
  – José Carlos Santos
  Jul 21 at 18:40
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! I came across this same argument for part 2 shortly after writing this question. I like the explanation, though I would be curious about how to construct an explicit function $phi$ as hinted. It seems such a map could only be defined in a neighborhood of the identity of $mathfrak g$...
  – D Ford
  Jul 21 at 18:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  My guess is that $phi$ is defined as follows: first, you find, for each $kin1,ldots,n$, a $Y_kinmathfrak h$ such that $tmapstoexp(tY_k)$ is equal to $tmapsto Fbigl(exp(tX_k)bigr)$. Then you define $phi$ as$$exp(tX_1)ldotsexp(tX_n)mapstoexp(tY_1)ldotsexp(tY_n).$$But this only occurred to me only after answering your question.
  – José Carlos Santos
  Jul 21 at 18:40
  
  

  
  
  
  

Thanks! I came across this same argument for part 2 shortly after writing this question. I like the explanation, though I would be curious about how to construct an explicit function $phi$ as hinted. It seems such a map could only be defined in a neighborhood of the identity of $mathfrak g$...
– D Ford
Jul 21 at 18:30

Thanks! I came across this same argument for part 2 shortly after writing this question. I like the explanation, though I would be curious about how to construct an explicit function $phi$ as hinted. It seems such a map could only be defined in a neighborhood of the identity of $mathfrak g$...
– D Ford
Jul 21 at 18:30

My guess is that $phi$ is defined as follows: first, you find, for each $kin1,ldots,n$, a $Y_kinmathfrak h$ such that $tmapstoexp(tY_k)$ is equal to $tmapsto Fbigl(exp(tX_k)bigr)$. Then you define $phi$ as$$exp(tX_1)ldotsexp(tX_n)mapstoexp(tY_1)ldotsexp(tY_n).$$But this only occurred to me only after answering your question.
– José Carlos Santos
Jul 21 at 18:40

My guess is that $phi$ is defined as follows: first, you find, for each $kin1,ldots,n$, a $Y_kinmathfrak h$ such that $tmapstoexp(tY_k)$ is equal to $tmapsto Fbigl(exp(tX_k)bigr)$. Then you define $phi$ as$$exp(tX_1)ldotsexp(tX_n)mapstoexp(tY_1)ldotsexp(tY_n).$$But this only occurred to me only after answering your question.
– José Carlos Santos
Jul 21 at 18:40

add a comment | 

 

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