Definition of a left-invariant vector field
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In DoCarmo’s Riemannian Geometry book, a smooth vector field $X$ on a Lie group $G$ is called left invariant if $d(L_x)_yX=X$ for all $x,yin G$.
The left side maps each $yin G$ to the tangent space $T_{xy}G$ while the right side maps each $y$ to $T_yG$. So in what sense are they equal?
differential-geometry lie-groups riemannian-geometry smooth-manifolds
asked Nov 18 at 20:17
Selflearner
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In DoCarmo’s Riemannian Geometry book, a smooth vector field $X$ on a Lie group $G$ is called left invariant if $d(L_x)_yX=X$ for all $x,yin G$.
The left side maps each $yin G$ to the tangent space $T_{xy}G$ while the right side maps each $y$ to $T_yG$. So in what sense are they equal?
differential-geometry lie-groups riemannian-geometry smooth-manifolds
asked Nov 18 at 20:17
Selflearner
377214
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
In DoCarmo’s Riemannian Geometry book, a smooth vector field $X$ on a Lie group $G$ is called left invariant if $d(L_x)_yX=X$ for all $x,yin G$.
The left side maps each $yin G$ to the tangent space $T_{xy}G$ while the right side maps each $y$ to $T_yG$. So in what sense are they equal?
differential-geometry lie-groups riemannian-geometry smooth-manifolds
asked Nov 18 at 20:17
Selflearner
377214
In DoCarmo’s Riemannian Geometry book, a smooth vector field $X$ on a Lie group $G$ is called left invariant if $d(L_x)_yX=X$ for all $x,yin G$.
The left side maps each $yin G$ to the tangent space $T_{xy}G$ while the right side maps each $y$ to $T_yG$. So in what sense are they equal?
differential-geometry lie-groups riemannian-geometry smooth-manifolds
differential-geometry lie-groups riemannian-geometry smooth-manifolds
asked Nov 18 at 20:17
Selflearner
377214
asked Nov 18 at 20:17
Selflearner
377214
asked Nov 18 at 20:17
Selflearner
377214
asked Nov 18 at 20:17
Selflearner
377214
asked Nov 18 at 20:17
Selflearner
377214
377214
add a comment |
add a comment |
1 Answer
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Let $G$ be a Lie group and $X$ a vector field. For $gin G$, let $L_g:Gto G$ denote left-multiplication, i.e., $L_g(p)=gp$. Then we say that $X$ is left-invariant if
$$(L_g)_*X=X.$$
Note that the pushforward makes sense on vector fields since $L_g$ is a diffeomorphism. Making this explicit at a point, say at $pin G$ yields
$$X_{L_g(p)}=((L_g)_*X)_p$$
(this is saying $X$ is $L_g$-related with itself). Or rewritten, we have that
$$X_{gp}=d(L_g)_p(X_p).$$
answered Nov 18 at 20:32
Matt
39218
Thanks, I should use such pushforwards to make sense of such expressions
– Selflearner
Nov 18 at 20:34
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Let $G$ be a Lie group and $X$ a vector field. For $gin G$, let $L_g:Gto G$ denote left-multiplication, i.e., $L_g(p)=gp$. Then we say that $X$ is left-invariant if
$$(L_g)_*X=X.$$
Note that the pushforward makes sense on vector fields since $L_g$ is a diffeomorphism. Making this explicit at a point, say at $pin G$ yields
$$X_{L_g(p)}=((L_g)_*X)_p$$
(this is saying $X$ is $L_g$-related with itself). Or rewritten, we have that
$$X_{gp}=d(L_g)_p(X_p).$$
answered Nov 18 at 20:32
Matt
39218
Thanks, I should use such pushforwards to make sense of such expressions
– Selflearner
Nov 18 at 20:34
add a comment |
up vote
0
down vote
accepted
Let $G$ be a Lie group and $X$ a vector field. For $gin G$, let $L_g:Gto G$ denote left-multiplication, i.e., $L_g(p)=gp$. Then we say that $X$ is left-invariant if
$$(L_g)_*X=X.$$
Note that the pushforward makes sense on vector fields since $L_g$ is a diffeomorphism. Making this explicit at a point, say at $pin G$ yields
$$X_{L_g(p)}=((L_g)_*X)_p$$
(this is saying $X$ is $L_g$-related with itself). Or rewritten, we have that
$$X_{gp}=d(L_g)_p(X_p).$$
answered Nov 18 at 20:32
Matt
39218
Thanks, I should use such pushforwards to make sense of such expressions
– Selflearner
Nov 18 at 20:34
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Let $G$ be a Lie group and $X$ a vector field. For $gin G$, let $L_g:Gto G$ denote left-multiplication, i.e., $L_g(p)=gp$. Then we say that $X$ is left-invariant if
$$(L_g)_*X=X.$$
Note that the pushforward makes sense on vector fields since $L_g$ is a diffeomorphism. Making this explicit at a point, say at $pin G$ yields
$$X_{L_g(p)}=((L_g)_*X)_p$$
(this is saying $X$ is $L_g$-related with itself). Or rewritten, we have that
$$X_{gp}=d(L_g)_p(X_p).$$
answered Nov 18 at 20:32
Matt
39218
Let $G$ be a Lie group and $X$ a vector field. For $gin G$, let $L_g:Gto G$ denote left-multiplication, i.e., $L_g(p)=gp$. Then we say that $X$ is left-invariant if
$$(L_g)_*X=X.$$
Note that the pushforward makes sense on vector fields since $L_g$ is a diffeomorphism. Making this explicit at a point, say at $pin G$ yields
$$X_{L_g(p)}=((L_g)_*X)_p$$
(this is saying $X$ is $L_g$-related with itself). Or rewritten, we have that
$$X_{gp}=d(L_g)_p(X_p).$$
answered Nov 18 at 20:32
Matt
39218
answered Nov 18 at 20:32
Matt
39218
answered Nov 18 at 20:32
Matt
39218
answered Nov 18 at 20:32
Matt
39218
39218
Thanks, I should use such pushforwards to make sense of such expressions
– Selflearner
Nov 18 at 20:34
add a comment |
Thanks, I should use such pushforwards to make sense of such expressions
– Selflearner
Nov 18 at 20:34
Thanks, I should use such pushforwards to make sense of such expressions
– Selflearner
Nov 18 at 20:34
Thanks, I should use such pushforwards to make sense of such expressions
– Selflearner
Nov 18 at 20:34
add a comment |
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