A relation between left-invariant vector fields
If $G$ is a Lie group with a bivariant metric and if $U,V,X$ are left invariant vector fields, I wish to prove that $langle[U,X],Vrangle=-langle U,[V,X]rangle$.
Following the proof of Do Carmo’s Riemannian Geometry book, I was able to understand the proof as far as the fact that $langle U,Vrangle=langle dx_tU,dx_tVrangle$.
Now DoCarmo says differentiating with respect to $t$ at $t=0$ gives the result using the fact that $$[Y,X]=lim_{trightarrow 0}frac{dx_tY-Y}t.$$ But I do not know how to do this last step. How should I proceed?
riemannian-geometry smooth-manifolds vector-fields
edited Nov 21 '18 at 20:46
John B
12.2k51840
asked Nov 21 '18 at 20:31
User12239
412216
add a comment |
If $G$ is a Lie group with a bivariant metric and if $U,V,X$ are left invariant vector fields, I wish to prove that $langle[U,X],Vrangle=-langle U,[V,X]rangle$.
Following the proof of Do Carmo’s Riemannian Geometry book, I was able to understand the proof as far as the fact that $langle U,Vrangle=langle dx_tU,dx_tVrangle$.
Now DoCarmo says differentiating with respect to $t$ at $t=0$ gives the result using the fact that $$[Y,X]=lim_{trightarrow 0}frac{dx_tY-Y}t.$$ But I do not know how to do this last step. How should I proceed?
riemannian-geometry smooth-manifolds vector-fields
edited Nov 21 '18 at 20:46
John B
12.2k51840
asked Nov 21 '18 at 20:31
User12239
412216
add a comment |
If $G$ is a Lie group with a bivariant metric and if $U,V,X$ are left invariant vector fields, I wish to prove that $langle[U,X],Vrangle=-langle U,[V,X]rangle$.
Following the proof of Do Carmo’s Riemannian Geometry book, I was able to understand the proof as far as the fact that $langle U,Vrangle=langle dx_tU,dx_tVrangle$.
Now DoCarmo says differentiating with respect to $t$ at $t=0$ gives the result using the fact that $$[Y,X]=lim_{trightarrow 0}frac{dx_tY-Y}t.$$ But I do not know how to do this last step. How should I proceed?
riemannian-geometry smooth-manifolds vector-fields
edited Nov 21 '18 at 20:46
John B
12.2k51840
asked Nov 21 '18 at 20:31
User12239
412216
If $G$ is a Lie group with a bivariant metric and if $U,V,X$ are left invariant vector fields, I wish to prove that $langle[U,X],Vrangle=-langle U,[V,X]rangle$.
Following the proof of Do Carmo’s Riemannian Geometry book, I was able to understand the proof as far as the fact that $langle U,Vrangle=langle dx_tU,dx_tVrangle$.
Now DoCarmo says differentiating with respect to $t$ at $t=0$ gives the result using the fact that $$[Y,X]=lim_{trightarrow 0}frac{dx_tY-Y}t.$$ But I do not know how to do this last step. How should I proceed?
riemannian-geometry smooth-manifolds vector-fields
riemannian-geometry smooth-manifolds vector-fields
edited Nov 21 '18 at 20:46
John B
12.2k51840
asked Nov 21 '18 at 20:31
User12239
412216
edited Nov 21 '18 at 20:46
John B
12.2k51840
asked Nov 21 '18 at 20:31
User12239
412216
edited Nov 21 '18 at 20:46
John B
12.2k51840
edited Nov 21 '18 at 20:46
John B
12.2k51840
edited Nov 21 '18 at 20:46
John B
12.2k51840
12.2k51840
asked Nov 21 '18 at 20:31
User12239
412216
asked Nov 21 '18 at 20:31
User12239
412216
asked Nov 21 '18 at 20:31
User12239
412216
412216
add a comment |
add a comment |
1 Answer
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Subtracting $langle U, dx_t Vrangle$ from $langle U,Vrangle=langle dx_tU,dx_tVrangle$ gives
$$
langle U,V-dx_t Vrangle=langle dx_tU-U,dx_tVrangle.
$$
Hence, taking the limit when $tto0$ yields the identity
$$
-langle U,[V,X]rangle=langle [U,X],Vrangle.
$$
answered Nov 21 '18 at 20:44
John B
12.2k51840
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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active
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votes
Subtracting $langle U, dx_t Vrangle$ from $langle U,Vrangle=langle dx_tU,dx_tVrangle$ gives
$$
langle U,V-dx_t Vrangle=langle dx_tU-U,dx_tVrangle.
$$
Hence, taking the limit when $tto0$ yields the identity
$$
-langle U,[V,X]rangle=langle [U,X],Vrangle.
$$
answered Nov 21 '18 at 20:44
John B
12.2k51840
add a comment |
Subtracting $langle U, dx_t Vrangle$ from $langle U,Vrangle=langle dx_tU,dx_tVrangle$ gives
$$
langle U,V-dx_t Vrangle=langle dx_tU-U,dx_tVrangle.
$$
Hence, taking the limit when $tto0$ yields the identity
$$
-langle U,[V,X]rangle=langle [U,X],Vrangle.
$$
answered Nov 21 '18 at 20:44
John B
12.2k51840
add a comment |
Subtracting $langle U, dx_t Vrangle$ from $langle U,Vrangle=langle dx_tU,dx_tVrangle$ gives
$$
langle U,V-dx_t Vrangle=langle dx_tU-U,dx_tVrangle.
$$
Hence, taking the limit when $tto0$ yields the identity
$$
-langle U,[V,X]rangle=langle [U,X],Vrangle.
$$
answered Nov 21 '18 at 20:44
John B
12.2k51840
Subtracting $langle U, dx_t Vrangle$ from $langle U,Vrangle=langle dx_tU,dx_tVrangle$ gives
$$
langle U,V-dx_t Vrangle=langle dx_tU-U,dx_tVrangle.
$$
Hence, taking the limit when $tto0$ yields the identity
$$
-langle U,[V,X]rangle=langle [U,X],Vrangle.
$$
answered Nov 21 '18 at 20:44
John B
12.2k51840
answered Nov 21 '18 at 20:44
John B
12.2k51840
answered Nov 21 '18 at 20:44
John B
12.2k51840
answered Nov 21 '18 at 20:44
John B
12.2k51840
12.2k51840
add a comment |
add a comment |
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