Moving frame method with non-matrix Lie group
$begingroup$
I am trying to understand the modern formulation of the moving frame method for Lie group acting on a manifold.
I know the following theorem
Let be $M$ a manifold, $G$ a Lie group and $omega$ the Maurer-Cartan form of $G$. If $f_1, f_2: M to G$ are two functions, than $f_1^*omega = f_2^*omega$ if and only if $f_1 = gf_2$ for a fixed $g in G$.
I have always seen this theorem proved for matrix Lie group. In this case that's easy because $G$ can play directly with his Lie algebra $frak{g}$ (for example we have $omega_g = g^{-1}dg$). My question is:
Is this theorem true for a general Lie group? How is this proved?
Thanks in advance.
differential-geometry lie-groups differential-forms
asked Dec 3 '18 at 18:31
Marco All-in NervoMarco All-in Nervo
36118
$endgroup$
add a comment |
$begingroup$
I am trying to understand the modern formulation of the moving frame method for Lie group acting on a manifold.
I know the following theorem
Let be $M$ a manifold, $G$ a Lie group and $omega$ the Maurer-Cartan form of $G$. If $f_1, f_2: M to G$ are two functions, than $f_1^*omega = f_2^*omega$ if and only if $f_1 = gf_2$ for a fixed $g in G$.
I have always seen this theorem proved for matrix Lie group. In this case that's easy because $G$ can play directly with his Lie algebra $frak{g}$ (for example we have $omega_g = g^{-1}dg$). My question is:
Is this theorem true for a general Lie group? How is this proved?
Thanks in advance.
differential-geometry lie-groups differential-forms
asked Dec 3 '18 at 18:31
Marco All-in NervoMarco All-in Nervo
36118
$endgroup$
add a comment |
$begingroup$
I am trying to understand the modern formulation of the moving frame method for Lie group acting on a manifold.
I know the following theorem
Let be $M$ a manifold, $G$ a Lie group and $omega$ the Maurer-Cartan form of $G$. If $f_1, f_2: M to G$ are two functions, than $f_1^*omega = f_2^*omega$ if and only if $f_1 = gf_2$ for a fixed $g in G$.
I have always seen this theorem proved for matrix Lie group. In this case that's easy because $G$ can play directly with his Lie algebra $frak{g}$ (for example we have $omega_g = g^{-1}dg$). My question is:
Is this theorem true for a general Lie group? How is this proved?
Thanks in advance.
differential-geometry lie-groups differential-forms
asked Dec 3 '18 at 18:31
Marco All-in NervoMarco All-in Nervo
36118
$endgroup$
I am trying to understand the modern formulation of the moving frame method for Lie group acting on a manifold.
I know the following theorem
Let be $M$ a manifold, $G$ a Lie group and $omega$ the Maurer-Cartan form of $G$. If $f_1, f_2: M to G$ are two functions, than $f_1^*omega = f_2^*omega$ if and only if $f_1 = gf_2$ for a fixed $g in G$.
I have always seen this theorem proved for matrix Lie group. In this case that's easy because $G$ can play directly with his Lie algebra $frak{g}$ (for example we have $omega_g = g^{-1}dg$). My question is:
Is this theorem true for a general Lie group? How is this proved?
Thanks in advance.
differential-geometry lie-groups differential-forms
differential-geometry lie-groups differential-forms
asked Dec 3 '18 at 18:31
Marco All-in NervoMarco All-in Nervo
36118
asked Dec 3 '18 at 18:31
Marco All-in NervoMarco All-in Nervo
36118
asked Dec 3 '18 at 18:31
Marco All-in NervoMarco All-in Nervo
36118
asked Dec 3 '18 at 18:31
Marco All-in NervoMarco All-in Nervo
36118
asked Dec 3 '18 at 18:31
Marco All-in NervoMarco All-in Nervo
36118
36118
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
There are various proofs in textbooks. My favorite proof is to use E. Cartan's graph trick and the Frobenius Theorem.
As you stated it, things aren't quite right. You need $M$ connected.
If $omega^1,dots,omega^n$ are $n$ basis left-invariant forms (so we pick a basis for $mathfrak g^*$ and pull back by $L_g$), consider the differential ideal generated by the $1$-forms
$$eta^i = f_1^*omega^i-f_2^*omega^i.$$
Conceptually, we're looking at the map $F=(f_1,f_2)colon Mto Gtimes G$ and pulling back the forms $phi^i=pi_1^*omega^i - pi_2^*omega^i$ by the product map. The differential system $phi^1=dots=phi^n=0$ is completely integrable, since
$$dphi = pi_1^*[omega,omega] - pi_2^*[omega,omega] = [phi,pi_1^*omega] + [pi_2^*omega,phi] equiv 0 pmodphi.$$
Indeed, integral manifolds of $phi^i=0$ give left cosets of the diagonal subgroup $Deltasubset Gtimes G$. Since $F^*phi^i = 0$ by hypothesis, and since $M$ is connected the image of $F$ must be contained in one of those integral manifolds, which says that $f_1=gf_2$ for some $gin G$.
answered Dec 3 '18 at 21:02
Ted ShifrinTed Shifrin
63.2k44489
$endgroup$
$begingroup$
Thanks, as always, for your answer! Is necessary to use Frobenius theorem? I have always seen it used for the existence part. And which textbooks do you have in mind?
$endgroup$
– Marco All-in Nervo
Dec 3 '18 at 23:18
$begingroup$
Well, in this case you can get the integral manifolds by inspection. I was thinking of the more interesting parallel result that you construct $fcolon Mto G$ if you have a system of $1$-forms on $M$ satisfying the Maurer-Cartan equations of $G$. ... This stuff is in Warner and Spivak (volume 1). There's also a beautiful paper of Griffiths on Lie Groups and Moving Frames (Duke, 41, no. 4, pp. 775-814). Also, see Chern/Chen/Lam, pp. 198 ff.
$endgroup$
– Ted Shifrin
Dec 3 '18 at 23:47
$begingroup$
Yes, I got your answer and your proof is really beautiful. I was looking for a simpler proof (as in Spivak, thank you!) with instruments that I can manage with my little experience. The paper of Griffiths seems really ...wow. Thank you again, next time I will start my question with "Dear Shifrin, here my new doubts about moving frames..."
$endgroup$
– Marco All-in Nervo
Dec 4 '18 at 20:20
add a comment |
$begingroup$
Here Spivak's proof from A Comprehensive Introduction to Differential Geometry, Vol. 1
answered Dec 4 '18 at 20:26
Marco All-in NervoMarco All-in Nervo
36118
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
There are various proofs in textbooks. My favorite proof is to use E. Cartan's graph trick and the Frobenius Theorem.
As you stated it, things aren't quite right. You need $M$ connected.
If $omega^1,dots,omega^n$ are $n$ basis left-invariant forms (so we pick a basis for $mathfrak g^*$ and pull back by $L_g$), consider the differential ideal generated by the $1$-forms
$$eta^i = f_1^*omega^i-f_2^*omega^i.$$
Conceptually, we're looking at the map $F=(f_1,f_2)colon Mto Gtimes G$ and pulling back the forms $phi^i=pi_1^*omega^i - pi_2^*omega^i$ by the product map. The differential system $phi^1=dots=phi^n=0$ is completely integrable, since
$$dphi = pi_1^*[omega,omega] - pi_2^*[omega,omega] = [phi,pi_1^*omega] + [pi_2^*omega,phi] equiv 0 pmodphi.$$
Indeed, integral manifolds of $phi^i=0$ give left cosets of the diagonal subgroup $Deltasubset Gtimes G$. Since $F^*phi^i = 0$ by hypothesis, and since $M$ is connected the image of $F$ must be contained in one of those integral manifolds, which says that $f_1=gf_2$ for some $gin G$.
answered Dec 3 '18 at 21:02
Ted ShifrinTed Shifrin
63.2k44489
$endgroup$
$begingroup$
Thanks, as always, for your answer! Is necessary to use Frobenius theorem? I have always seen it used for the existence part. And which textbooks do you have in mind?
$endgroup$
– Marco All-in Nervo
Dec 3 '18 at 23:18
$begingroup$
Well, in this case you can get the integral manifolds by inspection. I was thinking of the more interesting parallel result that you construct $fcolon Mto G$ if you have a system of $1$-forms on $M$ satisfying the Maurer-Cartan equations of $G$. ... This stuff is in Warner and Spivak (volume 1). There's also a beautiful paper of Griffiths on Lie Groups and Moving Frames (Duke, 41, no. 4, pp. 775-814). Also, see Chern/Chen/Lam, pp. 198 ff.
$endgroup$
– Ted Shifrin
Dec 3 '18 at 23:47
$begingroup$
Yes, I got your answer and your proof is really beautiful. I was looking for a simpler proof (as in Spivak, thank you!) with instruments that I can manage with my little experience. The paper of Griffiths seems really ...wow. Thank you again, next time I will start my question with "Dear Shifrin, here my new doubts about moving frames..."
$endgroup$
– Marco All-in Nervo
Dec 4 '18 at 20:20
add a comment |
$begingroup$
There are various proofs in textbooks. My favorite proof is to use E. Cartan's graph trick and the Frobenius Theorem.
As you stated it, things aren't quite right. You need $M$ connected.
If $omega^1,dots,omega^n$ are $n$ basis left-invariant forms (so we pick a basis for $mathfrak g^*$ and pull back by $L_g$), consider the differential ideal generated by the $1$-forms
$$eta^i = f_1^*omega^i-f_2^*omega^i.$$
Conceptually, we're looking at the map $F=(f_1,f_2)colon Mto Gtimes G$ and pulling back the forms $phi^i=pi_1^*omega^i - pi_2^*omega^i$ by the product map. The differential system $phi^1=dots=phi^n=0$ is completely integrable, since
$$dphi = pi_1^*[omega,omega] - pi_2^*[omega,omega] = [phi,pi_1^*omega] + [pi_2^*omega,phi] equiv 0 pmodphi.$$
Indeed, integral manifolds of $phi^i=0$ give left cosets of the diagonal subgroup $Deltasubset Gtimes G$. Since $F^*phi^i = 0$ by hypothesis, and since $M$ is connected the image of $F$ must be contained in one of those integral manifolds, which says that $f_1=gf_2$ for some $gin G$.
answered Dec 3 '18 at 21:02
Ted ShifrinTed Shifrin
63.2k44489
$endgroup$
$begingroup$
Thanks, as always, for your answer! Is necessary to use Frobenius theorem? I have always seen it used for the existence part. And which textbooks do you have in mind?
$endgroup$
– Marco All-in Nervo
Dec 3 '18 at 23:18
$begingroup$
Well, in this case you can get the integral manifolds by inspection. I was thinking of the more interesting parallel result that you construct $fcolon Mto G$ if you have a system of $1$-forms on $M$ satisfying the Maurer-Cartan equations of $G$. ... This stuff is in Warner and Spivak (volume 1). There's also a beautiful paper of Griffiths on Lie Groups and Moving Frames (Duke, 41, no. 4, pp. 775-814). Also, see Chern/Chen/Lam, pp. 198 ff.
$endgroup$
– Ted Shifrin
Dec 3 '18 at 23:47
$begingroup$
Yes, I got your answer and your proof is really beautiful. I was looking for a simpler proof (as in Spivak, thank you!) with instruments that I can manage with my little experience. The paper of Griffiths seems really ...wow. Thank you again, next time I will start my question with "Dear Shifrin, here my new doubts about moving frames..."
$endgroup$
– Marco All-in Nervo
Dec 4 '18 at 20:20
add a comment |
$begingroup$
There are various proofs in textbooks. My favorite proof is to use E. Cartan's graph trick and the Frobenius Theorem.
As you stated it, things aren't quite right. You need $M$ connected.
If $omega^1,dots,omega^n$ are $n$ basis left-invariant forms (so we pick a basis for $mathfrak g^*$ and pull back by $L_g$), consider the differential ideal generated by the $1$-forms
$$eta^i = f_1^*omega^i-f_2^*omega^i.$$
Conceptually, we're looking at the map $F=(f_1,f_2)colon Mto Gtimes G$ and pulling back the forms $phi^i=pi_1^*omega^i - pi_2^*omega^i$ by the product map. The differential system $phi^1=dots=phi^n=0$ is completely integrable, since
$$dphi = pi_1^*[omega,omega] - pi_2^*[omega,omega] = [phi,pi_1^*omega] + [pi_2^*omega,phi] equiv 0 pmodphi.$$
Indeed, integral manifolds of $phi^i=0$ give left cosets of the diagonal subgroup $Deltasubset Gtimes G$. Since $F^*phi^i = 0$ by hypothesis, and since $M$ is connected the image of $F$ must be contained in one of those integral manifolds, which says that $f_1=gf_2$ for some $gin G$.
answered Dec 3 '18 at 21:02
Ted ShifrinTed Shifrin
63.2k44489
$endgroup$
There are various proofs in textbooks. My favorite proof is to use E. Cartan's graph trick and the Frobenius Theorem.
As you stated it, things aren't quite right. You need $M$ connected.
If $omega^1,dots,omega^n$ are $n$ basis left-invariant forms (so we pick a basis for $mathfrak g^*$ and pull back by $L_g$), consider the differential ideal generated by the $1$-forms
$$eta^i = f_1^*omega^i-f_2^*omega^i.$$
Conceptually, we're looking at the map $F=(f_1,f_2)colon Mto Gtimes G$ and pulling back the forms $phi^i=pi_1^*omega^i - pi_2^*omega^i$ by the product map. The differential system $phi^1=dots=phi^n=0$ is completely integrable, since
$$dphi = pi_1^*[omega,omega] - pi_2^*[omega,omega] = [phi,pi_1^*omega] + [pi_2^*omega,phi] equiv 0 pmodphi.$$
Indeed, integral manifolds of $phi^i=0$ give left cosets of the diagonal subgroup $Deltasubset Gtimes G$. Since $F^*phi^i = 0$ by hypothesis, and since $M$ is connected the image of $F$ must be contained in one of those integral manifolds, which says that $f_1=gf_2$ for some $gin G$.
answered Dec 3 '18 at 21:02
Ted ShifrinTed Shifrin
63.2k44489
answered Dec 3 '18 at 21:02
Ted ShifrinTed Shifrin
63.2k44489
answered Dec 3 '18 at 21:02
Ted ShifrinTed Shifrin
63.2k44489
answered Dec 3 '18 at 21:02
Ted ShifrinTed Shifrin
63.2k44489
63.2k44489
$begingroup$
Thanks, as always, for your answer! Is necessary to use Frobenius theorem? I have always seen it used for the existence part. And which textbooks do you have in mind?
$endgroup$
– Marco All-in Nervo
Dec 3 '18 at 23:18
$begingroup$
Well, in this case you can get the integral manifolds by inspection. I was thinking of the more interesting parallel result that you construct $fcolon Mto G$ if you have a system of $1$-forms on $M$ satisfying the Maurer-Cartan equations of $G$. ... This stuff is in Warner and Spivak (volume 1). There's also a beautiful paper of Griffiths on Lie Groups and Moving Frames (Duke, 41, no. 4, pp. 775-814). Also, see Chern/Chen/Lam, pp. 198 ff.
$endgroup$
– Ted Shifrin
Dec 3 '18 at 23:47
$begingroup$
Yes, I got your answer and your proof is really beautiful. I was looking for a simpler proof (as in Spivak, thank you!) with instruments that I can manage with my little experience. The paper of Griffiths seems really ...wow. Thank you again, next time I will start my question with "Dear Shifrin, here my new doubts about moving frames..."
$endgroup$
– Marco All-in Nervo
Dec 4 '18 at 20:20
add a comment |
$begingroup$
Thanks, as always, for your answer! Is necessary to use Frobenius theorem? I have always seen it used for the existence part. And which textbooks do you have in mind?
$endgroup$
– Marco All-in Nervo
Dec 3 '18 at 23:18
$begingroup$
Well, in this case you can get the integral manifolds by inspection. I was thinking of the more interesting parallel result that you construct $fcolon Mto G$ if you have a system of $1$-forms on $M$ satisfying the Maurer-Cartan equations of $G$. ... This stuff is in Warner and Spivak (volume 1). There's also a beautiful paper of Griffiths on Lie Groups and Moving Frames (Duke, 41, no. 4, pp. 775-814). Also, see Chern/Chen/Lam, pp. 198 ff.
$endgroup$
– Ted Shifrin
Dec 3 '18 at 23:47
$begingroup$
Yes, I got your answer and your proof is really beautiful. I was looking for a simpler proof (as in Spivak, thank you!) with instruments that I can manage with my little experience. The paper of Griffiths seems really ...wow. Thank you again, next time I will start my question with "Dear Shifrin, here my new doubts about moving frames..."
$endgroup$
– Marco All-in Nervo
Dec 4 '18 at 20:20
$begingroup$
Thanks, as always, for your answer! Is necessary to use Frobenius theorem? I have always seen it used for the existence part. And which textbooks do you have in mind?
$endgroup$
– Marco All-in Nervo
Dec 3 '18 at 23:18
$begingroup$
Thanks, as always, for your answer! Is necessary to use Frobenius theorem? I have always seen it used for the existence part. And which textbooks do you have in mind?
$endgroup$
– Marco All-in Nervo
Dec 3 '18 at 23:18
$begingroup$
Well, in this case you can get the integral manifolds by inspection. I was thinking of the more interesting parallel result that you construct $fcolon Mto G$ if you have a system of $1$-forms on $M$ satisfying the Maurer-Cartan equations of $G$. ... This stuff is in Warner and Spivak (volume 1). There's also a beautiful paper of Griffiths on Lie Groups and Moving Frames (Duke, 41, no. 4, pp. 775-814). Also, see Chern/Chen/Lam, pp. 198 ff.
$endgroup$
– Ted Shifrin
Dec 3 '18 at 23:47
$begingroup$
Well, in this case you can get the integral manifolds by inspection. I was thinking of the more interesting parallel result that you construct $fcolon Mto G$ if you have a system of $1$-forms on $M$ satisfying the Maurer-Cartan equations of $G$. ... This stuff is in Warner and Spivak (volume 1). There's also a beautiful paper of Griffiths on Lie Groups and Moving Frames (Duke, 41, no. 4, pp. 775-814). Also, see Chern/Chen/Lam, pp. 198 ff.
$endgroup$
– Ted Shifrin
Dec 3 '18 at 23:47
$begingroup$
Yes, I got your answer and your proof is really beautiful. I was looking for a simpler proof (as in Spivak, thank you!) with instruments that I can manage with my little experience. The paper of Griffiths seems really ...wow. Thank you again, next time I will start my question with "Dear Shifrin, here my new doubts about moving frames..."
$endgroup$
– Marco All-in Nervo
Dec 4 '18 at 20:20
$begingroup$
Yes, I got your answer and your proof is really beautiful. I was looking for a simpler proof (as in Spivak, thank you!) with instruments that I can manage with my little experience. The paper of Griffiths seems really ...wow. Thank you again, next time I will start my question with "Dear Shifrin, here my new doubts about moving frames..."
$endgroup$
– Marco All-in Nervo
Dec 4 '18 at 20:20
add a comment |
$begingroup$
Here Spivak's proof from A Comprehensive Introduction to Differential Geometry, Vol. 1
answered Dec 4 '18 at 20:26
Marco All-in NervoMarco All-in Nervo
36118
$endgroup$
add a comment |
$begingroup$
Here Spivak's proof from A Comprehensive Introduction to Differential Geometry, Vol. 1
answered Dec 4 '18 at 20:26
Marco All-in NervoMarco All-in Nervo
36118
$endgroup$
add a comment |
$begingroup$
Here Spivak's proof from A Comprehensive Introduction to Differential Geometry, Vol. 1
answered Dec 4 '18 at 20:26
Marco All-in NervoMarco All-in Nervo
36118
$endgroup$
Here Spivak's proof from A Comprehensive Introduction to Differential Geometry, Vol. 1
answered Dec 4 '18 at 20:26
Marco All-in NervoMarco All-in Nervo
36118
answered Dec 4 '18 at 20:26
Marco All-in NervoMarco All-in Nervo
36118
answered Dec 4 '18 at 20:26
Marco All-in NervoMarco All-in Nervo
36118
answered Dec 4 '18 at 20:26
Marco All-in NervoMarco All-in Nervo
36118
36118
add a comment |
add a comment |
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