Extending to a local frame that agrees with given orientation
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Suppose that $(e_1, ldots, e_k)$ is an oriented basis for $T_pM$ where $M$ is an oriented Riemannian manifold. In general, we know that we can extend to a smooth local frame $(X_1, ldots, X_k)$ on $Uni p$ such that $X_irvert_p = e_i$ for each $i$.
But can we further stipulate that $(X_1rvert_q, ldots, X_krvert_q)$ is oriented for each $qin U$?
I have tried thinking of ways to shrink $U$ in a suitable way. I have played around with using the orientation form $omega$ on $M$ and with first applying Gram-Schmidt to $(X_1, ldots, X_k)$, but no progress.
Any help is much appreciated.
linear-algebra differential-geometry differential-topology riemannian-geometry smooth-manifolds
asked Dec 4 at 1:01
CuriousKid7
1,650717
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up vote
0
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Suppose that $(e_1, ldots, e_k)$ is an oriented basis for $T_pM$ where $M$ is an oriented Riemannian manifold. In general, we know that we can extend to a smooth local frame $(X_1, ldots, X_k)$ on $Uni p$ such that $X_irvert_p = e_i$ for each $i$.
But can we further stipulate that $(X_1rvert_q, ldots, X_krvert_q)$ is oriented for each $qin U$?
I have tried thinking of ways to shrink $U$ in a suitable way. I have played around with using the orientation form $omega$ on $M$ and with first applying Gram-Schmidt to $(X_1, ldots, X_k)$, but no progress.
Any help is much appreciated.
linear-algebra differential-geometry differential-topology riemannian-geometry smooth-manifolds
asked Dec 4 at 1:01
CuriousKid7
1,650717
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose that $(e_1, ldots, e_k)$ is an oriented basis for $T_pM$ where $M$ is an oriented Riemannian manifold. In general, we know that we can extend to a smooth local frame $(X_1, ldots, X_k)$ on $Uni p$ such that $X_irvert_p = e_i$ for each $i$.
But can we further stipulate that $(X_1rvert_q, ldots, X_krvert_q)$ is oriented for each $qin U$?
I have tried thinking of ways to shrink $U$ in a suitable way. I have played around with using the orientation form $omega$ on $M$ and with first applying Gram-Schmidt to $(X_1, ldots, X_k)$, but no progress.
Any help is much appreciated.
linear-algebra differential-geometry differential-topology riemannian-geometry smooth-manifolds
asked Dec 4 at 1:01
CuriousKid7
1,650717
Suppose that $(e_1, ldots, e_k)$ is an oriented basis for $T_pM$ where $M$ is an oriented Riemannian manifold. In general, we know that we can extend to a smooth local frame $(X_1, ldots, X_k)$ on $Uni p$ such that $X_irvert_p = e_i$ for each $i$.
But can we further stipulate that $(X_1rvert_q, ldots, X_krvert_q)$ is oriented for each $qin U$?
I have tried thinking of ways to shrink $U$ in a suitable way. I have played around with using the orientation form $omega$ on $M$ and with first applying Gram-Schmidt to $(X_1, ldots, X_k)$, but no progress.
Any help is much appreciated.
linear-algebra differential-geometry differential-topology riemannian-geometry smooth-manifolds
linear-algebra differential-geometry differential-topology riemannian-geometry smooth-manifolds
asked Dec 4 at 1:01
CuriousKid7
1,650717
asked Dec 4 at 1:01
CuriousKid7
1,650717
edited Dec 4 at 18:25
asked Dec 4 at 1:01
CuriousKid7
1,650717
asked Dec 4 at 1:01
CuriousKid7
1,650717
asked Dec 4 at 1:01
CuriousKid7
1,650717
1,650717
add a comment |
add a comment |
1 Answer
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The local frame $(X_i)$ will agree with the chosen orientation form: if $omega(e_1,dots,e_k)>0$, then $omega(X_1,dots,X_k)>0$ in $U$, because $omega(X_1,dots,X_k)|_q=0$ would imply that the $X_i's$ do not form a basis of $T_qM$.
answered Dec 4 at 18:47
Federico
4,097512
This solution shows how to restrict $U$ appropriately, but I don't see why vanishing at $omega rvert_q$ would imply that the $X_i$ don't form a basis. It's possible that an alternating tensor vanishes at a linearly independent tuple, right?
– CuriousKid7
Dec 5 at 5:01
You don't need to restrict $U$ in any way. $M$ is orientable and you have fixed an orientation form $omega$. What I'm saying is that if $(X_i)$ is a frame inside $U$, then $omega(X_1,dots,X_k)$ doesn't change sign inside $U$, because if it does there is a point at which it must vanish, and at that point $(X_i)$ would not be a frame
– Federico
Dec 5 at 13:51
1
"It's possible that an alternating tensor vanishes at a linearly independent tuple, right?" No, unless $omega=0$ at that point. If it vanishes at a linearly independent tuple, it vanishes on any tuple.
– Federico
Dec 5 at 13:52
1
Assume $omega(e_1,dots,e_k)=0$ for some basis $(e_i)$. Take any $k$-uple $v_j=a^i_je_i$. Then $omega(v_1,dots,v_k)=det(A)omega(e_1,dots,e_k)=0$, where $A$ is the matrix with entries $a^i_j$.
– Federico
Dec 5 at 13:54
Got it, thanks for clarifying!
– CuriousKid7
Dec 5 at 16:27
add a comment |
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1 Answer
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active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The local frame $(X_i)$ will agree with the chosen orientation form: if $omega(e_1,dots,e_k)>0$, then $omega(X_1,dots,X_k)>0$ in $U$, because $omega(X_1,dots,X_k)|_q=0$ would imply that the $X_i's$ do not form a basis of $T_qM$.
answered Dec 4 at 18:47
Federico
4,097512
This solution shows how to restrict $U$ appropriately, but I don't see why vanishing at $omega rvert_q$ would imply that the $X_i$ don't form a basis. It's possible that an alternating tensor vanishes at a linearly independent tuple, right?
– CuriousKid7
Dec 5 at 5:01
You don't need to restrict $U$ in any way. $M$ is orientable and you have fixed an orientation form $omega$. What I'm saying is that if $(X_i)$ is a frame inside $U$, then $omega(X_1,dots,X_k)$ doesn't change sign inside $U$, because if it does there is a point at which it must vanish, and at that point $(X_i)$ would not be a frame
– Federico
Dec 5 at 13:51
1
"It's possible that an alternating tensor vanishes at a linearly independent tuple, right?" No, unless $omega=0$ at that point. If it vanishes at a linearly independent tuple, it vanishes on any tuple.
– Federico
Dec 5 at 13:52
1
Assume $omega(e_1,dots,e_k)=0$ for some basis $(e_i)$. Take any $k$-uple $v_j=a^i_je_i$. Then $omega(v_1,dots,v_k)=det(A)omega(e_1,dots,e_k)=0$, where $A$ is the matrix with entries $a^i_j$.
– Federico
Dec 5 at 13:54
Got it, thanks for clarifying!
– CuriousKid7
Dec 5 at 16:27
add a comment |
up vote
1
down vote
accepted
The local frame $(X_i)$ will agree with the chosen orientation form: if $omega(e_1,dots,e_k)>0$, then $omega(X_1,dots,X_k)>0$ in $U$, because $omega(X_1,dots,X_k)|_q=0$ would imply that the $X_i's$ do not form a basis of $T_qM$.
answered Dec 4 at 18:47
Federico
4,097512
This solution shows how to restrict $U$ appropriately, but I don't see why vanishing at $omega rvert_q$ would imply that the $X_i$ don't form a basis. It's possible that an alternating tensor vanishes at a linearly independent tuple, right?
– CuriousKid7
Dec 5 at 5:01
You don't need to restrict $U$ in any way. $M$ is orientable and you have fixed an orientation form $omega$. What I'm saying is that if $(X_i)$ is a frame inside $U$, then $omega(X_1,dots,X_k)$ doesn't change sign inside $U$, because if it does there is a point at which it must vanish, and at that point $(X_i)$ would not be a frame
– Federico
Dec 5 at 13:51
1
"It's possible that an alternating tensor vanishes at a linearly independent tuple, right?" No, unless $omega=0$ at that point. If it vanishes at a linearly independent tuple, it vanishes on any tuple.
– Federico
Dec 5 at 13:52
1
Assume $omega(e_1,dots,e_k)=0$ for some basis $(e_i)$. Take any $k$-uple $v_j=a^i_je_i$. Then $omega(v_1,dots,v_k)=det(A)omega(e_1,dots,e_k)=0$, where $A$ is the matrix with entries $a^i_j$.
– Federico
Dec 5 at 13:54
Got it, thanks for clarifying!
– CuriousKid7
Dec 5 at 16:27
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The local frame $(X_i)$ will agree with the chosen orientation form: if $omega(e_1,dots,e_k)>0$, then $omega(X_1,dots,X_k)>0$ in $U$, because $omega(X_1,dots,X_k)|_q=0$ would imply that the $X_i's$ do not form a basis of $T_qM$.
answered Dec 4 at 18:47
Federico
4,097512
The local frame $(X_i)$ will agree with the chosen orientation form: if $omega(e_1,dots,e_k)>0$, then $omega(X_1,dots,X_k)>0$ in $U$, because $omega(X_1,dots,X_k)|_q=0$ would imply that the $X_i's$ do not form a basis of $T_qM$.
answered Dec 4 at 18:47
Federico
4,097512
answered Dec 4 at 18:47
Federico
4,097512
answered Dec 4 at 18:47
Federico
4,097512
answered Dec 4 at 18:47
Federico
4,097512
4,097512
This solution shows how to restrict $U$ appropriately, but I don't see why vanishing at $omega rvert_q$ would imply that the $X_i$ don't form a basis. It's possible that an alternating tensor vanishes at a linearly independent tuple, right?
– CuriousKid7
Dec 5 at 5:01
You don't need to restrict $U$ in any way. $M$ is orientable and you have fixed an orientation form $omega$. What I'm saying is that if $(X_i)$ is a frame inside $U$, then $omega(X_1,dots,X_k)$ doesn't change sign inside $U$, because if it does there is a point at which it must vanish, and at that point $(X_i)$ would not be a frame
– Federico
Dec 5 at 13:51
1
"It's possible that an alternating tensor vanishes at a linearly independent tuple, right?" No, unless $omega=0$ at that point. If it vanishes at a linearly independent tuple, it vanishes on any tuple.
– Federico
Dec 5 at 13:52
1
Assume $omega(e_1,dots,e_k)=0$ for some basis $(e_i)$. Take any $k$-uple $v_j=a^i_je_i$. Then $omega(v_1,dots,v_k)=det(A)omega(e_1,dots,e_k)=0$, where $A$ is the matrix with entries $a^i_j$.
– Federico
Dec 5 at 13:54
Got it, thanks for clarifying!
– CuriousKid7
Dec 5 at 16:27
add a comment |
This solution shows how to restrict $U$ appropriately, but I don't see why vanishing at $omega rvert_q$ would imply that the $X_i$ don't form a basis. It's possible that an alternating tensor vanishes at a linearly independent tuple, right?
– CuriousKid7
Dec 5 at 5:01
You don't need to restrict $U$ in any way. $M$ is orientable and you have fixed an orientation form $omega$. What I'm saying is that if $(X_i)$ is a frame inside $U$, then $omega(X_1,dots,X_k)$ doesn't change sign inside $U$, because if it does there is a point at which it must vanish, and at that point $(X_i)$ would not be a frame
– Federico
Dec 5 at 13:51
1
"It's possible that an alternating tensor vanishes at a linearly independent tuple, right?" No, unless $omega=0$ at that point. If it vanishes at a linearly independent tuple, it vanishes on any tuple.
– Federico
Dec 5 at 13:52
1
Assume $omega(e_1,dots,e_k)=0$ for some basis $(e_i)$. Take any $k$-uple $v_j=a^i_je_i$. Then $omega(v_1,dots,v_k)=det(A)omega(e_1,dots,e_k)=0$, where $A$ is the matrix with entries $a^i_j$.
– Federico
Dec 5 at 13:54
Got it, thanks for clarifying!
– CuriousKid7
Dec 5 at 16:27
This solution shows how to restrict $U$ appropriately, but I don't see why vanishing at $omega rvert_q$ would imply that the $X_i$ don't form a basis. It's possible that an alternating tensor vanishes at a linearly independent tuple, right?
– CuriousKid7
Dec 5 at 5:01
This solution shows how to restrict $U$ appropriately, but I don't see why vanishing at $omega rvert_q$ would imply that the $X_i$ don't form a basis. It's possible that an alternating tensor vanishes at a linearly independent tuple, right?
– CuriousKid7
Dec 5 at 5:01
You don't need to restrict $U$ in any way. $M$ is orientable and you have fixed an orientation form $omega$. What I'm saying is that if $(X_i)$ is a frame inside $U$, then $omega(X_1,dots,X_k)$ doesn't change sign inside $U$, because if it does there is a point at which it must vanish, and at that point $(X_i)$ would not be a frame
– Federico
Dec 5 at 13:51
You don't need to restrict $U$ in any way. $M$ is orientable and you have fixed an orientation form $omega$. What I'm saying is that if $(X_i)$ is a frame inside $U$, then $omega(X_1,dots,X_k)$ doesn't change sign inside $U$, because if it does there is a point at which it must vanish, and at that point $(X_i)$ would not be a frame
– Federico
Dec 5 at 13:51
1
1
"It's possible that an alternating tensor vanishes at a linearly independent tuple, right?" No, unless $omega=0$ at that point. If it vanishes at a linearly independent tuple, it vanishes on any tuple.
– Federico
Dec 5 at 13:52
"It's possible that an alternating tensor vanishes at a linearly independent tuple, right?" No, unless $omega=0$ at that point. If it vanishes at a linearly independent tuple, it vanishes on any tuple.
– Federico
Dec 5 at 13:52
1
1
Assume $omega(e_1,dots,e_k)=0$ for some basis $(e_i)$. Take any $k$-uple $v_j=a^i_je_i$. Then $omega(v_1,dots,v_k)=det(A)omega(e_1,dots,e_k)=0$, where $A$ is the matrix with entries $a^i_j$.
– Federico
Dec 5 at 13:54
Assume $omega(e_1,dots,e_k)=0$ for some basis $(e_i)$. Take any $k$-uple $v_j=a^i_je_i$. Then $omega(v_1,dots,v_k)=det(A)omega(e_1,dots,e_k)=0$, where $A$ is the matrix with entries $a^i_j$.
– Federico
Dec 5 at 13:54
Got it, thanks for clarifying!
– CuriousKid7
Dec 5 at 16:27
Got it, thanks for clarifying!
– CuriousKid7
Dec 5 at 16:27
add a comment |
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