What does the Jacobian matrix of the projection mapping for Normal bundle look like? (2.3.14 G&P)
I want to solve this question:
I feel like the previous question is similar to the one given in this link:
Natural projection of tangent bundle is submersion
Am I correct? but what does the Jacobian matrix look like in our situation here?
Thanks.
general-topology differential-geometry differential-topology transversality
asked Nov 26 at 14:41
Idonotknow
1347
add a comment |
I want to solve this question:
I feel like the previous question is similar to the one given in this link:
Natural projection of tangent bundle is submersion
Am I correct? but what does the Jacobian matrix look like in our situation here?
Thanks.
general-topology differential-geometry differential-topology transversality
asked Nov 26 at 14:41
Idonotknow
1347
add a comment |
I want to solve this question:
I feel like the previous question is similar to the one given in this link:
Natural projection of tangent bundle is submersion
Am I correct? but what does the Jacobian matrix look like in our situation here?
Thanks.
general-topology differential-geometry differential-topology transversality
asked Nov 26 at 14:41
Idonotknow
1347
I want to solve this question:
I feel like the previous question is similar to the one given in this link:
Natural projection of tangent bundle is submersion
Am I correct? but what does the Jacobian matrix look like in our situation here?
Thanks.
general-topology differential-geometry differential-topology transversality
general-topology differential-geometry differential-topology transversality
asked Nov 26 at 14:41
Idonotknow
1347
asked Nov 26 at 14:41
Idonotknow
1347
asked Nov 26 at 14:41
Idonotknow
1347
asked Nov 26 at 14:41
Idonotknow
1347
asked Nov 26 at 14:41
Idonotknow
1347
1347
add a comment |
add a comment |
1 Answer
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I don't recall all the notation from G&P so I'll try to just explain why the projection from the normal bundle of a manifold $M$ to the manifold is a submersion. Let's write this as $pi:NMto M$. A trivializing open neighborhood for $NMto M$ is an open neighborhood $Usubseteq M$ so that $pi^{-1}(U)cong Utimes mathbb{R}^k$. If $U$ is a trivializing open neighborhood for $NM$ with $U$ also a coordinate chart neighborhood on $M$ with coordinates $(x^1,ldots, x^m)$, then on $pi^{-1}(U)$ the bundle is parametrized by $(x^1,ldots, x^m, v^1,ldots, v^k)$.
If we think about the projection $pi^{-1}(U)to U$ and put it in coordinates, we are just forgetting about the last $k$ data above, so the map is locally $(x^1,ldots, x^m, v^1,ldots, v^k)to (x^1,ldots, x^m).$ The associated Jacobian matrix is of the form $J=[I_m|0_k]$ where $I_m$ is the $mtimes m$ identity matrix and $0_k$ is the $ktimes k$ zero matrix. Notice that $J$ is a surjective linear map, so that $pi: NMto M$ is a submersion.
answered Nov 26 at 15:34
Antonios-Alexandros Robotis
9,15041640
why $J$ is a surjective linear map?
– Idonotknow
Nov 27 at 16:39
Because its rank equals the dimension of the target space.
– Antonios-Alexandros Robotis
Nov 27 at 16:40
I understood you are saying that our map gives us the first coordinate only and the second is not included ..... this is why the jacobian is the identity for the first coordinates while zero for the last ..... is my understanding correct?
– Idonotknow
Nov 27 at 18:20
What about the question of what specifically is the preimage ?
– Intuition
Nov 27 at 18:36
what is the dimension of the normal bundle (our domain)?
– Idonotknow
Nov 28 at 19:51
|
show 2 more comments
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1 Answer
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I don't recall all the notation from G&P so I'll try to just explain why the projection from the normal bundle of a manifold $M$ to the manifold is a submersion. Let's write this as $pi:NMto M$. A trivializing open neighborhood for $NMto M$ is an open neighborhood $Usubseteq M$ so that $pi^{-1}(U)cong Utimes mathbb{R}^k$. If $U$ is a trivializing open neighborhood for $NM$ with $U$ also a coordinate chart neighborhood on $M$ with coordinates $(x^1,ldots, x^m)$, then on $pi^{-1}(U)$ the bundle is parametrized by $(x^1,ldots, x^m, v^1,ldots, v^k)$.
If we think about the projection $pi^{-1}(U)to U$ and put it in coordinates, we are just forgetting about the last $k$ data above, so the map is locally $(x^1,ldots, x^m, v^1,ldots, v^k)to (x^1,ldots, x^m).$ The associated Jacobian matrix is of the form $J=[I_m|0_k]$ where $I_m$ is the $mtimes m$ identity matrix and $0_k$ is the $ktimes k$ zero matrix. Notice that $J$ is a surjective linear map, so that $pi: NMto M$ is a submersion.
answered Nov 26 at 15:34
Antonios-Alexandros Robotis
9,15041640
why $J$ is a surjective linear map?
– Idonotknow
Nov 27 at 16:39
Because its rank equals the dimension of the target space.
– Antonios-Alexandros Robotis
Nov 27 at 16:40
I understood you are saying that our map gives us the first coordinate only and the second is not included ..... this is why the jacobian is the identity for the first coordinates while zero for the last ..... is my understanding correct?
– Idonotknow
Nov 27 at 18:20
What about the question of what specifically is the preimage ?
– Intuition
Nov 27 at 18:36
what is the dimension of the normal bundle (our domain)?
– Idonotknow
Nov 28 at 19:51
|
show 2 more comments
I don't recall all the notation from G&P so I'll try to just explain why the projection from the normal bundle of a manifold $M$ to the manifold is a submersion. Let's write this as $pi:NMto M$. A trivializing open neighborhood for $NMto M$ is an open neighborhood $Usubseteq M$ so that $pi^{-1}(U)cong Utimes mathbb{R}^k$. If $U$ is a trivializing open neighborhood for $NM$ with $U$ also a coordinate chart neighborhood on $M$ with coordinates $(x^1,ldots, x^m)$, then on $pi^{-1}(U)$ the bundle is parametrized by $(x^1,ldots, x^m, v^1,ldots, v^k)$.
If we think about the projection $pi^{-1}(U)to U$ and put it in coordinates, we are just forgetting about the last $k$ data above, so the map is locally $(x^1,ldots, x^m, v^1,ldots, v^k)to (x^1,ldots, x^m).$ The associated Jacobian matrix is of the form $J=[I_m|0_k]$ where $I_m$ is the $mtimes m$ identity matrix and $0_k$ is the $ktimes k$ zero matrix. Notice that $J$ is a surjective linear map, so that $pi: NMto M$ is a submersion.
answered Nov 26 at 15:34
Antonios-Alexandros Robotis
9,15041640
why $J$ is a surjective linear map?
– Idonotknow
Nov 27 at 16:39
Because its rank equals the dimension of the target space.
– Antonios-Alexandros Robotis
Nov 27 at 16:40
I understood you are saying that our map gives us the first coordinate only and the second is not included ..... this is why the jacobian is the identity for the first coordinates while zero for the last ..... is my understanding correct?
– Idonotknow
Nov 27 at 18:20
What about the question of what specifically is the preimage ?
– Intuition
Nov 27 at 18:36
what is the dimension of the normal bundle (our domain)?
– Idonotknow
Nov 28 at 19:51
|
show 2 more comments
I don't recall all the notation from G&P so I'll try to just explain why the projection from the normal bundle of a manifold $M$ to the manifold is a submersion. Let's write this as $pi:NMto M$. A trivializing open neighborhood for $NMto M$ is an open neighborhood $Usubseteq M$ so that $pi^{-1}(U)cong Utimes mathbb{R}^k$. If $U$ is a trivializing open neighborhood for $NM$ with $U$ also a coordinate chart neighborhood on $M$ with coordinates $(x^1,ldots, x^m)$, then on $pi^{-1}(U)$ the bundle is parametrized by $(x^1,ldots, x^m, v^1,ldots, v^k)$.
If we think about the projection $pi^{-1}(U)to U$ and put it in coordinates, we are just forgetting about the last $k$ data above, so the map is locally $(x^1,ldots, x^m, v^1,ldots, v^k)to (x^1,ldots, x^m).$ The associated Jacobian matrix is of the form $J=[I_m|0_k]$ where $I_m$ is the $mtimes m$ identity matrix and $0_k$ is the $ktimes k$ zero matrix. Notice that $J$ is a surjective linear map, so that $pi: NMto M$ is a submersion.
answered Nov 26 at 15:34
Antonios-Alexandros Robotis
9,15041640
I don't recall all the notation from G&P so I'll try to just explain why the projection from the normal bundle of a manifold $M$ to the manifold is a submersion. Let's write this as $pi:NMto M$. A trivializing open neighborhood for $NMto M$ is an open neighborhood $Usubseteq M$ so that $pi^{-1}(U)cong Utimes mathbb{R}^k$. If $U$ is a trivializing open neighborhood for $NM$ with $U$ also a coordinate chart neighborhood on $M$ with coordinates $(x^1,ldots, x^m)$, then on $pi^{-1}(U)$ the bundle is parametrized by $(x^1,ldots, x^m, v^1,ldots, v^k)$.
If we think about the projection $pi^{-1}(U)to U$ and put it in coordinates, we are just forgetting about the last $k$ data above, so the map is locally $(x^1,ldots, x^m, v^1,ldots, v^k)to (x^1,ldots, x^m).$ The associated Jacobian matrix is of the form $J=[I_m|0_k]$ where $I_m$ is the $mtimes m$ identity matrix and $0_k$ is the $ktimes k$ zero matrix. Notice that $J$ is a surjective linear map, so that $pi: NMto M$ is a submersion.
answered Nov 26 at 15:34
Antonios-Alexandros Robotis
9,15041640
answered Nov 26 at 15:34
Antonios-Alexandros Robotis
9,15041640
answered Nov 26 at 15:34
Antonios-Alexandros Robotis
9,15041640
answered Nov 26 at 15:34
Antonios-Alexandros Robotis
9,15041640
9,15041640
why $J$ is a surjective linear map?
– Idonotknow
Nov 27 at 16:39
Because its rank equals the dimension of the target space.
– Antonios-Alexandros Robotis
Nov 27 at 16:40
I understood you are saying that our map gives us the first coordinate only and the second is not included ..... this is why the jacobian is the identity for the first coordinates while zero for the last ..... is my understanding correct?
– Idonotknow
Nov 27 at 18:20
What about the question of what specifically is the preimage ?
– Intuition
Nov 27 at 18:36
what is the dimension of the normal bundle (our domain)?
– Idonotknow
Nov 28 at 19:51
|
show 2 more comments
why $J$ is a surjective linear map?
– Idonotknow
Nov 27 at 16:39
Because its rank equals the dimension of the target space.
– Antonios-Alexandros Robotis
Nov 27 at 16:40
I understood you are saying that our map gives us the first coordinate only and the second is not included ..... this is why the jacobian is the identity for the first coordinates while zero for the last ..... is my understanding correct?
– Idonotknow
Nov 27 at 18:20
What about the question of what specifically is the preimage ?
– Intuition
Nov 27 at 18:36
what is the dimension of the normal bundle (our domain)?
– Idonotknow
Nov 28 at 19:51
why $J$ is a surjective linear map?
– Idonotknow
Nov 27 at 16:39
why $J$ is a surjective linear map?
– Idonotknow
Nov 27 at 16:39
Because its rank equals the dimension of the target space.
– Antonios-Alexandros Robotis
Nov 27 at 16:40
Because its rank equals the dimension of the target space.
– Antonios-Alexandros Robotis
Nov 27 at 16:40
I understood you are saying that our map gives us the first coordinate only and the second is not included ..... this is why the jacobian is the identity for the first coordinates while zero for the last ..... is my understanding correct?
– Idonotknow
Nov 27 at 18:20
I understood you are saying that our map gives us the first coordinate only and the second is not included ..... this is why the jacobian is the identity for the first coordinates while zero for the last ..... is my understanding correct?
– Idonotknow
Nov 27 at 18:20
What about the question of what specifically is the preimage ?
– Intuition
Nov 27 at 18:36
What about the question of what specifically is the preimage ?
– Intuition
Nov 27 at 18:36
what is the dimension of the normal bundle (our domain)?
– Idonotknow
Nov 28 at 19:51
what is the dimension of the normal bundle (our domain)?
– Idonotknow
Nov 28 at 19:51
|
show 2 more comments
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