Mixing
Understanding k- jets
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K-jets can be realized as space of (up to equivalence classes) of Taylor polynomials of order k. There is a statement in V. Arnold's lectures in PDE stating that 1-jets of functions in the plane can be viewed a 5-dimensional space. Why is that?
1) Can someone clarify the concept?
2)How are jets defined for Riemannian manifolds?
differential-geometry pde
asked Jan 25 at 3:42
BigMBigM
2,56911530
$endgroup$
add a comment |
$begingroup$
K-jets can be realized as space of (up to equivalence classes) of Taylor polynomials of order k. There is a statement in V. Arnold's lectures in PDE stating that 1-jets of functions in the plane can be viewed a 5-dimensional space. Why is that?
1) Can someone clarify the concept?
2)How are jets defined for Riemannian manifolds?
differential-geometry pde
asked Jan 25 at 3:42
BigMBigM
2,56911530
$endgroup$
add a comment |
$begingroup$
K-jets can be realized as space of (up to equivalence classes) of Taylor polynomials of order k. There is a statement in V. Arnold's lectures in PDE stating that 1-jets of functions in the plane can be viewed a 5-dimensional space. Why is that?
1) Can someone clarify the concept?
2)How are jets defined for Riemannian manifolds?
differential-geometry pde
asked Jan 25 at 3:42
BigMBigM
2,56911530
$endgroup$
K-jets can be realized as space of (up to equivalence classes) of Taylor polynomials of order k. There is a statement in V. Arnold's lectures in PDE stating that 1-jets of functions in the plane can be viewed a 5-dimensional space. Why is that?
1) Can someone clarify the concept?
2)How are jets defined for Riemannian manifolds?
differential-geometry pde
differential-geometry pde
asked Jan 25 at 3:42
BigMBigM
2,56911530
asked Jan 25 at 3:42
BigMBigM
2,56911530
asked Jan 25 at 3:42
BigMBigM
2,56911530
asked Jan 25 at 3:42
BigMBigM
2,56911530
asked Jan 25 at 3:42
BigMBigM
2,56911530
2,56911530
add a comment |
add a comment |
1 Answer
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At each point in the plane, a $1$-jet is determined by three numbers, since
a Taylor series at point $(x_0,y_0)$ up to the linear terms has the form $a+b(x-x_0)+c(y-y_0)$. As a point then, the $1$-jets form a three-dimensional vector space. Then the $1$-jets in the plane are a rank three vector bundle over the
plane, so the total space has dimension $3+2=5$.
All this works just as well over a smooth manifold. Taking a coordinate
patch one can define jets as Taylor expansions in the local coordinates there.
Then one checks that these behave well when we transition between coordinate
patches. It's the same sort of argument that's used to show that the tangent
bundle of a manifold makes sense.
answered Jan 25 at 3:58
Lord Shark the UnknownLord Shark the Unknown
106k1161133
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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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$begingroup$
At each point in the plane, a $1$-jet is determined by three numbers, since
a Taylor series at point $(x_0,y_0)$ up to the linear terms has the form $a+b(x-x_0)+c(y-y_0)$. As a point then, the $1$-jets form a three-dimensional vector space. Then the $1$-jets in the plane are a rank three vector bundle over the
plane, so the total space has dimension $3+2=5$.
All this works just as well over a smooth manifold. Taking a coordinate
patch one can define jets as Taylor expansions in the local coordinates there.
Then one checks that these behave well when we transition between coordinate
patches. It's the same sort of argument that's used to show that the tangent
bundle of a manifold makes sense.
answered Jan 25 at 3:58
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
add a comment |
$begingroup$
At each point in the plane, a $1$-jet is determined by three numbers, since
a Taylor series at point $(x_0,y_0)$ up to the linear terms has the form $a+b(x-x_0)+c(y-y_0)$. As a point then, the $1$-jets form a three-dimensional vector space. Then the $1$-jets in the plane are a rank three vector bundle over the
plane, so the total space has dimension $3+2=5$.
All this works just as well over a smooth manifold. Taking a coordinate
patch one can define jets as Taylor expansions in the local coordinates there.
Then one checks that these behave well when we transition between coordinate
patches. It's the same sort of argument that's used to show that the tangent
bundle of a manifold makes sense.
answered Jan 25 at 3:58
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
add a comment |
$begingroup$
At each point in the plane, a $1$-jet is determined by three numbers, since
a Taylor series at point $(x_0,y_0)$ up to the linear terms has the form $a+b(x-x_0)+c(y-y_0)$. As a point then, the $1$-jets form a three-dimensional vector space. Then the $1$-jets in the plane are a rank three vector bundle over the
plane, so the total space has dimension $3+2=5$.
All this works just as well over a smooth manifold. Taking a coordinate
patch one can define jets as Taylor expansions in the local coordinates there.
Then one checks that these behave well when we transition between coordinate
patches. It's the same sort of argument that's used to show that the tangent
bundle of a manifold makes sense.
answered Jan 25 at 3:58
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
At each point in the plane, a $1$-jet is determined by three numbers, since
a Taylor series at point $(x_0,y_0)$ up to the linear terms has the form $a+b(x-x_0)+c(y-y_0)$. As a point then, the $1$-jets form a three-dimensional vector space. Then the $1$-jets in the plane are a rank three vector bundle over the
plane, so the total space has dimension $3+2=5$.
All this works just as well over a smooth manifold. Taking a coordinate
patch one can define jets as Taylor expansions in the local coordinates there.
Then one checks that these behave well when we transition between coordinate
patches. It's the same sort of argument that's used to show that the tangent
bundle of a manifold makes sense.
answered Jan 25 at 3:58
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 25 at 3:58
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 25 at 3:58
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Jan 25 at 3:58
Lord Shark the UnknownLord Shark the Unknown
106k1161133
106k1161133
add a comment |
add a comment |
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