Most general diffeomorphisms of a sphere.
$begingroup$
Given a n dimensional vector $V$, such that $|V|=1$, how can one write a general diffeomorphism which preserves it's length as an orthogonal matrix $M(V)$ which acts on V?
diffeomorphism
asked Dec 7 '18 at 11:53
zoobyzooby
985616
$endgroup$
add a comment |
$begingroup$
Given a n dimensional vector $V$, such that $|V|=1$, how can one write a general diffeomorphism which preserves it's length as an orthogonal matrix $M(V)$ which acts on V?
diffeomorphism
asked Dec 7 '18 at 11:53
zoobyzooby
985616
$endgroup$
2
$begingroup$
Your question doesn't make sense. $V$ need not have an inner product to start with, and a general diffeomorphism need not be linear.
$endgroup$
– user10354138
Dec 7 '18 at 12:38
$begingroup$
Read the question
$endgroup$
– zooby
Dec 7 '18 at 20:10
add a comment |
$begingroup$
Given a n dimensional vector $V$, such that $|V|=1$, how can one write a general diffeomorphism which preserves it's length as an orthogonal matrix $M(V)$ which acts on V?
diffeomorphism
asked Dec 7 '18 at 11:53
zoobyzooby
985616
$endgroup$
Given a n dimensional vector $V$, such that $|V|=1$, how can one write a general diffeomorphism which preserves it's length as an orthogonal matrix $M(V)$ which acts on V?
diffeomorphism
diffeomorphism
asked Dec 7 '18 at 11:53
zoobyzooby
985616
asked Dec 7 '18 at 11:53
zoobyzooby
985616
asked Dec 7 '18 at 11:53
zoobyzooby
985616
asked Dec 7 '18 at 11:53
zoobyzooby
985616
asked Dec 7 '18 at 11:53
zoobyzooby
985616
985616
2
$begingroup$
Your question doesn't make sense. $V$ need not have an inner product to start with, and a general diffeomorphism need not be linear.
$endgroup$
– user10354138
Dec 7 '18 at 12:38
$begingroup$
Read the question
$endgroup$
– zooby
Dec 7 '18 at 20:10
add a comment |
2
$begingroup$
Your question doesn't make sense. $V$ need not have an inner product to start with, and a general diffeomorphism need not be linear.
$endgroup$
– user10354138
Dec 7 '18 at 12:38
$begingroup$
Read the question
$endgroup$
– zooby
Dec 7 '18 at 20:10
2
2
$begingroup$
Your question doesn't make sense. $V$ need not have an inner product to start with, and a general diffeomorphism need not be linear.
$endgroup$
– user10354138
Dec 7 '18 at 12:38
$begingroup$
Your question doesn't make sense. $V$ need not have an inner product to start with, and a general diffeomorphism need not be linear.
$endgroup$
– user10354138
Dec 7 '18 at 12:38
$begingroup$
Read the question
$endgroup$
– zooby
Dec 7 '18 at 20:10
$begingroup$
Read the question
$endgroup$
– zooby
Dec 7 '18 at 20:10
add a comment |
1 Answer
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$begingroup$
I don't know what answer you are after, but the orientation preserving diffeomorphisms of $mathbb{R}^3$ that preserve the unit two-sphere $mathbb{S}^2$ can be 'represented by' all time-dependent vector fields defined on the unit sphere $mathbb{S}^2$ (time independent vector fields are a special case of time dependent ones). Equivalently, you can think of time dependent vector fields on $mathbb{S}^2$ as the usual vector fields $frac{partial}{partial t} + X(t,x)$ on $(-epsilon, 1+epsilon) times mathbb{S}^2$. Have in mind that two different vector fields can produce the same diffeomorphism, but any orientation preserving diffeomorphism of the unit sphere can be the time-one phase flow of a (time dependent) vector field. Furthermore, any time dependent vector field on the unit sphere extends to a time dependent vector field of $mathbb{R}^3$ and its time-one flow is the extension of the corresponding unit sphere diffeomorphism to a diffeomorphism of $mathbb{R}^3$. The extension can be achieved by either vector field extension from the sphere to the three space, or by some smooth version of Alexander's trick. If you compose all these time-one flow diffeomorphisms of the unit sphere with the antipodal map on the sphere, then you get all diffeomorphisms. This is due to the fact that the mapping class group of the sphere is $mathbb{Z}_2$.
answered Dec 27 '18 at 5:54
FuturologistFuturologist
7,3062520
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1 Answer
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1 Answer
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$begingroup$
I don't know what answer you are after, but the orientation preserving diffeomorphisms of $mathbb{R}^3$ that preserve the unit two-sphere $mathbb{S}^2$ can be 'represented by' all time-dependent vector fields defined on the unit sphere $mathbb{S}^2$ (time independent vector fields are a special case of time dependent ones). Equivalently, you can think of time dependent vector fields on $mathbb{S}^2$ as the usual vector fields $frac{partial}{partial t} + X(t,x)$ on $(-epsilon, 1+epsilon) times mathbb{S}^2$. Have in mind that two different vector fields can produce the same diffeomorphism, but any orientation preserving diffeomorphism of the unit sphere can be the time-one phase flow of a (time dependent) vector field. Furthermore, any time dependent vector field on the unit sphere extends to a time dependent vector field of $mathbb{R}^3$ and its time-one flow is the extension of the corresponding unit sphere diffeomorphism to a diffeomorphism of $mathbb{R}^3$. The extension can be achieved by either vector field extension from the sphere to the three space, or by some smooth version of Alexander's trick. If you compose all these time-one flow diffeomorphisms of the unit sphere with the antipodal map on the sphere, then you get all diffeomorphisms. This is due to the fact that the mapping class group of the sphere is $mathbb{Z}_2$.
answered Dec 27 '18 at 5:54
FuturologistFuturologist
7,3062520
$endgroup$
add a comment |
$begingroup$
I don't know what answer you are after, but the orientation preserving diffeomorphisms of $mathbb{R}^3$ that preserve the unit two-sphere $mathbb{S}^2$ can be 'represented by' all time-dependent vector fields defined on the unit sphere $mathbb{S}^2$ (time independent vector fields are a special case of time dependent ones). Equivalently, you can think of time dependent vector fields on $mathbb{S}^2$ as the usual vector fields $frac{partial}{partial t} + X(t,x)$ on $(-epsilon, 1+epsilon) times mathbb{S}^2$. Have in mind that two different vector fields can produce the same diffeomorphism, but any orientation preserving diffeomorphism of the unit sphere can be the time-one phase flow of a (time dependent) vector field. Furthermore, any time dependent vector field on the unit sphere extends to a time dependent vector field of $mathbb{R}^3$ and its time-one flow is the extension of the corresponding unit sphere diffeomorphism to a diffeomorphism of $mathbb{R}^3$. The extension can be achieved by either vector field extension from the sphere to the three space, or by some smooth version of Alexander's trick. If you compose all these time-one flow diffeomorphisms of the unit sphere with the antipodal map on the sphere, then you get all diffeomorphisms. This is due to the fact that the mapping class group of the sphere is $mathbb{Z}_2$.
answered Dec 27 '18 at 5:54
FuturologistFuturologist
7,3062520
$endgroup$
add a comment |
$begingroup$
I don't know what answer you are after, but the orientation preserving diffeomorphisms of $mathbb{R}^3$ that preserve the unit two-sphere $mathbb{S}^2$ can be 'represented by' all time-dependent vector fields defined on the unit sphere $mathbb{S}^2$ (time independent vector fields are a special case of time dependent ones). Equivalently, you can think of time dependent vector fields on $mathbb{S}^2$ as the usual vector fields $frac{partial}{partial t} + X(t,x)$ on $(-epsilon, 1+epsilon) times mathbb{S}^2$. Have in mind that two different vector fields can produce the same diffeomorphism, but any orientation preserving diffeomorphism of the unit sphere can be the time-one phase flow of a (time dependent) vector field. Furthermore, any time dependent vector field on the unit sphere extends to a time dependent vector field of $mathbb{R}^3$ and its time-one flow is the extension of the corresponding unit sphere diffeomorphism to a diffeomorphism of $mathbb{R}^3$. The extension can be achieved by either vector field extension from the sphere to the three space, or by some smooth version of Alexander's trick. If you compose all these time-one flow diffeomorphisms of the unit sphere with the antipodal map on the sphere, then you get all diffeomorphisms. This is due to the fact that the mapping class group of the sphere is $mathbb{Z}_2$.
answered Dec 27 '18 at 5:54
FuturologistFuturologist
7,3062520
$endgroup$
I don't know what answer you are after, but the orientation preserving diffeomorphisms of $mathbb{R}^3$ that preserve the unit two-sphere $mathbb{S}^2$ can be 'represented by' all time-dependent vector fields defined on the unit sphere $mathbb{S}^2$ (time independent vector fields are a special case of time dependent ones). Equivalently, you can think of time dependent vector fields on $mathbb{S}^2$ as the usual vector fields $frac{partial}{partial t} + X(t,x)$ on $(-epsilon, 1+epsilon) times mathbb{S}^2$. Have in mind that two different vector fields can produce the same diffeomorphism, but any orientation preserving diffeomorphism of the unit sphere can be the time-one phase flow of a (time dependent) vector field. Furthermore, any time dependent vector field on the unit sphere extends to a time dependent vector field of $mathbb{R}^3$ and its time-one flow is the extension of the corresponding unit sphere diffeomorphism to a diffeomorphism of $mathbb{R}^3$. The extension can be achieved by either vector field extension from the sphere to the three space, or by some smooth version of Alexander's trick. If you compose all these time-one flow diffeomorphisms of the unit sphere with the antipodal map on the sphere, then you get all diffeomorphisms. This is due to the fact that the mapping class group of the sphere is $mathbb{Z}_2$.
answered Dec 27 '18 at 5:54
FuturologistFuturologist
7,3062520
answered Dec 27 '18 at 5:54
FuturologistFuturologist
7,3062520
answered Dec 27 '18 at 5:54
FuturologistFuturologist
7,3062520
answered Dec 27 '18 at 5:54
FuturologistFuturologist
7,3062520
7,3062520
add a comment |
add a comment |
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2
$begingroup$
Your question doesn't make sense. $V$ need not have an inner product to start with, and a general diffeomorphism need not be linear.
$endgroup$
– user10354138
Dec 7 '18 at 12:38
$begingroup$
Read the question
$endgroup$
– zooby
Dec 7 '18 at 20:10