Is the orthogonal distribution of a Killing vector field always involutive?
$begingroup$
Let $M, g$ be a semi-Riemannian manifold and $K$ a Killing vector field on $M$. Let $D = {X in Gamma(TM) | g(X, K) = 0}$ be the orthogonal distribution to $K$. Does $D$ have to be involutive, i. e. $forall X, Y in D: [X, Y] in D$ (or equivalently by Frobenius' theorem $D$ is integrable).
I believe this not true in general but haven't been able to come up with a counterexample. (apparently a counterexample would need at least 4 dimensions since in dimension $leq 3$ you can always find locally orthogonal coordinates.)
differential-geometry riemannian-geometry vector-fields general-relativity
asked Jan 10 at 20:59
0x5390x539
1,450518
$endgroup$
add a comment |
$begingroup$
Let $M, g$ be a semi-Riemannian manifold and $K$ a Killing vector field on $M$. Let $D = {X in Gamma(TM) | g(X, K) = 0}$ be the orthogonal distribution to $K$. Does $D$ have to be involutive, i. e. $forall X, Y in D: [X, Y] in D$ (or equivalently by Frobenius' theorem $D$ is integrable).
I believe this not true in general but haven't been able to come up with a counterexample. (apparently a counterexample would need at least 4 dimensions since in dimension $leq 3$ you can always find locally orthogonal coordinates.)
differential-geometry riemannian-geometry vector-fields general-relativity
asked Jan 10 at 20:59
0x5390x539
1,450518
$endgroup$
$begingroup$
en.wikipedia.org/wiki/Sasakian_manifold
$endgroup$
– Travis
Jan 10 at 22:19
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@Travis could you elaborate on why this is relevant here?
$endgroup$
– 0x539
Jan 10 at 22:28
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Sure: One can define a Sasaki structure to be a triple $(M, g, k)$ where $g$ is a metric and $k$ is a Killing field satisfying a particular second-order equation (this second-order equation is exactly what makes the metric cone over M a Kahler manifold). It follows from the definition that for any Sasaki structure ${k}^{perp}$ is a contact structure and hence nonintegrable. Thus, every Sasaki structure is a counterexample.
$endgroup$
– Travis
Jan 10 at 22:46
$begingroup$
By the way, my answer below, which gives a particular Killing field on $S^3$, is perhaps the simplest example of a Sasaki manifold. In that case, the metric cone can be identified with $Bbb C^2 - { 0 }$ equipped with its usual Kahler structure.
$endgroup$
– Travis
Jan 10 at 22:47
add a comment |
$begingroup$
Let $M, g$ be a semi-Riemannian manifold and $K$ a Killing vector field on $M$. Let $D = {X in Gamma(TM) | g(X, K) = 0}$ be the orthogonal distribution to $K$. Does $D$ have to be involutive, i. e. $forall X, Y in D: [X, Y] in D$ (or equivalently by Frobenius' theorem $D$ is integrable).
I believe this not true in general but haven't been able to come up with a counterexample. (apparently a counterexample would need at least 4 dimensions since in dimension $leq 3$ you can always find locally orthogonal coordinates.)
differential-geometry riemannian-geometry vector-fields general-relativity
asked Jan 10 at 20:59
0x5390x539
1,450518
$endgroup$
Let $M, g$ be a semi-Riemannian manifold and $K$ a Killing vector field on $M$. Let $D = {X in Gamma(TM) | g(X, K) = 0}$ be the orthogonal distribution to $K$. Does $D$ have to be involutive, i. e. $forall X, Y in D: [X, Y] in D$ (or equivalently by Frobenius' theorem $D$ is integrable).
I believe this not true in general but haven't been able to come up with a counterexample. (apparently a counterexample would need at least 4 dimensions since in dimension $leq 3$ you can always find locally orthogonal coordinates.)
differential-geometry riemannian-geometry vector-fields general-relativity
differential-geometry riemannian-geometry vector-fields general-relativity
asked Jan 10 at 20:59
0x5390x539
1,450518
asked Jan 10 at 20:59
0x5390x539
1,450518
edited Jan 10 at 22:20
0x539
asked Jan 10 at 20:59
0x5390x539
1,450518
asked Jan 10 at 20:59
0x5390x539
1,450518
asked Jan 10 at 20:59
0x5390x539
1,450518
1,450518
$begingroup$
en.wikipedia.org/wiki/Sasakian_manifold
$endgroup$
– Travis
Jan 10 at 22:19
$begingroup$
@Travis could you elaborate on why this is relevant here?
$endgroup$
– 0x539
Jan 10 at 22:28
$begingroup$
Sure: One can define a Sasaki structure to be a triple $(M, g, k)$ where $g$ is a metric and $k$ is a Killing field satisfying a particular second-order equation (this second-order equation is exactly what makes the metric cone over M a Kahler manifold). It follows from the definition that for any Sasaki structure ${k}^{perp}$ is a contact structure and hence nonintegrable. Thus, every Sasaki structure is a counterexample.
$endgroup$
– Travis
Jan 10 at 22:46
$begingroup$
By the way, my answer below, which gives a particular Killing field on $S^3$, is perhaps the simplest example of a Sasaki manifold. In that case, the metric cone can be identified with $Bbb C^2 - { 0 }$ equipped with its usual Kahler structure.
$endgroup$
– Travis
Jan 10 at 22:47
add a comment |
$begingroup$
en.wikipedia.org/wiki/Sasakian_manifold
$endgroup$
– Travis
Jan 10 at 22:19
$begingroup$
@Travis could you elaborate on why this is relevant here?
$endgroup$
– 0x539
Jan 10 at 22:28
$begingroup$
Sure: One can define a Sasaki structure to be a triple $(M, g, k)$ where $g$ is a metric and $k$ is a Killing field satisfying a particular second-order equation (this second-order equation is exactly what makes the metric cone over M a Kahler manifold). It follows from the definition that for any Sasaki structure ${k}^{perp}$ is a contact structure and hence nonintegrable. Thus, every Sasaki structure is a counterexample.
$endgroup$
– Travis
Jan 10 at 22:46
$begingroup$
By the way, my answer below, which gives a particular Killing field on $S^3$, is perhaps the simplest example of a Sasaki manifold. In that case, the metric cone can be identified with $Bbb C^2 - { 0 }$ equipped with its usual Kahler structure.
$endgroup$
– Travis
Jan 10 at 22:47
$begingroup$
en.wikipedia.org/wiki/Sasakian_manifold
$endgroup$
– Travis
Jan 10 at 22:19
$begingroup$
en.wikipedia.org/wiki/Sasakian_manifold
$endgroup$
– Travis
Jan 10 at 22:19
$begingroup$
@Travis could you elaborate on why this is relevant here?
$endgroup$
– 0x539
Jan 10 at 22:28
$begingroup$
@Travis could you elaborate on why this is relevant here?
$endgroup$
– 0x539
Jan 10 at 22:28
$begingroup$
Sure: One can define a Sasaki structure to be a triple $(M, g, k)$ where $g$ is a metric and $k$ is a Killing field satisfying a particular second-order equation (this second-order equation is exactly what makes the metric cone over M a Kahler manifold). It follows from the definition that for any Sasaki structure ${k}^{perp}$ is a contact structure and hence nonintegrable. Thus, every Sasaki structure is a counterexample.
$endgroup$
– Travis
Jan 10 at 22:46
$begingroup$
Sure: One can define a Sasaki structure to be a triple $(M, g, k)$ where $g$ is a metric and $k$ is a Killing field satisfying a particular second-order equation (this second-order equation is exactly what makes the metric cone over M a Kahler manifold). It follows from the definition that for any Sasaki structure ${k}^{perp}$ is a contact structure and hence nonintegrable. Thus, every Sasaki structure is a counterexample.
$endgroup$
– Travis
Jan 10 at 22:46
$begingroup$
By the way, my answer below, which gives a particular Killing field on $S^3$, is perhaps the simplest example of a Sasaki manifold. In that case, the metric cone can be identified with $Bbb C^2 - { 0 }$ equipped with its usual Kahler structure.
$endgroup$
– Travis
Jan 10 at 22:47
$begingroup$
By the way, my answer below, which gives a particular Killing field on $S^3$, is perhaps the simplest example of a Sasaki manifold. In that case, the metric cone can be identified with $Bbb C^2 - { 0 }$ equipped with its usual Kahler structure.
$endgroup$
– Travis
Jan 10 at 22:47
add a comment |
2 Answers
2
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In $mathbb R^3$, $K=partial_z + xpartial_y - ypartial_x$, $D$ is the kernel of $alpha=g(K,cdot)=dz + x,dy-y,dx$, and we have $alphawedge,dalpha=2dzwedge dxwedge dyneq0$, so $D$ is not involutive.
answered Jan 10 at 21:16
user8268user8268
17k12947
$endgroup$
$begingroup$
is it correct that $K$ has no $partial_x$-component?
$endgroup$
– 0x539
Jan 10 at 21:19
$begingroup$
@0x539 oops, a typo
$endgroup$
– user8268
Jan 10 at 21:20
add a comment |
$begingroup$
On $S^3 subset Bbb C^2$, the vector field $X$ defined by $$X_{(z, w)} = i(z, w) = (iz, iw)$$ is a Killing field for the standard metric, but its orthogonal distribution $${X}^{perp} = operatorname{span}{(w, -z), (iw, -iz)}$$ is the standard contact distribution on $S^3$ and in particular is nonintegrable.
answered Jan 10 at 21:48
TravisTravis
64.1k769151
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In $mathbb R^3$, $K=partial_z + xpartial_y - ypartial_x$, $D$ is the kernel of $alpha=g(K,cdot)=dz + x,dy-y,dx$, and we have $alphawedge,dalpha=2dzwedge dxwedge dyneq0$, so $D$ is not involutive.
answered Jan 10 at 21:16
user8268user8268
17k12947
$endgroup$
$begingroup$
is it correct that $K$ has no $partial_x$-component?
$endgroup$
– 0x539
Jan 10 at 21:19
$begingroup$
@0x539 oops, a typo
$endgroup$
– user8268
Jan 10 at 21:20
add a comment |
$begingroup$
In $mathbb R^3$, $K=partial_z + xpartial_y - ypartial_x$, $D$ is the kernel of $alpha=g(K,cdot)=dz + x,dy-y,dx$, and we have $alphawedge,dalpha=2dzwedge dxwedge dyneq0$, so $D$ is not involutive.
answered Jan 10 at 21:16
user8268user8268
17k12947
$endgroup$
$begingroup$
is it correct that $K$ has no $partial_x$-component?
$endgroup$
– 0x539
Jan 10 at 21:19
$begingroup$
@0x539 oops, a typo
$endgroup$
– user8268
Jan 10 at 21:20
add a comment |
$begingroup$
In $mathbb R^3$, $K=partial_z + xpartial_y - ypartial_x$, $D$ is the kernel of $alpha=g(K,cdot)=dz + x,dy-y,dx$, and we have $alphawedge,dalpha=2dzwedge dxwedge dyneq0$, so $D$ is not involutive.
answered Jan 10 at 21:16
user8268user8268
17k12947
$endgroup$
In $mathbb R^3$, $K=partial_z + xpartial_y - ypartial_x$, $D$ is the kernel of $alpha=g(K,cdot)=dz + x,dy-y,dx$, and we have $alphawedge,dalpha=2dzwedge dxwedge dyneq0$, so $D$ is not involutive.
answered Jan 10 at 21:16
user8268user8268
17k12947
edited Jan 10 at 21:25
answered Jan 10 at 21:16
user8268user8268
17k12947
answered Jan 10 at 21:16
user8268user8268
17k12947
answered Jan 10 at 21:16
user8268user8268
17k12947
17k12947
$begingroup$
is it correct that $K$ has no $partial_x$-component?
$endgroup$
– 0x539
Jan 10 at 21:19
$begingroup$
@0x539 oops, a typo
$endgroup$
– user8268
Jan 10 at 21:20
add a comment |
$begingroup$
is it correct that $K$ has no $partial_x$-component?
$endgroup$
– 0x539
Jan 10 at 21:19
$begingroup$
@0x539 oops, a typo
$endgroup$
– user8268
Jan 10 at 21:20
$begingroup$
is it correct that $K$ has no $partial_x$-component?
$endgroup$
– 0x539
Jan 10 at 21:19
$begingroup$
is it correct that $K$ has no $partial_x$-component?
$endgroup$
– 0x539
Jan 10 at 21:19
$begingroup$
@0x539 oops, a typo
$endgroup$
– user8268
Jan 10 at 21:20
$begingroup$
@0x539 oops, a typo
$endgroup$
– user8268
Jan 10 at 21:20
add a comment |
$begingroup$
On $S^3 subset Bbb C^2$, the vector field $X$ defined by $$X_{(z, w)} = i(z, w) = (iz, iw)$$ is a Killing field for the standard metric, but its orthogonal distribution $${X}^{perp} = operatorname{span}{(w, -z), (iw, -iz)}$$ is the standard contact distribution on $S^3$ and in particular is nonintegrable.
answered Jan 10 at 21:48
TravisTravis
64.1k769151
$endgroup$
add a comment |
$begingroup$
On $S^3 subset Bbb C^2$, the vector field $X$ defined by $$X_{(z, w)} = i(z, w) = (iz, iw)$$ is a Killing field for the standard metric, but its orthogonal distribution $${X}^{perp} = operatorname{span}{(w, -z), (iw, -iz)}$$ is the standard contact distribution on $S^3$ and in particular is nonintegrable.
answered Jan 10 at 21:48
TravisTravis
64.1k769151
$endgroup$
add a comment |
$begingroup$
On $S^3 subset Bbb C^2$, the vector field $X$ defined by $$X_{(z, w)} = i(z, w) = (iz, iw)$$ is a Killing field for the standard metric, but its orthogonal distribution $${X}^{perp} = operatorname{span}{(w, -z), (iw, -iz)}$$ is the standard contact distribution on $S^3$ and in particular is nonintegrable.
answered Jan 10 at 21:48
TravisTravis
64.1k769151
$endgroup$
On $S^3 subset Bbb C^2$, the vector field $X$ defined by $$X_{(z, w)} = i(z, w) = (iz, iw)$$ is a Killing field for the standard metric, but its orthogonal distribution $${X}^{perp} = operatorname{span}{(w, -z), (iw, -iz)}$$ is the standard contact distribution on $S^3$ and in particular is nonintegrable.
answered Jan 10 at 21:48
TravisTravis
64.1k769151
edited Jan 10 at 22:39
answered Jan 10 at 21:48
TravisTravis
64.1k769151
answered Jan 10 at 21:48
TravisTravis
64.1k769151
answered Jan 10 at 21:48
TravisTravis
64.1k769151
64.1k769151
add a comment |
add a comment |
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$begingroup$
en.wikipedia.org/wiki/Sasakian_manifold
$endgroup$
– Travis
Jan 10 at 22:19
$begingroup$
@Travis could you elaborate on why this is relevant here?
$endgroup$
– 0x539
Jan 10 at 22:28
$begingroup$
Sure: One can define a Sasaki structure to be a triple $(M, g, k)$ where $g$ is a metric and $k$ is a Killing field satisfying a particular second-order equation (this second-order equation is exactly what makes the metric cone over M a Kahler manifold). It follows from the definition that for any Sasaki structure ${k}^{perp}$ is a contact structure and hence nonintegrable. Thus, every Sasaki structure is a counterexample.
$endgroup$
– Travis
Jan 10 at 22:46
$begingroup$
By the way, my answer below, which gives a particular Killing field on $S^3$, is perhaps the simplest example of a Sasaki manifold. In that case, the metric cone can be identified with $Bbb C^2 - { 0 }$ equipped with its usual Kahler structure.
$endgroup$
– Travis
Jan 10 at 22:47