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What is the range of the Levi-Civita connection?
Clash Royale CLAN TAG#URR8PPP
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Let $E$ a distribution in $TQ$, $Q$ a riemannian manifold. Let $X,Y$ vectorfields
tangent to $E$.
Question: is it true that $nabla_X Y in E + [E,E]$.
If not give a counterexample.
I would be content to know true or false for X=Y. This is motivated by
a short paper by Cartan n nonholonomic mechanics at the ICM 1928. The arguments
he uses seem to imply this conclusion, although in a roundabout way. Cartan shows that one can change the metric at will outside E+[E,E] without changing the NH geodesics.
By looking at Cartan's arguments, it seems that the result holds for any torsion free connection in TQ, no need to be compatible with a metric.
One chooses a frame with 3 types of vectorfields,
i) first ones tangent to E, ii) second group forming a complement of E in
E + [E,E] , iii) complete to TQ . One considers the dual coframe and writes the structure equations. The third group in the coframe is the derived ideal of
the ideal of forms that annihilate E.
I believe that writing the covariant derivative nabla_X Y for X,Y in E using the structure forms, one gets the desired result.
differential-geometry
asked Jul 15 at 11:44
Jair Koiller
162
add a comment |Â
up vote
2
down vote
favorite
Let $E$ a distribution in $TQ$, $Q$ a riemannian manifold. Let $X,Y$ vectorfields
tangent to $E$.
Question: is it true that $nabla_X Y in E + [E,E]$.
If not give a counterexample.
I would be content to know true or false for X=Y. This is motivated by
a short paper by Cartan n nonholonomic mechanics at the ICM 1928. The arguments
he uses seem to imply this conclusion, although in a roundabout way. Cartan shows that one can change the metric at will outside E+[E,E] without changing the NH geodesics.
By looking at Cartan's arguments, it seems that the result holds for any torsion free connection in TQ, no need to be compatible with a metric.
One chooses a frame with 3 types of vectorfields,
i) first ones tangent to E, ii) second group forming a complement of E in
E + [E,E] , iii) complete to TQ . One considers the dual coframe and writes the structure equations. The third group in the coframe is the derived ideal of
the ideal of forms that annihilate E.
I believe that writing the covariant derivative nabla_X Y for X,Y in E using the structure forms, one gets the desired result.
differential-geometry
asked Jul 15 at 11:44
Jair Koiller
162
This would extend the well know result for integrable distributions.
– Jair Koiller
Jul 15 at 11:45
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $E$ a distribution in $TQ$, $Q$ a riemannian manifold. Let $X,Y$ vectorfields
tangent to $E$.
Question: is it true that $nabla_X Y in E + [E,E]$.
If not give a counterexample.
I would be content to know true or false for X=Y. This is motivated by
a short paper by Cartan n nonholonomic mechanics at the ICM 1928. The arguments
he uses seem to imply this conclusion, although in a roundabout way. Cartan shows that one can change the metric at will outside E+[E,E] without changing the NH geodesics.
By looking at Cartan's arguments, it seems that the result holds for any torsion free connection in TQ, no need to be compatible with a metric.
One chooses a frame with 3 types of vectorfields,
i) first ones tangent to E, ii) second group forming a complement of E in
E + [E,E] , iii) complete to TQ . One considers the dual coframe and writes the structure equations. The third group in the coframe is the derived ideal of
the ideal of forms that annihilate E.
I believe that writing the covariant derivative nabla_X Y for X,Y in E using the structure forms, one gets the desired result.
differential-geometry
asked Jul 15 at 11:44
Jair Koiller
162
Let $E$ a distribution in $TQ$, $Q$ a riemannian manifold. Let $X,Y$ vectorfields
tangent to $E$.
Question: is it true that $nabla_X Y in E + [E,E]$.
If not give a counterexample.
I would be content to know true or false for X=Y. This is motivated by
a short paper by Cartan n nonholonomic mechanics at the ICM 1928. The arguments
he uses seem to imply this conclusion, although in a roundabout way. Cartan shows that one can change the metric at will outside E+[E,E] without changing the NH geodesics.
By looking at Cartan's arguments, it seems that the result holds for any torsion free connection in TQ, no need to be compatible with a metric.
One chooses a frame with 3 types of vectorfields,
i) first ones tangent to E, ii) second group forming a complement of E in
E + [E,E] , iii) complete to TQ . One considers the dual coframe and writes the structure equations. The third group in the coframe is the derived ideal of
the ideal of forms that annihilate E.
I believe that writing the covariant derivative nabla_X Y for X,Y in E using the structure forms, one gets the desired result.
differential-geometry
asked Jul 15 at 11:44
Jair Koiller
162
edited Jul 17 at 16:22
asked Jul 15 at 11:44
Jair Koiller
162
asked Jul 15 at 11:44
Jair Koiller
162
asked Jul 15 at 11:44
Jair Koiller
162
162
This would extend the well know result for integrable distributions.
– Jair Koiller
Jul 15 at 11:45
add a comment |Â
This would extend the well know result for integrable distributions.
– Jair Koiller
Jul 15 at 11:45
This would extend the well know result for integrable distributions.
– Jair Koiller
Jul 15 at 11:45
This would extend the well know result for integrable distributions.
– Jair Koiller
Jul 15 at 11:45
add a comment |Â
1 Answer
1
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0
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Let $Q = SU(2)$, and consider the diagonal maximal torus $U(1)$ as acting freely and isometrically on $SU(2)$ by multiplication from the right, which therefore gives rise to the Hopf fibration $SU(2) to SU(2)/U(1) = mathbbCP^1$. Let $E = VSU(2)$ be the vertical tangent bundle of this submersion, which is integrable, so that $E + [E,E] = E$. Finally, let $g_0$ be the $operatornameAd$-invariant Riemannian metric on $SU(2)$ induced by the Killing form, let $ell in C^infty(SU(2),(0,+infty))^U(1)$ be non-constant, and let $g = ell g_0$ be the resulting $U(1)$-invariant Riemannian metric on $Q = SU(2)$, so that $SU(2) to mathbbCP^1$ becomes a Riemannian submersion. Then for any unit vector $X in mathfraksu(2)$ with respect to the Killing form, if $X_P$ is the corresponding left-invariant vector field on $SU(2)$, then the orthogonal projection of $nabla_X_PX_P$ onto $E^perp = (E+[E,E])^perp$ is precisely the mean curvature vector field
$$
H = -frac12operatornamegradell neq 0
$$
of the Riemannian submersion $SU(2) to mathbbCP^1$. More generally, you'll run into trouble if $E$ is the tangent bundle to a foliation of $Q$ by some locally free isometric action of a connected Lie group, in which case, the obstruction to your claim is the orbit-wise second fundamental form (i.e., O'Neill's $T$-tensor for the Riemannian foliation $E$).
answered Jul 17 at 18:07
Branimir Ćaćić
9,69221846
Many thanks Branimir, will think about your example.
– Jair Koiller
Jul 17 at 18:13
1
For Branimir's example, any foliation works, with E the associated integrable distribution. Jair's condition on $nabla$ would imply that the leaves of E are totally geodesic. But typically, they are not. To get perhaps the simplest example possible, take a surface of revolution, with leaves being the circles -the orbits of rotation. The metric can be written $dr^2 + f(r)^2 d theta^2$. The leaves are the level sets of $r$. Only the circles corresponding to critical points of f are (totally) geodesic.
– Richard Montgomery
Jul 18 at 16:14
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $Q = SU(2)$, and consider the diagonal maximal torus $U(1)$ as acting freely and isometrically on $SU(2)$ by multiplication from the right, which therefore gives rise to the Hopf fibration $SU(2) to SU(2)/U(1) = mathbbCP^1$. Let $E = VSU(2)$ be the vertical tangent bundle of this submersion, which is integrable, so that $E + [E,E] = E$. Finally, let $g_0$ be the $operatornameAd$-invariant Riemannian metric on $SU(2)$ induced by the Killing form, let $ell in C^infty(SU(2),(0,+infty))^U(1)$ be non-constant, and let $g = ell g_0$ be the resulting $U(1)$-invariant Riemannian metric on $Q = SU(2)$, so that $SU(2) to mathbbCP^1$ becomes a Riemannian submersion. Then for any unit vector $X in mathfraksu(2)$ with respect to the Killing form, if $X_P$ is the corresponding left-invariant vector field on $SU(2)$, then the orthogonal projection of $nabla_X_PX_P$ onto $E^perp = (E+[E,E])^perp$ is precisely the mean curvature vector field
$$
H = -frac12operatornamegradell neq 0
$$
of the Riemannian submersion $SU(2) to mathbbCP^1$. More generally, you'll run into trouble if $E$ is the tangent bundle to a foliation of $Q$ by some locally free isometric action of a connected Lie group, in which case, the obstruction to your claim is the orbit-wise second fundamental form (i.e., O'Neill's $T$-tensor for the Riemannian foliation $E$).
answered Jul 17 at 18:07
Branimir Ćaćić
9,69221846
Many thanks Branimir, will think about your example.
– Jair Koiller
Jul 17 at 18:13
1
For Branimir's example, any foliation works, with E the associated integrable distribution. Jair's condition on $nabla$ would imply that the leaves of E are totally geodesic. But typically, they are not. To get perhaps the simplest example possible, take a surface of revolution, with leaves being the circles -the orbits of rotation. The metric can be written $dr^2 + f(r)^2 d theta^2$. The leaves are the level sets of $r$. Only the circles corresponding to critical points of f are (totally) geodesic.
– Richard Montgomery
Jul 18 at 16:14
add a comment |Â
up vote
0
down vote
Let $Q = SU(2)$, and consider the diagonal maximal torus $U(1)$ as acting freely and isometrically on $SU(2)$ by multiplication from the right, which therefore gives rise to the Hopf fibration $SU(2) to SU(2)/U(1) = mathbbCP^1$. Let $E = VSU(2)$ be the vertical tangent bundle of this submersion, which is integrable, so that $E + [E,E] = E$. Finally, let $g_0$ be the $operatornameAd$-invariant Riemannian metric on $SU(2)$ induced by the Killing form, let $ell in C^infty(SU(2),(0,+infty))^U(1)$ be non-constant, and let $g = ell g_0$ be the resulting $U(1)$-invariant Riemannian metric on $Q = SU(2)$, so that $SU(2) to mathbbCP^1$ becomes a Riemannian submersion. Then for any unit vector $X in mathfraksu(2)$ with respect to the Killing form, if $X_P$ is the corresponding left-invariant vector field on $SU(2)$, then the orthogonal projection of $nabla_X_PX_P$ onto $E^perp = (E+[E,E])^perp$ is precisely the mean curvature vector field
$$
H = -frac12operatornamegradell neq 0
$$
of the Riemannian submersion $SU(2) to mathbbCP^1$. More generally, you'll run into trouble if $E$ is the tangent bundle to a foliation of $Q$ by some locally free isometric action of a connected Lie group, in which case, the obstruction to your claim is the orbit-wise second fundamental form (i.e., O'Neill's $T$-tensor for the Riemannian foliation $E$).
answered Jul 17 at 18:07
Branimir Ćaćić
9,69221846
Many thanks Branimir, will think about your example.
– Jair Koiller
Jul 17 at 18:13
1
For Branimir's example, any foliation works, with E the associated integrable distribution. Jair's condition on $nabla$ would imply that the leaves of E are totally geodesic. But typically, they are not. To get perhaps the simplest example possible, take a surface of revolution, with leaves being the circles -the orbits of rotation. The metric can be written $dr^2 + f(r)^2 d theta^2$. The leaves are the level sets of $r$. Only the circles corresponding to critical points of f are (totally) geodesic.
– Richard Montgomery
Jul 18 at 16:14
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $Q = SU(2)$, and consider the diagonal maximal torus $U(1)$ as acting freely and isometrically on $SU(2)$ by multiplication from the right, which therefore gives rise to the Hopf fibration $SU(2) to SU(2)/U(1) = mathbbCP^1$. Let $E = VSU(2)$ be the vertical tangent bundle of this submersion, which is integrable, so that $E + [E,E] = E$. Finally, let $g_0$ be the $operatornameAd$-invariant Riemannian metric on $SU(2)$ induced by the Killing form, let $ell in C^infty(SU(2),(0,+infty))^U(1)$ be non-constant, and let $g = ell g_0$ be the resulting $U(1)$-invariant Riemannian metric on $Q = SU(2)$, so that $SU(2) to mathbbCP^1$ becomes a Riemannian submersion. Then for any unit vector $X in mathfraksu(2)$ with respect to the Killing form, if $X_P$ is the corresponding left-invariant vector field on $SU(2)$, then the orthogonal projection of $nabla_X_PX_P$ onto $E^perp = (E+[E,E])^perp$ is precisely the mean curvature vector field
$$
H = -frac12operatornamegradell neq 0
$$
of the Riemannian submersion $SU(2) to mathbbCP^1$. More generally, you'll run into trouble if $E$ is the tangent bundle to a foliation of $Q$ by some locally free isometric action of a connected Lie group, in which case, the obstruction to your claim is the orbit-wise second fundamental form (i.e., O'Neill's $T$-tensor for the Riemannian foliation $E$).
answered Jul 17 at 18:07
Branimir Ćaćić
9,69221846
Let $Q = SU(2)$, and consider the diagonal maximal torus $U(1)$ as acting freely and isometrically on $SU(2)$ by multiplication from the right, which therefore gives rise to the Hopf fibration $SU(2) to SU(2)/U(1) = mathbbCP^1$. Let $E = VSU(2)$ be the vertical tangent bundle of this submersion, which is integrable, so that $E + [E,E] = E$. Finally, let $g_0$ be the $operatornameAd$-invariant Riemannian metric on $SU(2)$ induced by the Killing form, let $ell in C^infty(SU(2),(0,+infty))^U(1)$ be non-constant, and let $g = ell g_0$ be the resulting $U(1)$-invariant Riemannian metric on $Q = SU(2)$, so that $SU(2) to mathbbCP^1$ becomes a Riemannian submersion. Then for any unit vector $X in mathfraksu(2)$ with respect to the Killing form, if $X_P$ is the corresponding left-invariant vector field on $SU(2)$, then the orthogonal projection of $nabla_X_PX_P$ onto $E^perp = (E+[E,E])^perp$ is precisely the mean curvature vector field
$$
H = -frac12operatornamegradell neq 0
$$
of the Riemannian submersion $SU(2) to mathbbCP^1$. More generally, you'll run into trouble if $E$ is the tangent bundle to a foliation of $Q$ by some locally free isometric action of a connected Lie group, in which case, the obstruction to your claim is the orbit-wise second fundamental form (i.e., O'Neill's $T$-tensor for the Riemannian foliation $E$).
answered Jul 17 at 18:07
Branimir Ćaćić
9,69221846
answered Jul 17 at 18:07
Branimir Ćaćić
9,69221846
answered Jul 17 at 18:07
Branimir Ćaćić
9,69221846
answered Jul 17 at 18:07
Branimir Ćaćić
9,69221846
9,69221846
Many thanks Branimir, will think about your example.
– Jair Koiller
Jul 17 at 18:13
1
For Branimir's example, any foliation works, with E the associated integrable distribution. Jair's condition on $nabla$ would imply that the leaves of E are totally geodesic. But typically, they are not. To get perhaps the simplest example possible, take a surface of revolution, with leaves being the circles -the orbits of rotation. The metric can be written $dr^2 + f(r)^2 d theta^2$. The leaves are the level sets of $r$. Only the circles corresponding to critical points of f are (totally) geodesic.
– Richard Montgomery
Jul 18 at 16:14
add a comment |Â
Many thanks Branimir, will think about your example.
– Jair Koiller
Jul 17 at 18:13
1
For Branimir's example, any foliation works, with E the associated integrable distribution. Jair's condition on $nabla$ would imply that the leaves of E are totally geodesic. But typically, they are not. To get perhaps the simplest example possible, take a surface of revolution, with leaves being the circles -the orbits of rotation. The metric can be written $dr^2 + f(r)^2 d theta^2$. The leaves are the level sets of $r$. Only the circles corresponding to critical points of f are (totally) geodesic.
– Richard Montgomery
Jul 18 at 16:14
Many thanks Branimir, will think about your example.
– Jair Koiller
Jul 17 at 18:13
Many thanks Branimir, will think about your example.
– Jair Koiller
Jul 17 at 18:13
1
1
For Branimir's example, any foliation works, with E the associated integrable distribution. Jair's condition on $nabla$ would imply that the leaves of E are totally geodesic. But typically, they are not. To get perhaps the simplest example possible, take a surface of revolution, with leaves being the circles -the orbits of rotation. The metric can be written $dr^2 + f(r)^2 d theta^2$. The leaves are the level sets of $r$. Only the circles corresponding to critical points of f are (totally) geodesic.
– Richard Montgomery
Jul 18 at 16:14
For Branimir's example, any foliation works, with E the associated integrable distribution. Jair's condition on $nabla$ would imply that the leaves of E are totally geodesic. But typically, they are not. To get perhaps the simplest example possible, take a surface of revolution, with leaves being the circles -the orbits of rotation. The metric can be written $dr^2 + f(r)^2 d theta^2$. The leaves are the level sets of $r$. Only the circles corresponding to critical points of f are (totally) geodesic.
– Richard Montgomery
Jul 18 at 16:14
add a comment |Â
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This would extend the well know result for integrable distributions.
– Jair Koiller
Jul 15 at 11:45