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Let $ M $ be the Moebius band. In other words, the total space of the nontrivial line bundle over the circle. Can we equip $ M $ with a metric such the the isometry group acts transitively?

My intuition is that yes we can because $ M $ is the total space of a vector bundle over a compact Riemannian homogenous space (the circle).

However thanks to Ben McKay for the argument that $ M $ does not have a transitive isometry group. Also this fact is stated in "Isometries of 2-Dimensional Riemannian Manifolds into Themselves" by Sumner Byron Myers.

More generally let $ E $ be the total space of a vector bundle with base space $ B $ where $ B $ is a compact Riemannian homogeneous manifold. When does there exist a metric on $ E $ such that $ E $ is Riemannian homogeneous? Does the bundle have to be equivariant with respect to the action of the isometry group of $ B $?

Let $ M $ be the Moebius band. In other words, the total space of the nontrivial line bundle over the circle. Can we equip $ M $ with a metric such the the isometry group acts transitively?

My intuition is that yes we can because $ M $ is the total space of a vector bundle over a compact Riemannian homogenous space (the circle).

However thanks to Ben McKay for the argument that $ M $ does not have a transitive isometry group. Also this fact is stated in "Isometries of 2-Dimensional Riemannian Manifolds into Themselves" by Sumner Byron Myers.

More generally let $ E $ be the total space of a vector bundle with base space $ B $ where $ B $ is a compact Riemannian homogeneous manifold. When does there exist a metric on $ E $ such that $ E $ is Riemannian homogeneous? Does the bundle have to be equivariant with respect to the action of the isometry group of $ B $?

Let $ M $ be the Moebius band. In other words, the total space of the nontrivial line bundle over the circle. Can we equip $ M $ with a metric such the the isometry group acts transitively?

My intuition is that yes we can because $ M $ is the total space of a vector bundle over a compact Riemannian homogenous space (the circle).

However thanks to Ben McKay for the argument that $ M $ does not have a transitive isometry group. Also this fact is stated in "Isometries of 2-Dimensional Riemannian Manifolds into Themselves" by Sumner Byron Myers.

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Let $ M $ be the Moebius band. In other words, the total space of the nontrivial line bundle over the circle. Can we equip $ M $ with a metric such the the isometry group acts transitively?

My intuition is that yes we can because $ M $ is the total space of a vector bundle over a compact Riemannian homogenous space (the circle).

However thanks to Ben McKay for the argument that $ M $ does not have a transitive isometry group. Also this fact is stated in "Isometries of 2-Dimensional Riemannian Manifolds into Themselves" by Sumner Byron Myers.

More generally let $ E $ be the total space of a vector bundle with base space $ B $ where $ B $ is a compact Riemannian homogeneous manifold. DoesWhen does there always exist a metric on $ E $ such that $ E $ is Riemannian homogeneous?

If not is there some condition under which this is true? Like does Does the bundle have to be equivariant with respect to the action of the isometry group of $ B $?

Let $ M $ be the Moebius band. In other words, the total space of the nontrivial line bundle over the circle. Can we equip $ M $ with a metric such the the isometry group acts transitively?

My intuition is that yes we can because $ M $ is the total space of a vector bundle over a compact Riemannian homogenous space (the circle).

More generally let $ E $ be the total space of a vector bundle with base space $ B $ where $ B $ is a compact Riemannian homogeneous manifold. Does there always exist a metric on $ E $ such that $ E $ is Riemannian homogeneous?

If not is there some condition under which this is true? Like does the bundle have to be equivariant with respect to the action of the isometry group of $ B $?

Let $ M $ be the Moebius band. In other words, the total space of the nontrivial line bundle over the circle. Can we equip $ M $ with a metric such the the isometry group acts transitively?

My intuition is that yes we can because $ M $ is the total space of a vector bundle over a compact Riemannian homogenous space (the circle).

However thanks to Ben McKay for the argument that $ M $ does not have a transitive isometry group. Also this fact is stated in "Isometries of 2-Dimensional Riemannian Manifolds into Themselves" by Sumner Byron Myers.

More generally let $ E $ be the total space of a vector bundle with base space $ B $ where $ B $ is a compact Riemannian homogeneous manifold. When does there exist a metric on $ E $ such that $ E $ is Riemannian homogeneous? Does the bundle have to be equivariant with respect to the action of the isometry group of $ B $?

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Is the Moebius strip ReimannianRiemannian homogeneous?

Let $ M $ be the Moebius band. In other words, the total space of the nontrivial line bundle over the circle. Can we equip $ M $ with a metric such the the isometry group acts transitively?

My intuition is that yes we can because $ M $ is the total space of a vector bundle over a compact Riemannian homogenous space (the circle).

More generally let $ E $ be the total space of a vector bundle with base space $ B $ where $ B $ is a compact Riemannian homogeneous manifold. Does there always exist a metric on $ E $ such that $ E $ is Riemannian homogeneous?

If not is there some condition under which this is true? Like does the bundle have to be equivariant with respect to the action of the isometry group of $ B $  ?

Is the Moebius strip Reimannian homogeneous?

Let $ M $ be the Moebius band. In other words, the total space of the nontrivial line bundle over the circle. Can we equip $ M $ with a metric such the the isometry group acts transitively?

My intuition is that yes we can because $ M $ is the total space of a vector bundle over a compact Riemannian homogenous space (the circle).

More generally let $ E $ be the total space of a vector bundle with base space $ B $ where $ B $ is a compact Riemannian homogeneous manifold. Does there always exist a metric on $ E $ such that $ E $ is Riemannian homogeneous?

If not is there some condition under which this is true? Like does the bundle have to be equivariant with respect to the action of the isometry group of $ B $  ?

Is the Moebius strip Riemannian homogeneous?

Let $ M $ be the Moebius band. In other words, the total space of the nontrivial line bundle over the circle. Can we equip $ M $ with a metric such the the isometry group acts transitively?

My intuition is that yes we can because $ M $ is the total space of a vector bundle over a compact Riemannian homogenous space (the circle).

More generally let $ E $ be the total space of a vector bundle with base space $ B $ where $ B $ is a compact Riemannian homogeneous manifold. Does there always exist a metric on $ E $ such that $ E $ is Riemannian homogeneous?

If not is there some condition under which this is true? Like does the bundle have to be equivariant with respect to the action of the isometry group of $ B $?

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