Are Bianchi IX Models Truly Homogeneous Yet Not Isotropic?

[befj0001]

In a Bianchi IX universe the metric must be invariant under the SO(3) group acting on the 3-sphere. Hence, the metric must be translation invariant in the spatial parts, where t=constant. This implies that the metric must take the form such that:

ds^2 = dt^2 - g_ij(t)(x^i)(x^j), where g is a function of t alone. Am I right about all this?

What concerns me is that someone told me that the metric:

ds^2 = -dt^2 + a^2(t)(dx)^2 + b^2(t)(dy)^2 + (b^2sin^2y+a^2cos^2y)(dz)^2 - 2a^2cosydxdz

belong to the Bianchi IX models. But this doesn't seem right?!

Am I right about the Biachi IX models being homogeneous but not necesseraly isotropic?

 

[Bill_K]

What concerns me is that someone told me that the metric:

ds^2 = -dt^2 + a^2(t)(dx)^2 + b^2(t)(dy)^2 + (b^2sin^2y+a^2cos^2y)(dz)^2 - 2a^2cosydxdz

belong to the Bianchi IX models. But this doesn't seem right?!

Why don't you just write down a few of the Killing vectors and calculate their commutators?

 

[befj0001]

Why don't you just write down a few of the Killing vectors and calculate their commutators?

Yes, the Lie algebra is isomorphic to SO(3). But what does it imply about the geometry of the universe? The spatial part being homogeneous under the action of SO(3) doesn't say anything to me!

I think like this: Since the underlying space of SO(3) is a 3-sphere, and then if the spatial part of the universe is a 3-sphere, the metric would be invariant under ordinary spatial translations just like the Minkowski space is invariant under ordinary spatial translations. But since Minkowski space is flat and noncompact, translation invariance is instead asociated by the Euclidean group of translations not SO(3). Am I on the right track?

My concern about the above metric was that I thought the x,y,z coordinates were cartesian coordiantes locally on the 3-sphere. Instead they are coordinates on the perpendicular coordinate axis witch span the entire four-manifold. Right?

 

[befj0001]

Any thoughts?

 

[Bill_K]

Any thoughts?

Well the first thing that should hit you in the face about this metric is that it is independent of both x and z. Which tells you there are two obvious Killing vectors, and furthermore that they commute. So the isometry group is not SO(3)! There may be four Killing vectors, not just three.