Are Bianchi IX Models Truly Homogeneous Yet Not Isotropic?

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Discussion Overview

The discussion revolves around the properties of Bianchi IX models in cosmology, specifically addressing whether these models can be considered homogeneous while not necessarily being isotropic. Participants explore the implications of the metric forms and the role of symmetry in defining the geometry of the universe.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant asserts that the Bianchi IX metric must be invariant under the SO(3) group acting on the 3-sphere, suggesting a specific form for the metric.
  • Another participant questions the validity of a proposed metric for Bianchi IX models, expressing confusion about its homogeneity and isotropy properties.
  • A different participant proposes calculating Killing vectors and their commutators to analyze the metric's properties, indicating that the Lie algebra is isomorphic to SO(3) but questioning the implications for the universe's geometry.
  • Concerns are raised about the interpretation of coordinates in the metric, with a participant suggesting that the coordinates may not be Cartesian and instead span the entire four-manifold.
  • Another participant points out that the proposed metric is independent of certain coordinates, indicating the presence of multiple Killing vectors and suggesting that the isometry group may not be SO(3) as initially thought.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the Bianchi IX models, particularly regarding the relationship between homogeneity and isotropy. There is no consensus on the correctness of the proposed metric or its implications for the geometry of the universe.

Contextual Notes

Participants highlight potential limitations in understanding the metric's properties, including assumptions about the coordinates and the implications of symmetry groups. The discussion remains open-ended with unresolved mathematical interpretations.

befj0001
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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?
 
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befj0001 said:
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?
 
Bill_K said:
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?
 
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Any thoughts?
 
befj0001 said:
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.
 
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