# Isotropy and homogeneity

## Main Question or Discussion Point

http://camoo.freeshell.org/27.16wrong.pdf" [Broken]

Mistake by the author?

Laura

Latex source below for quoting purposes but the .pdf may've been edited since then.

Exercise 27.16 asks you to show why a connected 3-space can't be
isotropic about 2 distinct points without being homogeneous.

Counterexample, though. Suppose the space is $S^3$, the
3-dimensional sphere. You could think of it as the equation for
$x^2+y^2+z^2+w^2=1$.

Then let the 2 separate points be antipodal points on the sphere. For
example $x=(1,0,0,0)$ and $-x=(-1,0,0,0)$.

You could have a matter distribution that was radially symmetric
around both of these points, because a rotation around x is also a
rotation around $-x$! But it doesn't have to be homogeneous. The
matter density could go up with distance from x or $-x$, up to the
"equator" $y^2+z^2+w^2=1$.

Am I missing something, or is this exercise just wrong?
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However if the universe were isotropic about 3 separate points, then it has to be homogeneous!

Laura

However if the universe were isotropic about 3 separate points, then it has to be homogeneous!

Laura
Do you know a proof for this?

Christian

One trivial way is to constrain the space to be flat, but thats not what your getting at here . I'm still trying to visualize your example but failing. I can see a 2-sphere being isotropic around two poles and not homogeneous, but I'm having trouble with the 3-sphere. Maybe it would be fruitful to try to come up with an explicit expression for a scalar field on the 3-sphere that would satisfy isotropy around two points without being homogeneous and see if you can do it. That might either prove you right or the problem right. Im not sure if a rotation about (-1,0,0,0) is the same as a rotation about (1,0,0,0).

In the two sphere picture the rotations are the same because you are basically spinning a globe on its axis so the rotation isn't a general rotation about a point but one around an axis.