# Isotropy and homogeneity

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

Laura

knobelc
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

Allday
One trivial way is to constrain the space to be flat, but that's 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. I am 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.