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Let us assume that the universe is isotropic. In other words, it looks the same in all directions. This link says that in that case, there will be spherical symmetry, which again implies that the metric looks like

$$\mathrm dr^2+f^2(r)(\mathrm d\theta^2+ \sin^2\theta\, \mathrm d\phi^2)$$

If this is indeed a spherically symmetric matrix, shouldn't the coefficients all be independent of spherical coordinates- $\theta$ and $\phi$ namely?

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  • $\begingroup$ @ChiralAnomaly- Yes I'll make the correction $\endgroup$ – Anju George May 10 at 20:23
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In your question you show the spatial part of a general isotropic and homogeneous metric.

Even though in such a space there exists spherical symmetry, you are free to express the same metric in, e.g., Cartesian coordinates. The spatial part of a flat such metric then reads \begin{align} dx^2 + dy^2 + dz^2. \end{align} This might look more familiar to you and no direction (none of $x$, $y$, or $z$) are special wrt the other. If no direction is special we have isotropy.

Expressing the Cartesian coordinates in terms of spherical coordinates \begin{align} x &= r\cos(\phi)\sin(\theta\\ y &= r\sin(\phi)\sin(\theta)\\ x &= r\cos(\theta) \end{align} and computing the differentials, leads to the form you quote in your question (with $f(r)=r$; rescaling $r$ or replacing it by a real $f(r)$ can lead to non-flat spaces).

That you find the coefficient of $d\phi^2$ to carry $\theta$ dependence simply is a result of your coordinate transformation.

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  • $\begingroup$ I guess this might be a sort of stupid question, but why are some coordinate systems more misleading than others? For instance, allowing for the fact that $f(r)=1$, we have a $\theta$ dependance for the coefficient of $d\phi^2$, which clearly suggests a lack of spherical symmetry. However, writing everything down in cartesian coordinates clears it all up. $\endgroup$ – Anju George May 10 at 20:35
  • $\begingroup$ @AnjuGeorge You can derive spherical symmetry irrespective of the coordinate system. Find a basis for the Killing vectors (solutions to the Killing equation for the metric), and see which Lie algebra they generate. In the end you can use any absurd coordinate system and you will always find the group of rotations. $\endgroup$ – JamalS May 10 at 20:50
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If all the components of the metric would be independent of spherical coordinates, then for any curve $r,\theta=const.$ or $r,\phi=const.$ you would get circles with circumference independent of both spherical coordinates.

But in f.e. flat space, it is obvious from the meaning of the coordinates that circumference of circles $r,\theta=const.$ (circles of given longitude) , actually depends on the value of coordinate $\theta$. Thus metric coeficient $g_{\phi\phi}$ must too.

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