# What is the $r$ coordinate in a $\mathbb{S}^3$ FLRW spactime?

I'm having trouble understanding what the $$r$$ reduced-circumference coordinate really is in a 3-sphere $$\mathbb{S}^3$$ context.

Let's start with the unit 3-sphere metric in hyperspherical $$(\psi, \theta,\phi)$$ coordinates:

$$ds^2 = h_{a b} dx^a dx^b= d \psi^2 + \sin(\psi)^2 \big(d \theta ^2 + \sin(\theta)^2 d \phi^2 \big)$$

here $$\psi \in (0,\pi)$$ covers the whole 3-sphere, i.e we foliate the 3-sphere with topological 2-spheres $$\mathbb{S}^3 = (0,\pi) \times \mathbb{S}^2$$ . To be extra fancy, we might also define a sort of normalized extrinsic curvature trace $$K^{*} \sim \partial_{\psi}\log\big(\gamma\big)$$ as a function to measure the 2-surface area (this is secretly a Lie-derivative along the normal to the 2-surfaces, but that's not important now), where $$\gamma$$ is the determinant of the induced 2-metric

$$\gamma_{A B} dx^A dx^B = \sin(\psi)^2 \big(d \theta ^2 + \sin(\theta)^2 d \phi^2 \big)$$.

This $$K^{*}$$ behaves like one would expect it to: it vanishes at the $$\psi = \pi / 2$$ "equator", where the 2-surface area is the greatest, and has asymptotes at $$\psi = 0, \pi$$, where the 2-surface areas go to zero.

My problem is the following: when one employs the usual $$r = \sin(\psi)$$ coordinate transformation and puts the metric into the "standard" FLRW form

$$ds^2 = \dfrac{dr^2}{1-r^2} + r^2 \big(d \theta ^2 + \sin(\theta)^2 d \phi^2 \big)$$

the $$r \in (0,1)$$ coordinate now only covers half-of the 3-sphere, say between $$\psi \in (0,\pi/2)$$.We should expect the then defined $$K^{*} \sim \partial_{r}\log\big(\gamma\big)$$ quantity to vanish at the $$r=1$$ "equator", since that is what corresponds to the $$\psi = \pi/2$$ 2-surface.

However,what we instead find is that $$K^{*} \sim \frac{1}{r}$$, which indicates the equator is now located at $$r=\infty$$, even tho the $$r$$ coordinate only runs between 0 and 1.

What is going on here? What is precisely the range and meaning of this $$r$$ coordinate? How come this indicates that the equator is at $$r=\infty$$, if that is not in the original coordinate range? Is this just a case of a badly chosen coordinate chart?

EDIT: Allow me to elaborate on $$K^{*}$$ :

Let the foliation of $$\mathbb{S}^3 = I \times \mathbb{S}^2$$ for some interval $$I$$, and let the 2-surfaces be the level sets of some smooth Morse function $$\rho : \mathbb{S}^3 \rightarrow \mathbb{R}$$ .

Let $$\rho^i$$ (excuse the abuse of notation) be a smooth vector field on $$\mathbb{S}^3$$, normalized such that $$\rho^i \; \partial_i \rho =1$$ throughout the 3-sphere. We can decompose the unit norm vector field $$\widehat{n}^i$$ of the foliation into a "lapse" $$\widehat{N}$$ and "shift" $$\widehat{N}^i$$ like

$$\widehat{n}^i = \widehat{N}^{-1}\; \big(\;\rho^i - \widehat{N}^i \big)$$

(which is analogous to the 3+1 ADM decomposition; also, the shift $$\widehat{N}^i$$ vanishes for these $$\mathbb{S}^3$$ calculations. )

Then finally we can define the "normalized extrinsic curvature" tensor $$K^{*}_{i j}$$ as follows:

$$K^{*}_{i j} = \dfrac{\widehat{N}}{2}\; \mathcal{L}_{\widehat{n}} \; \gamma_{i j} = \dfrac{1}{2} \mathcal{L}_{\rho} \;\gamma_{i,j} \,-\, D_{(i} \widehat{N}_{j)}$$

Where $$\gamma_{i j}$$ is the induced 2-metric with covariant derivative $$D_i$$.

The trace of this extrinsic curvature is the precisely the quantity above:

$$K^{*}(\rho) = \gamma^{i j} K^{*}_{i j} = \dfrac{1}{2} \;\gamma^{i j} \; \mathcal{L}_{\rho} \;\gamma_{i,j} = \dfrac{1}{2} \partial_{\rho} \log(\gamma)$$

Where I have used the vanishing of the "shift" $$\widehat{N}^i$$. This (with the substitution $$\rho = \psi$$ and $$\rho = r$$) yields the above claimed values:

$$K^{*}(\psi) = 2 \cot(\psi) \quad \quad K^{*}(r) = \frac{2}{r}$$

I believe this is a well defined geometric construction, so the divergent / zero values of this $$K^{*}$$ should indicate the minimum / maximum 2-surfaces areas. It is clear that in the $$\psi$$ case this makes sense, but less so in the $$r$$ case .

• kindv'e a segway, but I really encourage using the orthonormal tetrad formulation for this space, then spatial bases belong to the Lie algebra $\mathfrak{su(2)}$ and you can think of the spatial part of the metric in terms of the Maurer-Cartan form. Aug 26 at 21:11

The $$r$$-coordinate would be known as the circumference or areal radius. You can verify this by computing $$C_{\mathbb{S}^2}\oint \sqrt{\gamma_{\phi\phi}} d \phi\Big|_{\theta = \pi/2} = 2 \pi r \,,$$ $$A_{\mathbb{S}^2} = \oint \sqrt{\gamma} d \phi d\theta = 4 \pi r^2$$ That is, you foliate you $$\mathbb{S}^3$$ into 2-spheres and label them according to their circumference/areal radius.
Can one slice the 3-sphere so that there is a unique 2-sphere for every given areal radius in $$\mathbb{S}^3$$ you ask? No, one cannot, there are always at least two, similarly to the case of circumference radius and slicing up the 2-sphere into circles (that is, 1-spheres). So you are going to get a coordinate singularity when labelling this foliation, which appears at $$r=1$$. Fortunately, in this case you can resolve this "double-cover" by choosing $$r \in (-1,1)$$ or $$\psi \in (-\pi/2,\pi/2)$$. Then you see that since $$\psi = \pm \pi$$ are topologically identified, we also topologically identify $$r= \sin(\pm\pi/2)=\pm 1$$ as the same spheres.
(I do not quite understand your computations and arguments around $$K^*$$. It is a geometrically defined scalar, its behaviour should depend only on the folitation, not on coordinates. I believe you may be forgetting the normalization of the normal when using the $$r$$ coordinate.)
• I've edited the post with the construction of $K^{*}$. Apr 16 at 11:37