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 .

  • $\begingroup$ 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. $\endgroup$
    – R. Rankin
    Aug 26 at 21:11

1 Answer 1


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.)

  • $\begingroup$ I've edited the post with the construction of $K^{*}$. $\endgroup$
    – Johnny
    Apr 16 at 11:37

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