Principal and Gaussian curvature of the FRW metric

[shinobi20]

TL;DR

I'd like to confirm my calculations of the principal and Gaussian curvature of the spatial part of the Friedmann-Robertson-Walker (FRW) metric. Specifically, if the spatial part has negative curvature.

I would like to calculate the principal and Gaussian curvature of the spatial part of the Friedmann-Robertson-Walker (FRW) metric; specifically, the negative Gaussian curvature $k=-1$. The FRW metric is,

\begin{equation*}
ds^2 = -dt^2 + R(t)^2 \left( \frac{dr^2}{1-k r^2} + r^2 d\Omega^2 \right)
\end{equation*}

For a certain time slice $t = t_0$, the spatial part is,

\begin{align*}
& d\sigma^2 = R_0^2 \left( \frac{dr^2}{1-k r^2} + r^2 d\Omega^2 \right) \qquad R_0 \equiv R(t_0) \\
& d\sigma^2 = R_0^2 \left( \frac{dr^2}{1 + r^2} + r^2 d\Omega^2 \right), \qquad \text{k=-1}
\end{align*}

If we set $r = \sinh \psi$, then we get the metric (also called the first fundamental form) for the 3d hyperboloid in angular coordinates,

\begin{align*}
& d\sigma^2 = R_0^2 \left( d\psi^2 + \sinh^2 \psi d\theta^2 + \sinh^2 \psi \sin^2\theta d\phi^2 \right)\\
& d\sigma^2 = h_{ab} dx^a dx^b
\end{align*}

\begin{equation*}
h_{ab} =
\begin{bmatrix}
R_0^2 & 0 & 0\\
0 & R_0^2 \sinh^2 \psi & 0\\
0 & 0 & R_0^2 \sinh^2 \psi \sin^2 \theta
\end{bmatrix}
\end{equation*}

The 3d hyperboloid can be embedded in 4d Minkowski space. Let $g_{\mu\nu} = \text{diag}(-1,1,1,1)$ be the Minkowski metric and $h_{ab}$ be the 3d hyperboloid metric in angular coordinates. The 3d hyperboloid in $(x,y,z,w)$ coordinates is described by the equation,

\begin{align*}
& x^2 + y^2 + z^2 -w^2 = -R_0^2\\
& f(x,y,z,w) = x^2 + y^2 + z^2 -w^2 + R_0^2
\end{align*}

The parameterization is,

\begin{equation*}
x = R_0 \sinh \psi \sin\theta \cos \phi, \quad y = R_0 \sinh \psi \sin\theta \sin \phi, \quad z = R_0 \sinh \psi \cos\theta, \quad w = R_0 \cosh \psi
\end{equation*}

Given the point $\mathbf{v}^a = (x,y,z,w)$ on the 3d hyperboloid, the tangent vectors $\mathbf{e}_a$ are,

\begin{align*}
& \mathbf{e}_\psi = R_0 \left( \cosh \psi \sin\theta \cos \phi, \cosh \psi \sin\theta \sin \phi, \cosh \psi \cos\theta, \sinh \psi \right)\\
& \mathbf{e}_\theta = R_0 \left( \sinh \psi \cos\theta \cos \phi, \sinh \psi \cos\theta \sin \phi, -\sinh \psi \sin\theta, 0 \right)\\
& \mathbf{e}_\phi = R_0 \left( -\sinh \psi \sin\theta \sin \phi, \sinh \psi \sin\theta \cos \phi, 0 , 0 \right)
\end{align*}

The normal vector $N_\mu$ on the 3d hyperboloid can be calculated as,

\begin{equation*}
N_\mu = -\nabla f(x,y,z,w)
\end{equation*}

The normalized normal vector $n_\mu$ is,

\begin{align*}
& n_\mu = \frac{N_\mu}{\sqrt{ | g^{\mu\nu} N_\mu N_\nu | }}\\
& n_\mu = \left( -\sinh \psi \sin\theta \cos \phi, -\sinh \psi \sin\theta \sin \phi, -\sinh \psi \cos\theta, \cosh \psi \right)
\end{align*}

We also need the derivative of the tangent vectors $\partial_b \mathbf{e}_a$. The second fundamental form $L_{ab}$ can be calculated as,

\begin{equation*}
L_{ab} = g^{\mu\nu} (\partial_b \mathbf{e}_a)_\mu n_\nu, \quad
L_{ab} =
\begin{bmatrix}
-R_0 \cosh 2\psi & 0 & 0\\
0 & R_0 \sinh^2 \psi & 0\\
0 & 0 & R_0 \sinh^2 \psi \sin^2 \theta
\end{bmatrix}
\end{equation*}

The principal curvatures $k_i$ for $i=1,2,3$ are the eigenvalues of the matrix $h^{-1} L$,

\begin{equation*}
k_1 = \frac{1}{R_0}, \quad k_2 = \frac{1}{R_0}, \quad k_3 = -\frac{\cosh 2\psi}{R_0}
\end{equation*}

The Gaussian curvature is $k = k_1 k_2 k_3$.

Questions:
(1) Are my calculations correct?
(2) Are the answers for the principal curvatures correct?
(3) I'm wondering about $k_3$. As I know the Gaussian curvature should be $k=-1$ for $R_0 = 1$, but $k_3 = -1$ only for $\psi = 0$. Can anyone help clarify and explain more about the principal curvatures of the 3d hyperboloid?
(4) Another thing I'm unsure of is the sign of the normal vector, if I remove the minus sign then $k_1,k_2$ becomes negative and $k_3$ becomes positive while maintaining the same magnitude.

 

[ergospherical]

The second fundamental form looks a bit suspect (i.e. the cosh in the to left component). The 3d hyperboloid (H³) is maximally symmetric and should have constant intrinsic curvature everywhere. You should then find that $L \propto h$, specifically $L = (1/R_0) h$. And you should get constant principal curvatures.

 

[shinobi20]

The second fundamental form looks a bit suspect (i.e. the cosh in the to left component). The 3d hyperboloid (H³) is maximally symmetric and should have constant intrinsic curvature everywhere. You should then find that $L \propto h$, specifically $L = (1/R_0) h$. And you should get constant principal curvatures.

Indeed, if I remove $\cosh 2\psi$ then I get $k_3 = -\frac{1}{R_0}$. However, I'm not yet sure where I went wrong. The necessary quantities to calculate $L$ are listed above, although I did not explicitly type out the vectors $\partial_b \mathbf{e}_a$, but that is just the derivative of the tangent vectors; I've used Mathematica to calculate those quantities. Is my normal vector correct?

The issue is with the $(1,1)$ component $L_{11}$. The vector $\partial_1 \mathbf{e}_1$ is given by,

\begin{equation*}
\partial_1 \mathbf{e}_1 = \partial_\psi \mathbf{e}_\psi = (R_0 \sinh \psi \sin \theta \cos \phi, R_0 \sinh \psi \sin \theta \sin \phi, R_0 \sinh \psi \cos \theta, R_0 \cosh \psi )
\end{equation*}

Note the minus sign in the metric $g$ is for the coordinate $w$. $L_{11}$ is calculated as,

\begin{align*}
g^{\mu\nu} (\partial_1 \mathbf{e}_1)_\mu n_\nu & = -(\partial_1 \mathbf{e}_1)_1 n_1 + (\partial_1 \mathbf{e}_1)_2 n_2 + (\partial_1 \mathbf{e}_1)_3 n_3 + (\partial_1 \mathbf{e}_1)_4 n_4\\
& = -R_0 \cosh^2 \psi - R_0 \sinh^2 \psi \cos^2 \theta - R_0 \sinh^2 \psi \sin^2 \theta \cos^2 \phi - R_0 \sinh^2 \psi \sin^2 \theta \sin^2 \phi\\
& = -R_0 ( \cosh^2 \psi + \sinh^2 \psi )\\
& = -R_0 \cosh 2 \psi
\end{align*}

So, the issue is the term $\cosh^2 \psi + \sinh^2 \psi$.

 

[ergospherical]

\begin{align*}
L_{\psi \psi} = n_{\mu} \frac{\partial^2 x^{\mu}}{\partial \psi^2} &= - \frac{x}{R_0} R_0 \sinh{\psi} \sin{\theta} \cos{\phi} \\
&\quad - \frac{y}{R_0} R_0 \sinh{\psi} \sin{\theta} \sin{\phi} \\
&\quad - \frac{z}{R_0} R_0 \sinh{\psi} \cos{\theta} \\
&\quad + \frac{w}{R_0} R_0 \cosh{\psi}
\end{align*}
Reinserting the parameterizations of $x^{\mu}(\psi, \theta, \phi)$ gives $L_{\psi \psi} = R_0$, which is indeed $h_{\psi \psi} / R_0$.

 

[shinobi20]

\begin{align*}
L_{\psi \psi} = n_{\mu} \frac{\partial^2 x^{\mu}}{\partial \psi^2} &= - \frac{x}{R_0} R_0 \sinh{\psi} \sin{\theta} \cos{\phi} \\
&\quad - \frac{y}{R_0} R_0 \sinh{\psi} \sin{\theta} \sin{\phi} \\
&\quad - \frac{z}{R_0} R_0 \sinh{\psi} \cos{\theta} \\
&\quad + \frac{w}{R_0} R_0 \cosh{\psi}
\end{align*}
Reinserting the parameterizations of $x^{\mu}(\psi, \theta, \phi)$ gives $L_{\psi \psi} = R_0$, which is indeed $h_{\psi \psi} / R_0$.

Your last term should be $-\frac{w}{R_0} R_0 \cosh \psi$ due to the metric $g=diag(-1,1,1,1)$ right? The minus sign in $g$ is what makes it minus. That's the reason the issue comes out.

 

[ergospherical]

I already account for the signs. Albeit maybe I had chosen the other signature to you. With your signature convention, $f(x) := x^2 + y^2 + z^2 - w^2 + R_0^2 = 0$ gives $N_{\mu} = \partial_{\mu} f = (2x,2y, 2z, -2w)$ and $||N_{\mu}|| = 2R_0$ gives $n_{\mu} = \tfrac{1}{R_0}(x,y,z,-w)$. Then take $L_{ab} = n_{\mu} \tfrac{\partial^2 x^{\mu}}{\partial y^a \partial y^b}$, and that Einstein sum has a positive sign for the first three terms and a negative sign for the last term.

 

[shinobi20]

I already account for the signs. Albeit maybe I had chosen the other signature to you. With your signature convention, $f(x) := x^2 + y^2 + z^2 - w^2 + R_0^2 = 0$ gives $N_{\mu} = \partial_{\mu} f = (2x,2y, 2z, -2w)$ and $||N_{\mu}|| = 2R_0$ gives $n_{\mu} = \tfrac{1}{R_0}(x,y,z,-w)$. Then take $L_{ab} = n_{\mu} \tfrac{\partial^2 x^{\mu}}{\partial y^a \partial y^b}$, and that Einstein sum has a positive sign for the first three terms and a negative sign for the last term.

Actually, I made a mistake interpreting the vectors $\partial_b \mathbf{e}_a$ as lower index, i.e., $(\partial_b \mathbf{e}_a)_\mu$. Actually, they are upper index vectors $(\partial_b \mathbf{e}_a)^\mu$. So when computing $L_{\psi \psi}$, it is as you did, i.e., $L_{\psi \psi} = n_\mu (\partial_b \mathbf{e}_a)^\mu$.

My normal vector is,

\begin{align*}
& n_\mu = \frac{N_\mu}{\sqrt{ | g^{\mu\nu} N_\mu N_\nu | }}\\
& n_\mu = \left( -\sinh \psi \sin\theta \cos \phi, -\sinh \psi \sin\theta \sin \phi, -\sinh \psi \cos\theta, \cosh \psi \right)
\end{align*}

and

\begin{equation*}
\partial_\psi \mathbf{e}_\psi = (R_0 \sinh \psi \sin \theta \cos \phi, R_0 \sinh \psi \sin \theta \sin \phi, R_0 \sinh \psi \cos \theta, R_0 \cosh \psi )
\end{equation*}

When you take the inner product it indeed gives $R_0$. However, if this is the case then all $k_1, k_2, k_3$ have the same sign. If my normal vector has a minus sign, then $k_1 = 1/R_0, k_2 = 1/R_0, k_3 = 1/R_0$. If I remove the minus sign, then $k_1 = -1/R_0, k_2 = -1/R_0, k_3 = -1/R_0$. Shouldn't one of the principal curvatures have a different sign?

I believe the 3d hyperboloid is hyperbolic in some direction and spherical in another direction. As an example, for 2d hyperboloid the principal curvatures should be $k_1 = 1/R_0, k_2 = -1/R_0$.

The only way for this to happen is if $L_{\psi \psi} = -R_0$ not $R_0$.

 

[ergospherical]

The spatial isotropy forces sectional curvature to be equal in any direction at a point (and the homogeneity means they are also the same for any point).

 

[shinobi20]

The spatial isotropy forces sectional curvature to be equal in direction at a point (and the homogeneity means they are also the same for any point).

Oh yeah! So the correct principal curvatures are $k_1 = -1/R_0, k_2 = -1/R_0, k_3 = -1/R_0$?

If this is the case then the Gaussian curvature will be $k=k_1 k_2 k_3 = -1/R_0^3$ which has the correct sign. If $R_0 = 1$ then $k=-1$ which is the assumption from the start.