I'm trying to follow this:

"The metric on de Sitter space is the metric induced from the ambient Minkowski metric."


What I have in mind is that a 5D Minkowski metric:

$\begin{bmatrix}-c^2 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 1 \end{bmatrix}$

Somehow turns into a 4D de Sitter space, which I think is an FLRW dark energy model ($\Omega_\Lambda =1, \Omega_M=0$), which should have a metric like this:

$\begin{bmatrix}-c^2 & 0 & 0 & 0 \\ 0 & a(t)^2 & 0 & 0 \\ 0 & 0 & a(t)^2 & 0 \\ 0 & 0 & 0 & a(t)^2 \end{bmatrix}$

Where $a(t) = e^{H_0t}$.

Now maybe that's not what it's saying at all. So I probably could have stopped right there.

I sort of gathered that you take the 3D surface of a 4D dimensional hyperboloid. I try to simplify this concept to drawing a hyperbola in 2 dimensions, and get back one "hyperbolic" dimension. I don't think this is supposed to work. But I'm curious how close it gets.

I thought, I want to measure the distance along the curve from (1, 0), because that would have to be the origin of the resulting dimension. The arc length should be $l = R\theta$, and the length of a chord $L$ from those points in the 2d space should be:

$L^2 = (R \cosh( \frac{l}{R}) - R)^2 + (R \sinh(\frac{l}{R}))^2$

To go the other direction I think you can do:

$l = R \cosh^{-1} \left( \frac{1}{2} \sqrt{ \frac{2L^2}{R^2} + 1} +\frac{1}{2} \right)$

That gets something I think that could be considered in the same realm as $e^{H_0t}$, but it's not it.

Considering this formula:

$g_{ab}=\partial _{a}X^{\mu }\partial _{b}X^{\nu }g_{\mu \nu } $


If the goal (which is perhaps foolish in itself) is get a 1 dimensional metric that equals $(e^{H_0t})^2$, then it seems I can get "there" is by doing something like defining:

$L = R \cosh( \frac{l}{R}) + R \sinh(\frac{l}{R}) - 1$

Or something similar. But I'm not sure if that's "allowed"?

To recapp.

Can you literally get the 4D metric with the scale factor from the 5D flat metric like that?

Is my method of trying to get there valid?

Do the distances involved need to be arc lengths and chord lengths? Can I make up a somewhat arbitrary one, and induce a metric from that?

Thank you.


1 Answer 1


Consider the five-dimensional Minkowski space captured by $M =(\mathbb{R}^{5}, \eta)$, where $\eta$ is the metric you wrote down.

To get four-dimensional de Sitter space from $M$, you would need to identify a submanifold of $M$ with codimension $1$, call it $N$, such that the induced metric on $N$ from $\eta$ is the de Sitter metric.

The right manifold turns out to be a one-sheeted hyperboloid \begin{align} -x_{0}^{2} + \sum_{i=1}^{4} x_{i}^{2} = \alpha^{2}, \end{align} where $\alpha$ will be identified with the cosmological constant.

If you take the coordinates $x_{\mu}$ to be \begin{align} x_{0} &=\alpha \sinh \frac{t}{\alpha} +\frac{1}{\alpha}r^{2}e^{\frac{t}{\alpha}}, \\ x_{1} &=\alpha \sinh \frac{t}{\alpha} -\frac{1}{\alpha}r^{2}e^{\frac{t}{\alpha}}, \\ x_{i} &=e^{\frac{t}{\alpha}} y_{i}, \end{align} where $y_{i}$ are coordinates on the remaining flat plane of $M$, the induced metric on $N$ is simply given by \begin{align} ds^{2} = -dt^{2} + e^{\frac{2}{\alpha}t} dy^{2}. \end{align} There are different choices of coordinates you can make to describe the hyperboloid but this is one such slicing, showcasing the procedure of inducing a metric on a submanifold.

  • $\begingroup$ Thank you! So is 𝑖 in the second to last equation 2, 3, or 4? I'm also confused of where 𝑟 comes from, you have 𝛼 as the radius of the hyperboloid. And can you also maybe add why the $x_\mu$ coordinates are the sinh function? Is that a convenient choice or a geometric consequence? $\endgroup$ Jun 11, 2023 at 18:56
  • $\begingroup$ Yes, $i$ is for the remaining coordinates on $\mathbb{R}^{3} \subset M$. It's a convenient choice. A particular submanifold could have multiple embeddings in a larger space, and this is one choice. $\endgroup$ Jun 11, 2023 at 19:35
  • $\begingroup$ Just curious how this would work, if I wanted the induced metric to be $ds^2 =- e^{\frac{2}{\alpha}t} dt^2 + dy^2$, so there's a time scale factor instead, would I start with 2 timelike dimensions and 1 spacelike dimension? $\endgroup$ Jun 11, 2023 at 21:06

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