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Visualization of de Sitter spaces

+ 2 like - 0 dislike

When I was reading about de Sitter it said that the we looked in 5-dimensional Minkowski metric and that we had immersed in it a hyperboloid, and with some coordinate transformation we get some metric that we see resembles a known metric.

But how to visualize this? And why did we took the hyperboloid in the first place? None of the books I read on the subject really gave some simple explanation to that :\

I'm kinda puzzled about this :\ How do I 'see' things with given metric?

For instance, I have, for de Sitter:

$$ds^2~=~-dt^2+l^2 \cosh^2(t/l) d\Omega_3^2$$

And this describes a space tri-sphere that shrinks to a minimum at t=0, and then expands.

Now, this means that the $d\Omega_3^2$ is a three-sphere, contracts because of the hyperbolic cosine, and time is just normal like in Minkowski. Am I right in interpreting this?

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user dingo_d
asked Nov 9, 2012 in Theoretical Physics by dingo_d (110 points) [ no revision ]

2 Answers

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The motivation for this construction is explained here:

He first gives the example of a space of constant positive curvature - the 3-sphere, given by taking a flat Euclidean space of one higher dimension (4) and restricting to the subspace $(x_1, x_2, x_3, x_4)$ s.t. $$x_1^2+x_2^2+x_3^2+x_4^2=a^2$$ for some $a$. The metric on the sphere is just the induced metric from the embedding.

If, instead, we start in Minkowski space, and look at the subspace $(x_0, x_1, x_2, x_3)$ s.t. $$x_0^2-x_1^2-x_2^2-x_3^2=a^2$$ then we end up with a space of constant negative curvature, which we can draw as one sheet of a hyperboloid. Again the metric is just the induced metric.

De Sitter space is just exactly the same thing but starting in 5 dimensional flat Minkowski space instead of 4.

Your interpretation is right, but it's important to notice that time as defined here (the one in which the spatial 3-sphere contracts and expands) isn't a Killing vector of the spacetime, i.e. isn't a symmetry like in Minkowski space.

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user twistor59
answered Nov 9, 2012 by twistor59 (2,490 points) [ no revision ]
Thank you for the explanation, I had some troubles with this. I am going to look at the pdf you provided :)

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user dingo_d
+ 2 like - 0 dislike

I find it helpful to visualize de Sitter space-time using Felix Klein's approach to geometry; begin with projective space and pick a polarity that transforms trivially under the congruence group of the geometry. To get 3+1 de Sitter space-time one starts with a 4-d projective space that is modelled as the rays of a 5-d vector space $V_{5}$; a point in de Sitter space is a ray vector $\lambda x^{i}\in V_{5}$ where $\lambda$ is a weight with no physical significance. In Kleinian language, a polarity is a symmetric tensor $I_{ij}$ that maps points $x^{i}$ into hyperplanes $x_{i}=I_{ij}x^{j}$. The congruence group of de Sitter space is SO(4,1) and so the polarity $I_{ij}$ has to transform trivially (be invariant) under SO(4,1). In Kleinian language $I_{ij}$ is called the absolute polarity. A natural hypersurface is $x^{i}I_{ij}x^{j}=0$. The physical points of space are rays in projective space so that $\lambda x^{i}I_{ij}\lambda x^{j}=0$ must be the same hypersurface; that is why the RHS must be zero in order to define a hypersurface. If one transforms everything with a SO(4,1) matrix $[D(g)]^{i}_{ j}$, the points go $x'^{i}=[D(g)]^{i}_{ j}x^{j}$ and the absolute polarity remains unchanged so that, $$ x'^{i}I_{ij}x'^{j}=x^{i}I_{ij}x^{j}=0 $$ which shows that all "inertial" observers agree on the natural hypersurface. In this context inertial means observers connected by SO(4,1) transformations. In Kleinian language, the hypersurface is called the absolute quadric. The polarity is a symmetric tensor so it can be diagonalized; with a natural choice of the unit of length it is diag(I)=[1,1,1,1,-1]. In projective space the absolute quadric hypersurface is , $$ (x^{1})^{2}+(x^{2})^{2}+(x^{3})^{2}+(x^{4})^{2}-(x^{5})^{2}=0. $$ In the 5-d projective space the quadric hypersurface looks like a cone symmetrical about the $x^{5}$ axis. (Think of 1+1 de Sitter space-time in which the projective space is 3-d and the quadric hypersurface is the cone $x^2+y^2-z^2=0$.) However, instead of visualizing points as rays in the projective space, it is better to set up an arbitrary hyperplane and take the representation of the point as the point at which the ray intersects the hyperplane. In this way the geometry becomes a sketch on this hyperplane used as an aid to visualization. In particular, the absolute quadric becomes a conic section. The graphic shows the absolute quadric as an ellipse for 1+1 de Sitter space-time. The absolute quadric divides the space into two parts . The points inside the quadric are called ordinary points and the points outside the quadric are called ultra-infinite points. The ordinary points make up hyperbolic geometry. The ultra-infinite points make up de Sitter space-time. In the graphic the observer is shown at point p along with the observer's future and past infinity and the light cone. I got a lot of this from reading Coxeter's book "Non-Euclidean Geometry". The picture of de Sitter space here is what Schrodinger calls the elliptic view in his book "Expanding Universes"; the compact space-like slices are elliptic spaces and not spheres.

Kleinian view of 1+1 de Sitter  space-time

This post imported from StackExchange Physics at 2014-03-22 17:08 (UCT), posted by SE-user Stephen Blake
answered Nov 11, 2012 by Stephen Blake (70 points) [ no revision ]

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