<p>The AdS space is the sphere in <span class="math-container">$\mathbb{R}^{p,q+1}$</span>, not <span class="math-container">$\mathbb{R}^{p,q+1}$</span> itself. I think you may be confusing the two.</p>
<p>It's similar to the model of <a href="https://en.wikipedia.org/wiki/Elliptic_geometry" rel="nofollow noreferrer">elliptic geometry</a> as a sphere (with antipodal points identified) in 3D Euclidean space. The existence of that model doesn't mean that elliptic geometry is Euclidean. (And it isn't Euclidean, since it violates the parallel postulate.)</p>
<blockquote>
<p>Does this mean that a sphere in anti ADS space has a negative diameter? (-R²)?</p>
</blockquote>
<p>No. For one thing, it's not a sphere in AdS space, it's the entire AdS space. For another thing, you could just as well write the defining formula as <span class="math-container">$\sum_{j=1}^{q+1}t_j^2 - \sum^q_{i=1}x_i^2 = R^2$</span>.</p>
<p>On the other hand, there is nothing wrong with saying that AdS space has a negative radius-squared, as long as you make it clear that you're adopting that convention.</p>
<blockquote>
<p>Doesn't Einstein's spacetime interval metric <span class="math-container">$d=\sqrt{x^2 + y^2 + z^2 - ct^2}$</span> meet this definition of ADS space, where p=3 and q=1?</p>
</blockquote>
<p>No. Compare this to the sphere model of elliptic geometry. If you use <span class="math-container">$d=\sqrt{x^2 + y^2 + z^2}$</span> to define the distance between points on the sphere, you are measuring distances along straight lines in the background space that don't lie in the elliptic plane (sphere), and have nothing to do with elliptic geometry per se. They only exist &quot;by accident&quot; in that particular model. The reasonable way to define distance in elliptic geometry is along the elliptic plane's own lines (great circles), and that gives a different result.</p>

CC BY-SA 4.0
2024-07-19T00:50:30.173Z
2024-07-19T00:50:30.173Z
benrg
<p>Regarding the second question:</p>
<p>In Minkowski space, the spacetime interval is given by</p>
<p><span class="math-container">$$ds^2 = dx^2+dy^2+dz^2-c^2dt^2.$$</span></p>
<p>This just defines how you measure distances in spacetime. Note that <span class="math-container">$ds^2$</span> is not fixed to a certain value. (For example, <span class="math-container">$ds$</span> is zero for photons and nonzero for massive particles).</p>
<p>From this 4-D spacetime, we can now pick out all the points which satisfy</p>
<p><span class="math-container">$$x^2+y^2-z^2-c^2dt^2 = -R^2$$</span></p>
<p>for some value <span class="math-container">$R\neq 0$</span>. Not all points in Minkowski spacetime satisfy this condition, but some do. We define the set of points that satisfy this condition as anti de Sitter spacetime. In this case, the anti de Sitter spacetime is a subset of the underlying Minkowski spacetime (To drop some fancy math lingo, it is a submanifold.)</p>
<p>To make it easier, let's consider ordinary euclidian 2-D space. It is defined as the set of points <span class="math-container">$(x,y)$</span> where <span class="math-container">$x$</span> and <span class="math-container">$y$</span> are real numbers. If we now demand that <span class="math-container">$x^2+y^2=1$</span>, we get the set of all points which lie on the unit circle, which is a subset of the 2-D euclidian space.</p>
<p>If we demand <span class="math-container">$x^2-y^2=1$</span> or <span class="math-container">$x^2-y^2=-1$</span>, the subspace is hyperbolic (i.e. the points lie on hyperbolas.)</p>
<p>Regarding the first question, it sort of depends on what you mean by a sphere. A sphere is defined as the set of points whose distance to the center is some fixed value <span class="math-container">$r$</span>. That obviously depends on what you mean by distance, i.e. it depends on the underlying metric. In minkowski coordinates, the condition for a sphere would be
<span class="math-container">$$x^2 + y^2 + z^2 - c^2t^2 = R^2$$</span>
which is different to the condition for a hyperbolic space. So if you want to view anti de sitter space as a submanifold of an underlying minkowski space, you shouldn't think of <span class="math-container">$R$</span> as a radius.</p>

CC BY-SA 4.0
2024-07-18T20:14:13.770Z
2024-07-18T20:14:13.770Z
Lenard Kasselmann
<p><a href="https://ncatlab.org/nlab/show/anti+de+Sitter+spacetime#definition" rel="nofollow noreferrer">N-Lab says:</a></p>
<blockquote>
<p>We may define the anti de Sitter space of signature <span class="math-container">$(p,q)$</span> as isometrically embeded in Space <span class="math-container">$\mathbb{R}^{p,q+1}$</span> with coordinates <span class="math-container">$(x_1, \dots, x_p, t_1,\dots,t_{q+1})$</span> as the sphere <span class="math-container">$\sum^q_{i=1}x_i^2 - \sum_{j=1}^{q+1}t_j^2 = -R^2$</span>.</p>
</blockquote>
<p>TWO QUESTIONS:</p>
<ol>
<li><p>Does this mean that a sphere in anti ADS space has a negative diameter? (-R²)?</p>
</li>
<li><p>Doesn't Einstein's spacetime interval metric...</p>
</li>
</ol>
<p><span class="math-container">$$d=\sqrt{x^2 + y^2 + z^2 - ct^2}$$</span></p>
<p>...meet this definition of ADS space, where p=3 and q=1?</p>
<p>In fact, why aren't all Lorentzian manifolds anti-desitter spaces?</p>

CC BY-SA 4.0
2024-07-18T14:34:16.433Z
Perhaps better fit for Physics stack?
antlersoft
CC BY-SA 4.0
2024-07-18T14:45:55.843Z
@antlersoft why? Cosmology is part of astronomy, also explicitly as by the list of things on-topic in here in this site's help center.
planetmaker
<quantum-field-theory>
Does the spacetime interval metric define spacetime as necessarily anti desitter?
Miss Understands
# Does the spacetime interval metric define spacetime as necessarily anti desitter?

(Asked by: username_0 on 2024-07-18)

[N-Lab says:](https://ncatlab.org/nlab/show/anti+de+Sitter+spacetime#definition)

> We may define the anti de Sitter space of signature $(p,q)$ as isometrically embeded in Space $\mathbb{R}^{p,q+1}$ with coordinates $(x\_1, \dots, x\_p, t\_1,\dots,t\_{q+1})$ as the sphere $\sum^q\_{i=1}x\_i^2 - \sum\_{j=1}^{q+1}t\_j^2 = -R^2$.

TWO QUESTIONS:

1. Does this mean that a sphere in anti ADS space has a negative diameter? (-R²)?
2. Doesn't Einstein's spacetime interval metric...

$$d=\sqrt{x^2 + y^2 + z^2 - ct^2}$$

...meet this definition of ADS space, where p=3 and q=1?

In fact, why aren't all Lorentzian manifolds anti-desitter spaces?

**Tags:** quantum-field-theory

> **Comments:**
> **username_1**: Perhaps better fit for Physics stack?  
> **username_2**: @username_1 why? Cosmology is part of astronomy, also explicitly as by the list of things on-topic in here in this site's help center.  

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## 2 answers

### Answer by username_3 on 2024-07-19. Score: 4 | Accepted answer

The AdS space is the sphere in $\mathbb{R}^{p,q+1}$, not $\mathbb{R}^{p,q+1}$ itself. I think you may be confusing the two.

It's similar to the model of [elliptic geometry](https://en.wikipedia.org/wiki/Elliptic_geometry) as a sphere (with antipodal points identified) in 3D Euclidean space. The existence of that model doesn't mean that elliptic geometry is Euclidean. (And it isn't Euclidean, since it violates the parallel postulate.)

> Does this mean that a sphere in anti ADS space has a negative diameter? (-R²)?

No. For one thing, it's not a sphere in AdS space, it's the entire AdS space. For another thing, you could just as well write the defining formula as $\sum\_{j=1}^{q+1}t\_j^2 - \sum^q\_{i=1}x\_i^2 = R^2$.

On the other hand, there is nothing wrong with saying that AdS space has a negative radius-squared, as long as you make it clear that you're adopting that convention.

> Doesn't Einstein's spacetime interval metric $d=\sqrt{x^2 + y^2 + z^2 - ct^2}$ meet this definition of ADS space, where p=3 and q=1?

No. Compare this to the sphere model of elliptic geometry. If you use $d=\sqrt{x^2 + y^2 + z^2}$ to define the distance between points on the sphere, you are measuring distances along straight lines in the background space that don't lie in the elliptic plane (sphere), and have nothing to do with elliptic geometry per se. They only exist "by accident" in that particular model. The reasonable way to define distance in elliptic geometry is along the elliptic plane's own lines (great circles), and that gives a different result.


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### Answer by username_4 on 2024-07-18. Score: 3

Regarding the second question:

In Minkowski space, the spacetime interval is given by

$$ds^2 = dx^2+dy^2+dz^2-c^2dt^2.$$

This just defines how you measure distances in spacetime. Note that $ds^2$ is not fixed to a certain value. (For example, $ds$ is zero for photons and nonzero for massive particles).

From this 4-D spacetime, we can now pick out all the points which satisfy

$$x^2+y^2-z^2-c^2dt^2 = -R^2$$

for some value $R\neq 0$. Not all points in Minkowski spacetime satisfy this condition, but some do. We define the set of points that satisfy this condition as anti de Sitter spacetime. In this case, the anti de Sitter spacetime is a subset of the underlying Minkowski spacetime (To drop some fancy math lingo, it is a submanifold.)

To make it easier, let's consider ordinary euclidian 2-D space. It is defined as the set of points $(x,y)$ where $x$ and $y$ are real numbers. If we now demand that $x^2+y^2=1$, we get the set of all points which lie on the unit circle, which is a subset of the 2-D euclidian space.

If we demand $x^2-y^2=1$ or $x^2-y^2=-1$, the subspace is hyperbolic (i.e. the points lie on hyperbolas.)

Regarding the first question, it sort of depends on what you mean by a sphere. A sphere is defined as the set of points whose distance to the center is some fixed value $r$. That obviously depends on what you mean by distance, i.e. it depends on the underlying metric. In minkowski coordinates, the condition for a sphere would be
$$x^2 + y^2 + z^2 - c^2t^2 = R^2$$
which is different to the condition for a hyperbolic space. So if you want to view anti de sitter space as a submanifold of an underlying minkowski space, you shouldn't think of $R$ as a radius.


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