# I Maximally symmetric spaces

1. Jun 30, 2017

### davidge

I have a question regarding the FLRW metric used for cosmological analysis in S & G Relativity.

Let the coordinates of a point in the space time be $(t,r,\theta,\varphi)$. For constant $t, \theta$ and $\varphi$ we have the metric $$d \tau^2 = \frac{dr^2}{1 - kr^2}$$
My doubt is about this form of the metric. We know that it can describe a paraboloid, a 1-dim sphere (circle) or a plane, depending on the sign of $k$.

If $k = 1$ or $k = -1$ we have a maximally symmetric space, correct? (Namely the circle or the paraboloid, respectively.) Now are these the only possible one dimensional spaces of max symmetry? To put it another way, that form above of the metric is the unique possible form for one dimension if it is to describe a maximally symmetric space? (By one dimension I mean of course, that we need only one coordinate ($r$ in this case) to describe the space.)

2. Jun 30, 2017

### Staff: Mentor

No, we don't, because a paraboloid and a plane are 2-dimensional manifolds, while a circle is a 1-dimensional manifold. And for 1-dimensional manifolds, the concept of "maximally symmetric" is trivial. So you need to have at least two (spatial) dimensions, which means you need to include at least one of the angular coordinates.

3. Jun 30, 2017

### davidge

Is not a paraboloid defined in $\mathbb{R}^2$ as being all points $(x,y)$ such that $x^2-y^2 = r^2$? The same way, is not a circle defined as all points $(x,y)$ in $\mathbb{R}^2$ such that $x^2 + y^2 = r^2$? If the latter is intrinsically a one dimensional manifold, why can't the former be also a one dimensional manifold?

4. Jun 30, 2017

### Staff: Mentor

No. That's a parabola (a curve), not a paraboloid (a surface).

The more important point is that, as I said, the concept of "maximally symmetric" is trivial for 1-dimensional manifolds. So you need to add back one of the angular coordinates and look at those cases.

5. Jun 30, 2017

### davidge

Ok
I have noticed that $$\frac{dr^2}{1 + r^2}$$ actually corresponds to a sphere with negative radius! For $x^2 + y^2 = -1$ give that metric. Now why would we want a sphere with negative radius (that is, a imaginary radius) describing our own space?

6. Jun 30, 2017

### Staff: Mentor

No, it doesn't. The metric you wrote down is for a one-dimensional manifold. A sphere is a 2-dimensional manifold. Just because the name of the coordinate on your one-dimensional manifold is $r$ does not mean it's the "radius" of anything.

No, it doesn't. Please go back and read my previous posts, carefully.

7. Jun 30, 2017

### davidge

By sphere I mean the one dimensional sphere $\mathbb{S}^1$... the circle.
Why not? See

$\mathbb{S_{-}}^1:${$(x,y) \in \text{"some approp. 2 dim. spc."} \ | \ x^2 + y^2 = -1$}

$y^2 = - (1+x^2)$.
This implies that a parametrization of $\mathbb{C}$ into $\mathbb{S_{-}}^1$ is possible (this also shows that $\mathbb{S_{-}}^1$ is one dimensional).

$f^{(1)}_{\pm}: \mathbb{C} \longrightarrow \mathbb{S_{-}}^1 \\ t \longmapsto (t, \pm i \sqrt{1+t^2})$ and

$f^{(2)}_{\pm}: \mathbb{C} \longrightarrow \mathbb{S_{-}}^1 \\ t \longmapsto (\pm i \sqrt{1+t^2},t)$ will cover $\mathbb{S_{-}}^1$.

Now
$$dx = dt; dy = \pm \frac{it}{\sqrt{1+t^2}} \\ \Rightarrow dx^2 + dy^2 = dt^2 \bigg(1 - \frac{t^2}{1+t^2} \bigg) = \frac{dt^2}{1+t^2}$$

Last edited: Jun 30, 2017
8. Jun 30, 2017

### Staff: Mentor

There is no such thing as a circle with negative radius. If you really want to try interpreting the metric you wrote down as a curve in $\mathbb{R}^2$ such that $x^2 + y^2 = - 1$, try drawing that curve and see what you get.

9. Jun 30, 2017

### davidge

I'm not saying this is in $\mathbb{R}^2$. Please see above.

10. Jun 30, 2017

### Staff: Mentor

Because the manifold you think you are describing by $x^2 + y^2 = -1$ is not $\mathbb{S}^1$. Your "parameterization" requires complex numbers; that's not a valid parameterization (you have to use real numbers only).

11. Jun 30, 2017

### davidge

So whyyyyy the negative value of $k$ is allowed in $dr^2 / (1 - kr^2)$ if this describes a non real manifold. I dont understand it.

12. Jun 30, 2017

### Staff: Mentor

I'm sorry, you are still confused, and you don't seem to understand how to properly use the tools you are trying to use. I haven't even fully addressed your confusion yet, because I haven't had the time to look at your actual math in detail; I've just been pointing out obvious misuses of language, as well as the key point, which you still apparently have not paid attention to, that the concept of "maximally symmetric" is trivial for a 1-dimensional manifold. That makes everything you have posted so far, in which you have continued to insist on looking at 1-dimensional manifolds, a fool's errand regardless of anything else.

However, now that I have some more time, let's go through the misunderstandings of the tools in more detail. First, the expression you wrote down in the OP:

$$ds^2 = \frac{dr^2}{1 - k r^2}$$

(Note that I have written $ds^2$ instead of $d\tau^2$ because there is no need to bring in the complications of a non-positive-definite metric; the metrics we are talking about here are all positive definite.) You say that this "metric" describes a one-dimensional manifold. You also say that, for the case $k = +1$, this manifold is $\mathbb{S}^1$, the circle.

Now let's ask a simple question: in the metric as written above, what is the coordinate on the one-dimensional manifold? It is obviously $r$, correct? (I hope that's obvious.) But now let's look at your suggested parameterization of this circle in terms of $x$ and $y$: it is all points $(x, y)$ in $\mathbb{R}^2$ such that $x^2 + y^2 = r^2$. That makes $r$ the radius of the circle.

Do you see the problem yet? You are claiming that the radius of a circle is a coordinate along the circle. That is obvious nonsense. If we are going to parameterize the circle as a subset of $\mathbb{R}^2$, with $x$ and $y$ being the usual Cartesian coordinates on $\mathbb{R}^2$, then the coordinate on the circle is $\theta$, not $r$. The radius $r$ for a given circle does not change: all points on the circle have the same radius, So you obviously can't use $r$ as the coordinate on the circle.

Now let's look at your "circle with a negative radius", i.e., the case $\kappa = - 1$. Here we cannot parameterize the "circle" as a subset of $\mathbb{R}^2$ with $x^2 + y^2 = - r^2$ (note the minus sign on the right, corresponding to $k = - 1$). It should be obvious why not, but I'll say it anyway: the above expression has no solutions for real $x$, $y$, and $r$, and all three of those are real numbers.

Having done all that, let's now go back and do what I suggested several posts ago, and put back one of the angular coordinates that you left out in the OP. Here is what the metric looks like then:

$$ds^2 = \frac{dr^2}{1 - k r^2} + r^2 d\theta^2$$

This is obviously a metric on a 2-dimensional manifold which is symmetric about an "origin" $r = 0$. (The technical way of saying this is to note that neither of the metric coefficients depends on $\theta$, and therefore $\partial / \partial \theta$ is a Killing vector.) Furthermore, we can see that each of the possible values of $k$ gives a 2-dimensional manifold that we can easily recognize:

The case $k = 0$ obviously makes the manifold a Euclidean plane (the metric is just the usual flat metric in polar coordinates).

The case $k = +1$ makes the manifold a 2-sphere: the "origin" $r = 0$ can be thought of as the "North Pole" of the 2-sphere, and $r$ as the distance South along meridians, while $\theta$ is the "longitude" of the particular meridian you are on. Note that for $r = 1$ the denominator of $g_{rr}$ vanishes; this is a "coordinate singularity", which if we go through a lot of complicated math to deal with it, will end up telling is that that point $r = 1$ corresponds to the "South Pole" of the 2-sphere. Higher values of $r$ are not possible, so this manifold does not extend infinitely as the plane does. (The technical term is that this manifold is "compact", whereas the plane--and the hyperboloid, see below, are not.)

The case $k = -1$ makes the manifold something called "hyperbolic space" in 2 dimensions. This can be thought of as a hyperboloid, whose "origin" $r = 0$ is the apex of the hyperboloid; see, for example, here:

https://en.wikipedia.org/wiki/Hyperboloid_model

Note that for this case, as for the plane, there are no coordinate singularities ($g_{rr}$ is positive for all values of $r$), and the manifold extends infinitely.

Last edited: Jun 30, 2017
13. Jun 30, 2017

### Staff: Mentor

Read my previous post (#12). As you've written it it doesn't describe anything valid. Written correctly, as I do in post #12, you can see there what it describes.

Last edited: Jun 30, 2017
14. Jul 1, 2017

### davidge

I'm sorry. I should have said that I was using the same symbol $r$ in two different cases. So I was not using the radius as a coordinate. When I talked about parametrizing the circle, that $r$ comes from an open interval of $\mathbb{R}$, and when I gave the definition of the circle as all $(x,y) \in \mathbb{R}^2$ such that $x^2 + y^2 = r^2$, the $r^2$ here is a constant, the radius of the circle.

I'm sorry, but I still don't see how adding one more coordinate (how making it a two-manifold) changes the situation. In other words, I still don't see how we can derive $$ds^2 = \frac{dr^2}{1+r^2} + r^2 d \theta^2$$

15. Jul 1, 2017

### davidge

For instance, in derivation of Wikipidea, they choose $x^2+y^2+z^2+w^2 = \kappa^{-1}R^2$, where $\kappa$ can be equal to $-1$. In this case, we have again that problem you pointed out earlier, namely we have a non real situtation, where the summ of the squares of real numbers results in a negative number.

16. Jul 1, 2017

### Staff: Mentor

You're still missing the point. You wrote this in the OP:

$$ds^2 = \frac{dr^2}{1 - k r^2}$$

Here $dr$ is a coordinate differential of the coordinate $r$. You don't have a choice about that, because you obtained this expression from the standard FRW metric by setting $dt = d\theta = d\phi = 0$. That means you have to give $dr$ and $r$ the meanings they have in that metric, which are what I said. And that is all that is needed to justify the criticism that I made: that this metric, by itself, cannot describe a circle of radius $r$, because you can't use the radius of a circle as the coordinate along the circle, and $r$ is the coordinate in the 1-dimensional manifold defined by the above metric.

Um, by taking the standard FRW metric, and just setting $dt = d\phi = 0$? (I.e., by leaving one angular coordinate free.)

Let's step back even further, since it still seems like you don't understand how to properly use the tools you are trying to use. Let's just set $dt = 0$ in the standard FRW metric:

$$ds^2 = \frac{dr^2}{1 - k r^2} + r^2 \left( d\theta^2 + \sin^2 \theta d\phi^2 \right)$$

What is this the metric of? It's the metric of a 3-dimensional manifold that corresponds to "space" in the FRW spacetime at some constant value of coordinate time $t$. Depending on the value of $k$, this "space" is either a 3-sphere, flat Euclidean 3-space, or hyperbolic 3-space.

Visualizing these (at least the curved versions) is difficult, so let's drop one dimension by setting $d\phi = 0$; i.e., we are now considering a 2-dimensional surface in this 3-dimensional space defined by picking out a particular value of $\phi$ (for concreteness, let's say it's $\phi = 0$). That gives us the metric I wrote that appears to be confusing you:

$$ds^2 = \frac{dr^2}{1 - k r^2} + r^2 d\theta^2$$

Now, to make it perfectly clear what this is describing, let's first consider the easiest case to visualize: the case $k = 0$, where this metric is just describing a flat Euclidean plane. Suppose I want to pick out a circle in this plane centered on the origin. How will that circle be described? Obviously we set $r$ to some constant value (the radius of the circle), and leave $\theta$ free. In other words, the metric on this circle will be:

$$ds^2 = r^2 d\theta^2$$

Now for the key question: what if we take $k = + 1$ or $k = - 1$? What will the metric on a circle centered on the origin look like in those cases? The answer is, exactly the same as what I just wrote down. The fact that there is an extra factor in the $dr^2$ term in the metric of the 2-dimensional surface makes no difference when we pick out a circle centered on the origin and write its metric.

So what difference does the value of $k$ make? It tells us the relationship between the coordinate radius of the circle, namely $r$, and the physical radius of the circle, which will be given by the integral

$$R = \int_0^r \frac{dr'}{\sqrt{1 - k r'^2}}$$

where the integration variable is now $r'$ to make it clear that it is distinct from the coordinate value $r$ that is the upper limit of integration. For the case $k = 0$, the Euclidean plane, we have $R = r$; but for the other cases, we don't. If you work it out, you will see that $k = +1$ gives $R > r$, and the case $k = -1$ gives $R < r$. If you work things out further, you will see that $k = +1$ describes the intrinsic geometry of a 2-sphere, while the case $k = -1$ describes the intrinsic geometry of hyperbolic 2-space.

Note, however, that this physical radius $R$ makes no difference to the metric on the circle itself--that metric has $r$ in it, not $R$, which means that a unit increment of $\theta$ (i.e., one radian) gives a physical arc length along the circle of $r$. This is true regardless of the value of $k$.

So what 1-dimensional manifold do we get if we set $d\theta = 0$, so we just have the metric you wrote down in your OP? That metric, once again, is

$$ds^2 = \frac{dr^2}{1 - k r^2}$$

Let's first consider the case of the Euclidean plane, $k = 0$. The above metric then just describes a straight line through the origin (which line depends on which constant value of $\theta$ we pick when we set $d\theta = 0$--for concreteness, again, let's assume that we pick $\theta = 0$).

The case $k = -1$ is similar, except that the "line" now does not have a constant increment of physical distance for a constant increment of the coordinate $r$--each increment of $r$ gives a smaller increment of physical distance. If we look at an embedding of this manifold in a higher-dimensional Euclidean space, we will see that the manifold looks like a hyperbola (in fact it is just one hyperbolic "grid line" on the hyperbolic 2-space we obtained earlier), and the coordinate distance $r$ is arc length along the hyperbola, while the physical distance $R$ (which will be given by the same integral I wrote above) is the length along a straight line which is perpendicular to the axis of the hyperbola.

The case $k = +1$ has an extra complication, which is that the integral is undefined for $r \ge 1$. Within that limitation, we see that each constant increment of $r$ gives a larger increment of physical distance; and if we take the limit as $r \rightarrow 1$, we see that the physical distance comes out to $\pi$. If we look at an embedding of this manifold in a higher-dimensional Euclidean space, we see that it is an arc of a great circle on the 2-sphere we obtained earlier. But the radius of this great circle is not $r$ (it can't be, because $r$ is a coordinate and the radius of the great circle is a constant). Nor is it $R$ (which also varies with $r$). Nor is it even $1$, the limiting value of $r$, or $\pi$, the physical distance in the limit $r \rightarrow 1$. The actual radius of the great circle turns out to be $1/2$.

If we look at an embedding of this manifold in a higher-dimensional Euclidean space, we see that it is half of a "meridian" of a 2-sphere, starting at the North Pole, which corresponds to $r = 0$, and where the limit $r \rightarrow 1$ corresponds to the South Pole. (The actual South Pole is technically not part of the manifold since the metric is singular there.) The physical distance $R$ corresponds to arc length along this meridian. The coordinate distance $r$ corresponds to distance along the diameter of the 2-sphere that passes through the North and South Poles; but a constant increment of $r$ does not correspond to a constant increment of distance along this diameter. So it is not even true, for example, that the coordinate value $r = 1/2$ corresponds to the center of the 2-sphere (which you might erroneously think should be the case since the radius of the sphere is $1/2$, as above).

17. Jul 1, 2017

### Staff: Mentor

Since I said earlier that this was the real key point, let me explain how this works when it is done properly.

First, if we look at the three 1-dimensional manifolds (line, arc of great circle, or hyperbola) that I obtained in my previous post by gradually restricting coordinate values in the standard FRW metric, it makes no sense to ask whether they are "maximally symmetric" or not, because, as I said, the concept of "maximally symmetric" is trivial for 1-dimensional manifolds. Why? To answer that, we have to look at what "maximally symmetric" actually means.

For an example of the proper definition of "maximally symmetric", see Carroll's lecture notes, Chapter 8:

https://ned.ipac.caltech.edu/level5/March01/Carroll3/Carroll8.html

As you can see there, a maximally symmetric space is defined as a space which has the maximum possible number of Killing vectors. For a space of $n$ dimensions, this maximum possible number is $n (n + 1) / 2$. So for a 1-dimensional manifold, the maximum possible number of Killing vectors is $1$.

Now, what is a Killing vector? For our purposes right now, we don't need to go into the full-blown technical definition. It is enough to know that, if we can choose coordinates on a manifold such that the metric coefficients are independent of some coordinate $x$, then the coordinate basis vector $\partial / \partial x$ is a Killing vector. And it should be obvious that we can always choose a coordinate on any 1-dimensional manifold such that the metric coefficient (there is only one for this case) is independent of that coordinate: just choose the arc length $s$ as the coordinate, so that the metric is simply $ds^2 = ds^2$. So, as I said, the concept of "maximally symmetric" is trivial for a 1-dimensional manifold.

Now let's consider the case $n = 2$, i.e., 2-dimensional manifolds. Here the maximum possible number of Killing vectors is $3$. The canonical example of a maximally symmetric space is flat Euclidean space, so we expect flat Euclidean space to have 3 Killing vectors. And it does: 2 translations and one rotation. (More precisely, since we are talking about continuous manifolds, we would say that flat Euclidean space has a 3-parameter group of Killing vectors, of which two parameters describe the subgroup of translations--think of these as being able to put the origin at any point we want--and one parameter describes the subgroup of rotations--think of these as being able to rotate our axes about our chosen origin any way we want.)

It takes some more work to show that the 2-sphere and hyperbolic 2-space are maximally symmetric, but it can be done. (For the 2-sphere, think of the 3-parameter group of Killing vectors as being able to rotate the sphere independently about 3 mutually perpendicular axes.)

18. Jul 1, 2017

### Staff: Mentor

Yes, which means they are not doing rigorous math, they are just waving their hands. With negative $\kappa$ this equation does not describe a valid embedding of a 3-dimensional manifold in 4-dimensional Euclidean space. There are ways of making what they are doing mathematically rigorous, but doing so requires giving up the ability to call the result an embedding of the 3-dimensional manifold in 4-dimensional Euclidean space.

19. Jul 1, 2017

### davidge

First, thank you for your time in giving a complete answer to my questions. Incidentally, I'm going to give you the like number 3,000 .

About post #16: I get what you say there. The only thing I'm thinking of is about that integral you give for $R$. Wouldn't it be giving the circunference of the circle instead of its radius?

About post #17: Now I see why it's trivial to talk about max symmetry in one dimension.

Wouldn't it be wrong to do that? For instance, relativity theory don't require our universe embedded in a higher Euclidean Space.
So I'm not confortable with the need of an embedding in a one-higher dimensional space to derive the FLRW metric.

20. Jul 1, 2017

### Staff: Mentor

No. Why would you think that? Do you understand what the integral represents?

Wrong to do what? I don't understand.

There is no such need. Why do you think there is?

21. Jul 1, 2017

### davidge

That integral is the same that appears if we take an open interval of $\mathbb{R}$ and map it into $\mathbb{S}^1$. The metric will be $dt^2 / (1-t^2)$. And the circunference will be two times that integral. Note: I'm talking about a unit circle; $t$ is a element of $(-1,1)$ and this works only for $k= 1$, but it's just an example, of course.
Wrong to embbed the space in a higher dimensional space.
I don't know. You said that

22. Jul 1, 2017

### Staff: Mentor

No, it's the same that appears if you map an open interval of $\mathbb{R}$ into an open interval of $\mathbb{S}^1$. There is no mapping of an open interval of $\mathbb{R}$ into all of $\mathbb{S}^1$; that's topologically impossible. And mapping an open interval of $\mathbb{R}$ into an open interval of $\mathbb{S}^1$ is indistinguishable, just from the mapping and intervals alone, from mapping an open interval of $\mathbb{R}$ into an open interval of $\mathbb{R}$. The only way to tell the difference is if you have embeddings of both intervals into some higher dimensional space; but even then the embedding is extra structure that is not present in the mapping itself.

Where? Please give a specific reference.

23. Jul 1, 2017

### Staff: Mentor

If what you quoted just after is your specific reference, I have no idea why you think it says the same thing as you said here:

Just to be clear about the actual facts of the matter: the fact that it is possible to construct an embedding of a particular 3-dimensional space into a higher-dimensional space does not mean that such an embedding is logically necessary in order to define the 3-dimensional space. When various textbooks make use of embeddings of constant-time slices of FRW spacetime, it is for pedagogical purposes; that's all. It is perfectly possible to derive the FRW metric without making use of any embeddings; for example, see Chapter 8 of Carroll's lecture notes.

24. Jul 1, 2017

### davidge

Yes, that is what I was refering to.
Oh, I thought it was a necessary step.
You're correct. The mapping I've chosen would map the circle minus one of its points.

25. Jul 1, 2017

### Staff: Mentor

Which, if we don't make use of any embedding of either manifold in a higher dimensional space, is indistinguishable from mapping $\mathbb{R}$ into $\mathbb{R}$, as I said before.

As far as the FRW spacetime goes, the fact that we are only mapping into an open region means that, in the chart we have been using all along in this thread, all we can cover is an open region of "space" which, topologically, is $\mathbb{R}^3$. For the $k = +1$ case, in particular, we cannot cover with this chart all of the $\mathbb{S}^3$ that is the actual manifold corresponding to "space" at a constant time. (In fact, it turns out that with this chart we can't even cover all but the "South Pole" of the $\mathbb{S}^3$, i.e., we can't even cover all but one point; we can only cover an open region centered on the "North Pole" and approaching, but never reaching, the "equator". I think I misstated the analogue of this for the $\mathbb{S}^2$ case in an earlier post.)