a maximally symmetric space


by physlad
Tags: maximally, space, symmetric
physlad
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#1
May15-09, 03:38 AM
P: 22
when a space (or spacetime) is said to be maximally symmetric, does this mean that it is homogeneous?
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VKint
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#2
May17-09, 10:21 PM
P: 111
A manifold is said to be maximally symmetric if it has the same number of symmetries as ordinary Euclidean space. More formally, a space is maximally symmetric if it has [tex] \displaystyle \frac{1}{2} n(n+1) [/tex] linearly independent Killing vectors, where [tex] n [/tex] is the dimension (the same number as Euclidean space).There are a number of equivalent ways of characterizing these types of spaces; one of them involves proving that a space is maximally symmetric if and only if it is both homogeneous and isotropic. (Essentially, homogeneity is "invariance under translations," and isotropy is "invariance under rotations.")

It is also possible to prove that a manifold is maximally symmetric if and only if the Riemann tensor obeys the relation
[tex] \displaystyle R_{abcd} = \frac{R}{n(n-1)} (g_{ac} g_{bd} - g_{ad} g_{bc}) \textrm{,} [/tex]
where [tex] g_{\mu \nu} [/tex] is the metric and [tex] R [/tex] is the curvature scalar.
apeiron
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#3
May26-09, 06:04 PM
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Two quick questions:

Does this mean a flat Euclidean space is more symmetric than spaces with closed and open curvature (hyperspheric and hyperbolic)?

And second, does this relate to an axis of scale symmetry? I mean a euclidean space looks the same (homogenous) on any scale, whereas step back far enough, and a hypersphere reveals its closed loop paths, a hyperbolic space reveals its divergences.

This would seem to be a third kind of invariance as it is not about translations or rotations but the self-similarity of geometric expansions and contractions.

VKint
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#4
May28-09, 12:29 AM
P: 111

a maximally symmetric space


Does this mean a flat Euclidean space is more symmetric than spaces with closed and open curvature (hyperspheric and hyperbolic)?
It depends on what you mean by "symmetry." Generally, in this context, a "symmetry" is taken to be an isometry of the metric, i.e., a diffeomorphism which pulls back the metric to itself. If this is taken as the definition, then, as I said above, all maximally symmetric spaces share the same number of symmetries. This includes the maximally symmetric spaces of both closed and open varieties, i.e., hyperboloids and hyperspheres (in general relativity, the analogous spaces are the de Sitter and anti-de Sitter cosmologies).

However, if you also include changes of scale in the definition of symmetry, the answer is different. A "change of scale" of the type you have described is usually called a conformal mapping; more precisely, two metrics [tex] g [/tex] and [tex] \tilde{g} [/tex] are conformally related if there exists a (nonvanishing, smooth) function [tex] \omega [/tex] of the coordinate functions such that [tex] \tilde{g} = \omega^2 g [/tex]. The idea of "symmetry under changes of scale" can then be articulated in terms of the existence of a diffeomorphism [tex] \varphi : M \to M [/tex] (where [tex] M [/tex] is the manifold) such that [tex] \varphi^{*} g [/tex] (i.e., the pullback of the metric by [tex] \varphi [/tex]) is conformally related to [tex] g [/tex]. Thus, a space is said to have a conformal symmetry if there is a diffeomorphism [tex] \varphi [/tex] and a smooth nonvanishing function [tex] \omega [/tex] such that [tex] \varphi^{*} g = \omega^2 g [/tex]. The map [tex] \varphi [/tex] is called a conformal isometry, by analogy with regular isometries.

Another way to phrase this is in terms of Killing fields. You may be familiar with ordinary Killing vector fields, which are defined as the generators of isometries and obey the equation [tex] \mathcal{L}_K g = 0 [/tex], where [tex] K [/tex] is the Killing field (i.e., [tex] K [/tex] Lie-transports the metric [tex] g [/tex]). In coordinates, this reduces to [tex] \nabla_{(a} K_{b)} = 0 [/tex]. Similarly, a conformal Killing vector [tex] V [/tex] is defined as the generator of a conformal isometry: [tex] \mathcal{L}_V g = \omega^2 g [/tex]. We can evaluate what [tex] \omega^2 [/tex] must be explicitly by expanding this in coordinates and taking a trace. In the end, we get
[tex]
\nabla_{(a} V_{b)} = \frac{2}{n} (\nabla^c V_c) g_{ab} \textrm{,}
[/tex]
which is called the conformal Killing equation. Since [tex] \nabla [/tex] is just the ordinary partial derivative operator on Euclidean space, any constant vector field on [tex] \mathbb{R}^n [/tex] is a conformal Killing field (and, in fact, an ordinary Killing field, too, though these are not the only Killing fields in this case). Thus, in addition to a full complement of ordinary Killing vectors, Euclidean space is also blessed with a maximal collection of conformal Killing vectors, making it "uber-maximally symmetric," unlike hyperspheres and hyperboloids. So, indeed, there is a "third kind of invariance," as you say, having to do with changes of scale, that Euclidean space possesses but other so-called maximally symmetric spaces do not.
apeiron
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#5
May29-09, 04:19 AM
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Quote Quote by VKint View Post
Thus, in addition to a full complement of ordinary Killing vectors, Euclidean space is also blessed with a maximal collection of conformal Killing vectors, making it "uber-maximally symmetric," unlike hyperspheres and hyperboloids. So, indeed, there is a "third kind of invariance," as you say, having to do with changes of scale, that Euclidean space possesses but other so-called maximally symmetric spaces do not.
Thanks for a quick introduction to the jargon in this area.

It seems significant that flat euclidean geometry is picked out in this regard.

Noether would say for every symmetry there is an inertia or conservation law, so what is law that results from scale symmetry?
VKint
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#6
May31-09, 06:15 PM
P: 111
Noether would say for every symmetry there is an inertia or conservation law, so what is law that results from scale symmetry?
There are a few problems with attempting to invoke Noether's Theorem to produce a symmetry from conformal Killing vectors. First, in order for Noether's Theorem to apply, you need a differentiable symmetry of an action or Lagrangian of a physical system. Thus, Killing vectors and conformal Killing vectors do not give rise to conserved quantities for just any path through a manifold, but they do if the path can be described via an action proportional to the metric, such as the action defining geodesics:
[tex]
I = \frac{1}{2} \int g_{ab} \frac{dx^a}{d\tau} \frac{dx^b}{d\tau} \textrm{.}
[/tex]
Furthermore, the type of "symmetry" indicated by the existence of conformal Killing vectors isn't exactly a "symmetry" in the sense of Noether's Theorem, since the conformal Killing equation doesn't quite give the invariance of the metric that Noether's Theorem requires. Instead, a conformal symmetry is defined as a diffeomorphism which changes the metric, but only in a certain allowed fashion, i.e., [tex] \varphi^{*} g = \omega^2 g [/tex]. In order to apply Noether's Theorem, it is my (admittedly limited) understanding that you would need something like [tex] \varphi^{*} g = g [/tex], which defines an ordinary Killing vector.

Despite all this, the presence of conformal Killing vectors is in fact enough to guarantee that certain quantities will be conserved, but only for a very special class of paths, i.e., null geodesics. (Obviously, this only applies to Lorentzian manifolds. I have no idea whether or not you can use a conformal Killing vector to construct an invariant in an ordinary Riemannian setting.) First, note that for an ordinary Killing vector [tex] K^a [/tex], if [tex] T^b [/tex] denotes the tangent vector to a unit-speed geodesic, we have
[tex]
T^a \nabla_a (T^b K_b) = T^a T^b K_{b;a} + K_b T^a T \indices{^b_{;a}} = 0
[/tex]
(by Killing's equation and the geodesic equation). Thus, the quantity [tex] L = T^b K_b [/tex] is conserved along geodesics. This holds for all Riemannian or pseudo-Riemannian metrics, and gives rise to such classical laws as the conservation of momentum, energy, and angular momentum in flat space.

A similar calculation for conformal geodesics [tex] Y^b [/tex], using the equation I gave in the last post, gives
[tex]
T^a \nabla_a (T^b Y_b) = \frac{1}{n} (\nabla^c Y_c) T^a T_a \textrm{.}
[/tex]
Thus, [tex] L [/tex] is not conserved for conformal Killing fields. However, if [tex] T [/tex] is the tangent to a null geodesic, then [tex] T^a T_a = 0 [/tex], so [tex] L [/tex] is again conserved. I think this is about as close to a Noether current as you can get from a conformal Killing vector. I haven't seen this applied in any specific examples, so I can't at the moment describe in physical terms precisely what this indicates for null paths of interest. I'll get back to you if I have any flashes of inspiration.
apeiron
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#7
May31-09, 06:57 PM
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Thanks again for a lengthy reply. I'm afraid I can only get about a fifth of the maths and will have to read up on conformal killing fields. I'm coming at this question from a familiarity with fractals, powerlaws and thermodynamics.

The physical relevance of the question is the problem of the flatness of the expanding universe. Usually it is taken that it is a surprise the universe is (asymptotically) flat rather than curved positively/negatively. This seems an unnaturally poised situation.

But the alternative view would be that flatness is instead the self-selecting outcome for some reason. Because it has the highest symmetry (having scale symmetry as well) it is the trough state, the equilibrium balance.

Putting it simply, the other noether symmetries are you can go left or right, spin clockwise or anticlockwise, and go backwards or forwards in time (a little questionable this one).

Then in a fractal realm, you would be free to shrink or expand. Go either direction in this sense with no change. An observer moving across scale would still look out and see the same world, the event would still have the same relationship to the context.

A light cone would seem to have this kind of "inertia", this smooth expansion in scale. The intensity of a force wanes with an inverse square law, but the total force is a conserved quantity. So there seems a connection there.


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