Confused by DeSitter again!

On Thu, 7 Aug 2003, Serenus Zeitblom wrote:

I think it is pretty well-known by now that DeSitter space can be
sliced in lots of different ways, eg by flat slices or by
spherical ones.

Indeed, families of hyperslices with the geometry of S^3, E^3, and H^3
respectively are all present and valuable for various purposes.

What I don't understand is this: why does anyone take the flat slicing
seriously? I mean, it is obviously going to lead to a geodesically
incomplete spacetime---you can have timelike worldlines coming in from
"beyond the universe"! That is pretty ridiculous

It would indeed be absurd to say that by changing to a different
coordinate chart on M, you can alter an intrinsic property (geodesic
completeness). Of course the answer is that you -can't- do this; you must
be somehow confusing the notion of a coordinate chart on M (which need not
cover all of M) with M itself.

In this case, M is the de Sitter manifold H^(1,3). Examples of intrinsic
properties include its topology (M is homeomorphic to the topological
manifold RxS^3), the geodesic completeness property, and its conformal
structure, which can be diagrammed
______
|\ /|
| \ / |
| \/ |
| /\ |
| / \ |
|/____\|

(This crude ASCII sketch is a stand-in for a similar diagram in which the
diagonals have slope +/-1.)

For details concerning what I just said, see

author = {S. W. Hawking and G. F. R. Ellis},
title = {The Large Scale Structure of Space-Time},
publisher = {Cambridge University Press},
year = 1973}

In this book you will also find details of several charts; as an exercise
you can find transformations from any of those to any in the following
list which are not discussed in HE.

1. Comoving with an irrotational timelike geodesic congruence (integral
curves of a vector field X) everywhere orthogonal to a (contracting then
re-expanding) family of hyperslices with S^3 geometry:

ds^2 = -dt^2 + a^2 cosh(t/a)^2 [cos(r)^2 dz^2 + dr^2 + sin(r)^2 du^2 ],

-infty t infty, 0 r pi/2, -pi z,u pi

This chart covers almost all of M. Note that the coordinate lines z = z0,
r = r0, u = u0 correspond to the world lines of a certain family of
inertial observers, i.e. a certain family of timelike geodesics on M.
Because the geodesics "fill up an open set without intersection"-- or
better yet, because they are the integral curves of a vector field X on
M-- we call it a "congruence". Because this congruence of timelike
geodesics is irrotational (i.e. the vector field X has vanishing vorticity
tensor), it defines an orthogonal family of spatial hyperslices t = t0.
(See Hawking & Ellis for congruences and vorticity.) From the form of the
line element it is obvious that these hyperslices are three-spheres of
"radius" rho = a cosh(t/a), i.e. on -infty t 0 they contract to a
minimal size reached at t = t0, and then expand again.

It is very important to understand that this chart actually corresponds to
a whole family of different congruences. To understand this, consider the
usual embedding as a hyperboloid of one sheet of the two dimensional de
Sitter space H^(1,1). Here the "latitude circles" correspond to the
slices t = t0, with the "equatorial circle" corresponding to the "minimal
size" circle t = 0. Note that as t increases, the latitude circles shrink
to radius a and then re-expand. (This is obviously very closely analogous
to what we saw for our irrotational congruence of inertial observers
orthogonal to S^3 hyperslices above.) But now imagine applying a boost in
the embedding space E^(1,3) to our H^(1,1), moving the original latitude
circles to a new family of "tilted" circles on H^(1,1). The same boost
moves all but two of the world lines of our observers to new hyperbolas,
so we have a new irrotational congruence of timelike geodesics with a new
family of orthogonal slices (circles). Similarly on H^(1,3).

If this seems puzzling, it might help to note that one can define on E^3
not just one "cylindrical coordinate chart"

ds^2 = dz^2 + dr^2 + r^2 du^2,

-infty z infty, 0 r infty, -pi u pi

but a whole -family- of similar charts. To obtain the others: rotate the
omitted half plane u = pi, or less trivially, translate/rotate the axis of
symmetry r = 0 to a new line.

Similarly, by choosing various pairs of antipodal points on S^2 (embedded
in E^3 in the usual way) we obtain different families of "latitude
circles", so we have a whole family of "polar spherical charts" on S^2.
Given any one we can obtain the others by rotating the semicircle
("International Date Line" u = 0) or, less trivially, by choosing a new
pair of antipodal points ("North Pole" and "South Pole").

Note that the freedom to define a whole family of charts such that the
metric tensor takes the same form in terms of the various coordinates is
closely connected to the existence of symmetries (Killing vector fields)
for these familiar Riemannian manifolds.

Pedantic aside: I said above that the domain of the chart above covers
-almost- all of H^(1,3). Only "almost" because we need to delete certain
"world sheets" corresponding on t = t0 to a pair of linked great circles
(the degenerate Hopf tori at r = 0 and r = pi/2) as well as "cuts" on
each Hopf torus t = t0, r = r0 along u = 0 and along z = 0. IOW, the
failure of this chart to cover all of M is due only to the failure of the
Hopf chart to cover all of S^3, which is closely analogous to the failure
of an ordinary polar spherical chart to cover all of S^2 (which omits the
two poles), or of a cylindrical chart to cover all of E^3 (which omits a
half plane with boundary the axis of cylindrical symmetry). So this
failure to cover all of M is as harmless as the failure of a cylindrical
coordinate chart to cover all of E^3. Another way to understand this is
to observe that our vector field X is defined on all of H^(1,3), and by
moving around the omitted surfaces on each S^3 hyperslice we obtain
"trivially different" charts which taken together cover M, and are each
associated with one and the same vector field X on M.

2. Comoving with irrotational timelike geodesic congruence (integral curves
of vector field Y) everywhere orthogonal to an expanding family of spatial
hyperslices with E^3 geometry:

ds^2 = -dt^2 + exp(-2t/a) [ dx^2 + dy^2 + dz^2 ],

-infty t,x,y,z infty

(Not the same t coordinate an in (1)!) The domain of this chart covers the
region

______
|\*****|
| \****|
| \***|
| \**|
| \*|
|_____\|

Note well: even though the coordinates range over all of R^4, the chart
only covers half of M! This doesn't mean that the other half has
vanished, only that it is not represented in this chart. In particular,
even though our new irrotational timelike geodesic congruence (integral
curves of vector field Y) is defined only on half of M, our original
vector field X is of course still defined, and as the diagram suggests
half of each integral curve of X appears in our new chart.

To understand the nature of Y, look again at H^(1,1), embedded in E^(1,2)
in the usual way. This is a ruled surface with rulings comprising two
families of null geodesics in E^(1,2) (straight lines in the embedding
space), which are precisely the null geodesics on H^(1,1). On "the"
equatorial circle of H^(1,1) choose two antipodal points, and choose a
pair of "parallel" rulings. These define the boundary of the analogous
chart on H^(1,1). Note that H^(1,1) can be covered "piecewise" using
different examples of such charts (change the defining circle and
antipodal point pair).

Exercise: does each Y define a unique chart of the above form? Compare
with X vs. the S^3 slices above.

(See the discussion in Hawking and Ellis for a nice picture of the
analogous slices in H^(1,1).)

Readers familiar with hyperbolic geometry should note that these vanishing
(three-dimensional) intrinsic curvature slices of H^(1,3) are analogous to
"horospheres" in H^3 and "horocycles" in H^2, which also have vanishing
intrinsic curvature.

Exercise: Use the method of geodesic Lagrangians to find the geodesic
equations for our chart. Use the method of effective potentials to find
the following first integrals (asterisk = d/ds, s the parameter of the
geodesic to be found, e = -1,0,1 for timelike, null and spacelike
geodesics respectively):

x* = A exp(-2t/a) y* = B exp(-2t/a) z* = C exp(-2t/a)

t*^2 = e - (A^2 + B^2 + C^2) exp(-2t/a)

Explain why this shows at once that the coordinate lines mentioned above
are indeed timelike geodesics. Can you write down the complete solution
of the geodesic equations?

Exercise: Read off the obvious coframe o^1 , o^2, o^3, o^4 from the line
element (dual respectively to one timelike and three spacelike vector
fields). Take the exterior derivatives of these one-forms and guess the
connection one-forms

o^1_2 = exp(t/a) dx/a

o^1_3 = exp(t/a) dy/a

o^1_4 = exp(t/a) dz/a

Compute the covariant derivative D_X X and verify again that the
coordinate lines above are geodesics.

Exercise: Use the connection one-forms above to compute the curvature
two-forms and read off the components of the Riemann tensor wrt the given
coframe. (You should get R^1_(212) = 1/a^2 etc.; this corresponds to
geodesic divergence of our congruence, i.e. uniform negative curvature, as
expected. Note that again a appears as a "size" parameter analogous to
the "radius" of a sphere.)

Exercise: Verify that Y is irrotational and show that it has constant
expansion, to wit, in our coframe we have 1/a diag(1,1,1). Compute

E_(ab) = R_(abcd) Y^b Y^d

and verify that in our coframe this is -1/a^2 diag(1,1,1). Interpret these
results in terms of physical observations by the family of observers
corresponding to Y. Does the scaling with parameter a make sense?
(Hint: think "dilation"!)

Exercise: Look up past posts on finding the wave operator on a Lorentzian
manifold, starting with a coframe o^1, o^2, o^3, o^4. Show that the wave
equation on H^(1,3) in our chart takes the simple form

h_(tt) + 3/a h_t = exp(-2t/a) [ h_(xx) + h_(yy) + h_(zz) ]

Separate the variables, finding h(t,x,y,z) = P(t) F(x) G(y) H(z) where

F(x) = a1 exp(-A x) + a2 exp(A x)

G(y) = b1 exp(-B y) + b2 exp(B y)

H(z) = c1 sin(q z) + c2 exp(q y), q^2 = (A^2 + B^2) - E^2/a

P(t) = e1 [ exp((-t+a^(3/2) E exp(-t/a))/a) a E
- a^(1/2) exp(a^(1/2) E exp(-t/a)) ]
+ e2 [ a^(1/2) exp(-a^(1/2) E exp(-t/a))
+ exp(-(t+a^(3/2) E exp(-t/a))/a) a E ]

Here, A,B,C,D,a1,a2,b1,b2,c1,c2,e1,e2 are constants. Are they redundant?
Can you write down the general solution of the wave equation on H^(1,3)?
Next, look up past posts on computing symmetry groups of PDEs; can you
find the symmetry group of the wave equation in the form above? Choose
unidimensional subgroups and find the corresponding solutions. Explain
how these results are related to what you found using separation of
variables.

Exercise: Verify that

k = exp(-t/a) (e_1 + e_2)

defines a null geodesic congruence with expansion scalar exp(-t/a)/a and
vanishing shear and twist scalars. Interpret physically.

Exercise: Find a transformation to another chart based on Y,

ds^2 = a^2/w^2 [-dw^2 + 2 dw dx + dx^2 + dy^2 + dz^2 ],

0 w infty, -infty x,y,z infty

Analyze this following the model of the preceding exercises.

Exercise: Find a transformation to a third chart based on Y,

ds^2 = -dT^2 + (dr + r/a dT)^2 + r^2 (du^2 + sin(u)^2 dv^2),

-infty T infty, 0 r infty, 0 u pi, -pi v pi

Sketch some null geodesics and integral curves of Y. What happens to Y at
the sphere r = a? Is this a coordinate singularity? Interpret
physically. (Hint: this chart is the "zero mass" case of a Painleve chart
for the Schwarzschild-de Sitter spacetime.)

3. Comoving with irrotational timelike geodesic congruence (integral
curves of vector field Z) everywhere orthogonal to a family of spatial
hyperslices with H^3 geometry:

ds^2 = -dt^2 + a sinh(t/a)^2 [ (dx^2 + dy^2 + dz^2)/z^2 ],

0 t, z infty, -infty x, y infty

(Not the same t coordinate as in (1),(2)!) The domain of this chart covers
the region
______
|\****/|
| \**/ |
| \/ |
| /\ |
| / \ |
|/____\|

Note that the hyperslices t = t0 are each -complete- hyperbolic spaces
H^3. IOW, note that in our first slicing, the slices were compact but in
the second two they are noncompact. The discussion above, especially if
read with reference to the pictures/discussion in Hawking and Ellis,
should make it clear why this is not at all a contradiction.

We don't normally do that kind of thing; geodesic incompleteness is
tolerated only when the curvature diverges or something like that. So
why should we tolerate it in the case of DeSitter space?

Geodesic completeness is an intrinsic property enjoyed by H^(1,3). This
property holds irrespective of what coordinate charts we use to represent
this manifold. If we choose a chart whose domain does not cover the
entire manifold, we will be able to find some geodesics which reach a
"boundary" of that chart (or have coordinates which diverge) after finite
parameter lapse, but this just means we can't study the entire geodesic in
question using the given chart. To follow it outside the domain of our
chart we must switch to another chart.

A homely example: consider an ordinary polar coordinate chart

ds^2 = du^2 + sin(u)^2 dv^2,

0 u pi, -pi v pi

Note that the domain omits the North pole (u = 0), the South pole (u =
pi), and the International Date Line (the semicircular arc v = 0, aka v =
pi). Now, every latitude circle (including the equator u = pi/2, which is
a geodesic) will encounter the Date Line. This doesn't mean that S^2 is
geodesically incomplete, only that to follow latitudes across the Date
Line, we need to change to another chart, e.g. by setting v' = v + pi/2,
which gives a new polar chart in which the Date Line is just the
semicircular arc v' = pi/2, which lies in the domain of validity of the
new chart. Similarly, to follow longitudes through the North Pole, we
need to change to a new polar chart defined by a different pair of
antipodal points; e.g. choose such a pair lying on the equator of our
original chart.

Compare this example with a polar chart for H^2, and then with H^(1,1) as
above. It should be clear that whether our chart omits a "set of volume
zero" or "half the manifold", in each case we can follow geodesics or
other curves outside a given chart simply by changing to another chart
partially overlapping the first, but covering at least some of the region
omitted by it.

The only special thing there is that, being empty of matter, DeSitter is
extremely symmetric, so you can hide the incompleteness with a nice
choice of coordinates.

Both H^(1,3) and S^2 are geodesically complete. This is an intrinsic
property which is not affected by the failings of particular charts.

This is just like the Milne Universe, a so-called cosmology which is
really just a chunk of Minkowski space.

The Milne chart is analogous to chart with H^3 hyperslices above. The
fact that these charts don't cover all of E^(1,3) or H^(1,3) respectively
do not contradict the geodesic completeness of these manifolds.

The fact is, however, that the full DeSitter has spheres as slices; the
flat slicing is just a mathematical trick of no real importance,
^^^^^^^^^^

How about -utility-?

Exercise: use the method of geodesic Lagrangians to find the geodesic
equations in a chart you may be more familiar with:

ds^2 = -dt^2

+ a^2 cosh(t/a)^2 [dR^2 + sin(R)^2 (dU^2 + sin(U)^2 dV^2) ],

-infty t infty, 0 R pi, 0 U pi, -pi V pi

Compare with the equations found in the exercises above. Which would you
rather try to solve?

In short, DeSitter is REALLY a cosmology with finite spherical space.
Right?

I hope it is now abundantly clear that H^(1,3) has multiple slicings with
any of S^3, E^3 or H^3 geometry. REALLY! :-/ Each of these slicings gives
different geometrical insight into the nature of the geometry of H^(1,3),
but in various circumstances one may be more convenient or natural than
the others.

On Thu, 7 Aug 2003, Serenus Zeitblom wrote:

I think it is pretty well-known by now that DeSitter space can be
sliced in lots of different ways, eg by flat slices or by
spherical ones.

Indeed, families of hyperslices with the geometry of S^3, E^3, and H^3
respectively are all present and valuable for various purposes.

What I don't understand is this: why does anyone take the flat slicing
seriously? I mean, it is obviously going to lead to a geodesically
incomplete spacetime---you can have timelike worldlines coming in from
"beyond the universe"! That is pretty ridiculous

It would indeed be absurd to say that by changing to a different
coordinate chart on M, you can alter an intrinsic property (geodesic
completeness). Of course the answer is that you -can't- do this; you must
be somehow confusing the notion of a coordinate chart on M (which need not
cover all of M) with M itself.

In this case, M is the de Sitter manifold H^(1,3). Examples of intrinsic
properties include its topology (M is homeomorphic to the topological
manifold RxS^3), the geodesic completeness property, and its conformal
structure, which can be diagrammed
______
|\ /|
| \ / |
| \/ |
| /\ |
| / \ |
|/____\|

(This crude ASCII sketch is a stand-in for a similar diagram in which the
diagonals have slope +/-1.)

For details concerning what I just said, see

author = {S. W. Hawking and G. F. R. Ellis},
title = {The Large Scale Structure of Space-Time},
publisher = {Cambridge University Press},
year = 1973}

In this book you will also find details of several charts; as an exercise
you can find transformations from any of those to any in the following
list which are not discussed in HE.

1. Comoving with an irrotational timelike geodesic congruence (integral
curves of a vector field X) everywhere orthogonal to a (contracting then
re-expanding) family of hyperslices with S^3 geometry:

ds^2 = -dt^2 + a^2 cosh(t/a)^2 [cos(r)^2 dz^2 + dr^2 + sin(r)^2 du^2 ],

-infty t infty, 0 r pi/2, -pi z,u pi

This chart covers almost all of M. Note that the coordinate lines z = z0,
r = r0, u = u0 correspond to the world lines of a certain family of
inertial observers, i.e. a certain family of timelike geodesics on M.
Because the geodesics "fill up an open set without intersection"-- or
better yet, because they are the integral curves of a vector field X on
M-- we call it a "congruence". Because this congruence of timelike
geodesics is irrotational (i.e. the vector field X has vanishing vorticity
tensor), it defines an orthogonal family of spatial hyperslices t = t0.
(See Hawking & Ellis for congruences and vorticity.) From the form of the
line element it is obvious that these hyperslices are three-spheres of
"radius" rho = a cosh(t/a), i.e. on -infty t 0 they contract to a
minimal size reached at t = t0, and then expand again.

It is very important to understand that this chart actually corresponds to
a whole family of different congruences. To understand this, consider the
usual embedding as a hyperboloid of one sheet of the two dimensional de
Sitter space H^(1,1). Here the "latitude circles" correspond to the
slices t = t0, with the "equatorial circle" corresponding to the "minimal
size" circle t = 0. Note that as t increases, the latitude circles shrink
to radius a and then re-expand. (This is obviously very closely analogous
to what we saw for our irrotational congruence of inertial observers
orthogonal to S^3 hyperslices above.) But now imagine applying a boost in
the embedding space E^(1,3) to our H^(1,1), moving the original latitude
circles to a new family of "tilted" circles on H^(1,1). The same boost
moves all but two of the world lines of our observers to new hyperbolas,
so we have a new irrotational congruence of timelike geodesics with a new
family of orthogonal slices (circles). Similarly on H^(1,3).

If this seems puzzling, it might help to note that one can define on E^3
not just one "cylindrical coordinate chart"

ds^2 = dz^2 + dr^2 + r^2 du^2,

-infty z infty, 0 r infty, -pi u pi

but a whole -family- of similar charts. To obtain the others: rotate the
omitted half plane u = pi, or less trivially, translate/rotate the axis of
symmetry r = 0 to a new line.

Similarly, by choosing various pairs of antipodal points on S^2 (embedded
in E^3 in the usual way) we obtain different families of "latitude
circles", so we have a whole family of "polar spherical charts" on S^2.
Given any one we can obtain the others by rotating the semicircle
("International Date Line" u = 0) or, less trivially, by choosing a new
pair of antipodal points ("North Pole" and "South Pole").

Note that the freedom to define a whole family of charts such that the
metric tensor takes the same form in terms of the various coordinates is
closely connected to the existence of symmetries (Killing vector fields)
for these familiar Riemannian manifolds.

Pedantic aside: I said above that the domain of the chart above covers
-almost- all of H^(1,3). Only "almost" because we need to delete certain
"world sheets" corresponding on t = t0 to a pair of linked great circles
(the degenerate Hopf tori at r = 0 and r = pi/2) as well as "cuts" on
each Hopf torus t = t0, r = r0 along u = 0 and along z = 0. IOW, the
failure of this chart to cover all of M is due only to the failure of the
Hopf chart to cover all of S^3, which is closely analogous to the failure
of an ordinary polar spherical chart to cover all of S^2 (which omits the
two poles), or of a cylindrical chart to cover all of E^3 (which omits a
half plane with boundary the axis of cylindrical symmetry). So this
failure to cover all of M is as harmless as the failure of a cylindrical
coordinate chart to cover all of E^3. Another way to understand this is
to observe that our vector field X is defined on all of H^(1,3), and by
moving around the omitted surfaces on each S^3 hyperslice we obtain
"trivially different" charts which taken together cover M, and are each
associated with one and the same vector field X on M.

2. Comoving with irrotational timelike geodesic congruence (integral curves
of vector field Y) everywhere orthogonal to an expanding family of spatial
hyperslices with E^3 geometry:

ds^2 = -dt^2 + exp(-2t/a) [ dx^2 + dy^2 + dz^2 ],

-infty t,x,y,z infty

(Not the same t coordinate an in (1)!) The domain of this chart covers the
region

______
|\*****|
| \****|
| \***|
| \**|
| \*|
|_____\|

Note well: even though the coordinates range over all of R^4, the chart
only covers half of M! This doesn't mean that the other half has
vanished, only that it is not represented in this chart. In particular,
even though our new irrotational timelike geodesic congruence (integral
curves of vector field Y) is defined only on half of M, our original
vector field X is of course still defined, and as the diagram suggests
half of each integral curve of X appears in our new chart.

To understand the nature of Y, look again at H^(1,1), embedded in E^(1,2)
in the usual way. This is a ruled surface with rulings comprising two
families of null geodesics in E^(1,2) (straight lines in the embedding
space), which are precisely the null geodesics on H^(1,1). On "the"
equatorial circle of H^(1,1) choose two antipodal points, and choose a
pair of "parallel" rulings. These define the boundary of the analogous
chart on H^(1,1). Note that H^(1,1) can be covered "piecewise" using
different examples of such charts (change the defining circle and
antipodal point pair).

Exercise: does each Y define a unique chart of the above form? Compare
with X vs. the S^3 slices above.

(See the discussion in Hawking and Ellis for a nice picture of the
analogous slices in H^(1,1).)

Readers familiar with hyperbolic geometry should note that these vanishing
(three-dimensional) intrinsic curvature slices of H^(1,3) are analogous to
"horospheres" in H^3 and "horocycles" in H^2, which also have vanishing
intrinsic curvature.

Exercise: Use the method of geodesic Lagrangians to find the geodesic
equations for our chart. Use the method of effective potentials to find
the following first integrals (asterisk = d/ds, s the parameter of the
geodesic to be found, e = -1,0,1 for timelike, null and spacelike
geodesics respectively):

x* = A exp(-2t/a) y* = B exp(-2t/a) z* = C exp(-2t/a)

t*^2 = e - (A^2 + B^2 + C^2) exp(-2t/a)

Explain why this shows at once that the coordinate lines mentioned above
are indeed timelike geodesics. Can you write down the complete solution
of the geodesic equations?

Exercise: Read off the obvious coframe o^1 , o^2, o^3, o^4 from the line
element (dual respectively to one timelike and three spacelike vector
fields). Take the exterior derivatives of these one-forms and guess the
connection one-forms

o^1_2 = exp(t/a) dx/a

o^1_3 = exp(t/a) dy/a

o^1_4 = exp(t/a) dz/a

Compute the covariant derivative D_X X and verify again that the
coordinate lines above are geodesics.

Exercise: Use the connection one-forms above to compute the curvature
two-forms and read off the components of the Riemann tensor wrt the given
coframe. (You should get R^1_(212) = 1/a^2 etc.; this corresponds to
geodesic divergence of our congruence, i.e. uniform negative curvature, as
expected. Note that again a appears as a "size" parameter analogous to
the "radius" of a sphere.)

Exercise: Verify that Y is irrotational and show that it has constant
expansion, to wit, in our coframe we have 1/a diag(1,1,1). Compute

E_(ab) = R_(abcd) Y^b Y^d

and verify that in our coframe this is -1/a^2 diag(1,1,1). Interpret these
results in terms of physical observations by the family of observers
corresponding to Y. Does the scaling with parameter a make sense?
(Hint: think "dilation"!)

Exercise: Look up past posts on finding the wave operator on a Lorentzian
manifold, starting with a coframe o^1, o^2, o^3, o^4. Show that the wave
equation on H^(1,3) in our chart takes the simple form

h_(tt) + 3/a h_t = exp(-2t/a) [ h_(xx) + h_(yy) + h_(zz) ]

Separate the variables, finding h(t,x,y,z) = P(t) F(x) G(y) H(z) where

F(x) = a1 exp(-A x) + a2 exp(A x)

G(y) = b1 exp(-B y) + b2 exp(B y)

H(z) = c1 sin(q z) + c2 exp(q y), q^2 = (A^2 + B^2) - E^2/a

P(t) = e1 [ exp((-t+a^(3/2) E exp(-t/a))/a) a E
- a^(1/2) exp(a^(1/2) E exp(-t/a)) ]
+ e2 [ a^(1/2) exp(-a^(1/2) E exp(-t/a))
+ exp(-(t+a^(3/2) E exp(-t/a))/a) a E ]

Here, A,B,C,D,a1,a2,b1,b2,c1,c2,e1,e2 are constants. Are they redundant?
Can you write down the general solution of the wave equation on H^(1,3)?
Next, look up past posts on computing symmetry groups of PDEs; can you
find the symmetry group of the wave equation in the form above? Choose
unidimensional subgroups and find the corresponding solutions. Explain
how these results are related to what you found using separation of
variables.

Exercise: Verify that

k = exp(-t/a) (e_1 + e_2)

defines a null geodesic congruence with expansion scalar exp(-t/a)/a and
vanishing shear and twist scalars. Interpret physically.

Exercise: Find a transformation to another chart based on Y,

ds^2 = a^2/w^2 [-dw^2 + 2 dw dx + dx^2 + dy^2 + dz^2 ],

0 w infty, -infty x,y,z infty

Analyze this following the model of the preceding exercises.

Exercise: Find a transformation to a third chart based on Y,

ds^2 = -dT^2 + (dr + r/a dT)^2 + r^2 (du^2 + sin(u)^2 dv^2),

-infty T infty, 0 r infty, 0 u pi, -pi v pi

Sketch some null geodesics and integral curves of Y. What happens to Y at
the sphere r = a? Is this a coordinate singularity? Interpret
physically. (Hint: this chart is the "zero mass" case of a Painleve chart
for the Schwarzschild-de Sitter spacetime.)

3. Comoving with irrotational timelike geodesic congruence (integral
curves of vector field Z) everywhere orthogonal to a family of spatial
hyperslices with H^3 geometry:

ds^2 = -dt^2 + a sinh(t/a)^2 [ (dx^2 + dy^2 + dz^2)/z^2 ],

0 t, z infty, -infty x, y infty

(Not the same t coordinate as in (1),(2)!) The domain of this chart covers
the region
______
|\****/|
| \**/ |
| \/ |
| /\ |
| / \ |
|/____\|

Note that the hyperslices t = t0 are each -complete- hyperbolic spaces
H^3. IOW, note that in our first slicing, the slices were compact but in
the second two they are noncompact. The discussion above, especially if
read with reference to the pictures/discussion in Hawking and Ellis,
should make it clear why this is not at all a contradiction.

We don't normally do that kind of thing; geodesic incompleteness is
tolerated only when the curvature diverges or something like that. So
why should we tolerate it in the case of DeSitter space?

Geodesic completeness is an intrinsic property enjoyed by H^(1,3). This
property holds irrespective of what coordinate charts we use to represent
this manifold. If we choose a chart whose domain does not cover the
entire manifold, we will be able to find some geodesics which reach a
"boundary" of that chart (or have coordinates which diverge) after finite
parameter lapse, but this just means we can't study the entire geodesic in
question using the given chart. To follow it outside the domain of our
chart we must switch to another chart.

A homely example: consider an ordinary polar coordinate chart

ds^2 = du^2 + sin(u)^2 dv^2,

0 u pi, -pi v pi

Note that the domain omits the North pole (u = 0), the South pole (u =
pi), and the International Date Line (the semicircular arc v = 0, aka v =
pi). Now, every latitude circle (including the equator u = pi/2, which is
a geodesic) will encounter the Date Line. This doesn't mean that S^2 is
geodesically incomplete, only that to follow latitudes across the Date
Line, we need to change to another chart, e.g. by setting v' = v + pi/2,
which gives a new polar chart in which the Date Line is just the
semicircular arc v' = pi/2, which lies in the domain of validity of the
new chart. Similarly, to follow longitudes through the North Pole, we
need to change to a new polar chart defined by a different pair of
antipodal points; e.g. choose such a pair lying on the equator of our
original chart.

Compare this example with a polar chart for H^2, and then with H^(1,1) as
above. It should be clear that whether our chart omits a "set of volume
zero" or "half the manifold", in each case we can follow geodesics or
other curves outside a given chart simply by changing to another chart
partially overlapping the first, but covering at least some of the region
omitted by it.

The only special thing there is that, being empty of matter, DeSitter is
extremely symmetric, so you can hide the incompleteness with a nice
choice of coordinates.

Both H^(1,3) and S^2 are geodesically complete. This is an intrinsic
property which is not affected by the failings of particular charts.

This is just like the Milne Universe, a so-called cosmology which is
really just a chunk of Minkowski space.

The Milne chart is analogous to chart with H^3 hyperslices above. The
fact that these charts don't cover all of E^(1,3) or H^(1,3) respectively
do not contradict the geodesic completeness of these manifolds.

The fact is, however, that the full DeSitter has spheres as slices; the
flat slicing is just a mathematical trick of no real importance,
^^^^^^^^^^

How about -utility-?

Exercise: use the method of geodesic Lagrangians to find the geodesic
equations in a chart you may be more familiar with:

ds^2 = -dt^2

+ a^2 cosh(t/a)^2 [dR^2 + sin(R)^2 (dU^2 + sin(U)^2 dV^2) ],

-infty t infty, 0 R pi, 0 U pi, -pi V pi

Compare with the equations found in the exercises above. Which would you
rather try to solve?

In short, DeSitter is REALLY a cosmology with finite spherical space.
Right?

I hope it is now abundantly clear that H^(1,3) has multiple slicings with
any of S^3, E^3 or H^3 geometry. REALLY! :-/ Each of these slicings gives
different geometrical insight into the nature of the geometry of H^(1,3),
but in various circumstances one may be more convenient or natural than
the others.

On Thu, 7 Aug 2003, Serenus Zeitblom wrote:

I think it is pretty well-known by now that DeSitter space can be
sliced in lots of different ways, eg by flat slices or by
spherical ones.

Indeed, families of hyperslices with the geometry of S^3, E^3, and H^3
respectively are all present and valuable for various purposes.

What I don't understand is this: why does anyone take the flat slicing
seriously? I mean, it is obviously going to lead to a geodesically
incomplete spacetime---you can have timelike worldlines coming in from
"beyond the universe"! That is pretty ridiculous

It would indeed be absurd to say that by changing to a different
coordinate chart on M, you can alter an intrinsic property (geodesic
completeness). Of course the answer is that you -can't- do this; you must
be somehow confusing the notion of a coordinate chart on M (which need not
cover all of M) with M itself.

In this case, M is the de Sitter manifold H^(1,3). Examples of intrinsic
properties include its topology (M is homeomorphic to the topological
manifold RxS^3), the geodesic completeness property, and its conformal
structure, which can be diagrammed
______
|\ /|
| \ / |
| \/ |
| /\ |
| / \ |
|/____\|

(This crude ASCII sketch is a stand-in for a similar diagram in which the
diagonals have slope +/-1.)

For details concerning what I just said, see

author = {S. W. Hawking and G. F. R. Ellis},
title = {The Large Scale Structure of Space-Time},
publisher = {Cambridge University Press},
year = 1973}

In this book you will also find details of several charts; as an exercise
you can find transformations from any of those to any in the following
list which are not discussed in HE.

1. Comoving with an irrotational timelike geodesic congruence (integral
curves of a vector field X) everywhere orthogonal to a (contracting then
re-expanding) family of hyperslices with S^3 geometry:

ds^2 = -dt^2 + a^2 cosh(t/a)^2 [cos(r)^2 dz^2 + dr^2 + sin(r)^2 du^2 ],

-infty t infty, 0 r pi/2, -pi z,u pi

This chart covers almost all of M. Note that the coordinate lines z = z0,
r = r0, u = u0 correspond to the world lines of a certain family of
inertial observers, i.e. a certain family of timelike geodesics on M.
Because the geodesics "fill up an open set without intersection"-- or
better yet, because they are the integral curves of a vector field X on
M-- we call it a "congruence". Because this congruence of timelike
geodesics is irrotational (i.e. the vector field X has vanishing vorticity
tensor), it defines an orthogonal family of spatial hyperslices t = t0.
(See Hawking & Ellis for congruences and vorticity.) From the form of the
line element it is obvious that these hyperslices are three-spheres of
"radius" rho = a cosh(t/a), i.e. on -infty t 0 they contract to a
minimal size reached at t = t0, and then expand again.

It is very important to understand that this chart actually corresponds to
a whole family of different congruences. To understand this, consider the
usual embedding as a hyperboloid of one sheet of the two dimensional de
Sitter space H^(1,1). Here the "latitude circles" correspond to the
slices t = t0, with the "equatorial circle" corresponding to the "minimal
size" circle t = 0. Note that as t increases, the latitude circles shrink
to radius a and then re-expand. (This is obviously very closely analogous
to what we saw for our irrotational congruence of inertial observers
orthogonal to S^3 hyperslices above.) But now imagine applying a boost in
the embedding space E^(1,3) to our H^(1,1), moving the original latitude
circles to a new family of "tilted" circles on H^(1,1). The same boost
moves all but two of the world lines of our observers to new hyperbolas,
so we have a new irrotational congruence of timelike geodesics with a new
family of orthogonal slices (circles). Similarly on H^(1,3).

If this seems puzzling, it might help to note that one can define on E^3
not just one "cylindrical coordinate chart"

ds^2 = dz^2 + dr^2 + r^2 du^2,

-infty z infty, 0 r infty, -pi u pi

but a whole -family- of similar charts. To obtain the others: rotate the
omitted half plane u = pi, or less trivially, translate/rotate the axis of
symmetry r = 0 to a new line.

Similarly, by choosing various pairs of antipodal points on S^2 (embedded
in E^3 in the usual way) we obtain different families of "latitude
circles", so we have a whole family of "polar spherical charts" on S^2.
Given any one we can obtain the others by rotating the semicircle
("International Date Line" u = 0) or, less trivially, by choosing a new
pair of antipodal points ("North Pole" and "South Pole").

Note that the freedom to define a whole family of charts such that the
metric tensor takes the same form in terms of the various coordinates is
closely connected to the existence of symmetries (Killing vector fields)
for these familiar Riemannian manifolds.

Pedantic aside: I said above that the domain of the chart above covers
-almost- all of H^(1,3). Only "almost" because we need to delete certain
"world sheets" corresponding on t = t0 to a pair of linked great circles
(the degenerate Hopf tori at r = 0 and r = pi/2) as well as "cuts" on
each Hopf torus t = t0, r = r0 along u = 0 and along z = 0. IOW, the
failure of this chart to cover all of M is due only to the failure of the
Hopf chart to cover all of S^3, which is closely analogous to the failure
of an ordinary polar spherical chart to cover all of S^2 (which omits the
two poles), or of a cylindrical chart to cover all of E^3 (which omits a
half plane with boundary the axis of cylindrical symmetry). So this
failure to cover all of M is as harmless as the failure of a cylindrical
coordinate chart to cover all of E^3. Another way to understand this is
to observe that our vector field X is defined on all of H^(1,3), and by
moving around the omitted surfaces on each S^3 hyperslice we obtain
"trivially different" charts which taken together cover M, and are each
associated with one and the same vector field X on M.

2. Comoving with irrotational timelike geodesic congruence (integral curves
of vector field Y) everywhere orthogonal to an expanding family of spatial
hyperslices with E^3 geometry:

ds^2 = -dt^2 + exp(-2t/a) [ dx^2 + dy^2 + dz^2 ],

-infty t,x,y,z infty

(Not the same t coordinate an in (1)!) The domain of this chart covers the
region

______
|\*****|
| \****|
| \***|
| \**|
| \*|
|_____\|

Note well: even though the coordinates range over all of R^4, the chart
only covers half of M! This doesn't mean that the other half has
vanished, only that it is not represented in this chart. In particular,
even though our new irrotational timelike geodesic congruence (integral
curves of vector field Y) is defined only on half of M, our original
vector field X is of course still defined, and as the diagram suggests
half of each integral curve of X appears in our new chart.

To understand the nature of Y, look again at H^(1,1), embedded in E^(1,2)
in the usual way. This is a ruled surface with rulings comprising two
families of null geodesics in E^(1,2) (straight lines in the embedding
space), which are precisely the null geodesics on H^(1,1). On "the"
equatorial circle of H^(1,1) choose two antipodal points, and choose a
pair of "parallel" rulings. These define the boundary of the analogous
chart on H^(1,1). Note that H^(1,1) can be covered "piecewise" using
different examples of such charts (change the defining circle and
antipodal point pair).

Exercise: does each Y define a unique chart of the above form? Compare
with X vs. the S^3 slices above.

(See the discussion in Hawking and Ellis for a nice picture of the
analogous slices in H^(1,1).)

Readers familiar with hyperbolic geometry should note that these vanishing
(three-dimensional) intrinsic curvature slices of H^(1,3) are analogous to
"horospheres" in H^3 and "horocycles" in H^2, which also have vanishing
intrinsic curvature.

Exercise: Use the method of geodesic Lagrangians to find the geodesic
equations for our chart. Use the method of effective potentials to find
the following first integrals (asterisk = d/ds, s the parameter of the
geodesic to be found, e = -1,0,1 for timelike, null and spacelike
geodesics respectively):

x* = A exp(-2t/a) y* = B exp(-2t/a) z* = C exp(-2t/a)

t*^2 = e - (A^2 + B^2 + C^2) exp(-2t/a)

Explain why this shows at once that the coordinate lines mentioned above
are indeed timelike geodesics. Can you write down the complete solution
of the geodesic equations?

Exercise: Read off the obvious coframe o^1 , o^2, o^3, o^4 from the line
element (dual respectively to one timelike and three spacelike vector
fields). Take the exterior derivatives of these one-forms and guess the
connection one-forms

o^1_2 = exp(t/a) dx/a

o^1_3 = exp(t/a) dy/a

o^1_4 = exp(t/a) dz/a

Compute the covariant derivative D_X X and verify again that the
coordinate lines above are geodesics.

Exercise: Use the connection one-forms above to compute the curvature
two-forms and read off the components of the Riemann tensor wrt the given
coframe. (You should get R^1_(212) = 1/a^2 etc.; this corresponds to
geodesic divergence of our congruence, i.e. uniform negative curvature, as
expected. Note that again a appears as a "size" parameter analogous to
the "radius" of a sphere.)

Exercise: Verify that Y is irrotational and show that it has constant
expansion, to wit, in our coframe we have 1/a diag(1,1,1). Compute

E_(ab) = R_(abcd) Y^b Y^d

and verify that in our coframe this is -1/a^2 diag(1,1,1). Interpret these
results in terms of physical observations by the family of observers
corresponding to Y. Does the scaling with parameter a make sense?
(Hint: think "dilation"!)

Exercise: Look up past posts on finding the wave operator on a Lorentzian
manifold, starting with a coframe o^1, o^2, o^3, o^4. Show that the wave
equation on H^(1,3) in our chart takes the simple form

h_(tt) + 3/a h_t = exp(-2t/a) [ h_(xx) + h_(yy) + h_(zz) ]

Separate the variables, finding h(t,x,y,z) = P(t) F(x) G(y) H(z) where

F(x) = a1 exp(-A x) + a2 exp(A x)

G(y) = b1 exp(-B y) + b2 exp(B y)

H(z) = c1 sin(q z) + c2 exp(q y), q^2 = (A^2 + B^2) - E^2/a

P(t) = e1 [ exp((-t+a^(3/2) E exp(-t/a))/a) a E
- a^(1/2) exp(a^(1/2) E exp(-t/a)) ]
+ e2 [ a^(1/2) exp(-a^(1/2) E exp(-t/a))
+ exp(-(t+a^(3/2) E exp(-t/a))/a) a E ]

Here, A,B,C,D,a1,a2,b1,b2,c1,c2,e1,e2 are constants. Are they redundant?
Can you write down the general solution of the wave equation on H^(1,3)?
Next, look up past posts on computing symmetry groups of PDEs; can you
find the symmetry group of the wave equation in the form above? Choose
unidimensional subgroups and find the corresponding solutions. Explain
how these results are related to what you found using separation of
variables.

Exercise: Verify that

k = exp(-t/a) (e_1 + e_2)

defines a null geodesic congruence with expansion scalar exp(-t/a)/a and
vanishing shear and twist scalars. Interpret physically.

Exercise: Find a transformation to another chart based on Y,

ds^2 = a^2/w^2 [-dw^2 + 2 dw dx + dx^2 + dy^2 + dz^2 ],

0 w infty, -infty x,y,z infty

Analyze this following the model of the preceding exercises.

Exercise: Find a transformation to a third chart based on Y,

ds^2 = -dT^2 + (dr + r/a dT)^2 + r^2 (du^2 + sin(u)^2 dv^2),

-infty T infty, 0 r infty, 0 u pi, -pi v pi

Sketch some null geodesics and integral curves of Y. What happens to Y at
the sphere r = a? Is this a coordinate singularity? Interpret
physically. (Hint: this chart is the "zero mass" case of a Painleve chart
for the Schwarzschild-de Sitter spacetime.)

3. Comoving with irrotational timelike geodesic congruence (integral
curves of vector field Z) everywhere orthogonal to a family of spatial
hyperslices with H^3 geometry:

ds^2 = -dt^2 + a sinh(t/a)^2 [ (dx^2 + dy^2 + dz^2)/z^2 ],

0 t, z infty, -infty x, y infty

(Not the same t coordinate as in (1),(2)!) The domain of this chart covers
the region
______
|\****/|
| \**/ |
| \/ |
| /\ |
| / \ |
|/____\|

Note that the hyperslices t = t0 are each -complete- hyperbolic spaces
H^3. IOW, note that in our first slicing, the slices were compact but in
the second two they are noncompact. The discussion above, especially if
read with reference to the pictures/discussion in Hawking and Ellis,
should make it clear why this is not at all a contradiction.

We don't normally do that kind of thing; geodesic incompleteness is
tolerated only when the curvature diverges or something like that. So
why should we tolerate it in the case of DeSitter space?

Geodesic completeness is an intrinsic property enjoyed by H^(1,3). This
property holds irrespective of what coordinate charts we use to represent
this manifold. If we choose a chart whose domain does not cover the
entire manifold, we will be able to find some geodesics which reach a
"boundary" of that chart (or have coordinates which diverge) after finite
parameter lapse, but this just means we can't study the entire geodesic in
question using the given chart. To follow it outside the domain of our
chart we must switch to another chart.

A homely example: consider an ordinary polar coordinate chart

ds^2 = du^2 + sin(u)^2 dv^2,

0 u pi, -pi v pi

Note that the domain omits the North pole (u = 0), the South pole (u =
pi), and the International Date Line (the semicircular arc v = 0, aka v =
pi). Now, every latitude circle (including the equator u = pi/2, which is
a geodesic) will encounter the Date Line. This doesn't mean that S^2 is
geodesically incomplete, only that to follow latitudes across the Date
Line, we need to change to another chart, e.g. by setting v' = v + pi/2,
which gives a new polar chart in which the Date Line is just the
semicircular arc v' = pi/2, which lies in the domain of validity of the
new chart. Similarly, to follow longitudes through the North Pole, we
need to change to a new polar chart defined by a different pair of
antipodal points; e.g. choose such a pair lying on the equator of our
original chart.

Compare this example with a polar chart for H^2, and then with H^(1,1) as
above. It should be clear that whether our chart omits a "set of volume
zero" or "half the manifold", in each case we can follow geodesics or
other curves outside a given chart simply by changing to another chart
partially overlapping the first, but covering at least some of the region
omitted by it.

The only special thing there is that, being empty of matter, DeSitter is
extremely symmetric, so you can hide the incompleteness with a nice
choice of coordinates.

Both H^(1,3) and S^2 are geodesically complete. This is an intrinsic
property which is not affected by the failings of particular charts.

This is just like the Milne Universe, a so-called cosmology which is
really just a chunk of Minkowski space.

The Milne chart is analogous to chart with H^3 hyperslices above. The
fact that these charts don't cover all of E^(1,3) or H^(1,3) respectively
do not contradict the geodesic completeness of these manifolds.

The fact is, however, that the full DeSitter has spheres as slices; the
flat slicing is just a mathematical trick of no real importance,
^^^^^^^^^^

How about -utility-?

Exercise: use the method of geodesic Lagrangians to find the geodesic
equations in a chart you may be more familiar with:

ds^2 = -dt^2

+ a^2 cosh(t/a)^2 [dR^2 + sin(R)^2 (dU^2 + sin(U)^2 dV^2) ],

-infty t infty, 0 R pi, 0 U pi, -pi V pi

Compare with the equations found in the exercises above. Which would you
rather try to solve?

In short, DeSitter is REALLY a cosmology with finite spherical space.
Right?

I hope it is now abundantly clear that H^(1,3) has multiple slicings with
any of S^3, E^3 or H^3 geometry. REALLY! :-/ Each of these slicings gives
different geometrical insight into the nature of the geometry of H^(1,3),
but in various circumstances one may be more convenient or natural than
the others.

On Thu, 7 Aug 2003, Serenus Zeitblom wrote:

I think it is pretty well-known by now that DeSitter space can be
sliced in lots of different ways, eg by flat slices or by
spherical ones.

Indeed, families of hyperslices with the geometry of S^3, E^3, and H^3
respectively are all present and valuable for various purposes.

What I don't understand is this: why does anyone take the flat slicing
seriously? I mean, it is obviously going to lead to a geodesically
incomplete spacetime---you can have timelike worldlines coming in from
"beyond the universe"! That is pretty ridiculous

It would indeed be absurd to say that by changing to a different
coordinate chart on M, you can alter an intrinsic property (geodesic
completeness). Of course the answer is that you -can't- do this; you must
be somehow confusing the notion of a coordinate chart on M (which need not
cover all of M) with M itself.

In this case, M is the de Sitter manifold H^(1,3). Examples of intrinsic
properties include its topology (M is homeomorphic to the topological
manifold RxS^3), the geodesic completeness property, and its conformal
structure, which can be diagrammed
______
|\ /|
| \ / |
| \/ |
| /\ |
| / \ |
|/____\|

(This crude ASCII sketch is a stand-in for a similar diagram in which the
diagonals have slope +/-1.)

For details concerning what I just said, see

author = {S. W. Hawking and G. F. R. Ellis},
title = {The Large Scale Structure of Space-Time},
publisher = {Cambridge University Press},
year = 1973}

In this book you will also find details of several charts; as an exercise
you can find transformations from any of those to any in the following
list which are not discussed in HE.

1. Comoving with an irrotational timelike geodesic congruence (integral
curves of a vector field X) everywhere orthogonal to a (contracting then
re-expanding) family of hyperslices with S^3 geometry:

ds^2 = -dt^2 + a^2 cosh(t/a)^2 [cos(r)^2 dz^2 + dr^2 + sin(r)^2 du^2 ],

-infty t infty, 0 r pi/2, -pi z,u pi

This chart covers almost all of M. Note that the coordinate lines z = z0,
r = r0, u = u0 correspond to the world lines of a certain family of
inertial observers, i.e. a certain family of timelike geodesics on M.
Because the geodesics "fill up an open set without intersection"-- or
better yet, because they are the integral curves of a vector field X on
M-- we call it a "congruence". Because this congruence of timelike
geodesics is irrotational (i.e. the vector field X has vanishing vorticity
tensor), it defines an orthogonal family of spatial hyperslices t = t0.
(See Hawking & Ellis for congruences and vorticity.) From the form of the
line element it is obvious that these hyperslices are three-spheres of
"radius" rho = a cosh(t/a), i.e. on -infty t 0 they contract to a
minimal size reached at t = t0, and then expand again.

It is very important to understand that this chart actually corresponds to
a whole family of different congruences. To understand this, consider the
usual embedding as a hyperboloid of one sheet of the two dimensional de
Sitter space H^(1,1). Here the "latitude circles" correspond to the
slices t = t0, with the "equatorial circle" corresponding to the "minimal
size" circle t = 0. Note that as t increases, the latitude circles shrink
to radius a and then re-expand. (This is obviously very closely analogous
to what we saw for our irrotational congruence of inertial observers
orthogonal to S^3 hyperslices above.) But now imagine applying a boost in
the embedding space E^(1,3) to our H^(1,1), moving the original latitude
circles to a new family of "tilted" circles on H^(1,1). The same boost
moves all but two of the world lines of our observers to new hyperbolas,
so we have a new irrotational congruence of timelike geodesics with a new
family of orthogonal slices (circles). Similarly on H^(1,3).

If this seems puzzling, it might help to note that one can define on E^3
not just one "cylindrical coordinate chart"

ds^2 = dz^2 + dr^2 + r^2 du^2,

-infty z infty, 0 r infty, -pi u pi

but a whole -family- of similar charts. To obtain the others: rotate the
omitted half plane u = pi, or less trivially, translate/rotate the axis of
symmetry r = 0 to a new line.

Similarly, by choosing various pairs of antipodal points on S^2 (embedded
in E^3 in the usual way) we obtain different families of "latitude
circles", so we have a whole family of "polar spherical charts" on S^2.
Given any one we can obtain the others by rotating the semicircle
("International Date Line" u = 0) or, less trivially, by choosing a new
pair of antipodal points ("North Pole" and "South Pole").

Note that the freedom to define a whole family of charts such that the
metric tensor takes the same form in terms of the various coordinates is
closely connected to the existence of symmetries (Killing vector fields)
for these familiar Riemannian manifolds.

Pedantic aside: I said above that the domain of the chart above covers
-almost- all of H^(1,3). Only "almost" because we need to delete certain
"world sheets" corresponding on t = t0 to a pair of linked great circles
(the degenerate Hopf tori at r = 0 and r = pi/2) as well as "cuts" on
each Hopf torus t = t0, r = r0 along u = 0 and along z = 0. IOW, the
failure of this chart to cover all of M is due only to the failure of the
Hopf chart to cover all of S^3, which is closely analogous to the failure
of an ordinary polar spherical chart to cover all of S^2 (which omits the
two poles), or of a cylindrical chart to cover all of E^3 (which omits a
half plane with boundary the axis of cylindrical symmetry). So this
failure to cover all of M is as harmless as the failure of a cylindrical
coordinate chart to cover all of E^3. Another way to understand this is
to observe that our vector field X is defined on all of H^(1,3), and by
moving around the omitted surfaces on each S^3 hyperslice we obtain
"trivially different" charts which taken together cover M, and are each
associated with one and the same vector field X on M.

2. Comoving with irrotational timelike geodesic congruence (integral curves
of vector field Y) everywhere orthogonal to an expanding family of spatial
hyperslices with E^3 geometry:

ds^2 = -dt^2 + exp(-2t/a) [ dx^2 + dy^2 + dz^2 ],

-infty t,x,y,z infty

(Not the same t coordinate an in (1)!) The domain of this chart covers the
region

______
|\*****|
| \****|
| \***|
| \**|
| \*|
|_____\|

Note well: even though the coordinates range over all of R^4, the chart
only covers half of M! This doesn't mean that the other half has
vanished, only that it is not represented in this chart. In particular,
even though our new irrotational timelike geodesic congruence (integral
curves of vector field Y) is defined only on half of M, our original
vector field X is of course still defined, and as the diagram suggests
half of each integral curve of X appears in our new chart.

To understand the nature of Y, look again at H^(1,1), embedded in E^(1,2)
in the usual way. This is a ruled surface with rulings comprising two
families of null geodesics in E^(1,2) (straight lines in the embedding
space), which are precisely the null geodesics on H^(1,1). On "the"
equatorial circle of H^(1,1) choose two antipodal points, and choose a
pair of "parallel" rulings. These define the boundary of the analogous
chart on H^(1,1). Note that H^(1,1) can be covered "piecewise" using
different examples of such charts (change the defining circle and
antipodal point pair).

Exercise: does each Y define a unique chart of the above form? Compare
with X vs. the S^3 slices above.

(See the discussion in Hawking and Ellis for a nice picture of the
analogous slices in H^(1,1).)

Readers familiar with hyperbolic geometry should note that these vanishing
(three-dimensional) intrinsic curvature slices of H^(1,3) are analogous to
"horospheres" in H^3 and "horocycles" in H^2, which also have vanishing
intrinsic curvature.

Exercise: Use the method of geodesic Lagrangians to find the geodesic
equations for our chart. Use the method of effective potentials to find
the following first integrals (asterisk = d/ds, s the parameter of the
geodesic to be found, e = -1,0,1 for timelike, null and spacelike
geodesics respectively):

x* = A exp(-2t/a) y* = B exp(-2t/a) z* = C exp(-2t/a)

t*^2 = e - (A^2 + B^2 + C^2) exp(-2t/a)

Explain why this shows at once that the coordinate lines mentioned above
are indeed timelike geodesics. Can you write down the complete solution
of the geodesic equations?

Exercise: Read off the obvious coframe o^1 , o^2, o^3, o^4 from the line
element (dual respectively to one timelike and three spacelike vector
fields). Take the exterior derivatives of these one-forms and guess the
connection one-forms

o^1_2 = exp(t/a) dx/a

o^1_3 = exp(t/a) dy/a

o^1_4 = exp(t/a) dz/a

Compute the covariant derivative D_X X and verify again that the
coordinate lines above are geodesics.

Exercise: Use the connection one-forms above to compute the curvature
two-forms and read off the components of the Riemann tensor wrt the given
coframe. (You should get R^1_(212) = 1/a^2 etc.; this corresponds to
geodesic divergence of our congruence, i.e. uniform negative curvature, as
expected. Note that again a appears as a "size" parameter analogous to
the "radius" of a sphere.)

Exercise: Verify that Y is irrotational and show that it has constant
expansion, to wit, in our coframe we have 1/a diag(1,1,1). Compute

E_(ab) = R_(abcd) Y^b Y^d

and verify that in our coframe this is -1/a^2 diag(1,1,1). Interpret these
results in terms of physical observations by the family of observers
corresponding to Y. Does the scaling with parameter a make sense?
(Hint: think "dilation"!)

Exercise: Look up past posts on finding the wave operator on a Lorentzian
manifold, starting with a coframe o^1, o^2, o^3, o^4. Show that the wave
equation on H^(1,3) in our chart takes the simple form

h_(tt) + 3/a h_t = exp(-2t/a) [ h_(xx) + h_(yy) + h_(zz) ]

Separate the variables, finding h(t,x,y,z) = P(t) F(x) G(y) H(z) where

F(x) = a1 exp(-A x) + a2 exp(A x)

G(y) = b1 exp(-B y) + b2 exp(B y)

H(z) = c1 sin(q z) + c2 exp(q y), q^2 = (A^2 + B^2) - E^2/a

P(t) = e1 [ exp((-t+a^(3/2) E exp(-t/a))/a) a E
- a^(1/2) exp(a^(1/2) E exp(-t/a)) ]
+ e2 [ a^(1/2) exp(-a^(1/2) E exp(-t/a))
+ exp(-(t+a^(3/2) E exp(-t/a))/a) a E ]

Here, A,B,C,D,a1,a2,b1,b2,c1,c2,e1,e2 are constants. Are they redundant?
Can you write down the general solution of the wave equation on H^(1,3)?
Next, look up past posts on computing symmetry groups of PDEs; can you
find the symmetry group of the wave equation in the form above? Choose
unidimensional subgroups and find the corresponding solutions. Explain
how these results are related to what you found using separation of
variables.

Exercise: Verify that

k = exp(-t/a) (e_1 + e_2)

defines a null geodesic congruence with expansion scalar exp(-t/a)/a and
vanishing shear and twist scalars. Interpret physically.

Exercise: Find a transformation to another chart based on Y,

ds^2 = a^2/w^2 [-dw^2 + 2 dw dx + dx^2 + dy^2 + dz^2 ],

0 w infty, -infty x,y,z infty

Analyze this following the model of the preceding exercises.

Exercise: Find a transformation to a third chart based on Y,

ds^2 = -dT^2 + (dr + r/a dT)^2 + r^2 (du^2 + sin(u)^2 dv^2),

-infty T infty, 0 r infty, 0 u pi, -pi v pi

Sketch some null geodesics and integral curves of Y. What happens to Y at
the sphere r = a? Is this a coordinate singularity? Interpret
physically. (Hint: this chart is the "zero mass" case of a Painleve chart
for the Schwarzschild-de Sitter spacetime.)

3. Comoving with irrotational timelike geodesic congruence (integral
curves of vector field Z) everywhere orthogonal to a family of spatial
hyperslices with H^3 geometry:

ds^2 = -dt^2 + a sinh(t/a)^2 [ (dx^2 + dy^2 + dz^2)/z^2 ],

0 t, z infty, -infty x, y infty

(Not the same t coordinate as in (1),(2)!) The domain of this chart covers
the region
______
|\****/|
| \**/ |
| \/ |
| /\ |
| / \ |
|/____\|

Note that the hyperslices t = t0 are each -complete- hyperbolic spaces
H^3. IOW, note that in our first slicing, the slices were compact but in
the second two they are noncompact. The discussion above, especially if
read with reference to the pictures/discussion in Hawking and Ellis,
should make it clear why this is not at all a contradiction.

We don't normally do that kind of thing; geodesic incompleteness is
tolerated only when the curvature diverges or something like that. So
why should we tolerate it in the case of DeSitter space?

Geodesic completeness is an intrinsic property enjoyed by H^(1,3). This
property holds irrespective of what coordinate charts we use to represent
this manifold. If we choose a chart whose domain does not cover the
entire manifold, we will be able to find some geodesics which reach a
"boundary" of that chart (or have coordinates which diverge) after finite
parameter lapse, but this just means we can't study the entire geodesic in
question using the given chart. To follow it outside the domain of our
chart we must switch to another chart.

A homely example: consider an ordinary polar coordinate chart

ds^2 = du^2 + sin(u)^2 dv^2,

0 u pi, -pi v pi

Note that the domain omits the North pole (u = 0), the South pole (u =
pi), and the International Date Line (the semicircular arc v = 0, aka v =
pi). Now, every latitude circle (including the equator u = pi/2, which is
a geodesic) will encounter the Date Line. This doesn't mean that S^2 is
geodesically incomplete, only that to follow latitudes across the Date
Line, we need to change to another chart, e.g. by setting v' = v + pi/2,
which gives a new polar chart in which the Date Line is just the
semicircular arc v' = pi/2, which lies in the domain of validity of the
new chart. Similarly, to follow longitudes through the North Pole, we
need to change to a new polar chart defined by a different pair of
antipodal points; e.g. choose such a pair lying on the equator of our
original chart.

Compare this example with a polar chart for H^2, and then with H^(1,1) as
above. It should be clear that whether our chart omits a "set of volume
zero" or "half the manifold", in each case we can follow geodesics or
other curves outside a given chart simply by changing to another chart
partially overlapping the first, but covering at least some of the region
omitted by it.

The only special thing there is that, being empty of matter, DeSitter is
extremely symmetric, so you can hide the incompleteness with a nice
choice of coordinates.

Both H^(1,3) and S^2 are geodesically complete. This is an intrinsic
property which is not affected by the failings of particular charts.

This is just like the Milne Universe, a so-called cosmology which is
really just a chunk of Minkowski space.

The Milne chart is analogous to chart with H^3 hyperslices above. The
fact that these charts don't cover all of E^(1,3) or H^(1,3) respectively
do not contradict the geodesic completeness of these manifolds.

The fact is, however, that the full DeSitter has spheres as slices; the
flat slicing is just a mathematical trick of no real importance,
^^^^^^^^^^

How about -utility-?

Exercise: use the method of geodesic Lagrangians to find the geodesic
equations in a chart you may be more familiar with:

ds^2 = -dt^2

+ a^2 cosh(t/a)^2 [dR^2 + sin(R)^2 (dU^2 + sin(U)^2 dV^2) ],

-infty t infty, 0 R pi, 0 U pi, -pi V pi

Compare with the equations found in the exercises above. Which would you
rather try to solve?

In short, DeSitter is REALLY a cosmology with finite spherical space.
Right?

I hope it is now abundantly clear that H^(1,3) has multiple slicings with
any of S^3, E^3 or H^3 geometry. REALLY! :-/ Each of these slicings gives
different geometrical insight into the nature of the geometry of H^(1,3),
but in various circumstances one may be more convenient or natural than
the others.

[[Mod. note -- lines re-wrapped -- jt]]

"Chris Hillman (remoev -animal to reply)" wrote in message ...
On Thu, 7 Aug 2003, Serenus Zeitblom wrote:

I think it is pretty well-known by now that DeSitter space can be
sliced in lots of different ways, eg by flat slices or by
spherical ones.

Indeed, families of hyperslices with the geometry of S^3, E^3, and H^3
respectively are all present and valuable for various purposes.

What I don't understand is this: why does anyone take the flat slicing
seriously? I mean, it is obviously going to lead to a geodesically
incomplete spacetime---you can have timelike worldlines coming in from
"beyond the universe"! That is pretty ridiculous

It would indeed be absurd to say that by changing to a different
coordinate chart on M, you can alter an intrinsic property (geodesic
completeness). Of course the answer is that you -can't- do this; you must
be somehow confusing the notion of a coordinate chart on M (which need not
cover all of M) with M itself.

I wasn't, of course, suggesting that ALL OF deSitter can be made
geodesically incomplete just by changing coordinates. But of course any
space can be made geodesically incomplete by cutting parts of it out and
throwing them away. Sometimes, you can however do this and cover things
up by a clever choice of coordinates on the remaining piece.

What I mean is this:
I frequently hear and see people saying
[A] the WMAP data show that the spatial sections of our universe are
flat; and
[b] this, combined with the fact that the universe is accelerating,
means that,
to a first approximation [ignoring matter and radiation], the metric
of the universe can be written as

ds^2 = -dt^2 + exp(-2t/a) [ dx^2 + dy^2 + dz^2 ]

What I am objecting to is the fact that this is not a metric on
"deSitter space"
---it is a metric on a PART of deSItter space. If we cut away the
other half, then what is left over is of course geodesically
incomplete. And normally we don't do that kind of thing!

Let me put it to you this way. Suppose you believe that the spatial
sections of the universe are flat and simply connected. Then you might
observe that the above metric is a metric on R x R^3, and that it solves
the Einstein equations with a positive cosmological constant. It looks
ok in these coordinates, but IN FACT it is geodesically incomplete.
The easiest way to show this is to do as you did, draw the Penrose
diagram, and see that what you really have here is just half of the
diagram. Cutting away the other half is of course going to produce a
geodesically incomplete spacetime. R x R^3 endowed with the above
metric is geodesically incomplete, true or false?

What I'm driving at is this. If we believe that our accelerating
spacetime would be deSitter space if there were no matter or radiation
in it, then we are implicitly claiming that the topology of spacetime
is really R x S^3. Of course we can do things like take stereographic
projections, but only by cutting out and throwing away parts of the
space. Agreed?

In short, DeSitter is REALLY a cosmology with finite spherical space.
Right?

I hope it is now abundantly clear that H^(1,3) has multiple slicings with
any of S^3, E^3 or H^3 geometry. REALLY! :-/

I agree. I wasn't objecting to these well-known mathematical facts. I
was really objecting to the idea that "deSitter spacetime" can have flat
spatial sections. It can, but only if you cut half of it off and throw
it away. By the way, I am told that the Steady-State people were quite
willing to do this, precisely because they didn't believe in matter
conservation and actually welcomed the idea of stuff coming from
"outside the universe" [ie, really, from the half of deSitter they
were throwing away!] Not coincidentally, they always used the flat
slicing!

Each of these slicings gives
different geometrical insight into the nature of the geometry of H^(1,3),
but in various circumstances one may be more convenient or natural than
the others.

As long as nobody tries to tell me that PART OF deSitter space (either
the half covered by the flat slicing or the quarter covered by the
hyperbolic one) is a viable cosmology BY ITSELF, then of course I have
no objections.

[[Mod. note -- lines re-wrapped -- jt]]

"Chris Hillman (remoev -animal to reply)" wrote in message ...
On Thu, 7 Aug 2003, Serenus Zeitblom wrote:

I think it is pretty well-known by now that DeSitter space can be
sliced in lots of different ways, eg by flat slices or by
spherical ones.

Indeed, families of hyperslices with the geometry of S^3, E^3, and H^3
respectively are all present and valuable for various purposes.

What I don't understand is this: why does anyone take the flat slicing
seriously? I mean, it is obviously going to lead to a geodesically
incomplete spacetime---you can have timelike worldlines coming in from
"beyond the universe"! That is pretty ridiculous

It would indeed be absurd to say that by changing to a different
coordinate chart on M, you can alter an intrinsic property (geodesic
completeness). Of course the answer is that you -can't- do this; you must
be somehow confusing the notion of a coordinate chart on M (which need not
cover all of M) with M itself.

I wasn't, of course, suggesting that ALL OF deSitter can be made
geodesically incomplete just by changing coordinates. But of course any
space can be made geodesically incomplete by cutting parts of it out and
throwing them away. Sometimes, you can however do this and cover things
up by a clever choice of coordinates on the remaining piece.

What I mean is this:
I frequently hear and see people saying
[A] the WMAP data show that the spatial sections of our universe are
flat; and
[b] this, combined with the fact that the universe is accelerating,
means that,
to a first approximation [ignoring matter and radiation], the metric
of the universe can be written as

ds^2 = -dt^2 + exp(-2t/a) [ dx^2 + dy^2 + dz^2 ]

What I am objecting to is the fact that this is not a metric on
"deSitter space"
---it is a metric on a PART of deSItter space. If we cut away the
other half, then what is left over is of course geodesically
incomplete. And normally we don't do that kind of thing!

Let me put it to you this way. Suppose you believe that the spatial
sections of the universe are flat and simply connected. Then you might
observe that the above metric is a metric on R x R^3, and that it solves
the Einstein equations with a positive cosmological constant. It looks
ok in these coordinates, but IN FACT it is geodesically incomplete.
The easiest way to show this is to do as you did, draw the Penrose
diagram, and see that what you really have here is just half of the
diagram. Cutting away the other half is of course going to produce a
geodesically incomplete spacetime. R x R^3 endowed with the above
metric is geodesically incomplete, true or false?

What I'm driving at is this. If we believe that our accelerating
spacetime would be deSitter space if there were no matter or radiation
in it, then we are implicitly claiming that the topology of spacetime
is really R x S^3. Of course we can do things like take stereographic
projections, but only by cutting out and throwing away parts of the
space. Agreed?

In short, DeSitter is REALLY a cosmology with finite spherical space.
Right?

I hope it is now abundantly clear that H^(1,3) has multiple slicings with
any of S^3, E^3 or H^3 geometry. REALLY! :-/

I agree. I wasn't objecting to these well-known mathematical facts. I
was really objecting to the idea that "deSitter spacetime" can have flat
spatial sections. It can, but only if you cut half of it off and throw
it away. By the way, I am told that the Steady-State people were quite
willing to do this, precisely because they didn't believe in matter
conservation and actually welcomed the idea of stuff coming from
"outside the universe" [ie, really, from the half of deSitter they
were throwing away!] Not coincidentally, they always used the flat
slicing!

Each of these slicings gives
different geometrical insight into the nature of the geometry of H^(1,3),
but in various circumstances one may be more convenient or natural than
the others.

As long as nobody tries to tell me that PART OF deSitter space (either
the half covered by the flat slicing or the quarter covered by the
hyperbolic one) is a viable cosmology BY ITSELF, then of course I have
no objections.