Mixing
Can a Lorentzian manifold be compact (without time loops)?
$begingroup$
Consider a manifold with metric signature (+++-). Which means that it is a curved 4 dimensional surface but also locally is Minkowski and can be assigned light-cones.
Now, in the 3 space directions there is no problem with it being closed. e.g. it could be $mathbb{R}times S_3$ for example.
But can we also close the surface off in the `time' direction? Because there will be at least one point (possibly more) where we can't assign a light cone. e.g. at the south and north poles. (Assuming no time loops).
Is there a mathematical theory of closed Lorenzian manifolds?
(Hawking suggested bolting on a Euclidean space to close off the manifold using "imaginary time" but that doesn't really make sense to me.)
I would like to know what the mathematicians think of this.
metric-spaces manifolds
edited Jan 11 at 23:23
Paul Frost
10.3k3933
asked Jan 9 at 17:18
zoobyzooby
1,015616
$endgroup$
add a comment |
$begingroup$
Consider a manifold with metric signature (+++-). Which means that it is a curved 4 dimensional surface but also locally is Minkowski and can be assigned light-cones.
Now, in the 3 space directions there is no problem with it being closed. e.g. it could be $mathbb{R}times S_3$ for example.
But can we also close the surface off in the `time' direction? Because there will be at least one point (possibly more) where we can't assign a light cone. e.g. at the south and north poles. (Assuming no time loops).
Is there a mathematical theory of closed Lorenzian manifolds?
(Hawking suggested bolting on a Euclidean space to close off the manifold using "imaginary time" but that doesn't really make sense to me.)
I would like to know what the mathematicians think of this.
metric-spaces manifolds
edited Jan 11 at 23:23
Paul Frost
10.3k3933
asked Jan 9 at 17:18
zoobyzooby
1,015616
$endgroup$
2
$begingroup$
I think, you should rewrite your question. It appears that what you are really asking is: Does there exist a compact Lorentzian manifold which contains no time-like loops? The answer to this question is negative: Compactness implies existence of such loops. A proof is not hard, I am sure it is in O'Niel's book "Semi-Riemannian geometry". As for the literature on compact Lorentzian manifolds, it is quite large.
$endgroup$
– Moishe Cohen
Jan 11 at 18:34
add a comment |
$begingroup$
Consider a manifold with metric signature (+++-). Which means that it is a curved 4 dimensional surface but also locally is Minkowski and can be assigned light-cones.
Now, in the 3 space directions there is no problem with it being closed. e.g. it could be $mathbb{R}times S_3$ for example.
But can we also close the surface off in the `time' direction? Because there will be at least one point (possibly more) where we can't assign a light cone. e.g. at the south and north poles. (Assuming no time loops).
Is there a mathematical theory of closed Lorenzian manifolds?
(Hawking suggested bolting on a Euclidean space to close off the manifold using "imaginary time" but that doesn't really make sense to me.)
I would like to know what the mathematicians think of this.
metric-spaces manifolds
edited Jan 11 at 23:23
Paul Frost
10.3k3933
asked Jan 9 at 17:18
zoobyzooby
1,015616
$endgroup$
Consider a manifold with metric signature (+++-). Which means that it is a curved 4 dimensional surface but also locally is Minkowski and can be assigned light-cones.
Now, in the 3 space directions there is no problem with it being closed. e.g. it could be $mathbb{R}times S_3$ for example.
But can we also close the surface off in the `time' direction? Because there will be at least one point (possibly more) where we can't assign a light cone. e.g. at the south and north poles. (Assuming no time loops).
Is there a mathematical theory of closed Lorenzian manifolds?
(Hawking suggested bolting on a Euclidean space to close off the manifold using "imaginary time" but that doesn't really make sense to me.)
I would like to know what the mathematicians think of this.
metric-spaces manifolds
metric-spaces manifolds
edited Jan 11 at 23:23
Paul Frost
10.3k3933
asked Jan 9 at 17:18
zoobyzooby
1,015616
edited Jan 11 at 23:23
Paul Frost
10.3k3933
asked Jan 9 at 17:18
zoobyzooby
1,015616
edited Jan 11 at 23:23
Paul Frost
10.3k3933
edited Jan 11 at 23:23
Paul Frost
10.3k3933
edited Jan 11 at 23:23
Paul Frost
10.3k3933
10.3k3933
asked Jan 9 at 17:18
zoobyzooby
1,015616
asked Jan 9 at 17:18
zoobyzooby
1,015616
asked Jan 9 at 17:18
zoobyzooby
1,015616
1,015616
2
$begingroup$
I think, you should rewrite your question. It appears that what you are really asking is: Does there exist a compact Lorentzian manifold which contains no time-like loops? The answer to this question is negative: Compactness implies existence of such loops. A proof is not hard, I am sure it is in O'Niel's book "Semi-Riemannian geometry". As for the literature on compact Lorentzian manifolds, it is quite large.
$endgroup$
– Moishe Cohen
Jan 11 at 18:34
add a comment |
2
$begingroup$
I think, you should rewrite your question. It appears that what you are really asking is: Does there exist a compact Lorentzian manifold which contains no time-like loops? The answer to this question is negative: Compactness implies existence of such loops. A proof is not hard, I am sure it is in O'Niel's book "Semi-Riemannian geometry". As for the literature on compact Lorentzian manifolds, it is quite large.
$endgroup$
– Moishe Cohen
Jan 11 at 18:34
2
2
$begingroup$
I think, you should rewrite your question. It appears that what you are really asking is: Does there exist a compact Lorentzian manifold which contains no time-like loops? The answer to this question is negative: Compactness implies existence of such loops. A proof is not hard, I am sure it is in O'Niel's book "Semi-Riemannian geometry". As for the literature on compact Lorentzian manifolds, it is quite large.
$endgroup$
– Moishe Cohen
Jan 11 at 18:34
$begingroup$
I think, you should rewrite your question. It appears that what you are really asking is: Does there exist a compact Lorentzian manifold which contains no time-like loops? The answer to this question is negative: Compactness implies existence of such loops. A proof is not hard, I am sure it is in O'Niel's book "Semi-Riemannian geometry". As for the literature on compact Lorentzian manifolds, it is quite large.
$endgroup$
– Moishe Cohen
Jan 11 at 18:34
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It appears that you are asking the following question:
Does there exist a compact Lorentzian manifold which contains no time-like (casual) loops?
The answer to this question is negative: Compactness implies existence of such loops. A proof is not hard, see Lemma 10 on page 407 of
O'Neill's book "Semi-Riemannian geometry", which is the standard source for mathematical treatment of Lorentzian manifolds.
One can even prove that a compact Lorentzian manifold contains a casual loop which is "almost geodesic".
As for mathematical literature on compact Lorentzian manifolds, it is quite extensive. Take a look at a relatively recent paper
- S. Suhr, Closed geodesics in Lorentzian surfaces. Trans. Amer. Math. Soc. 365 (2013), no. 3, 1469–1486.
and references therein.
Here is just a couple of results concerning geodesic completeness of compact Lorentzian manifolds. Recall that, according to the Hopf-Ronow theorem, every compact Riemannian manifold is geodesically complete, i.e. every geodesic extends indefinitely. In the Lorentzian case
every compact Lorentzian manifold of constant sectional curvature is geodesically complete, see
- Y.Carrière, Autour de la conjecture de L. Markus sur les variétés affines. Invent. Math. 95 (1989), no. 3, 615–628. (zero curvature case)
and
- B. Klingler, Complétude des variétés lorentziennes à courbure constante. Math. Ann. 306 (1996), no. 2, 353–370. (general case)
Thus, every compact (locally) flat Lorentzian $(n+1)$-dimensional manifold is isometric to one of the form
$$
{mathbb R}^{n,1}/Gamma
$$
where ${mathbb R}^{n,1}$ is the standard Lorentzian space-time and $Gamma$ is a properly discontinuous (torsion-free) group of isometries of ${mathbb R}^{n,1}$. A great deal is known about the structure of such groups, for instance, it is known (W.Goldman) that $Gamma$ contains a polycyclic subgroup of finite index, i.e. is "close to" being commutative. An easy example is a flat Lorentzian metric on the $n+1$-dimensional torus (see Tsemo's answer). But there are more complex examples such that $Gamma$ contains no commutative subgroups of finite index.
In contrast, there are incomplete Lorentzian metrics on 2-dimensional tori, see this paper for a survey:
- M. Sánchez, An introduction to the completeness of compact semi-Riemannian manifolds. Séminaire de Théorie Spectrale et Géométrie, No. 13, Année 1994–1995, 37–53, Sémin. Théor. Spectr. Géom., 13, Univ. Grenoble I, Saint-Martin-d'Hères, 1995.
And here is an open problem:
Question. Is it true that every compact Lorentzian manifold contains a closed geodesic?
Lastly, because of the existence of casual loops, physicists tend to regard compact Lorentzian manifolds as non-physical. This I never understood: For all what we know, casual loops exist in "our" space-time, it's just to traverse them takes more than the life-time of our universe.
answered Jan 12 at 13:17
Moishe CohenMoishe Cohen
46.6k342108
$endgroup$
add a comment |
$begingroup$
The tranlations preserve the the flat $b$ metric of signature $(+,+,+,-)$ of $mathbb{R}^4$. $T^4$, the $4$-dimensional torus is the quotient of $mathbb{R}^4$ by four translations whose directions are independent, thus $b$ induces a Lorentzian metric on $T^4$.
answered Jan 9 at 17:23
Tsemo AristideTsemo Aristide
57.7k11445
$endgroup$
1
$begingroup$
Does that involve time loops? Are there any compact surfaces without time loops?
$endgroup$
– zooby
Jan 9 at 17:26
1
$begingroup$
@zooby Interesting new questions, but misplaced here in the comment space $ddotfrown;$ Make new posts out of it! $ddotsmile;$ Every space has its place ...
$endgroup$
– Hanno
Jan 9 at 18:49
1
$begingroup$
Well it does say that in the question.
$endgroup$
– zooby
Jan 9 at 19:32
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It appears that you are asking the following question:
Does there exist a compact Lorentzian manifold which contains no time-like (casual) loops?
The answer to this question is negative: Compactness implies existence of such loops. A proof is not hard, see Lemma 10 on page 407 of
O'Neill's book "Semi-Riemannian geometry", which is the standard source for mathematical treatment of Lorentzian manifolds.
One can even prove that a compact Lorentzian manifold contains a casual loop which is "almost geodesic".
As for mathematical literature on compact Lorentzian manifolds, it is quite extensive. Take a look at a relatively recent paper
- S. Suhr, Closed geodesics in Lorentzian surfaces. Trans. Amer. Math. Soc. 365 (2013), no. 3, 1469–1486.
and references therein.
Here is just a couple of results concerning geodesic completeness of compact Lorentzian manifolds. Recall that, according to the Hopf-Ronow theorem, every compact Riemannian manifold is geodesically complete, i.e. every geodesic extends indefinitely. In the Lorentzian case
every compact Lorentzian manifold of constant sectional curvature is geodesically complete, see
- Y.Carrière, Autour de la conjecture de L. Markus sur les variétés affines. Invent. Math. 95 (1989), no. 3, 615–628. (zero curvature case)
and
- B. Klingler, Complétude des variétés lorentziennes à courbure constante. Math. Ann. 306 (1996), no. 2, 353–370. (general case)
Thus, every compact (locally) flat Lorentzian $(n+1)$-dimensional manifold is isometric to one of the form
$$
{mathbb R}^{n,1}/Gamma
$$
where ${mathbb R}^{n,1}$ is the standard Lorentzian space-time and $Gamma$ is a properly discontinuous (torsion-free) group of isometries of ${mathbb R}^{n,1}$. A great deal is known about the structure of such groups, for instance, it is known (W.Goldman) that $Gamma$ contains a polycyclic subgroup of finite index, i.e. is "close to" being commutative. An easy example is a flat Lorentzian metric on the $n+1$-dimensional torus (see Tsemo's answer). But there are more complex examples such that $Gamma$ contains no commutative subgroups of finite index.
In contrast, there are incomplete Lorentzian metrics on 2-dimensional tori, see this paper for a survey:
- M. Sánchez, An introduction to the completeness of compact semi-Riemannian manifolds. Séminaire de Théorie Spectrale et Géométrie, No. 13, Année 1994–1995, 37–53, Sémin. Théor. Spectr. Géom., 13, Univ. Grenoble I, Saint-Martin-d'Hères, 1995.
And here is an open problem:
Question. Is it true that every compact Lorentzian manifold contains a closed geodesic?
Lastly, because of the existence of casual loops, physicists tend to regard compact Lorentzian manifolds as non-physical. This I never understood: For all what we know, casual loops exist in "our" space-time, it's just to traverse them takes more than the life-time of our universe.
answered Jan 12 at 13:17
Moishe CohenMoishe Cohen
46.6k342108
$endgroup$
add a comment |
$begingroup$
It appears that you are asking the following question:
Does there exist a compact Lorentzian manifold which contains no time-like (casual) loops?
The answer to this question is negative: Compactness implies existence of such loops. A proof is not hard, see Lemma 10 on page 407 of
O'Neill's book "Semi-Riemannian geometry", which is the standard source for mathematical treatment of Lorentzian manifolds.
One can even prove that a compact Lorentzian manifold contains a casual loop which is "almost geodesic".
As for mathematical literature on compact Lorentzian manifolds, it is quite extensive. Take a look at a relatively recent paper
- S. Suhr, Closed geodesics in Lorentzian surfaces. Trans. Amer. Math. Soc. 365 (2013), no. 3, 1469–1486.
and references therein.
Here is just a couple of results concerning geodesic completeness of compact Lorentzian manifolds. Recall that, according to the Hopf-Ronow theorem, every compact Riemannian manifold is geodesically complete, i.e. every geodesic extends indefinitely. In the Lorentzian case
every compact Lorentzian manifold of constant sectional curvature is geodesically complete, see
- Y.Carrière, Autour de la conjecture de L. Markus sur les variétés affines. Invent. Math. 95 (1989), no. 3, 615–628. (zero curvature case)
and
- B. Klingler, Complétude des variétés lorentziennes à courbure constante. Math. Ann. 306 (1996), no. 2, 353–370. (general case)
Thus, every compact (locally) flat Lorentzian $(n+1)$-dimensional manifold is isometric to one of the form
$$
{mathbb R}^{n,1}/Gamma
$$
where ${mathbb R}^{n,1}$ is the standard Lorentzian space-time and $Gamma$ is a properly discontinuous (torsion-free) group of isometries of ${mathbb R}^{n,1}$. A great deal is known about the structure of such groups, for instance, it is known (W.Goldman) that $Gamma$ contains a polycyclic subgroup of finite index, i.e. is "close to" being commutative. An easy example is a flat Lorentzian metric on the $n+1$-dimensional torus (see Tsemo's answer). But there are more complex examples such that $Gamma$ contains no commutative subgroups of finite index.
In contrast, there are incomplete Lorentzian metrics on 2-dimensional tori, see this paper for a survey:
- M. Sánchez, An introduction to the completeness of compact semi-Riemannian manifolds. Séminaire de Théorie Spectrale et Géométrie, No. 13, Année 1994–1995, 37–53, Sémin. Théor. Spectr. Géom., 13, Univ. Grenoble I, Saint-Martin-d'Hères, 1995.
And here is an open problem:
Question. Is it true that every compact Lorentzian manifold contains a closed geodesic?
Lastly, because of the existence of casual loops, physicists tend to regard compact Lorentzian manifolds as non-physical. This I never understood: For all what we know, casual loops exist in "our" space-time, it's just to traverse them takes more than the life-time of our universe.
answered Jan 12 at 13:17
Moishe CohenMoishe Cohen
46.6k342108
$endgroup$
add a comment |
$begingroup$
It appears that you are asking the following question:
Does there exist a compact Lorentzian manifold which contains no time-like (casual) loops?
The answer to this question is negative: Compactness implies existence of such loops. A proof is not hard, see Lemma 10 on page 407 of
O'Neill's book "Semi-Riemannian geometry", which is the standard source for mathematical treatment of Lorentzian manifolds.
One can even prove that a compact Lorentzian manifold contains a casual loop which is "almost geodesic".
As for mathematical literature on compact Lorentzian manifolds, it is quite extensive. Take a look at a relatively recent paper
- S. Suhr, Closed geodesics in Lorentzian surfaces. Trans. Amer. Math. Soc. 365 (2013), no. 3, 1469–1486.
and references therein.
Here is just a couple of results concerning geodesic completeness of compact Lorentzian manifolds. Recall that, according to the Hopf-Ronow theorem, every compact Riemannian manifold is geodesically complete, i.e. every geodesic extends indefinitely. In the Lorentzian case
every compact Lorentzian manifold of constant sectional curvature is geodesically complete, see
- Y.Carrière, Autour de la conjecture de L. Markus sur les variétés affines. Invent. Math. 95 (1989), no. 3, 615–628. (zero curvature case)
and
- B. Klingler, Complétude des variétés lorentziennes à courbure constante. Math. Ann. 306 (1996), no. 2, 353–370. (general case)
Thus, every compact (locally) flat Lorentzian $(n+1)$-dimensional manifold is isometric to one of the form
$$
{mathbb R}^{n,1}/Gamma
$$
where ${mathbb R}^{n,1}$ is the standard Lorentzian space-time and $Gamma$ is a properly discontinuous (torsion-free) group of isometries of ${mathbb R}^{n,1}$. A great deal is known about the structure of such groups, for instance, it is known (W.Goldman) that $Gamma$ contains a polycyclic subgroup of finite index, i.e. is "close to" being commutative. An easy example is a flat Lorentzian metric on the $n+1$-dimensional torus (see Tsemo's answer). But there are more complex examples such that $Gamma$ contains no commutative subgroups of finite index.
In contrast, there are incomplete Lorentzian metrics on 2-dimensional tori, see this paper for a survey:
- M. Sánchez, An introduction to the completeness of compact semi-Riemannian manifolds. Séminaire de Théorie Spectrale et Géométrie, No. 13, Année 1994–1995, 37–53, Sémin. Théor. Spectr. Géom., 13, Univ. Grenoble I, Saint-Martin-d'Hères, 1995.
And here is an open problem:
Question. Is it true that every compact Lorentzian manifold contains a closed geodesic?
Lastly, because of the existence of casual loops, physicists tend to regard compact Lorentzian manifolds as non-physical. This I never understood: For all what we know, casual loops exist in "our" space-time, it's just to traverse them takes more than the life-time of our universe.
answered Jan 12 at 13:17
Moishe CohenMoishe Cohen
46.6k342108
$endgroup$
It appears that you are asking the following question:
Does there exist a compact Lorentzian manifold which contains no time-like (casual) loops?
The answer to this question is negative: Compactness implies existence of such loops. A proof is not hard, see Lemma 10 on page 407 of
O'Neill's book "Semi-Riemannian geometry", which is the standard source for mathematical treatment of Lorentzian manifolds.
One can even prove that a compact Lorentzian manifold contains a casual loop which is "almost geodesic".
As for mathematical literature on compact Lorentzian manifolds, it is quite extensive. Take a look at a relatively recent paper
- S. Suhr, Closed geodesics in Lorentzian surfaces. Trans. Amer. Math. Soc. 365 (2013), no. 3, 1469–1486.
and references therein.
Here is just a couple of results concerning geodesic completeness of compact Lorentzian manifolds. Recall that, according to the Hopf-Ronow theorem, every compact Riemannian manifold is geodesically complete, i.e. every geodesic extends indefinitely. In the Lorentzian case
every compact Lorentzian manifold of constant sectional curvature is geodesically complete, see
- Y.Carrière, Autour de la conjecture de L. Markus sur les variétés affines. Invent. Math. 95 (1989), no. 3, 615–628. (zero curvature case)
and
- B. Klingler, Complétude des variétés lorentziennes à courbure constante. Math. Ann. 306 (1996), no. 2, 353–370. (general case)
Thus, every compact (locally) flat Lorentzian $(n+1)$-dimensional manifold is isometric to one of the form
$$
{mathbb R}^{n,1}/Gamma
$$
where ${mathbb R}^{n,1}$ is the standard Lorentzian space-time and $Gamma$ is a properly discontinuous (torsion-free) group of isometries of ${mathbb R}^{n,1}$. A great deal is known about the structure of such groups, for instance, it is known (W.Goldman) that $Gamma$ contains a polycyclic subgroup of finite index, i.e. is "close to" being commutative. An easy example is a flat Lorentzian metric on the $n+1$-dimensional torus (see Tsemo's answer). But there are more complex examples such that $Gamma$ contains no commutative subgroups of finite index.
In contrast, there are incomplete Lorentzian metrics on 2-dimensional tori, see this paper for a survey:
- M. Sánchez, An introduction to the completeness of compact semi-Riemannian manifolds. Séminaire de Théorie Spectrale et Géométrie, No. 13, Année 1994–1995, 37–53, Sémin. Théor. Spectr. Géom., 13, Univ. Grenoble I, Saint-Martin-d'Hères, 1995.
And here is an open problem:
Question. Is it true that every compact Lorentzian manifold contains a closed geodesic?
Lastly, because of the existence of casual loops, physicists tend to regard compact Lorentzian manifolds as non-physical. This I never understood: For all what we know, casual loops exist in "our" space-time, it's just to traverse them takes more than the life-time of our universe.
answered Jan 12 at 13:17
Moishe CohenMoishe Cohen
46.6k342108
edited Jan 12 at 18:38
answered Jan 12 at 13:17
Moishe CohenMoishe Cohen
46.6k342108
answered Jan 12 at 13:17
Moishe CohenMoishe Cohen
46.6k342108
answered Jan 12 at 13:17
Moishe CohenMoishe Cohen
46.6k342108
46.6k342108
add a comment |
add a comment |
$begingroup$
The tranlations preserve the the flat $b$ metric of signature $(+,+,+,-)$ of $mathbb{R}^4$. $T^4$, the $4$-dimensional torus is the quotient of $mathbb{R}^4$ by four translations whose directions are independent, thus $b$ induces a Lorentzian metric on $T^4$.
answered Jan 9 at 17:23
Tsemo AristideTsemo Aristide
57.7k11445
$endgroup$
1
$begingroup$
Does that involve time loops? Are there any compact surfaces without time loops?
$endgroup$
– zooby
Jan 9 at 17:26
1
$begingroup$
@zooby Interesting new questions, but misplaced here in the comment space $ddotfrown;$ Make new posts out of it! $ddotsmile;$ Every space has its place ...
$endgroup$
– Hanno
Jan 9 at 18:49
1
$begingroup$
Well it does say that in the question.
$endgroup$
– zooby
Jan 9 at 19:32
add a comment |
$begingroup$
The tranlations preserve the the flat $b$ metric of signature $(+,+,+,-)$ of $mathbb{R}^4$. $T^4$, the $4$-dimensional torus is the quotient of $mathbb{R}^4$ by four translations whose directions are independent, thus $b$ induces a Lorentzian metric on $T^4$.
answered Jan 9 at 17:23
Tsemo AristideTsemo Aristide
57.7k11445
$endgroup$
1
$begingroup$
Does that involve time loops? Are there any compact surfaces without time loops?
$endgroup$
– zooby
Jan 9 at 17:26
1
$begingroup$
@zooby Interesting new questions, but misplaced here in the comment space $ddotfrown;$ Make new posts out of it! $ddotsmile;$ Every space has its place ...
$endgroup$
– Hanno
Jan 9 at 18:49
1
$begingroup$
Well it does say that in the question.
$endgroup$
– zooby
Jan 9 at 19:32
add a comment |
$begingroup$
The tranlations preserve the the flat $b$ metric of signature $(+,+,+,-)$ of $mathbb{R}^4$. $T^4$, the $4$-dimensional torus is the quotient of $mathbb{R}^4$ by four translations whose directions are independent, thus $b$ induces a Lorentzian metric on $T^4$.
answered Jan 9 at 17:23
Tsemo AristideTsemo Aristide
57.7k11445
$endgroup$
The tranlations preserve the the flat $b$ metric of signature $(+,+,+,-)$ of $mathbb{R}^4$. $T^4$, the $4$-dimensional torus is the quotient of $mathbb{R}^4$ by four translations whose directions are independent, thus $b$ induces a Lorentzian metric on $T^4$.
answered Jan 9 at 17:23
Tsemo AristideTsemo Aristide
57.7k11445
answered Jan 9 at 17:23
Tsemo AristideTsemo Aristide
57.7k11445
answered Jan 9 at 17:23
Tsemo AristideTsemo Aristide
57.7k11445
answered Jan 9 at 17:23
Tsemo AristideTsemo Aristide
57.7k11445
57.7k11445
1
$begingroup$
Does that involve time loops? Are there any compact surfaces without time loops?
$endgroup$
– zooby
Jan 9 at 17:26
1
$begingroup$
@zooby Interesting new questions, but misplaced here in the comment space $ddotfrown;$ Make new posts out of it! $ddotsmile;$ Every space has its place ...
$endgroup$
– Hanno
Jan 9 at 18:49
1
$begingroup$
Well it does say that in the question.
$endgroup$
– zooby
Jan 9 at 19:32
add a comment |
1
$begingroup$
Does that involve time loops? Are there any compact surfaces without time loops?
$endgroup$
– zooby
Jan 9 at 17:26
1
$begingroup$
@zooby Interesting new questions, but misplaced here in the comment space $ddotfrown;$ Make new posts out of it! $ddotsmile;$ Every space has its place ...
$endgroup$
– Hanno
Jan 9 at 18:49
1
$begingroup$
Well it does say that in the question.
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– zooby
Jan 9 at 19:32
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Does that involve time loops? Are there any compact surfaces without time loops?
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– zooby
Jan 9 at 17:26
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Does that involve time loops? Are there any compact surfaces without time loops?
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– zooby
Jan 9 at 17:26
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@zooby Interesting new questions, but misplaced here in the comment space $ddotfrown;$ Make new posts out of it! $ddotsmile;$ Every space has its place ...
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– Hanno
Jan 9 at 18:49
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@zooby Interesting new questions, but misplaced here in the comment space $ddotfrown;$ Make new posts out of it! $ddotsmile;$ Every space has its place ...
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– Hanno
Jan 9 at 18:49
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Well it does say that in the question.
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– zooby
Jan 9 at 19:32
$begingroup$
Well it does say that in the question.
$endgroup$
– zooby
Jan 9 at 19:32
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I think, you should rewrite your question. It appears that what you are really asking is: Does there exist a compact Lorentzian manifold which contains no time-like loops? The answer to this question is negative: Compactness implies existence of such loops. A proof is not hard, I am sure it is in O'Niel's book "Semi-Riemannian geometry". As for the literature on compact Lorentzian manifolds, it is quite large.
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– Moishe Cohen
Jan 11 at 18:34