Topological consequences of negative and zero Einstein condition
$begingroup$
Let $(M,g)$ be a complete Riemannian manifold which is Einstein, i.e. $mathrm{Ric}=kg$ for some constant $kin mathbb{R}$.
1) If $k<0$, is $M$ then necessarily noncompact? If so, does the condition $k<0$ give any other topological restraints?
All examples of negative Einstein manifolds that I know are noncompact (e.g. hyperbolic space). Note that when $k>0$, $M$ must be compact by Myers' theorem. I know that any Riemannian manifold admits a complete metric with $mathrm{Ric}<0$, but I suspect that the Einstein condition is much more restrictive.
2) Does the condition $k=0$ give any topological information? For example, can $S^n$ admit a metric with $k=0$?
geometry differential-geometry riemannian-geometry smooth-manifolds
edited Jan 18 at 19:41
Bernard
121k740116
asked Jan 18 at 19:06
srpsrp
2728
$endgroup$
add a comment |
$begingroup$
Let $(M,g)$ be a complete Riemannian manifold which is Einstein, i.e. $mathrm{Ric}=kg$ for some constant $kin mathbb{R}$.
1) If $k<0$, is $M$ then necessarily noncompact? If so, does the condition $k<0$ give any other topological restraints?
All examples of negative Einstein manifolds that I know are noncompact (e.g. hyperbolic space). Note that when $k>0$, $M$ must be compact by Myers' theorem. I know that any Riemannian manifold admits a complete metric with $mathrm{Ric}<0$, but I suspect that the Einstein condition is much more restrictive.
2) Does the condition $k=0$ give any topological information? For example, can $S^n$ admit a metric with $k=0$?
geometry differential-geometry riemannian-geometry smooth-manifolds
edited Jan 18 at 19:41
Bernard
121k740116
asked Jan 18 at 19:06
srpsrp
2728
$endgroup$
$begingroup$
I think (2) is a famous open question.
$endgroup$
– Mike Miller
Jan 18 at 21:49
$begingroup$
@MikeMiller: Which part of (2) do you think is an open question, i.e. the part about $S^n$, or the first part?
$endgroup$
– srp
Jan 19 at 0:20
1
$begingroup$
I am quite confident at least that it is open for spheres. As of Besse, "Einstein manifolds", 1987, I may quote 6.3: "In dimensions greater than 4, we do not know any topological restriction for a manifold to be Einstein. It may very well be that any manifold with dimension greater than 4 admits a negative Einstein metric - or, that most manifolds do." I would need to search the literature to know if any partial progress has been made.
$endgroup$
– Mike Miller
Jan 19 at 0:40
1
$begingroup$
Oh, (1) is definitively false. See Aubin (and separately Yau) for the resolution of the Calabi conjecture for Kahler manifolds with $c_1(M, J)^n < 0$ (it is said that $c_1$ is negative). This provides Kahler-Einstein metrics with negative Einstein constant. This is Theorem 11.17 listed in Besse.
$endgroup$
– Mike Miller
Jan 19 at 0:51
add a comment |
$begingroup$
Let $(M,g)$ be a complete Riemannian manifold which is Einstein, i.e. $mathrm{Ric}=kg$ for some constant $kin mathbb{R}$.
1) If $k<0$, is $M$ then necessarily noncompact? If so, does the condition $k<0$ give any other topological restraints?
All examples of negative Einstein manifolds that I know are noncompact (e.g. hyperbolic space). Note that when $k>0$, $M$ must be compact by Myers' theorem. I know that any Riemannian manifold admits a complete metric with $mathrm{Ric}<0$, but I suspect that the Einstein condition is much more restrictive.
2) Does the condition $k=0$ give any topological information? For example, can $S^n$ admit a metric with $k=0$?
geometry differential-geometry riemannian-geometry smooth-manifolds
edited Jan 18 at 19:41
Bernard
121k740116
asked Jan 18 at 19:06
srpsrp
2728
$endgroup$
Let $(M,g)$ be a complete Riemannian manifold which is Einstein, i.e. $mathrm{Ric}=kg$ for some constant $kin mathbb{R}$.
1) If $k<0$, is $M$ then necessarily noncompact? If so, does the condition $k<0$ give any other topological restraints?
All examples of negative Einstein manifolds that I know are noncompact (e.g. hyperbolic space). Note that when $k>0$, $M$ must be compact by Myers' theorem. I know that any Riemannian manifold admits a complete metric with $mathrm{Ric}<0$, but I suspect that the Einstein condition is much more restrictive.
2) Does the condition $k=0$ give any topological information? For example, can $S^n$ admit a metric with $k=0$?
geometry differential-geometry riemannian-geometry smooth-manifolds
geometry differential-geometry riemannian-geometry smooth-manifolds
edited Jan 18 at 19:41
Bernard
121k740116
asked Jan 18 at 19:06
srpsrp
2728
edited Jan 18 at 19:41
Bernard
121k740116
asked Jan 18 at 19:06
srpsrp
2728
edited Jan 18 at 19:41
Bernard
121k740116
edited Jan 18 at 19:41
Bernard
121k740116
edited Jan 18 at 19:41
Bernard
121k740116
121k740116
asked Jan 18 at 19:06
srpsrp
2728
asked Jan 18 at 19:06
srpsrp
2728
asked Jan 18 at 19:06
srpsrp
2728
2728
$begingroup$
I think (2) is a famous open question.
$endgroup$
– Mike Miller
Jan 18 at 21:49
$begingroup$
@MikeMiller: Which part of (2) do you think is an open question, i.e. the part about $S^n$, or the first part?
$endgroup$
– srp
Jan 19 at 0:20
1
$begingroup$
I am quite confident at least that it is open for spheres. As of Besse, "Einstein manifolds", 1987, I may quote 6.3: "In dimensions greater than 4, we do not know any topological restriction for a manifold to be Einstein. It may very well be that any manifold with dimension greater than 4 admits a negative Einstein metric - or, that most manifolds do." I would need to search the literature to know if any partial progress has been made.
$endgroup$
– Mike Miller
Jan 19 at 0:40
1
$begingroup$
Oh, (1) is definitively false. See Aubin (and separately Yau) for the resolution of the Calabi conjecture for Kahler manifolds with $c_1(M, J)^n < 0$ (it is said that $c_1$ is negative). This provides Kahler-Einstein metrics with negative Einstein constant. This is Theorem 11.17 listed in Besse.
$endgroup$
– Mike Miller
Jan 19 at 0:51
add a comment |
$begingroup$
I think (2) is a famous open question.
$endgroup$
– Mike Miller
Jan 18 at 21:49
$begingroup$
@MikeMiller: Which part of (2) do you think is an open question, i.e. the part about $S^n$, or the first part?
$endgroup$
– srp
Jan 19 at 0:20
1
$begingroup$
I am quite confident at least that it is open for spheres. As of Besse, "Einstein manifolds", 1987, I may quote 6.3: "In dimensions greater than 4, we do not know any topological restriction for a manifold to be Einstein. It may very well be that any manifold with dimension greater than 4 admits a negative Einstein metric - or, that most manifolds do." I would need to search the literature to know if any partial progress has been made.
$endgroup$
– Mike Miller
Jan 19 at 0:40
1
$begingroup$
Oh, (1) is definitively false. See Aubin (and separately Yau) for the resolution of the Calabi conjecture for Kahler manifolds with $c_1(M, J)^n < 0$ (it is said that $c_1$ is negative). This provides Kahler-Einstein metrics with negative Einstein constant. This is Theorem 11.17 listed in Besse.
$endgroup$
– Mike Miller
Jan 19 at 0:51
$begingroup$
I think (2) is a famous open question.
$endgroup$
– Mike Miller
Jan 18 at 21:49
$begingroup$
I think (2) is a famous open question.
$endgroup$
– Mike Miller
Jan 18 at 21:49
$begingroup$
@MikeMiller: Which part of (2) do you think is an open question, i.e. the part about $S^n$, or the first part?
$endgroup$
– srp
Jan 19 at 0:20
$begingroup$
@MikeMiller: Which part of (2) do you think is an open question, i.e. the part about $S^n$, or the first part?
$endgroup$
– srp
Jan 19 at 0:20
1
1
$begingroup$
I am quite confident at least that it is open for spheres. As of Besse, "Einstein manifolds", 1987, I may quote 6.3: "In dimensions greater than 4, we do not know any topological restriction for a manifold to be Einstein. It may very well be that any manifold with dimension greater than 4 admits a negative Einstein metric - or, that most manifolds do." I would need to search the literature to know if any partial progress has been made.
$endgroup$
– Mike Miller
Jan 19 at 0:40
$begingroup$
I am quite confident at least that it is open for spheres. As of Besse, "Einstein manifolds", 1987, I may quote 6.3: "In dimensions greater than 4, we do not know any topological restriction for a manifold to be Einstein. It may very well be that any manifold with dimension greater than 4 admits a negative Einstein metric - or, that most manifolds do." I would need to search the literature to know if any partial progress has been made.
$endgroup$
– Mike Miller
Jan 19 at 0:40
1
1
$begingroup$
Oh, (1) is definitively false. See Aubin (and separately Yau) for the resolution of the Calabi conjecture for Kahler manifolds with $c_1(M, J)^n < 0$ (it is said that $c_1$ is negative). This provides Kahler-Einstein metrics with negative Einstein constant. This is Theorem 11.17 listed in Besse.
$endgroup$
– Mike Miller
Jan 19 at 0:51
$begingroup$
Oh, (1) is definitively false. See Aubin (and separately Yau) for the resolution of the Calabi conjecture for Kahler manifolds with $c_1(M, J)^n < 0$ (it is said that $c_1$ is negative). This provides Kahler-Einstein metrics with negative Einstein constant. This is Theorem 11.17 listed in Besse.
$endgroup$
– Mike Miller
Jan 19 at 0:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Everything here is from Besse's book "Einstein manifolds". If you are interested in the subject, you owe it to yourself to have a copy nearby.
(1) is false for a number of reasons.
In dimension 2, negative Ricci curvature is the same thing as negative scalar curvature (and Einstein is the same as constant scalar curvature), so every closed surface of genus $g > 1$ admits a negative Einstein metric by the uniformization theorem. Besse give a more elementary argument in terms of pants decompositions.
In dimension 3, Einstein is the same as constant sectional curvature; see eg here; the Riemann curvature tensor is determined by (and determines) the Ricci tensor. many hyperbolic manifolds in dimension 3; these are in fact the majority of 3-manifolds, in some sense. It is not true that negative Ricci curvature is the same as negative sectional curvature (that is, if we do not assume constancy). A counterexample is given by $S^2 times S^1$.
In dimensions larger than 3 constant sectional curvature manifolds are still Einstein manifolds (with the same sign). So an example of a closed hyperbolic manifold still gives you an example of a closed Einstein manifold with negative sign. These are less universal than in dimension 3, but many still exist. So there are counterexamples to your question.
There are even simply-connected closed Einstein manifolds; Aubin proved, before Yau's theorem on the existence of Kahler-Einstein metrics when $c_1 = 0$, that if $(M,J)$ is a Kahler manifold with $c_1^n < 0$, then $M$ admits a Kahler-Einstein metric of negative sign. Such manifolds exist in great supply in all even dimensions. (See Besse, chapter 11.) The existence of a Ricci-flat metric on the K3 surface is the simplest version of Yau's result on the $c_1 = 0$ case, and is very impressive.
For (2), I will simply quote to you what Besse have to say on the subject.
In dimensions greater than $4$, we do not know of any topological restriction for a manifold to be Einstein. It may very well be that any manifold with dimension greater than $4$ admits a negative Einstein metric - or, that most manifolds do.
I do not know what has changed on the subject, but I don't think much has. It is still open whether or not $S^n$ has an Einstein metric of negative Einstein constant for $n > 3$, I think. (By the discussion above it cannot for $n leq 3$.)
answered Jan 19 at 4:02
Mike MillerMike Miller
37.4k472139
$endgroup$
$begingroup$
Mike: This is almost right, but it is not true that in dimension 3 negative Ricci is equivalent to negative sectional curvature. But, of course, negative Einstein is the same as constant negative curvature. In fact, every manifold of dimension $ge 3$ admits a metric of negative Ricci curvature (not Einstein, of course). The problem of existence of Einstein metrics on compact manifolds of dimension $ge 5$ is wide-open. What is known comparing to Besse's book are more examples, but, still, no obstructions.
$endgroup$
– Moishe Cohen
Jan 19 at 15:47
$begingroup$
@MoisheCohen I have been carrying that 'fact' with me for a while now, so thanks for correcting it. I'll update with the correct statement shortly.
$endgroup$
– Mike Miller
Jan 19 at 17:11
$begingroup$
@Moishe After a moment's thought your comment was completely clear. A little embarrassing to have made that mistake; I appreciate your having commented.
$endgroup$
– Mike Miller
Jan 20 at 16:36
add a comment |
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$begingroup$
Everything here is from Besse's book "Einstein manifolds". If you are interested in the subject, you owe it to yourself to have a copy nearby.
(1) is false for a number of reasons.
In dimension 2, negative Ricci curvature is the same thing as negative scalar curvature (and Einstein is the same as constant scalar curvature), so every closed surface of genus $g > 1$ admits a negative Einstein metric by the uniformization theorem. Besse give a more elementary argument in terms of pants decompositions.
In dimension 3, Einstein is the same as constant sectional curvature; see eg here; the Riemann curvature tensor is determined by (and determines) the Ricci tensor. many hyperbolic manifolds in dimension 3; these are in fact the majority of 3-manifolds, in some sense. It is not true that negative Ricci curvature is the same as negative sectional curvature (that is, if we do not assume constancy). A counterexample is given by $S^2 times S^1$.
In dimensions larger than 3 constant sectional curvature manifolds are still Einstein manifolds (with the same sign). So an example of a closed hyperbolic manifold still gives you an example of a closed Einstein manifold with negative sign. These are less universal than in dimension 3, but many still exist. So there are counterexamples to your question.
There are even simply-connected closed Einstein manifolds; Aubin proved, before Yau's theorem on the existence of Kahler-Einstein metrics when $c_1 = 0$, that if $(M,J)$ is a Kahler manifold with $c_1^n < 0$, then $M$ admits a Kahler-Einstein metric of negative sign. Such manifolds exist in great supply in all even dimensions. (See Besse, chapter 11.) The existence of a Ricci-flat metric on the K3 surface is the simplest version of Yau's result on the $c_1 = 0$ case, and is very impressive.
For (2), I will simply quote to you what Besse have to say on the subject.
In dimensions greater than $4$, we do not know of any topological restriction for a manifold to be Einstein. It may very well be that any manifold with dimension greater than $4$ admits a negative Einstein metric - or, that most manifolds do.
I do not know what has changed on the subject, but I don't think much has. It is still open whether or not $S^n$ has an Einstein metric of negative Einstein constant for $n > 3$, I think. (By the discussion above it cannot for $n leq 3$.)
answered Jan 19 at 4:02
Mike MillerMike Miller
37.4k472139
$endgroup$
$begingroup$
Mike: This is almost right, but it is not true that in dimension 3 negative Ricci is equivalent to negative sectional curvature. But, of course, negative Einstein is the same as constant negative curvature. In fact, every manifold of dimension $ge 3$ admits a metric of negative Ricci curvature (not Einstein, of course). The problem of existence of Einstein metrics on compact manifolds of dimension $ge 5$ is wide-open. What is known comparing to Besse's book are more examples, but, still, no obstructions.
$endgroup$
– Moishe Cohen
Jan 19 at 15:47
$begingroup$
@MoisheCohen I have been carrying that 'fact' with me for a while now, so thanks for correcting it. I'll update with the correct statement shortly.
$endgroup$
– Mike Miller
Jan 19 at 17:11
$begingroup$
@Moishe After a moment's thought your comment was completely clear. A little embarrassing to have made that mistake; I appreciate your having commented.
$endgroup$
– Mike Miller
Jan 20 at 16:36
add a comment |
$begingroup$
Everything here is from Besse's book "Einstein manifolds". If you are interested in the subject, you owe it to yourself to have a copy nearby.
(1) is false for a number of reasons.
In dimension 2, negative Ricci curvature is the same thing as negative scalar curvature (and Einstein is the same as constant scalar curvature), so every closed surface of genus $g > 1$ admits a negative Einstein metric by the uniformization theorem. Besse give a more elementary argument in terms of pants decompositions.
In dimension 3, Einstein is the same as constant sectional curvature; see eg here; the Riemann curvature tensor is determined by (and determines) the Ricci tensor. many hyperbolic manifolds in dimension 3; these are in fact the majority of 3-manifolds, in some sense. It is not true that negative Ricci curvature is the same as negative sectional curvature (that is, if we do not assume constancy). A counterexample is given by $S^2 times S^1$.
In dimensions larger than 3 constant sectional curvature manifolds are still Einstein manifolds (with the same sign). So an example of a closed hyperbolic manifold still gives you an example of a closed Einstein manifold with negative sign. These are less universal than in dimension 3, but many still exist. So there are counterexamples to your question.
There are even simply-connected closed Einstein manifolds; Aubin proved, before Yau's theorem on the existence of Kahler-Einstein metrics when $c_1 = 0$, that if $(M,J)$ is a Kahler manifold with $c_1^n < 0$, then $M$ admits a Kahler-Einstein metric of negative sign. Such manifolds exist in great supply in all even dimensions. (See Besse, chapter 11.) The existence of a Ricci-flat metric on the K3 surface is the simplest version of Yau's result on the $c_1 = 0$ case, and is very impressive.
For (2), I will simply quote to you what Besse have to say on the subject.
In dimensions greater than $4$, we do not know of any topological restriction for a manifold to be Einstein. It may very well be that any manifold with dimension greater than $4$ admits a negative Einstein metric - or, that most manifolds do.
I do not know what has changed on the subject, but I don't think much has. It is still open whether or not $S^n$ has an Einstein metric of negative Einstein constant for $n > 3$, I think. (By the discussion above it cannot for $n leq 3$.)
answered Jan 19 at 4:02
Mike MillerMike Miller
37.4k472139
$endgroup$
$begingroup$
Mike: This is almost right, but it is not true that in dimension 3 negative Ricci is equivalent to negative sectional curvature. But, of course, negative Einstein is the same as constant negative curvature. In fact, every manifold of dimension $ge 3$ admits a metric of negative Ricci curvature (not Einstein, of course). The problem of existence of Einstein metrics on compact manifolds of dimension $ge 5$ is wide-open. What is known comparing to Besse's book are more examples, but, still, no obstructions.
$endgroup$
– Moishe Cohen
Jan 19 at 15:47
$begingroup$
@MoisheCohen I have been carrying that 'fact' with me for a while now, so thanks for correcting it. I'll update with the correct statement shortly.
$endgroup$
– Mike Miller
Jan 19 at 17:11
$begingroup$
@Moishe After a moment's thought your comment was completely clear. A little embarrassing to have made that mistake; I appreciate your having commented.
$endgroup$
– Mike Miller
Jan 20 at 16:36
add a comment |
$begingroup$
Everything here is from Besse's book "Einstein manifolds". If you are interested in the subject, you owe it to yourself to have a copy nearby.
(1) is false for a number of reasons.
In dimension 2, negative Ricci curvature is the same thing as negative scalar curvature (and Einstein is the same as constant scalar curvature), so every closed surface of genus $g > 1$ admits a negative Einstein metric by the uniformization theorem. Besse give a more elementary argument in terms of pants decompositions.
In dimension 3, Einstein is the same as constant sectional curvature; see eg here; the Riemann curvature tensor is determined by (and determines) the Ricci tensor. many hyperbolic manifolds in dimension 3; these are in fact the majority of 3-manifolds, in some sense. It is not true that negative Ricci curvature is the same as negative sectional curvature (that is, if we do not assume constancy). A counterexample is given by $S^2 times S^1$.
In dimensions larger than 3 constant sectional curvature manifolds are still Einstein manifolds (with the same sign). So an example of a closed hyperbolic manifold still gives you an example of a closed Einstein manifold with negative sign. These are less universal than in dimension 3, but many still exist. So there are counterexamples to your question.
There are even simply-connected closed Einstein manifolds; Aubin proved, before Yau's theorem on the existence of Kahler-Einstein metrics when $c_1 = 0$, that if $(M,J)$ is a Kahler manifold with $c_1^n < 0$, then $M$ admits a Kahler-Einstein metric of negative sign. Such manifolds exist in great supply in all even dimensions. (See Besse, chapter 11.) The existence of a Ricci-flat metric on the K3 surface is the simplest version of Yau's result on the $c_1 = 0$ case, and is very impressive.
For (2), I will simply quote to you what Besse have to say on the subject.
In dimensions greater than $4$, we do not know of any topological restriction for a manifold to be Einstein. It may very well be that any manifold with dimension greater than $4$ admits a negative Einstein metric - or, that most manifolds do.
I do not know what has changed on the subject, but I don't think much has. It is still open whether or not $S^n$ has an Einstein metric of negative Einstein constant for $n > 3$, I think. (By the discussion above it cannot for $n leq 3$.)
answered Jan 19 at 4:02
Mike MillerMike Miller
37.4k472139
$endgroup$
Everything here is from Besse's book "Einstein manifolds". If you are interested in the subject, you owe it to yourself to have a copy nearby.
(1) is false for a number of reasons.
In dimension 2, negative Ricci curvature is the same thing as negative scalar curvature (and Einstein is the same as constant scalar curvature), so every closed surface of genus $g > 1$ admits a negative Einstein metric by the uniformization theorem. Besse give a more elementary argument in terms of pants decompositions.
In dimension 3, Einstein is the same as constant sectional curvature; see eg here; the Riemann curvature tensor is determined by (and determines) the Ricci tensor. many hyperbolic manifolds in dimension 3; these are in fact the majority of 3-manifolds, in some sense. It is not true that negative Ricci curvature is the same as negative sectional curvature (that is, if we do not assume constancy). A counterexample is given by $S^2 times S^1$.
In dimensions larger than 3 constant sectional curvature manifolds are still Einstein manifolds (with the same sign). So an example of a closed hyperbolic manifold still gives you an example of a closed Einstein manifold with negative sign. These are less universal than in dimension 3, but many still exist. So there are counterexamples to your question.
There are even simply-connected closed Einstein manifolds; Aubin proved, before Yau's theorem on the existence of Kahler-Einstein metrics when $c_1 = 0$, that if $(M,J)$ is a Kahler manifold with $c_1^n < 0$, then $M$ admits a Kahler-Einstein metric of negative sign. Such manifolds exist in great supply in all even dimensions. (See Besse, chapter 11.) The existence of a Ricci-flat metric on the K3 surface is the simplest version of Yau's result on the $c_1 = 0$ case, and is very impressive.
For (2), I will simply quote to you what Besse have to say on the subject.
In dimensions greater than $4$, we do not know of any topological restriction for a manifold to be Einstein. It may very well be that any manifold with dimension greater than $4$ admits a negative Einstein metric - or, that most manifolds do.
I do not know what has changed on the subject, but I don't think much has. It is still open whether or not $S^n$ has an Einstein metric of negative Einstein constant for $n > 3$, I think. (By the discussion above it cannot for $n leq 3$.)
answered Jan 19 at 4:02
Mike MillerMike Miller
37.4k472139
edited Jan 20 at 16:35
answered Jan 19 at 4:02
Mike MillerMike Miller
37.4k472139
answered Jan 19 at 4:02
Mike MillerMike Miller
37.4k472139
answered Jan 19 at 4:02
Mike MillerMike Miller
37.4k472139
37.4k472139
$begingroup$
Mike: This is almost right, but it is not true that in dimension 3 negative Ricci is equivalent to negative sectional curvature. But, of course, negative Einstein is the same as constant negative curvature. In fact, every manifold of dimension $ge 3$ admits a metric of negative Ricci curvature (not Einstein, of course). The problem of existence of Einstein metrics on compact manifolds of dimension $ge 5$ is wide-open. What is known comparing to Besse's book are more examples, but, still, no obstructions.
$endgroup$
– Moishe Cohen
Jan 19 at 15:47
$begingroup$
@MoisheCohen I have been carrying that 'fact' with me for a while now, so thanks for correcting it. I'll update with the correct statement shortly.
$endgroup$
– Mike Miller
Jan 19 at 17:11
$begingroup$
@Moishe After a moment's thought your comment was completely clear. A little embarrassing to have made that mistake; I appreciate your having commented.
$endgroup$
– Mike Miller
Jan 20 at 16:36
add a comment |
$begingroup$
Mike: This is almost right, but it is not true that in dimension 3 negative Ricci is equivalent to negative sectional curvature. But, of course, negative Einstein is the same as constant negative curvature. In fact, every manifold of dimension $ge 3$ admits a metric of negative Ricci curvature (not Einstein, of course). The problem of existence of Einstein metrics on compact manifolds of dimension $ge 5$ is wide-open. What is known comparing to Besse's book are more examples, but, still, no obstructions.
$endgroup$
– Moishe Cohen
Jan 19 at 15:47
$begingroup$
@MoisheCohen I have been carrying that 'fact' with me for a while now, so thanks for correcting it. I'll update with the correct statement shortly.
$endgroup$
– Mike Miller
Jan 19 at 17:11
$begingroup$
@Moishe After a moment's thought your comment was completely clear. A little embarrassing to have made that mistake; I appreciate your having commented.
$endgroup$
– Mike Miller
Jan 20 at 16:36
$begingroup$
Mike: This is almost right, but it is not true that in dimension 3 negative Ricci is equivalent to negative sectional curvature. But, of course, negative Einstein is the same as constant negative curvature. In fact, every manifold of dimension $ge 3$ admits a metric of negative Ricci curvature (not Einstein, of course). The problem of existence of Einstein metrics on compact manifolds of dimension $ge 5$ is wide-open. What is known comparing to Besse's book are more examples, but, still, no obstructions.
$endgroup$
– Moishe Cohen
Jan 19 at 15:47
$begingroup$
Mike: This is almost right, but it is not true that in dimension 3 negative Ricci is equivalent to negative sectional curvature. But, of course, negative Einstein is the same as constant negative curvature. In fact, every manifold of dimension $ge 3$ admits a metric of negative Ricci curvature (not Einstein, of course). The problem of existence of Einstein metrics on compact manifolds of dimension $ge 5$ is wide-open. What is known comparing to Besse's book are more examples, but, still, no obstructions.
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– Moishe Cohen
Jan 19 at 15:47
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@MoisheCohen I have been carrying that 'fact' with me for a while now, so thanks for correcting it. I'll update with the correct statement shortly.
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– Mike Miller
Jan 19 at 17:11
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@MoisheCohen I have been carrying that 'fact' with me for a while now, so thanks for correcting it. I'll update with the correct statement shortly.
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– Mike Miller
Jan 19 at 17:11
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@Moishe After a moment's thought your comment was completely clear. A little embarrassing to have made that mistake; I appreciate your having commented.
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– Mike Miller
Jan 20 at 16:36
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@Moishe After a moment's thought your comment was completely clear. A little embarrassing to have made that mistake; I appreciate your having commented.
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– Mike Miller
Jan 20 at 16:36
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I think (2) is a famous open question.
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– Mike Miller
Jan 18 at 21:49
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@MikeMiller: Which part of (2) do you think is an open question, i.e. the part about $S^n$, or the first part?
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– srp
Jan 19 at 0:20
1
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I am quite confident at least that it is open for spheres. As of Besse, "Einstein manifolds", 1987, I may quote 6.3: "In dimensions greater than 4, we do not know any topological restriction for a manifold to be Einstein. It may very well be that any manifold with dimension greater than 4 admits a negative Einstein metric - or, that most manifolds do." I would need to search the literature to know if any partial progress has been made.
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– Mike Miller
Jan 19 at 0:40
1
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Oh, (1) is definitively false. See Aubin (and separately Yau) for the resolution of the Calabi conjecture for Kahler manifolds with $c_1(M, J)^n < 0$ (it is said that $c_1$ is negative). This provides Kahler-Einstein metrics with negative Einstein constant. This is Theorem 11.17 listed in Besse.
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– Mike Miller
Jan 19 at 0:51