Einstein metrics on spheres
$begingroup$
I've got a couple of quick questions that came up after reading a peculiar statement in some article. The sentence says something like "... is the $N$-dimensional sphere with constant Ricci curvature equal to $K$...", and the questions are something like:
For $(mathbb{S}^n,g)$ the sphere with its standard differential structure and $some$ Riemannian metric on it,
1.a. Does $g$ being an Einstein metric implies that it is actually the round metric (up to some normalization constant)?
1.b. Does the answer change if we change to an alternative differential structure (when possible)?
I guess this shouldn't be true, so in this case
2. Is there an intuitive way to understand how one could construct a metric which is Einstein but not of constant curvature?
Anyways, I thank you all in advance for sharing your knowledge.
differential-geometry riemannian-geometry
edited 12 hours ago
Michael Albanese
64.6k1599315
asked 12 hours ago
Bruce WayneBruce Wayne
453213
$endgroup$
add a comment |
$begingroup$
I've got a couple of quick questions that came up after reading a peculiar statement in some article. The sentence says something like "... is the $N$-dimensional sphere with constant Ricci curvature equal to $K$...", and the questions are something like:
For $(mathbb{S}^n,g)$ the sphere with its standard differential structure and $some$ Riemannian metric on it,
1.a. Does $g$ being an Einstein metric implies that it is actually the round metric (up to some normalization constant)?
1.b. Does the answer change if we change to an alternative differential structure (when possible)?
I guess this shouldn't be true, so in this case
2. Is there an intuitive way to understand how one could construct a metric which is Einstein but not of constant curvature?
Anyways, I thank you all in advance for sharing your knowledge.
differential-geometry riemannian-geometry
edited 12 hours ago
Michael Albanese
64.6k1599315
asked 12 hours ago
Bruce WayneBruce Wayne
453213
$endgroup$
add a comment |
$begingroup$
I've got a couple of quick questions that came up after reading a peculiar statement in some article. The sentence says something like "... is the $N$-dimensional sphere with constant Ricci curvature equal to $K$...", and the questions are something like:
For $(mathbb{S}^n,g)$ the sphere with its standard differential structure and $some$ Riemannian metric on it,
1.a. Does $g$ being an Einstein metric implies that it is actually the round metric (up to some normalization constant)?
1.b. Does the answer change if we change to an alternative differential structure (when possible)?
I guess this shouldn't be true, so in this case
2. Is there an intuitive way to understand how one could construct a metric which is Einstein but not of constant curvature?
Anyways, I thank you all in advance for sharing your knowledge.
differential-geometry riemannian-geometry
edited 12 hours ago
Michael Albanese
64.6k1599315
asked 12 hours ago
Bruce WayneBruce Wayne
453213
$endgroup$
I've got a couple of quick questions that came up after reading a peculiar statement in some article. The sentence says something like "... is the $N$-dimensional sphere with constant Ricci curvature equal to $K$...", and the questions are something like:
For $(mathbb{S}^n,g)$ the sphere with its standard differential structure and $some$ Riemannian metric on it,
1.a. Does $g$ being an Einstein metric implies that it is actually the round metric (up to some normalization constant)?
1.b. Does the answer change if we change to an alternative differential structure (when possible)?
I guess this shouldn't be true, so in this case
2. Is there an intuitive way to understand how one could construct a metric which is Einstein but not of constant curvature?
Anyways, I thank you all in advance for sharing your knowledge.
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
edited 12 hours ago
Michael Albanese
64.6k1599315
asked 12 hours ago
Bruce WayneBruce Wayne
453213
edited 12 hours ago
Michael Albanese
64.6k1599315
asked 12 hours ago
Bruce WayneBruce Wayne
453213
edited 12 hours ago
Michael Albanese
64.6k1599315
edited 12 hours ago
Michael Albanese
64.6k1599315
edited 12 hours ago
Michael Albanese
64.6k1599315
64.6k1599315
asked 12 hours ago
Bruce WayneBruce Wayne
453213
asked 12 hours ago
Bruce WayneBruce Wayne
453213
asked 12 hours ago
Bruce WayneBruce Wayne
453213
453213
add a comment |
add a comment |
1 Answer
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No, there are Einstein metrics on spheres which are not rescalings of the round metric. See the introduction of Einstein metrics on spheres by Boyer, Galicki, & Kollár for some constructions. However, as far as I am aware, there are no known examples of Einstein metrics with non-positive Einstein constant. In particular, it is an open question as to whether $S^n$ admits a Ricci-flat metric for $n geq 4$.
If we consider exotic spheres, they do not admit a 'round metric' or any metric of constant curvature, so I'm not sure what is meant by this. However, there are examples of Einstein metrics on exotic spheres, see Einstein Metrics on Exotic Spheres in Dimensions 7, 11, and 15 by Boyer, Galicki, Kollár, & Thomas for example. Note however that there are some exotic spheres which, if they admit Einstein metrics, must have negative Einstein constant.
Finding Einstein metrics which are not constant curvature is, in general, a hard thing to do and an area of active research.
answered 12 hours ago
Michael AlbaneseMichael Albanese
64.6k1599315
$endgroup$
$begingroup$
great, thanks! Yeah, of course you are right, question 1b doesn't make sense as stated. I wrote it fast, sorry!
$endgroup$
– Bruce Wayne
11 hours ago
add a comment |
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1 Answer
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1 Answer
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$begingroup$
No, there are Einstein metrics on spheres which are not rescalings of the round metric. See the introduction of Einstein metrics on spheres by Boyer, Galicki, & Kollár for some constructions. However, as far as I am aware, there are no known examples of Einstein metrics with non-positive Einstein constant. In particular, it is an open question as to whether $S^n$ admits a Ricci-flat metric for $n geq 4$.
If we consider exotic spheres, they do not admit a 'round metric' or any metric of constant curvature, so I'm not sure what is meant by this. However, there are examples of Einstein metrics on exotic spheres, see Einstein Metrics on Exotic Spheres in Dimensions 7, 11, and 15 by Boyer, Galicki, Kollár, & Thomas for example. Note however that there are some exotic spheres which, if they admit Einstein metrics, must have negative Einstein constant.
Finding Einstein metrics which are not constant curvature is, in general, a hard thing to do and an area of active research.
answered 12 hours ago
Michael AlbaneseMichael Albanese
64.6k1599315
$endgroup$
$begingroup$
great, thanks! Yeah, of course you are right, question 1b doesn't make sense as stated. I wrote it fast, sorry!
$endgroup$
– Bruce Wayne
11 hours ago
add a comment |
$begingroup$
No, there are Einstein metrics on spheres which are not rescalings of the round metric. See the introduction of Einstein metrics on spheres by Boyer, Galicki, & Kollár for some constructions. However, as far as I am aware, there are no known examples of Einstein metrics with non-positive Einstein constant. In particular, it is an open question as to whether $S^n$ admits a Ricci-flat metric for $n geq 4$.
If we consider exotic spheres, they do not admit a 'round metric' or any metric of constant curvature, so I'm not sure what is meant by this. However, there are examples of Einstein metrics on exotic spheres, see Einstein Metrics on Exotic Spheres in Dimensions 7, 11, and 15 by Boyer, Galicki, Kollár, & Thomas for example. Note however that there are some exotic spheres which, if they admit Einstein metrics, must have negative Einstein constant.
Finding Einstein metrics which are not constant curvature is, in general, a hard thing to do and an area of active research.
answered 12 hours ago
Michael AlbaneseMichael Albanese
64.6k1599315
$endgroup$
$begingroup$
great, thanks! Yeah, of course you are right, question 1b doesn't make sense as stated. I wrote it fast, sorry!
$endgroup$
– Bruce Wayne
11 hours ago
add a comment |
$begingroup$
No, there are Einstein metrics on spheres which are not rescalings of the round metric. See the introduction of Einstein metrics on spheres by Boyer, Galicki, & Kollár for some constructions. However, as far as I am aware, there are no known examples of Einstein metrics with non-positive Einstein constant. In particular, it is an open question as to whether $S^n$ admits a Ricci-flat metric for $n geq 4$.
If we consider exotic spheres, they do not admit a 'round metric' or any metric of constant curvature, so I'm not sure what is meant by this. However, there are examples of Einstein metrics on exotic spheres, see Einstein Metrics on Exotic Spheres in Dimensions 7, 11, and 15 by Boyer, Galicki, Kollár, & Thomas for example. Note however that there are some exotic spheres which, if they admit Einstein metrics, must have negative Einstein constant.
Finding Einstein metrics which are not constant curvature is, in general, a hard thing to do and an area of active research.
answered 12 hours ago
Michael AlbaneseMichael Albanese
64.6k1599315
$endgroup$
No, there are Einstein metrics on spheres which are not rescalings of the round metric. See the introduction of Einstein metrics on spheres by Boyer, Galicki, & Kollár for some constructions. However, as far as I am aware, there are no known examples of Einstein metrics with non-positive Einstein constant. In particular, it is an open question as to whether $S^n$ admits a Ricci-flat metric for $n geq 4$.
If we consider exotic spheres, they do not admit a 'round metric' or any metric of constant curvature, so I'm not sure what is meant by this. However, there are examples of Einstein metrics on exotic spheres, see Einstein Metrics on Exotic Spheres in Dimensions 7, 11, and 15 by Boyer, Galicki, Kollár, & Thomas for example. Note however that there are some exotic spheres which, if they admit Einstein metrics, must have negative Einstein constant.
Finding Einstein metrics which are not constant curvature is, in general, a hard thing to do and an area of active research.
answered 12 hours ago
Michael AlbaneseMichael Albanese
64.6k1599315
edited 11 hours ago
answered 12 hours ago
Michael AlbaneseMichael Albanese
64.6k1599315
answered 12 hours ago
Michael AlbaneseMichael Albanese
64.6k1599315
answered 12 hours ago
Michael AlbaneseMichael Albanese
64.6k1599315
64.6k1599315
$begingroup$
great, thanks! Yeah, of course you are right, question 1b doesn't make sense as stated. I wrote it fast, sorry!
$endgroup$
– Bruce Wayne
11 hours ago
add a comment |
$begingroup$
great, thanks! Yeah, of course you are right, question 1b doesn't make sense as stated. I wrote it fast, sorry!
$endgroup$
– Bruce Wayne
11 hours ago
$begingroup$
great, thanks! Yeah, of course you are right, question 1b doesn't make sense as stated. I wrote it fast, sorry!
$endgroup$
– Bruce Wayne
11 hours ago
$begingroup$
great, thanks! Yeah, of course you are right, question 1b doesn't make sense as stated. I wrote it fast, sorry!
$endgroup$
– Bruce Wayne
11 hours ago
add a comment |
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