Einstein metrics on spheresSmooth Poincaré Conjecturerescaled metric quantities on rescaling metricsReversing the Ricci flowThoughts about sectional curvatureEasy solution to Yamabe problem for surfacesShowing that Ricci curvature of round unit sphere $(S^n,g_0)$ is $Ric(g_0)=(n-1)g_0$Uniformization of metrics vs. uniformization of Riemann surfacesCounterexample to Gunther Theorem when assuming only a Ricci curvature upper boundPointwise conformal vs. conformally diffeomorphic metrics in dimension 2Upper volume bounds for submanifolds

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Einstein metrics on spheres

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Einstein metrics on spheres


Smooth Poincaré Conjecturerescaled metric quantities on rescaling metricsReversing the Ricci flowThoughts about sectional curvatureEasy solution to Yamabe problem for surfacesShowing that Ricci curvature of round unit sphere $(S^n,g_0)$ is $Ric(g_0)=(n-1)g_0$Uniformization of metrics vs. uniformization of Riemann surfacesCounterexample to Gunther Theorem when assuming only a Ricci curvature upper boundPointwise conformal vs. conformally diffeomorphic metrics in dimension 2Upper volume bounds for submanifolds













2












$begingroup$


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 $(mathbbS^n,g)$ 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.










share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    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 $(mathbbS^n,g)$ 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.










    share|cite|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      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 $(mathbbS^n,g)$ 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.










      share|cite|improve this question











      $endgroup$




      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 $(mathbbS^n,g)$ 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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 3 hours ago









      Michael Albanese

      64.6k1599315




      64.6k1599315










      asked 4 hours ago









      Bruce WayneBruce Wayne

      448213




      448213




















          1 Answer
          1






          active

          oldest

          votes


















          5












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            great, thanks! Yeah, of course you are right, question 1b doesn't make sense as stated. I wrote it fast, sorry!
            $endgroup$
            – Bruce Wayne
            2 hours ago











          Your Answer





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          1 Answer
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          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          5












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            great, thanks! Yeah, of course you are right, question 1b doesn't make sense as stated. I wrote it fast, sorry!
            $endgroup$
            – Bruce Wayne
            2 hours ago















          5












          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            great, thanks! Yeah, of course you are right, question 1b doesn't make sense as stated. I wrote it fast, sorry!
            $endgroup$
            – Bruce Wayne
            2 hours ago













          5












          5








          5





          $begingroup$

          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.






          share|cite|improve this answer











          $endgroup$



          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.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 3 hours ago

























          answered 3 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
            2 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
            2 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
          2 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
          2 hours ago

















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