curvature of $S^p times S^q$ is not constant

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From Petersen book Corollary 5.6.12:




If $M$ is closed simply connected manifold with constant curvature $k$
then $k>0$ and $M=S^n$. Thus, $S^p times S^q$, $CP^n$ do not admit any constant
curvature metrics.




How is the second part of the Corollary proved?



I mean with $S^2 times S^1$ which has $S^2 times R$ as a universal covering space i understand it, since $S^2 times R$ is not a space form $S^n_k$, but what about the other cases?




riemannian-geometry curvature




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  • $S^2times S^1$ is not simply connected.
    – Arnaud Mortier
    Jul 15 at 10:23














up vote
0
down vote

favorite












From Petersen book Corollary 5.6.12:




If $M$ is closed simply connected manifold with constant curvature $k$
then $k>0$ and $M=S^n$. Thus, $S^p times S^q$, $CP^n$ do not admit any constant
curvature metrics.




How is the second part of the Corollary proved?



I mean with $S^2 times S^1$ which has $S^2 times R$ as a universal covering space i understand it, since $S^2 times R$ is not a space form $S^n_k$, but what about the other cases?




riemannian-geometry curvature




share|cite|improve this question





















  • $S^2times S^1$ is not simply connected.
    – Arnaud Mortier
    Jul 15 at 10:23












up vote
0
down vote

favorite









up vote
0
down vote

favorite











From Petersen book Corollary 5.6.12:




If $M$ is closed simply connected manifold with constant curvature $k$
then $k>0$ and $M=S^n$. Thus, $S^p times S^q$, $CP^n$ do not admit any constant
curvature metrics.




How is the second part of the Corollary proved?



I mean with $S^2 times S^1$ which has $S^2 times R$ as a universal covering space i understand it, since $S^2 times R$ is not a space form $S^n_k$, but what about the other cases?




riemannian-geometry curvature




share|cite|improve this question













From Petersen book Corollary 5.6.12:




If $M$ is closed simply connected manifold with constant curvature $k$
then $k>0$ and $M=S^n$. Thus, $S^p times S^q$, $CP^n$ do not admit any constant
curvature metrics.




How is the second part of the Corollary proved?



I mean with $S^2 times S^1$ which has $S^2 times R$ as a universal covering space i understand it, since $S^2 times R$ is not a space form $S^n_k$, but what about the other cases?





riemannian-geometry curvature





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 15 at 10:31
























asked Jul 15 at 10:18









Allotrios

636




636











  • $S^2times S^1$ is not simply connected.
    – Arnaud Mortier
    Jul 15 at 10:23
















  • $S^2times S^1$ is not simply connected.
    – Arnaud Mortier
    Jul 15 at 10:23















$S^2times S^1$ is not simply connected.
– Arnaud Mortier
Jul 15 at 10:23




$S^2times S^1$ is not simply connected.
– Arnaud Mortier
Jul 15 at 10:23










1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










You're missing a part of the statement. This is the statement in the book (Petersen's Riemannian Geometry, third edition):




Corollary $5.6.12.$ If $M$ is a closed simply connected manifold with constant curvature $k$, then $k > 0$ and $M = S^n$. Thus, $S^ptimes S^q$, $mathbbCP^n$ do not admit any constant curvature metrics.




Now the conclusion follows immediately as $S^ptimes S^q$ (for $p, q > 1$) and $mathbbCP^n$ are closed simply connected manifolds which are not $S^n$.



If $M = S^1times S^1$, then $M$ admits a metric of constant curvature zero. For $M = S^ptimes S^q$ with $p = 1$ or $q = 1$ but not both, then as you said, the universal cover is not a space form so $M$ does not admit a constant curvature metric.






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  • Yes, my typo, but i still cannot get it.
    – Allotrios
    Jul 15 at 10:32










  • @Allotrios: I have expanded my answer.
    – Michael Albanese
    Jul 15 at 10:33










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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










You're missing a part of the statement. This is the statement in the book (Petersen's Riemannian Geometry, third edition):




Corollary $5.6.12.$ If $M$ is a closed simply connected manifold with constant curvature $k$, then $k > 0$ and $M = S^n$. Thus, $S^ptimes S^q$, $mathbbCP^n$ do not admit any constant curvature metrics.




Now the conclusion follows immediately as $S^ptimes S^q$ (for $p, q > 1$) and $mathbbCP^n$ are closed simply connected manifolds which are not $S^n$.



If $M = S^1times S^1$, then $M$ admits a metric of constant curvature zero. For $M = S^ptimes S^q$ with $p = 1$ or $q = 1$ but not both, then as you said, the universal cover is not a space form so $M$ does not admit a constant curvature metric.






share|cite|improve this answer























  • Yes, my typo, but i still cannot get it.
    – Allotrios
    Jul 15 at 10:32










  • @Allotrios: I have expanded my answer.
    – Michael Albanese
    Jul 15 at 10:33














up vote
1
down vote



accepted










You're missing a part of the statement. This is the statement in the book (Petersen's Riemannian Geometry, third edition):




Corollary $5.6.12.$ If $M$ is a closed simply connected manifold with constant curvature $k$, then $k > 0$ and $M = S^n$. Thus, $S^ptimes S^q$, $mathbbCP^n$ do not admit any constant curvature metrics.




Now the conclusion follows immediately as $S^ptimes S^q$ (for $p, q > 1$) and $mathbbCP^n$ are closed simply connected manifolds which are not $S^n$.



If $M = S^1times S^1$, then $M$ admits a metric of constant curvature zero. For $M = S^ptimes S^q$ with $p = 1$ or $q = 1$ but not both, then as you said, the universal cover is not a space form so $M$ does not admit a constant curvature metric.






share|cite|improve this answer























  • Yes, my typo, but i still cannot get it.
    – Allotrios
    Jul 15 at 10:32










  • @Allotrios: I have expanded my answer.
    – Michael Albanese
    Jul 15 at 10:33












up vote
1
down vote



accepted







up vote
1
down vote



accepted






You're missing a part of the statement. This is the statement in the book (Petersen's Riemannian Geometry, third edition):




Corollary $5.6.12.$ If $M$ is a closed simply connected manifold with constant curvature $k$, then $k > 0$ and $M = S^n$. Thus, $S^ptimes S^q$, $mathbbCP^n$ do not admit any constant curvature metrics.




Now the conclusion follows immediately as $S^ptimes S^q$ (for $p, q > 1$) and $mathbbCP^n$ are closed simply connected manifolds which are not $S^n$.



If $M = S^1times S^1$, then $M$ admits a metric of constant curvature zero. For $M = S^ptimes S^q$ with $p = 1$ or $q = 1$ but not both, then as you said, the universal cover is not a space form so $M$ does not admit a constant curvature metric.






share|cite|improve this answer















You're missing a part of the statement. This is the statement in the book (Petersen's Riemannian Geometry, third edition):




Corollary $5.6.12.$ If $M$ is a closed simply connected manifold with constant curvature $k$, then $k > 0$ and $M = S^n$. Thus, $S^ptimes S^q$, $mathbbCP^n$ do not admit any constant curvature metrics.




Now the conclusion follows immediately as $S^ptimes S^q$ (for $p, q > 1$) and $mathbbCP^n$ are closed simply connected manifolds which are not $S^n$.



If $M = S^1times S^1$, then $M$ admits a metric of constant curvature zero. For $M = S^ptimes S^q$ with $p = 1$ or $q = 1$ but not both, then as you said, the universal cover is not a space form so $M$ does not admit a constant curvature metric.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 15 at 10:46


























answered Jul 15 at 10:25









Michael Albanese

61.3k1591290




61.3k1591290











  • Yes, my typo, but i still cannot get it.
    – Allotrios
    Jul 15 at 10:32










  • @Allotrios: I have expanded my answer.
    – Michael Albanese
    Jul 15 at 10:33
















  • Yes, my typo, but i still cannot get it.
    – Allotrios
    Jul 15 at 10:32










  • @Allotrios: I have expanded my answer.
    – Michael Albanese
    Jul 15 at 10:33















Yes, my typo, but i still cannot get it.
– Allotrios
Jul 15 at 10:32




Yes, my typo, but i still cannot get it.
– Allotrios
Jul 15 at 10:32












@Allotrios: I have expanded my answer.
– Michael Albanese
Jul 15 at 10:33




@Allotrios: I have expanded my answer.
– Michael Albanese
Jul 15 at 10:33












 

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