Positive scalar curvature in dimension 4Does the curvature determine the metric?why positive scalar curvature manifoldsCorollary to Preissman's theoremWhen can a manifold be curvature free?Positive Ricci curvature vs sectional curvatureTrying to Understand Scalar Curvature – Send Help!Does the proof of Yamabe problem gives a method for finding metric of constant scalar curvature?Does positive Yamabe invariant imply every metric in that conformal class has positive scalar curvature?Positive scalar curvature metric on $S^4$Compact totally geodesic submanifolds in manifold with positive sectional curvature

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Positive scalar curvature in dimension 4


Does the curvature determine the metric?why positive scalar curvature manifoldsCorollary to Preissman's theoremWhen can a manifold be curvature free?Positive Ricci curvature vs sectional curvatureTrying to Understand Scalar Curvature – Send Help!Does the proof of Yamabe problem gives a method for finding metric of constant scalar curvature?Does positive Yamabe invariant imply every metric in that conformal class has positive scalar curvature?Positive scalar curvature metric on $S^4$Compact totally geodesic submanifolds in manifold with positive sectional curvature













5












$begingroup$


Let $M^n$ be a compact simply connected spin manifold.



Gromov, Lawson, and Stolz proved that if $ngeq 5$, then $M$ admits a metric of positive scalar curvature iff $alpha(M)=0$.



Question: What happens in dimension 4?



  1. Are there compact simply connected spin manifolds of dimension 4 which have a metric of positive scalar curvature and $alpha(M)neq 0$?

  2. Are there compact simply connected spin manifolds of dimension 4 which have $alpha(M)=0$ but no metric of positive scalar curvature?

  3. What are necessary and/or sufficient conditions for a compact simply connected spin 4-manifold to have a metric with positive scalar curvature?









share|cite|improve this question











$endgroup$
















    5












    $begingroup$


    Let $M^n$ be a compact simply connected spin manifold.



    Gromov, Lawson, and Stolz proved that if $ngeq 5$, then $M$ admits a metric of positive scalar curvature iff $alpha(M)=0$.



    Question: What happens in dimension 4?



    1. Are there compact simply connected spin manifolds of dimension 4 which have a metric of positive scalar curvature and $alpha(M)neq 0$?

    2. Are there compact simply connected spin manifolds of dimension 4 which have $alpha(M)=0$ but no metric of positive scalar curvature?

    3. What are necessary and/or sufficient conditions for a compact simply connected spin 4-manifold to have a metric with positive scalar curvature?









    share|cite|improve this question











    $endgroup$














      5












      5








      5


      2



      $begingroup$


      Let $M^n$ be a compact simply connected spin manifold.



      Gromov, Lawson, and Stolz proved that if $ngeq 5$, then $M$ admits a metric of positive scalar curvature iff $alpha(M)=0$.



      Question: What happens in dimension 4?



      1. Are there compact simply connected spin manifolds of dimension 4 which have a metric of positive scalar curvature and $alpha(M)neq 0$?

      2. Are there compact simply connected spin manifolds of dimension 4 which have $alpha(M)=0$ but no metric of positive scalar curvature?

      3. What are necessary and/or sufficient conditions for a compact simply connected spin 4-manifold to have a metric with positive scalar curvature?









      share|cite|improve this question











      $endgroup$




      Let $M^n$ be a compact simply connected spin manifold.



      Gromov, Lawson, and Stolz proved that if $ngeq 5$, then $M$ admits a metric of positive scalar curvature iff $alpha(M)=0$.



      Question: What happens in dimension 4?



      1. Are there compact simply connected spin manifolds of dimension 4 which have a metric of positive scalar curvature and $alpha(M)neq 0$?

      2. Are there compact simply connected spin manifolds of dimension 4 which have $alpha(M)=0$ but no metric of positive scalar curvature?

      3. What are necessary and/or sufficient conditions for a compact simply connected spin 4-manifold to have a metric with positive scalar curvature?






      differential-geometry riemannian-geometry 4-manifolds






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 21 '18 at 11:41









      Michael Albanese

      64.2k1599313




      64.2k1599313










      asked Mar 21 '18 at 4:05









      rpfrpf

      1,110513




      1,110513




















          1 Answer
          1






          active

          oldest

          votes


















          5












          $begingroup$

          1. Lichnerowicz proved that if $M$ is a closed spin manifold which admits a positive scalar curvature metric, then $hatA(M) = 0$. In dimensions $4k$, $alpha(M) = 2hatA(M)$, so it follows that $alpha(M) = 0$. In fact, Hitchin proved that $alpha(M) = 0$, regardless of dimension.



          2. In dimension four, $hatA(M) = -frac124p_1(M) = -frac18tau(M)$, so $alpha(M) = 0$ if and only if $tau(M) = 0$. If $M$ is a simply connected, closed, spin four-manifold with signature zero, then by Freedman's classification, there is an integer $k geq 0$ such that $M$ is homeomorphic to $k(S^2times S^2)$, the connected sum of $k$ copies of $S^2times S^2$; note, the connected sum of zero copies of $S^2times S^2$ is defined to be $S^4$. If $M$ is diffeomorphic to $k(S^2times S^2)$, then $M$ admits positive scalar curvature metrics. Said another way, the topological manifolds $k(S^2times S^2)$ all admit positive scalar curvature for their standard smooth structure.



            A closed four-manifold can potentially admit a countably infinite number of smooth structures, and the existence of a positive scalar curvature metric depends on which smooth structure you choose. So it is possible that for some choice of $k$ and some choice of smooth structure on $k(S^2times S^2)$, there does not admit a positive scalar curvature metric. Such examples exist. In Hirzebruch Surfaces: Degenerations,
            Related Braid Monodromy, Galois Covers
            , Teichner constructs examples of simply connected general-type complex surfaces which are spin and have signature zero; see Theorem $5.8$. These are homeomorphic to $k(S^2times S^2)$ for some $k$, but general type surfaces do not admit positive scalar curvature metrics.



          3. In dimension four, unlike in other dimensions, there are gauge theoretic obstructions to the existence of positive scalar curvature metrics. In particular, on a closed smooth four-manifold with $b^+ geq 2$, the existence of a positive scalar curvature metric implies that all of its Seiberg-Witten invariants vanish. This is a necessary condition, but it is not sufficient. Note that these invariants are usually too complicated to calculate, except in ideal situations such as when the four-manifold in question is the underlying four-manifold of a Kähler surface.






          share|cite|improve this answer











          $endgroup$












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            1 Answer
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            5












            $begingroup$

            1. Lichnerowicz proved that if $M$ is a closed spin manifold which admits a positive scalar curvature metric, then $hatA(M) = 0$. In dimensions $4k$, $alpha(M) = 2hatA(M)$, so it follows that $alpha(M) = 0$. In fact, Hitchin proved that $alpha(M) = 0$, regardless of dimension.



            2. In dimension four, $hatA(M) = -frac124p_1(M) = -frac18tau(M)$, so $alpha(M) = 0$ if and only if $tau(M) = 0$. If $M$ is a simply connected, closed, spin four-manifold with signature zero, then by Freedman's classification, there is an integer $k geq 0$ such that $M$ is homeomorphic to $k(S^2times S^2)$, the connected sum of $k$ copies of $S^2times S^2$; note, the connected sum of zero copies of $S^2times S^2$ is defined to be $S^4$. If $M$ is diffeomorphic to $k(S^2times S^2)$, then $M$ admits positive scalar curvature metrics. Said another way, the topological manifolds $k(S^2times S^2)$ all admit positive scalar curvature for their standard smooth structure.



              A closed four-manifold can potentially admit a countably infinite number of smooth structures, and the existence of a positive scalar curvature metric depends on which smooth structure you choose. So it is possible that for some choice of $k$ and some choice of smooth structure on $k(S^2times S^2)$, there does not admit a positive scalar curvature metric. Such examples exist. In Hirzebruch Surfaces: Degenerations,
              Related Braid Monodromy, Galois Covers
              , Teichner constructs examples of simply connected general-type complex surfaces which are spin and have signature zero; see Theorem $5.8$. These are homeomorphic to $k(S^2times S^2)$ for some $k$, but general type surfaces do not admit positive scalar curvature metrics.



            3. In dimension four, unlike in other dimensions, there are gauge theoretic obstructions to the existence of positive scalar curvature metrics. In particular, on a closed smooth four-manifold with $b^+ geq 2$, the existence of a positive scalar curvature metric implies that all of its Seiberg-Witten invariants vanish. This is a necessary condition, but it is not sufficient. Note that these invariants are usually too complicated to calculate, except in ideal situations such as when the four-manifold in question is the underlying four-manifold of a Kähler surface.






            share|cite|improve this answer











            $endgroup$

















              5












              $begingroup$

              1. Lichnerowicz proved that if $M$ is a closed spin manifold which admits a positive scalar curvature metric, then $hatA(M) = 0$. In dimensions $4k$, $alpha(M) = 2hatA(M)$, so it follows that $alpha(M) = 0$. In fact, Hitchin proved that $alpha(M) = 0$, regardless of dimension.



              2. In dimension four, $hatA(M) = -frac124p_1(M) = -frac18tau(M)$, so $alpha(M) = 0$ if and only if $tau(M) = 0$. If $M$ is a simply connected, closed, spin four-manifold with signature zero, then by Freedman's classification, there is an integer $k geq 0$ such that $M$ is homeomorphic to $k(S^2times S^2)$, the connected sum of $k$ copies of $S^2times S^2$; note, the connected sum of zero copies of $S^2times S^2$ is defined to be $S^4$. If $M$ is diffeomorphic to $k(S^2times S^2)$, then $M$ admits positive scalar curvature metrics. Said another way, the topological manifolds $k(S^2times S^2)$ all admit positive scalar curvature for their standard smooth structure.



                A closed four-manifold can potentially admit a countably infinite number of smooth structures, and the existence of a positive scalar curvature metric depends on which smooth structure you choose. So it is possible that for some choice of $k$ and some choice of smooth structure on $k(S^2times S^2)$, there does not admit a positive scalar curvature metric. Such examples exist. In Hirzebruch Surfaces: Degenerations,
                Related Braid Monodromy, Galois Covers
                , Teichner constructs examples of simply connected general-type complex surfaces which are spin and have signature zero; see Theorem $5.8$. These are homeomorphic to $k(S^2times S^2)$ for some $k$, but general type surfaces do not admit positive scalar curvature metrics.



              3. In dimension four, unlike in other dimensions, there are gauge theoretic obstructions to the existence of positive scalar curvature metrics. In particular, on a closed smooth four-manifold with $b^+ geq 2$, the existence of a positive scalar curvature metric implies that all of its Seiberg-Witten invariants vanish. This is a necessary condition, but it is not sufficient. Note that these invariants are usually too complicated to calculate, except in ideal situations such as when the four-manifold in question is the underlying four-manifold of a Kähler surface.






              share|cite|improve this answer











              $endgroup$















                5












                5








                5





                $begingroup$

                1. Lichnerowicz proved that if $M$ is a closed spin manifold which admits a positive scalar curvature metric, then $hatA(M) = 0$. In dimensions $4k$, $alpha(M) = 2hatA(M)$, so it follows that $alpha(M) = 0$. In fact, Hitchin proved that $alpha(M) = 0$, regardless of dimension.



                2. In dimension four, $hatA(M) = -frac124p_1(M) = -frac18tau(M)$, so $alpha(M) = 0$ if and only if $tau(M) = 0$. If $M$ is a simply connected, closed, spin four-manifold with signature zero, then by Freedman's classification, there is an integer $k geq 0$ such that $M$ is homeomorphic to $k(S^2times S^2)$, the connected sum of $k$ copies of $S^2times S^2$; note, the connected sum of zero copies of $S^2times S^2$ is defined to be $S^4$. If $M$ is diffeomorphic to $k(S^2times S^2)$, then $M$ admits positive scalar curvature metrics. Said another way, the topological manifolds $k(S^2times S^2)$ all admit positive scalar curvature for their standard smooth structure.



                  A closed four-manifold can potentially admit a countably infinite number of smooth structures, and the existence of a positive scalar curvature metric depends on which smooth structure you choose. So it is possible that for some choice of $k$ and some choice of smooth structure on $k(S^2times S^2)$, there does not admit a positive scalar curvature metric. Such examples exist. In Hirzebruch Surfaces: Degenerations,
                  Related Braid Monodromy, Galois Covers
                  , Teichner constructs examples of simply connected general-type complex surfaces which are spin and have signature zero; see Theorem $5.8$. These are homeomorphic to $k(S^2times S^2)$ for some $k$, but general type surfaces do not admit positive scalar curvature metrics.



                3. In dimension four, unlike in other dimensions, there are gauge theoretic obstructions to the existence of positive scalar curvature metrics. In particular, on a closed smooth four-manifold with $b^+ geq 2$, the existence of a positive scalar curvature metric implies that all of its Seiberg-Witten invariants vanish. This is a necessary condition, but it is not sufficient. Note that these invariants are usually too complicated to calculate, except in ideal situations such as when the four-manifold in question is the underlying four-manifold of a Kähler surface.






                share|cite|improve this answer











                $endgroup$



                1. Lichnerowicz proved that if $M$ is a closed spin manifold which admits a positive scalar curvature metric, then $hatA(M) = 0$. In dimensions $4k$, $alpha(M) = 2hatA(M)$, so it follows that $alpha(M) = 0$. In fact, Hitchin proved that $alpha(M) = 0$, regardless of dimension.



                2. In dimension four, $hatA(M) = -frac124p_1(M) = -frac18tau(M)$, so $alpha(M) = 0$ if and only if $tau(M) = 0$. If $M$ is a simply connected, closed, spin four-manifold with signature zero, then by Freedman's classification, there is an integer $k geq 0$ such that $M$ is homeomorphic to $k(S^2times S^2)$, the connected sum of $k$ copies of $S^2times S^2$; note, the connected sum of zero copies of $S^2times S^2$ is defined to be $S^4$. If $M$ is diffeomorphic to $k(S^2times S^2)$, then $M$ admits positive scalar curvature metrics. Said another way, the topological manifolds $k(S^2times S^2)$ all admit positive scalar curvature for their standard smooth structure.



                  A closed four-manifold can potentially admit a countably infinite number of smooth structures, and the existence of a positive scalar curvature metric depends on which smooth structure you choose. So it is possible that for some choice of $k$ and some choice of smooth structure on $k(S^2times S^2)$, there does not admit a positive scalar curvature metric. Such examples exist. In Hirzebruch Surfaces: Degenerations,
                  Related Braid Monodromy, Galois Covers
                  , Teichner constructs examples of simply connected general-type complex surfaces which are spin and have signature zero; see Theorem $5.8$. These are homeomorphic to $k(S^2times S^2)$ for some $k$, but general type surfaces do not admit positive scalar curvature metrics.



                3. In dimension four, unlike in other dimensions, there are gauge theoretic obstructions to the existence of positive scalar curvature metrics. In particular, on a closed smooth four-manifold with $b^+ geq 2$, the existence of a positive scalar curvature metric implies that all of its Seiberg-Witten invariants vanish. This is a necessary condition, but it is not sufficient. Note that these invariants are usually too complicated to calculate, except in ideal situations such as when the four-manifold in question is the underlying four-manifold of a Kähler surface.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Mar 16 at 21:08

























                answered Mar 21 '18 at 11:33









                Michael AlbaneseMichael Albanese

                64.2k1599313




                64.2k1599313



























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