Realizing cohomology classes by submanifoldsEquivalent singular chains and differential forms, as functionals on forms, on compact Riemannian manifoldsHomologically trivial submanifoldsPoincaré duality with boundary conditionsAn analytic subset as a singular homology class of a compact manifoldIs there a Hodge isomorphism theorem for part-tangential, part-normal, harmonic differential forms?Constructing a Sobolev space containing the differential k-forms of a Riemannian manifoldCohomology classes represented by submanifoldsTwisted calibrations and sufficient conditions on homology of sub-manifoldsA Generalized De Rham cohomologyGodbillon's proof of strong excision in de Rham cohomology with compact support

Realizing cohomology classes by submanifolds


Equivalent singular chains and differential forms, as functionals on forms, on compact Riemannian manifoldsHomologically trivial submanifoldsPoincaré duality with boundary conditionsAn analytic subset as a singular homology class of a compact manifoldIs there a Hodge isomorphism theorem for part-tangential, part-normal, harmonic differential forms?Constructing a Sobolev space containing the differential k-forms of a Riemannian manifoldCohomology classes represented by submanifoldsTwisted calibrations and sufficient conditions on homology of sub-manifoldsA Generalized De Rham cohomologyGodbillon's proof of strong excision in de Rham cohomology with compact support













11












$begingroup$


In "Quelques propriétés globales des variétés différentiables", Thom gives conditions for a class in singular homology of a compact manifold to be realized by a smooth oriented submanifold (see e.g. Theoreme II.27).



I have the following variation of this question:




Let $M$ be a finite-dimensional manifold. Given $lambda in H^k(M, mathbb R)^*$, does there exists a compact oriented submanifold $S$ of $M$ and a closed form $alpha$ on $S$ such that
$$lambda([beta]) = int_S beta wedge alpha,$$
where $beta$ is a closed $k$-form on $M$, and we used the de Rham isomorphism to identify de Rham cohomology with singular cohomology. Of course, the dimension of $S$ and the degree of $alpha$ need to satify $dim S = k + # alpha$ for this integral to make sense.




I'm interested in conditions on $M$, $k$ and/or $lambda$ that ensure such a represenation. Moreover, I suspect the following:




There exists a lattice in $H^k(M, mathbb R)^*$ whose elements can be realized as above such that $alpha$ has integral periods (i.e. the integrals of $alpha$ over cycles are integers).











share|cite|improve this question











$endgroup$
















    11












    $begingroup$


    In "Quelques propriétés globales des variétés différentiables", Thom gives conditions for a class in singular homology of a compact manifold to be realized by a smooth oriented submanifold (see e.g. Theoreme II.27).



    I have the following variation of this question:




    Let $M$ be a finite-dimensional manifold. Given $lambda in H^k(M, mathbb R)^*$, does there exists a compact oriented submanifold $S$ of $M$ and a closed form $alpha$ on $S$ such that
    $$lambda([beta]) = int_S beta wedge alpha,$$
    where $beta$ is a closed $k$-form on $M$, and we used the de Rham isomorphism to identify de Rham cohomology with singular cohomology. Of course, the dimension of $S$ and the degree of $alpha$ need to satify $dim S = k + # alpha$ for this integral to make sense.




    I'm interested in conditions on $M$, $k$ and/or $lambda$ that ensure such a represenation. Moreover, I suspect the following:




    There exists a lattice in $H^k(M, mathbb R)^*$ whose elements can be realized as above such that $alpha$ has integral periods (i.e. the integrals of $alpha$ over cycles are integers).











    share|cite|improve this question











    $endgroup$














      11












      11








      11


      2



      $begingroup$


      In "Quelques propriétés globales des variétés différentiables", Thom gives conditions for a class in singular homology of a compact manifold to be realized by a smooth oriented submanifold (see e.g. Theoreme II.27).



      I have the following variation of this question:




      Let $M$ be a finite-dimensional manifold. Given $lambda in H^k(M, mathbb R)^*$, does there exists a compact oriented submanifold $S$ of $M$ and a closed form $alpha$ on $S$ such that
      $$lambda([beta]) = int_S beta wedge alpha,$$
      where $beta$ is a closed $k$-form on $M$, and we used the de Rham isomorphism to identify de Rham cohomology with singular cohomology. Of course, the dimension of $S$ and the degree of $alpha$ need to satify $dim S = k + # alpha$ for this integral to make sense.




      I'm interested in conditions on $M$, $k$ and/or $lambda$ that ensure such a represenation. Moreover, I suspect the following:




      There exists a lattice in $H^k(M, mathbb R)^*$ whose elements can be realized as above such that $alpha$ has integral periods (i.e. the integrals of $alpha$ over cycles are integers).











      share|cite|improve this question











      $endgroup$




      In "Quelques propriétés globales des variétés différentiables", Thom gives conditions for a class in singular homology of a compact manifold to be realized by a smooth oriented submanifold (see e.g. Theoreme II.27).



      I have the following variation of this question:




      Let $M$ be a finite-dimensional manifold. Given $lambda in H^k(M, mathbb R)^*$, does there exists a compact oriented submanifold $S$ of $M$ and a closed form $alpha$ on $S$ such that
      $$lambda([beta]) = int_S beta wedge alpha,$$
      where $beta$ is a closed $k$-form on $M$, and we used the de Rham isomorphism to identify de Rham cohomology with singular cohomology. Of course, the dimension of $S$ and the degree of $alpha$ need to satify $dim S = k + # alpha$ for this integral to make sense.




      I'm interested in conditions on $M$, $k$ and/or $lambda$ that ensure such a represenation. Moreover, I suspect the following:




      There exists a lattice in $H^k(M, mathbb R)^*$ whose elements can be realized as above such that $alpha$ has integral periods (i.e. the integrals of $alpha$ over cycles are integers).








      dg.differential-geometry at.algebraic-topology differential-topology






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited May 24 at 20:23







      Tobias Diez

















      asked May 24 at 10:10









      Tobias DiezTobias Diez

      2,7281334




      2,7281334




















          2 Answers
          2






          active

          oldest

          votes


















          16












          $begingroup$

          Your question is just a reformulation of what Thom did, so the answer is always yes.



          Since the Stokes map from de~Rham cohomology to singular cohomology (with real coefficients) is an isomorphism, your problem is equivalent to that of understanding the degree to which the map
          $$
          Omega_bullet(M)otimes Bbb R to H^bullet_DR(M;Bbb R)^ast
          $$

          is surjective. Here, the source is oriented bordism of manifolds mapping to $M$ and the target is the linear dual of de Rham cohomology. The identification $H^bullet_DR(M;Bbb R)^ast cong H_bullet(M;Bbb R)$ (singular homology on the right) enables one to reformulate the problem as to that of studying the degree to which
          $$
          Omega_bullet(M) to H_bullet(M)
          $$

          is surjective modulo torsion. The latter homomorphism assigns to a representative of a bordism class $Sigmato M$ the image of the fundamental class $[Sigma]$ in $H_bullet(M)$.



          It is known that that real (or rational) homology classes are representable by oriented smooth manifolds. Here is a dumb reason: there is a Hurewicz map
          $$pi_bullet^textst(M)to Omega_bullet(M)$$
          from stable homotopy to oriented bordism. The composite
          $$pi_bullet^textst(M)to Omega_bullet(M)to H_bullet(M)
          $$
          is just the usual Hurewicz map.



          It follows from Serre's thesis that the composite is an isomorphism modulo torsion (more specifically, Serre showed that the map $Sto hBbb Z$ from the sphere to the Eilenberg Mac Lane spectrum is a rational homotopy equivalence. If we smash this map with $M_+$ and take homotopy groups we get the statement about the Hurewicz map for $M$). Then it follows that the map $Omega_bullet(M)to H_bullet(M)$ is a surjection modulo torsion.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Does this also address the last statement of the question?
            $endgroup$
            – მამუკა ჯიბლაძე
            May 25 at 14:48


















          1












          $begingroup$

          It seems to me, that the original question asked about the cokernel of the map $Omega_*(M) to H_*(M).$ You are right, Thom's theorem claims it is finite. But can one say more about it?






          share|cite|improve this answer









          $endgroup$













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            2 Answers
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            active

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            2 Answers
            2






            active

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            active

            oldest

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            active

            oldest

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            16












            $begingroup$

            Your question is just a reformulation of what Thom did, so the answer is always yes.



            Since the Stokes map from de~Rham cohomology to singular cohomology (with real coefficients) is an isomorphism, your problem is equivalent to that of understanding the degree to which the map
            $$
            Omega_bullet(M)otimes Bbb R to H^bullet_DR(M;Bbb R)^ast
            $$

            is surjective. Here, the source is oriented bordism of manifolds mapping to $M$ and the target is the linear dual of de Rham cohomology. The identification $H^bullet_DR(M;Bbb R)^ast cong H_bullet(M;Bbb R)$ (singular homology on the right) enables one to reformulate the problem as to that of studying the degree to which
            $$
            Omega_bullet(M) to H_bullet(M)
            $$

            is surjective modulo torsion. The latter homomorphism assigns to a representative of a bordism class $Sigmato M$ the image of the fundamental class $[Sigma]$ in $H_bullet(M)$.



            It is known that that real (or rational) homology classes are representable by oriented smooth manifolds. Here is a dumb reason: there is a Hurewicz map
            $$pi_bullet^textst(M)to Omega_bullet(M)$$
            from stable homotopy to oriented bordism. The composite
            $$pi_bullet^textst(M)to Omega_bullet(M)to H_bullet(M)
            $$
            is just the usual Hurewicz map.



            It follows from Serre's thesis that the composite is an isomorphism modulo torsion (more specifically, Serre showed that the map $Sto hBbb Z$ from the sphere to the Eilenberg Mac Lane spectrum is a rational homotopy equivalence. If we smash this map with $M_+$ and take homotopy groups we get the statement about the Hurewicz map for $M$). Then it follows that the map $Omega_bullet(M)to H_bullet(M)$ is a surjection modulo torsion.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Does this also address the last statement of the question?
              $endgroup$
              – მამუკა ჯიბლაძე
              May 25 at 14:48















            16












            $begingroup$

            Your question is just a reformulation of what Thom did, so the answer is always yes.



            Since the Stokes map from de~Rham cohomology to singular cohomology (with real coefficients) is an isomorphism, your problem is equivalent to that of understanding the degree to which the map
            $$
            Omega_bullet(M)otimes Bbb R to H^bullet_DR(M;Bbb R)^ast
            $$

            is surjective. Here, the source is oriented bordism of manifolds mapping to $M$ and the target is the linear dual of de Rham cohomology. The identification $H^bullet_DR(M;Bbb R)^ast cong H_bullet(M;Bbb R)$ (singular homology on the right) enables one to reformulate the problem as to that of studying the degree to which
            $$
            Omega_bullet(M) to H_bullet(M)
            $$

            is surjective modulo torsion. The latter homomorphism assigns to a representative of a bordism class $Sigmato M$ the image of the fundamental class $[Sigma]$ in $H_bullet(M)$.



            It is known that that real (or rational) homology classes are representable by oriented smooth manifolds. Here is a dumb reason: there is a Hurewicz map
            $$pi_bullet^textst(M)to Omega_bullet(M)$$
            from stable homotopy to oriented bordism. The composite
            $$pi_bullet^textst(M)to Omega_bullet(M)to H_bullet(M)
            $$
            is just the usual Hurewicz map.



            It follows from Serre's thesis that the composite is an isomorphism modulo torsion (more specifically, Serre showed that the map $Sto hBbb Z$ from the sphere to the Eilenberg Mac Lane spectrum is a rational homotopy equivalence. If we smash this map with $M_+$ and take homotopy groups we get the statement about the Hurewicz map for $M$). Then it follows that the map $Omega_bullet(M)to H_bullet(M)$ is a surjection modulo torsion.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Does this also address the last statement of the question?
              $endgroup$
              – მამუკა ჯიბლაძე
              May 25 at 14:48













            16












            16








            16





            $begingroup$

            Your question is just a reformulation of what Thom did, so the answer is always yes.



            Since the Stokes map from de~Rham cohomology to singular cohomology (with real coefficients) is an isomorphism, your problem is equivalent to that of understanding the degree to which the map
            $$
            Omega_bullet(M)otimes Bbb R to H^bullet_DR(M;Bbb R)^ast
            $$

            is surjective. Here, the source is oriented bordism of manifolds mapping to $M$ and the target is the linear dual of de Rham cohomology. The identification $H^bullet_DR(M;Bbb R)^ast cong H_bullet(M;Bbb R)$ (singular homology on the right) enables one to reformulate the problem as to that of studying the degree to which
            $$
            Omega_bullet(M) to H_bullet(M)
            $$

            is surjective modulo torsion. The latter homomorphism assigns to a representative of a bordism class $Sigmato M$ the image of the fundamental class $[Sigma]$ in $H_bullet(M)$.



            It is known that that real (or rational) homology classes are representable by oriented smooth manifolds. Here is a dumb reason: there is a Hurewicz map
            $$pi_bullet^textst(M)to Omega_bullet(M)$$
            from stable homotopy to oriented bordism. The composite
            $$pi_bullet^textst(M)to Omega_bullet(M)to H_bullet(M)
            $$
            is just the usual Hurewicz map.



            It follows from Serre's thesis that the composite is an isomorphism modulo torsion (more specifically, Serre showed that the map $Sto hBbb Z$ from the sphere to the Eilenberg Mac Lane spectrum is a rational homotopy equivalence. If we smash this map with $M_+$ and take homotopy groups we get the statement about the Hurewicz map for $M$). Then it follows that the map $Omega_bullet(M)to H_bullet(M)$ is a surjection modulo torsion.






            share|cite|improve this answer











            $endgroup$



            Your question is just a reformulation of what Thom did, so the answer is always yes.



            Since the Stokes map from de~Rham cohomology to singular cohomology (with real coefficients) is an isomorphism, your problem is equivalent to that of understanding the degree to which the map
            $$
            Omega_bullet(M)otimes Bbb R to H^bullet_DR(M;Bbb R)^ast
            $$

            is surjective. Here, the source is oriented bordism of manifolds mapping to $M$ and the target is the linear dual of de Rham cohomology. The identification $H^bullet_DR(M;Bbb R)^ast cong H_bullet(M;Bbb R)$ (singular homology on the right) enables one to reformulate the problem as to that of studying the degree to which
            $$
            Omega_bullet(M) to H_bullet(M)
            $$

            is surjective modulo torsion. The latter homomorphism assigns to a representative of a bordism class $Sigmato M$ the image of the fundamental class $[Sigma]$ in $H_bullet(M)$.



            It is known that that real (or rational) homology classes are representable by oriented smooth manifolds. Here is a dumb reason: there is a Hurewicz map
            $$pi_bullet^textst(M)to Omega_bullet(M)$$
            from stable homotopy to oriented bordism. The composite
            $$pi_bullet^textst(M)to Omega_bullet(M)to H_bullet(M)
            $$
            is just the usual Hurewicz map.



            It follows from Serre's thesis that the composite is an isomorphism modulo torsion (more specifically, Serre showed that the map $Sto hBbb Z$ from the sphere to the Eilenberg Mac Lane spectrum is a rational homotopy equivalence. If we smash this map with $M_+$ and take homotopy groups we get the statement about the Hurewicz map for $M$). Then it follows that the map $Omega_bullet(M)to H_bullet(M)$ is a surjection modulo torsion.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited May 24 at 18:25

























            answered May 24 at 17:51









            John KleinJohn Klein

            12.1k3680




            12.1k3680











            • $begingroup$
              Does this also address the last statement of the question?
              $endgroup$
              – მამუკა ჯიბლაძე
              May 25 at 14:48
















            • $begingroup$
              Does this also address the last statement of the question?
              $endgroup$
              – მამუკა ჯიბლაძე
              May 25 at 14:48















            $begingroup$
            Does this also address the last statement of the question?
            $endgroup$
            – მამუკა ჯიბლაძე
            May 25 at 14:48




            $begingroup$
            Does this also address the last statement of the question?
            $endgroup$
            – მამუკა ჯიბლაძე
            May 25 at 14:48











            1












            $begingroup$

            It seems to me, that the original question asked about the cokernel of the map $Omega_*(M) to H_*(M).$ You are right, Thom's theorem claims it is finite. But can one say more about it?






            share|cite|improve this answer









            $endgroup$

















              1












              $begingroup$

              It seems to me, that the original question asked about the cokernel of the map $Omega_*(M) to H_*(M).$ You are right, Thom's theorem claims it is finite. But can one say more about it?






              share|cite|improve this answer









              $endgroup$















                1












                1








                1





                $begingroup$

                It seems to me, that the original question asked about the cokernel of the map $Omega_*(M) to H_*(M).$ You are right, Thom's theorem claims it is finite. But can one say more about it?






                share|cite|improve this answer









                $endgroup$



                It seems to me, that the original question asked about the cokernel of the map $Omega_*(M) to H_*(M).$ You are right, Thom's theorem claims it is finite. But can one say more about it?







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered May 29 at 10:18









                András SzűcsAndrás Szűcs

                2,01911518




                2,01911518



























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