Wu formula for manifolds with boundaryWhat manifold has $mathbbHP^odd$ as a boundary?Measuring the failure of pushforward to commute with Steenrod squaresTopological invariance of Stiefel-Whitney classes for open smooth manifoldsWho discovered this definition of Stiefel-Whitney classes?Two set of axioms for Stiefel-Whitney classesDiscrete Pin structuresExample of a finite group $G$ with low dimensional cohomology not generated by Stiefel-Whitney classes of flat vector bundles over $BG$

Wu formula for manifolds with boundary


What manifold has $mathbbHP^odd$ as a boundary?Measuring the failure of pushforward to commute with Steenrod squaresTopological invariance of Stiefel-Whitney classes for open smooth manifoldsWho discovered this definition of Stiefel-Whitney classes?Two set of axioms for Stiefel-Whitney classesDiscrete Pin structuresExample of a finite group $G$ with low dimensional cohomology not generated by Stiefel-Whitney classes of flat vector bundles over $BG$













16












$begingroup$


The classical Wu formula claims that if $M$ is a smooth closed $n$-manifold with fundamental class $zin H_n(M;mathbbZ_2)$, then the total Stiefel-Whitney class $w(M)$ is equal to $Sq(v)$, where $v=sum v_iin H^*(M;mathbbZ_2)$ is the unique cohomology class such that
$$langle vcup x,zrangle=langle Sq(x),zrangle$$
for all $xin H^*(M;mathbbZ_2)$. Thus, for $kge0$, $v_kcup x=Sq^k(x)$ for all $xin H^n-k(M;mathbbZ_2)$, and
$$w_k(M)=sum_i+j=kSq^i(v_j).$$
Here the Poincare duality guarantees the existence and uniqueness of $v$.



My question: if $M$ is a smooth compact $n$-manifold with boundary, is there a similar Wu formula?
In this case, there is a fundamental class $zin H_n(M,partial M;mathbbZ_2)$ and the relative Poincare duality claims that capping with $z$ yields duality isomorphisms
$$D:H^p(M,partial M;mathbbZ_2)to H_n-p(M;mathbbZ_2)$$
and
$$D:H^p(M;mathbbZ_2)to H_n-p(M,partial M;mathbbZ_2).$$



Thank you!










share|cite|improve this question









$endgroup$


















    16












    $begingroup$


    The classical Wu formula claims that if $M$ is a smooth closed $n$-manifold with fundamental class $zin H_n(M;mathbbZ_2)$, then the total Stiefel-Whitney class $w(M)$ is equal to $Sq(v)$, where $v=sum v_iin H^*(M;mathbbZ_2)$ is the unique cohomology class such that
    $$langle vcup x,zrangle=langle Sq(x),zrangle$$
    for all $xin H^*(M;mathbbZ_2)$. Thus, for $kge0$, $v_kcup x=Sq^k(x)$ for all $xin H^n-k(M;mathbbZ_2)$, and
    $$w_k(M)=sum_i+j=kSq^i(v_j).$$
    Here the Poincare duality guarantees the existence and uniqueness of $v$.



    My question: if $M$ is a smooth compact $n$-manifold with boundary, is there a similar Wu formula?
    In this case, there is a fundamental class $zin H_n(M,partial M;mathbbZ_2)$ and the relative Poincare duality claims that capping with $z$ yields duality isomorphisms
    $$D:H^p(M,partial M;mathbbZ_2)to H_n-p(M;mathbbZ_2)$$
    and
    $$D:H^p(M;mathbbZ_2)to H_n-p(M,partial M;mathbbZ_2).$$



    Thank you!










    share|cite|improve this question









    $endgroup$
















      16












      16








      16


      2



      $begingroup$


      The classical Wu formula claims that if $M$ is a smooth closed $n$-manifold with fundamental class $zin H_n(M;mathbbZ_2)$, then the total Stiefel-Whitney class $w(M)$ is equal to $Sq(v)$, where $v=sum v_iin H^*(M;mathbbZ_2)$ is the unique cohomology class such that
      $$langle vcup x,zrangle=langle Sq(x),zrangle$$
      for all $xin H^*(M;mathbbZ_2)$. Thus, for $kge0$, $v_kcup x=Sq^k(x)$ for all $xin H^n-k(M;mathbbZ_2)$, and
      $$w_k(M)=sum_i+j=kSq^i(v_j).$$
      Here the Poincare duality guarantees the existence and uniqueness of $v$.



      My question: if $M$ is a smooth compact $n$-manifold with boundary, is there a similar Wu formula?
      In this case, there is a fundamental class $zin H_n(M,partial M;mathbbZ_2)$ and the relative Poincare duality claims that capping with $z$ yields duality isomorphisms
      $$D:H^p(M,partial M;mathbbZ_2)to H_n-p(M;mathbbZ_2)$$
      and
      $$D:H^p(M;mathbbZ_2)to H_n-p(M,partial M;mathbbZ_2).$$



      Thank you!










      share|cite|improve this question









      $endgroup$




      The classical Wu formula claims that if $M$ is a smooth closed $n$-manifold with fundamental class $zin H_n(M;mathbbZ_2)$, then the total Stiefel-Whitney class $w(M)$ is equal to $Sq(v)$, where $v=sum v_iin H^*(M;mathbbZ_2)$ is the unique cohomology class such that
      $$langle vcup x,zrangle=langle Sq(x),zrangle$$
      for all $xin H^*(M;mathbbZ_2)$. Thus, for $kge0$, $v_kcup x=Sq^k(x)$ for all $xin H^n-k(M;mathbbZ_2)$, and
      $$w_k(M)=sum_i+j=kSq^i(v_j).$$
      Here the Poincare duality guarantees the existence and uniqueness of $v$.



      My question: if $M$ is a smooth compact $n$-manifold with boundary, is there a similar Wu formula?
      In this case, there is a fundamental class $zin H_n(M,partial M;mathbbZ_2)$ and the relative Poincare duality claims that capping with $z$ yields duality isomorphisms
      $$D:H^p(M,partial M;mathbbZ_2)to H_n-p(M;mathbbZ_2)$$
      and
      $$D:H^p(M;mathbbZ_2)to H_n-p(M,partial M;mathbbZ_2).$$



      Thank you!







      at.algebraic-topology gt.geometric-topology cohomology smooth-manifolds characteristic-classes






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      asked Apr 17 at 17:05









      BorromeanBorromean

      6253 silver badges12 bronze badges




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          1 Answer
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          $begingroup$

          A relative Wu formula for manifolds with boundary is discussed in Section 7 of



          Kervaire, Michel A., Relative characteristic classes, Am. J. Math. 79, 517-558 (1957). ZBL0173.51201.



          In particular, there are relative Wu classes $U^qin H^q(M;mathbbZ/2)$ for $q=0,1,ldots , n$ defined by the property that
          $$
          Sq^q(x)=U^qcup x in H^n(M,partial M;mathbbZ/2)
          $$

          for all $xin H^n-q(M,partial M;mathbbZ/2)$. Kervaire deduces the relative Wu formula $w(M)=Sq(U)$ from the absolute Wu formula for the double $N=Mcup_partial M M$ (the closed manifold obtained by gluing two copies of $M$ along the identity map of $partial M$), using naturality with respect to the inclusion $i:Mhookrightarrow N$.






          share|cite|improve this answer









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            active

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            15














            $begingroup$

            A relative Wu formula for manifolds with boundary is discussed in Section 7 of



            Kervaire, Michel A., Relative characteristic classes, Am. J. Math. 79, 517-558 (1957). ZBL0173.51201.



            In particular, there are relative Wu classes $U^qin H^q(M;mathbbZ/2)$ for $q=0,1,ldots , n$ defined by the property that
            $$
            Sq^q(x)=U^qcup x in H^n(M,partial M;mathbbZ/2)
            $$

            for all $xin H^n-q(M,partial M;mathbbZ/2)$. Kervaire deduces the relative Wu formula $w(M)=Sq(U)$ from the absolute Wu formula for the double $N=Mcup_partial M M$ (the closed manifold obtained by gluing two copies of $M$ along the identity map of $partial M$), using naturality with respect to the inclusion $i:Mhookrightarrow N$.






            share|cite|improve this answer









            $endgroup$



















              15














              $begingroup$

              A relative Wu formula for manifolds with boundary is discussed in Section 7 of



              Kervaire, Michel A., Relative characteristic classes, Am. J. Math. 79, 517-558 (1957). ZBL0173.51201.



              In particular, there are relative Wu classes $U^qin H^q(M;mathbbZ/2)$ for $q=0,1,ldots , n$ defined by the property that
              $$
              Sq^q(x)=U^qcup x in H^n(M,partial M;mathbbZ/2)
              $$

              for all $xin H^n-q(M,partial M;mathbbZ/2)$. Kervaire deduces the relative Wu formula $w(M)=Sq(U)$ from the absolute Wu formula for the double $N=Mcup_partial M M$ (the closed manifold obtained by gluing two copies of $M$ along the identity map of $partial M$), using naturality with respect to the inclusion $i:Mhookrightarrow N$.






              share|cite|improve this answer









              $endgroup$

















                15














                15










                15







                $begingroup$

                A relative Wu formula for manifolds with boundary is discussed in Section 7 of



                Kervaire, Michel A., Relative characteristic classes, Am. J. Math. 79, 517-558 (1957). ZBL0173.51201.



                In particular, there are relative Wu classes $U^qin H^q(M;mathbbZ/2)$ for $q=0,1,ldots , n$ defined by the property that
                $$
                Sq^q(x)=U^qcup x in H^n(M,partial M;mathbbZ/2)
                $$

                for all $xin H^n-q(M,partial M;mathbbZ/2)$. Kervaire deduces the relative Wu formula $w(M)=Sq(U)$ from the absolute Wu formula for the double $N=Mcup_partial M M$ (the closed manifold obtained by gluing two copies of $M$ along the identity map of $partial M$), using naturality with respect to the inclusion $i:Mhookrightarrow N$.






                share|cite|improve this answer









                $endgroup$



                A relative Wu formula for manifolds with boundary is discussed in Section 7 of



                Kervaire, Michel A., Relative characteristic classes, Am. J. Math. 79, 517-558 (1957). ZBL0173.51201.



                In particular, there are relative Wu classes $U^qin H^q(M;mathbbZ/2)$ for $q=0,1,ldots , n$ defined by the property that
                $$
                Sq^q(x)=U^qcup x in H^n(M,partial M;mathbbZ/2)
                $$

                for all $xin H^n-q(M,partial M;mathbbZ/2)$. Kervaire deduces the relative Wu formula $w(M)=Sq(U)$ from the absolute Wu formula for the double $N=Mcup_partial M M$ (the closed manifold obtained by gluing two copies of $M$ along the identity map of $partial M$), using naturality with respect to the inclusion $i:Mhookrightarrow N$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Apr 17 at 18:47









                Mark GrantMark Grant

                24.3k6 gold badges68 silver badges146 bronze badges




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