What does it mean geometrically that the diagonal cohomology class is concentrated along the diagonal?












2












$begingroup$


In Milnor Stasheff's Characteristic Classes, the geometric interpretation of Lemma 11.8 on P125 is that the diagonal cohomology class is "concentrated along the diagonal". The lemma states that




For any cohomology class $a in H^*(M)$, the product $(a times 1) cup u''$ is equal to $(1 times a) cup u''$, in $H^*(M times M)$.




Here, $u''$ is the "diagonal cohomology class", defined as in https://mathoverflow.net/a/74199/42662, or alternatively the image of the fundamental class of $H^n(M times M, M times M - Delta(M)) to H^n(M times M)$.



How does the formal statement in the lemma give rise to the geometric "concentration" interpretation?






algebraic-topology homology-cohomology characteristic-classes







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    In Milnor Stasheff's Characteristic Classes, the geometric interpretation of Lemma 11.8 on P125 is that the diagonal cohomology class is "concentrated along the diagonal". The lemma states that




    For any cohomology class $a in H^*(M)$, the product $(a times 1) cup u''$ is equal to $(1 times a) cup u''$, in $H^*(M times M)$.




    Here, $u''$ is the "diagonal cohomology class", defined as in https://mathoverflow.net/a/74199/42662, or alternatively the image of the fundamental class of $H^n(M times M, M times M - Delta(M)) to H^n(M times M)$.



    How does the formal statement in the lemma give rise to the geometric "concentration" interpretation?






    algebraic-topology homology-cohomology characteristic-classes







    share|cite|improve this question











    $endgroup$















      2












      2








      2


      0



      $begingroup$


      In Milnor Stasheff's Characteristic Classes, the geometric interpretation of Lemma 11.8 on P125 is that the diagonal cohomology class is "concentrated along the diagonal". The lemma states that




      For any cohomology class $a in H^*(M)$, the product $(a times 1) cup u''$ is equal to $(1 times a) cup u''$, in $H^*(M times M)$.




      Here, $u''$ is the "diagonal cohomology class", defined as in https://mathoverflow.net/a/74199/42662, or alternatively the image of the fundamental class of $H^n(M times M, M times M - Delta(M)) to H^n(M times M)$.



      How does the formal statement in the lemma give rise to the geometric "concentration" interpretation?






      algebraic-topology homology-cohomology characteristic-classes







      share|cite|improve this question











      $endgroup$




      In Milnor Stasheff's Characteristic Classes, the geometric interpretation of Lemma 11.8 on P125 is that the diagonal cohomology class is "concentrated along the diagonal". The lemma states that




      For any cohomology class $a in H^*(M)$, the product $(a times 1) cup u''$ is equal to $(1 times a) cup u''$, in $H^*(M times M)$.




      Here, $u''$ is the "diagonal cohomology class", defined as in https://mathoverflow.net/a/74199/42662, or alternatively the image of the fundamental class of $H^n(M times M, M times M - Delta(M)) to H^n(M times M)$.



      How does the formal statement in the lemma give rise to the geometric "concentration" interpretation?






      algebraic-topology homology-cohomology characteristic-classes




      algebraic-topology homology-cohomology characteristic-classes






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited May 15 '17 at 21:36







      Kevin Yin

















      asked May 8 '17 at 19:23









      Kevin YinKevin Yin

      234110




      234110






















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          $begingroup$

          It means that if $alpha, beta in H^*(M times M)$ are two forms which are equal when restricted to the diagonal, then their pairing with the diagonal class $int_{M times M} alpha u'' = int_{M times M} beta u''$ are equal. So in a sense only the value of $u''$ when restricted to the diagonal counts. (You also need Poincaré duality to be able to say that knowing all pairings of $u''$ with all other classes is enough to know $u''$.)



          Indeed, by Künneth's formula, over a field you have $H^*(M times M) = H^*(M) otimes H^*(M)$. So you can decompose $alpha = sum_i alpha'_i otimes alpha''_i$ and $beta = sum_j beta'_j otimes beta''_j$. The restriction to the diagonal is just the cup product: if $delta : M to M times M$ is the inclusion of the diagonal ($delta(x) = (x,x)$) then $delta^*alpha = sum_i alpha'_i alpha''_i$ and similarly for $beta$. Suppose $delta^*alpha = delta^*beta$. Then you get:
          $$begin{align}
          int_{M times M} alpha u''
          & = sum_i int_{M times M} (alpha'_i otimes alpha''_i) u'' \
          & = int_{M times M} (sum_i alpha'_i alpha''_i otimes 1) u'' \
          & = int_{M times M} (sum_j beta'_j beta''_j otimes 1) u'' \
          & = int_{M times M} beta u''.
          end{align}$$
          Here I used $(a otimes b) u'' = (a otimes 1) (1 otimes b) u'') = (a otimes 1) (b otimes 1) u'' = (ab otimes 1) u''$.



          Or for something more "geometric", if you work with smooth manifolds and differential forms: if you choose any neighborhood of the diagonal, no matter how small, you will be able to find a representative of $u''$ which vanishes outside this neighborhood.






          share|cite|improve this answer











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            $begingroup$

            It means that if $alpha, beta in H^*(M times M)$ are two forms which are equal when restricted to the diagonal, then their pairing with the diagonal class $int_{M times M} alpha u'' = int_{M times M} beta u''$ are equal. So in a sense only the value of $u''$ when restricted to the diagonal counts. (You also need Poincaré duality to be able to say that knowing all pairings of $u''$ with all other classes is enough to know $u''$.)



            Indeed, by Künneth's formula, over a field you have $H^*(M times M) = H^*(M) otimes H^*(M)$. So you can decompose $alpha = sum_i alpha'_i otimes alpha''_i$ and $beta = sum_j beta'_j otimes beta''_j$. The restriction to the diagonal is just the cup product: if $delta : M to M times M$ is the inclusion of the diagonal ($delta(x) = (x,x)$) then $delta^*alpha = sum_i alpha'_i alpha''_i$ and similarly for $beta$. Suppose $delta^*alpha = delta^*beta$. Then you get:
            $$begin{align}
            int_{M times M} alpha u''
            & = sum_i int_{M times M} (alpha'_i otimes alpha''_i) u'' \
            & = int_{M times M} (sum_i alpha'_i alpha''_i otimes 1) u'' \
            & = int_{M times M} (sum_j beta'_j beta''_j otimes 1) u'' \
            & = int_{M times M} beta u''.
            end{align}$$
            Here I used $(a otimes b) u'' = (a otimes 1) (1 otimes b) u'') = (a otimes 1) (b otimes 1) u'' = (ab otimes 1) u''$.



            Or for something more "geometric", if you work with smooth manifolds and differential forms: if you choose any neighborhood of the diagonal, no matter how small, you will be able to find a representative of $u''$ which vanishes outside this neighborhood.






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              It means that if $alpha, beta in H^*(M times M)$ are two forms which are equal when restricted to the diagonal, then their pairing with the diagonal class $int_{M times M} alpha u'' = int_{M times M} beta u''$ are equal. So in a sense only the value of $u''$ when restricted to the diagonal counts. (You also need Poincaré duality to be able to say that knowing all pairings of $u''$ with all other classes is enough to know $u''$.)



              Indeed, by Künneth's formula, over a field you have $H^*(M times M) = H^*(M) otimes H^*(M)$. So you can decompose $alpha = sum_i alpha'_i otimes alpha''_i$ and $beta = sum_j beta'_j otimes beta''_j$. The restriction to the diagonal is just the cup product: if $delta : M to M times M$ is the inclusion of the diagonal ($delta(x) = (x,x)$) then $delta^*alpha = sum_i alpha'_i alpha''_i$ and similarly for $beta$. Suppose $delta^*alpha = delta^*beta$. Then you get:
              $$begin{align}
              int_{M times M} alpha u''
              & = sum_i int_{M times M} (alpha'_i otimes alpha''_i) u'' \
              & = int_{M times M} (sum_i alpha'_i alpha''_i otimes 1) u'' \
              & = int_{M times M} (sum_j beta'_j beta''_j otimes 1) u'' \
              & = int_{M times M} beta u''.
              end{align}$$
              Here I used $(a otimes b) u'' = (a otimes 1) (1 otimes b) u'') = (a otimes 1) (b otimes 1) u'' = (ab otimes 1) u''$.



              Or for something more "geometric", if you work with smooth manifolds and differential forms: if you choose any neighborhood of the diagonal, no matter how small, you will be able to find a representative of $u''$ which vanishes outside this neighborhood.






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                It means that if $alpha, beta in H^*(M times M)$ are two forms which are equal when restricted to the diagonal, then their pairing with the diagonal class $int_{M times M} alpha u'' = int_{M times M} beta u''$ are equal. So in a sense only the value of $u''$ when restricted to the diagonal counts. (You also need Poincaré duality to be able to say that knowing all pairings of $u''$ with all other classes is enough to know $u''$.)



                Indeed, by Künneth's formula, over a field you have $H^*(M times M) = H^*(M) otimes H^*(M)$. So you can decompose $alpha = sum_i alpha'_i otimes alpha''_i$ and $beta = sum_j beta'_j otimes beta''_j$. The restriction to the diagonal is just the cup product: if $delta : M to M times M$ is the inclusion of the diagonal ($delta(x) = (x,x)$) then $delta^*alpha = sum_i alpha'_i alpha''_i$ and similarly for $beta$. Suppose $delta^*alpha = delta^*beta$. Then you get:
                $$begin{align}
                int_{M times M} alpha u''
                & = sum_i int_{M times M} (alpha'_i otimes alpha''_i) u'' \
                & = int_{M times M} (sum_i alpha'_i alpha''_i otimes 1) u'' \
                & = int_{M times M} (sum_j beta'_j beta''_j otimes 1) u'' \
                & = int_{M times M} beta u''.
                end{align}$$
                Here I used $(a otimes b) u'' = (a otimes 1) (1 otimes b) u'') = (a otimes 1) (b otimes 1) u'' = (ab otimes 1) u''$.



                Or for something more "geometric", if you work with smooth manifolds and differential forms: if you choose any neighborhood of the diagonal, no matter how small, you will be able to find a representative of $u''$ which vanishes outside this neighborhood.






                share|cite|improve this answer











                $endgroup$



                It means that if $alpha, beta in H^*(M times M)$ are two forms which are equal when restricted to the diagonal, then their pairing with the diagonal class $int_{M times M} alpha u'' = int_{M times M} beta u''$ are equal. So in a sense only the value of $u''$ when restricted to the diagonal counts. (You also need Poincaré duality to be able to say that knowing all pairings of $u''$ with all other classes is enough to know $u''$.)



                Indeed, by Künneth's formula, over a field you have $H^*(M times M) = H^*(M) otimes H^*(M)$. So you can decompose $alpha = sum_i alpha'_i otimes alpha''_i$ and $beta = sum_j beta'_j otimes beta''_j$. The restriction to the diagonal is just the cup product: if $delta : M to M times M$ is the inclusion of the diagonal ($delta(x) = (x,x)$) then $delta^*alpha = sum_i alpha'_i alpha''_i$ and similarly for $beta$. Suppose $delta^*alpha = delta^*beta$. Then you get:
                $$begin{align}
                int_{M times M} alpha u''
                & = sum_i int_{M times M} (alpha'_i otimes alpha''_i) u'' \
                & = int_{M times M} (sum_i alpha'_i alpha''_i otimes 1) u'' \
                & = int_{M times M} (sum_j beta'_j beta''_j otimes 1) u'' \
                & = int_{M times M} beta u''.
                end{align}$$
                Here I used $(a otimes b) u'' = (a otimes 1) (1 otimes b) u'') = (a otimes 1) (b otimes 1) u'' = (ab otimes 1) u''$.



                Or for something more "geometric", if you work with smooth manifolds and differential forms: if you choose any neighborhood of the diagonal, no matter how small, you will be able to find a representative of $u''$ which vanishes outside this neighborhood.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Dec 19 '18 at 15:41

























                answered Dec 19 '18 at 15:32









                Najib IdrissiNajib Idrissi

                42k473143




                42k473143






























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