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






















          1 Answer
          1






          active

          oldest

          votes


















          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











          $endgroup$














            Your Answer





            StackExchange.ifUsing("editor", function () {
            return StackExchange.using("mathjaxEditing", function () {
            StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
            StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
            });
            });
            }, "mathjax-editing");

            StackExchange.ready(function() {
            var channelOptions = {
            tags: "".split(" "),
            id: "69"
            };
            initTagRenderer("".split(" "), "".split(" "), channelOptions);

            StackExchange.using("externalEditor", function() {
            // Have to fire editor after snippets, if snippets enabled
            if (StackExchange.settings.snippets.snippetsEnabled) {
            StackExchange.using("snippets", function() {
            createEditor();
            });
            }
            else {
            createEditor();
            }
            });

            function createEditor() {
            StackExchange.prepareEditor({
            heartbeatType: 'answer',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2271994%2fwhat-does-it-mean-geometrically-that-the-diagonal-cohomology-class-is-concentrat%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            1 Answer
            1






            active

            oldest

            votes








            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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












              $begingroup$

              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






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid …



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2271994%2fwhat-does-it-mean-geometrically-that-the-diagonal-cohomology-class-is-concentrat%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    -

                    Popular posts from this blog

                    Plaza Victoria

                    Image
                    Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                    Read more

                    Puebla de Zaragoza

                    Image
                    Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                    Read more

                    Musa

                    Image
                    Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                    Read more