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Question about Chern Character in Hatcher's book



The 2019 Stack Overflow Developer Survey Results Are InQuestion involving the Chern character from the book “Fibre Bundles”Chern Character Isomorphism for non-CW complexeschern character of wedge product of bundlesQuestion about Hatcher's book CW complexAbout homotopy fiber at Hatcher's bookUniqueness of the Chern characterHatcher's KTVB - Proposed Isomorphismchern character domain and codomainQuestion about proof of Kunneth theorem in Hatcher's bookquestion about a notation in Atiyah's book K-Theory










2












$begingroup$


I have a question about an argument from Allen Hatcher's script Vector Bundles and K-Theory in Cor. 4.4 (see page 110). Here the excerpt:



enter image description here



We consider a vector bundle $E to S^2n$, Then for Chern classes we know (by cosidering cohomology groups of $S^2n$) that $c_1(E) =... c_n-1(E)=0$.



Futhermore by definitionof Chern character we have $ch(E)= dim E + s_n(c_1,ldots, c_n)/n!$



My question is why holds



$$s_n(c_1,ldots, c_n)/n!=pm nc_n(E)/n!$$? (*)



The author refers to a recursion formula from page 63:



$s_n= sigma_1 s_n-1 - cdots +(-1)^n-1nsigma_n$.



where $sigma_k$ are the $k$-th symmetric polynomials.



What I don't understand is why $s_n(c_1,ldots, c_n)/n!=pm nc_n(E)/n!$ and not $s_n(c_1,..., c_n)/n!=c_n^n(E)/n!$?



Indeed, here the symmetric polynomials are considered in variebles $t_i:= c_i(E)$ therefore $sigma_1= sum c_i(E)= c_n(E)$ and $sigma_k=0$ for $k >1$ since all summands of $sigma_k$ the containa factor $c_j$ with $j neq n$. But this contracicts (*). Where is the error in my reasonings?



Thank you.










share|cite|improve this question











$endgroup$
















    2












    $begingroup$


    I have a question about an argument from Allen Hatcher's script Vector Bundles and K-Theory in Cor. 4.4 (see page 110). Here the excerpt:



    enter image description here



    We consider a vector bundle $E to S^2n$, Then for Chern classes we know (by cosidering cohomology groups of $S^2n$) that $c_1(E) =... c_n-1(E)=0$.



    Futhermore by definitionof Chern character we have $ch(E)= dim E + s_n(c_1,ldots, c_n)/n!$



    My question is why holds



    $$s_n(c_1,ldots, c_n)/n!=pm nc_n(E)/n!$$? (*)



    The author refers to a recursion formula from page 63:



    $s_n= sigma_1 s_n-1 - cdots +(-1)^n-1nsigma_n$.



    where $sigma_k$ are the $k$-th symmetric polynomials.



    What I don't understand is why $s_n(c_1,ldots, c_n)/n!=pm nc_n(E)/n!$ and not $s_n(c_1,..., c_n)/n!=c_n^n(E)/n!$?



    Indeed, here the symmetric polynomials are considered in variebles $t_i:= c_i(E)$ therefore $sigma_1= sum c_i(E)= c_n(E)$ and $sigma_k=0$ for $k >1$ since all summands of $sigma_k$ the containa factor $c_j$ with $j neq n$. But this contracicts (*). Where is the error in my reasonings?



    Thank you.










    share|cite|improve this question











    $endgroup$














      2












      2








      2





      $begingroup$


      I have a question about an argument from Allen Hatcher's script Vector Bundles and K-Theory in Cor. 4.4 (see page 110). Here the excerpt:



      enter image description here



      We consider a vector bundle $E to S^2n$, Then for Chern classes we know (by cosidering cohomology groups of $S^2n$) that $c_1(E) =... c_n-1(E)=0$.



      Futhermore by definitionof Chern character we have $ch(E)= dim E + s_n(c_1,ldots, c_n)/n!$



      My question is why holds



      $$s_n(c_1,ldots, c_n)/n!=pm nc_n(E)/n!$$? (*)



      The author refers to a recursion formula from page 63:



      $s_n= sigma_1 s_n-1 - cdots +(-1)^n-1nsigma_n$.



      where $sigma_k$ are the $k$-th symmetric polynomials.



      What I don't understand is why $s_n(c_1,ldots, c_n)/n!=pm nc_n(E)/n!$ and not $s_n(c_1,..., c_n)/n!=c_n^n(E)/n!$?



      Indeed, here the symmetric polynomials are considered in variebles $t_i:= c_i(E)$ therefore $sigma_1= sum c_i(E)= c_n(E)$ and $sigma_k=0$ for $k >1$ since all summands of $sigma_k$ the containa factor $c_j$ with $j neq n$. But this contracicts (*). Where is the error in my reasonings?



      Thank you.










      share|cite|improve this question











      $endgroup$




      I have a question about an argument from Allen Hatcher's script Vector Bundles and K-Theory in Cor. 4.4 (see page 110). Here the excerpt:



      enter image description here



      We consider a vector bundle $E to S^2n$, Then for Chern classes we know (by cosidering cohomology groups of $S^2n$) that $c_1(E) =... c_n-1(E)=0$.



      Futhermore by definitionof Chern character we have $ch(E)= dim E + s_n(c_1,ldots, c_n)/n!$



      My question is why holds



      $$s_n(c_1,ldots, c_n)/n!=pm nc_n(E)/n!$$? (*)



      The author refers to a recursion formula from page 63:



      $s_n= sigma_1 s_n-1 - cdots +(-1)^n-1nsigma_n$.



      where $sigma_k$ are the $k$-th symmetric polynomials.



      What I don't understand is why $s_n(c_1,ldots, c_n)/n!=pm nc_n(E)/n!$ and not $s_n(c_1,..., c_n)/n!=c_n^n(E)/n!$?



      Indeed, here the symmetric polynomials are considered in variebles $t_i:= c_i(E)$ therefore $sigma_1= sum c_i(E)= c_n(E)$ and $sigma_k=0$ for $k >1$ since all summands of $sigma_k$ the containa factor $c_j$ with $j neq n$. But this contracicts (*). Where is the error in my reasonings?



      Thank you.







      algebraic-topology vector-bundles topological-k-theory






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 24 at 5:02









      Andrews

      1,2812423




      1,2812423










      asked Mar 24 at 4:48









      KarlPeterKarlPeter

      6771416




      6771416




















          2 Answers
          2






          active

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          2












          $begingroup$

          I think you are confused about what $s_n$ means here. The notation $s_n(c_1,dots,c_n)$ does not mean we are substituting the $c_i$ for the variables $t_i$ in the symmetric polynomial $t_1^n+dots+t_n^n$. Rather, $s_n$ is defined as the polynomial which, when inputted the elementary symmetric polynomials in $t_1,dots,t_n$, outputs $t_1^n+dots+t_n^n$. That is, $$s_n(sigma_1(t_1,dots,t_n),dots,sigma_n(t_1,dots,t_n))=t_1^n+dots+t_n^n.$$ In particular, this means that when we apply the formula $$s_n= sigma_1 s_n-1 - dots +(-1)^n-1nsigma_n$$ to $s_n(c_1,dots,c_n)$, we are substituting $c_i$ for $sigma_i$, not for $t_i$. Since $c_i=0$ for $0<i<n$, we get just $$s_n(c_1,dots,c_n)=(-1)^n-1nc_n.$$






          share|cite|improve this answer









          $endgroup$




















            1












            $begingroup$

            I always find it helps to think about these things in terms of the Splitting Principle:




            For every complex vector bundle $E to X$ of rank $n$ (where maybe $X$ has to be paracompact) there is a "splitting space" $S(X)$ and a continuous function $fcolon S(X) to X$ such that $f^*(E) cong oplus_i=1 ^n L_i$ where $L_i to S(X)$ is a complex line bundle and $f^*colon H^*(X) to H^*(S(X))$ is injective.




            If we let $x_i = c_1(L_i)$ then $$f^*(c_k(E)) = sigma_k(x_1,dots,x_n)$$



            where $sigma_kin mathbbZ[x_1,dots,x_n]$ is the $k$-th elementary symmetric polynomial. In particular $sigma_1 = x_1 + dots + x_n$ doesn't correspond to $c_n(E)$, it's $c_1(E)$.



            The polynomial $x_1^k + dots + x_n^kin mathbbZ[x_1,dots,x_n]$ is symmetric so The Theory implies it can be expressed as a polynomial in the elementary symmetric poynomials, i.e. there is an $s_k in mathbbZ[y_1,dots, y_n]$ (called a "Netwon Polynomial") such that $s_k(sigma_1, dots, sigma_n) = x_1^k + dots + x_n^k$. Then if $c_1(E),dots,c_n-1(E)$ vanish so will $sigma_1, dots sigma_n-1in H^*(S(X))$, so the recursion formula reduces to



            $$ s_n = (-1)^n n sigma_n $$






            share|cite|improve this answer











            $endgroup$













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              2 Answers
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              2 Answers
              2






              active

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              active

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              active

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              2












              $begingroup$

              I think you are confused about what $s_n$ means here. The notation $s_n(c_1,dots,c_n)$ does not mean we are substituting the $c_i$ for the variables $t_i$ in the symmetric polynomial $t_1^n+dots+t_n^n$. Rather, $s_n$ is defined as the polynomial which, when inputted the elementary symmetric polynomials in $t_1,dots,t_n$, outputs $t_1^n+dots+t_n^n$. That is, $$s_n(sigma_1(t_1,dots,t_n),dots,sigma_n(t_1,dots,t_n))=t_1^n+dots+t_n^n.$$ In particular, this means that when we apply the formula $$s_n= sigma_1 s_n-1 - dots +(-1)^n-1nsigma_n$$ to $s_n(c_1,dots,c_n)$, we are substituting $c_i$ for $sigma_i$, not for $t_i$. Since $c_i=0$ for $0<i<n$, we get just $$s_n(c_1,dots,c_n)=(-1)^n-1nc_n.$$






              share|cite|improve this answer









              $endgroup$

















                2












                $begingroup$

                I think you are confused about what $s_n$ means here. The notation $s_n(c_1,dots,c_n)$ does not mean we are substituting the $c_i$ for the variables $t_i$ in the symmetric polynomial $t_1^n+dots+t_n^n$. Rather, $s_n$ is defined as the polynomial which, when inputted the elementary symmetric polynomials in $t_1,dots,t_n$, outputs $t_1^n+dots+t_n^n$. That is, $$s_n(sigma_1(t_1,dots,t_n),dots,sigma_n(t_1,dots,t_n))=t_1^n+dots+t_n^n.$$ In particular, this means that when we apply the formula $$s_n= sigma_1 s_n-1 - dots +(-1)^n-1nsigma_n$$ to $s_n(c_1,dots,c_n)$, we are substituting $c_i$ for $sigma_i$, not for $t_i$. Since $c_i=0$ for $0<i<n$, we get just $$s_n(c_1,dots,c_n)=(-1)^n-1nc_n.$$






                share|cite|improve this answer









                $endgroup$















                  2












                  2








                  2





                  $begingroup$

                  I think you are confused about what $s_n$ means here. The notation $s_n(c_1,dots,c_n)$ does not mean we are substituting the $c_i$ for the variables $t_i$ in the symmetric polynomial $t_1^n+dots+t_n^n$. Rather, $s_n$ is defined as the polynomial which, when inputted the elementary symmetric polynomials in $t_1,dots,t_n$, outputs $t_1^n+dots+t_n^n$. That is, $$s_n(sigma_1(t_1,dots,t_n),dots,sigma_n(t_1,dots,t_n))=t_1^n+dots+t_n^n.$$ In particular, this means that when we apply the formula $$s_n= sigma_1 s_n-1 - dots +(-1)^n-1nsigma_n$$ to $s_n(c_1,dots,c_n)$, we are substituting $c_i$ for $sigma_i$, not for $t_i$. Since $c_i=0$ for $0<i<n$, we get just $$s_n(c_1,dots,c_n)=(-1)^n-1nc_n.$$






                  share|cite|improve this answer









                  $endgroup$



                  I think you are confused about what $s_n$ means here. The notation $s_n(c_1,dots,c_n)$ does not mean we are substituting the $c_i$ for the variables $t_i$ in the symmetric polynomial $t_1^n+dots+t_n^n$. Rather, $s_n$ is defined as the polynomial which, when inputted the elementary symmetric polynomials in $t_1,dots,t_n$, outputs $t_1^n+dots+t_n^n$. That is, $$s_n(sigma_1(t_1,dots,t_n),dots,sigma_n(t_1,dots,t_n))=t_1^n+dots+t_n^n.$$ In particular, this means that when we apply the formula $$s_n= sigma_1 s_n-1 - dots +(-1)^n-1nsigma_n$$ to $s_n(c_1,dots,c_n)$, we are substituting $c_i$ for $sigma_i$, not for $t_i$. Since $c_i=0$ for $0<i<n$, we get just $$s_n(c_1,dots,c_n)=(-1)^n-1nc_n.$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Mar 24 at 5:36









                  Eric WofseyEric Wofsey

                  193k14220352




                  193k14220352





















                      1












                      $begingroup$

                      I always find it helps to think about these things in terms of the Splitting Principle:




                      For every complex vector bundle $E to X$ of rank $n$ (where maybe $X$ has to be paracompact) there is a "splitting space" $S(X)$ and a continuous function $fcolon S(X) to X$ such that $f^*(E) cong oplus_i=1 ^n L_i$ where $L_i to S(X)$ is a complex line bundle and $f^*colon H^*(X) to H^*(S(X))$ is injective.




                      If we let $x_i = c_1(L_i)$ then $$f^*(c_k(E)) = sigma_k(x_1,dots,x_n)$$



                      where $sigma_kin mathbbZ[x_1,dots,x_n]$ is the $k$-th elementary symmetric polynomial. In particular $sigma_1 = x_1 + dots + x_n$ doesn't correspond to $c_n(E)$, it's $c_1(E)$.



                      The polynomial $x_1^k + dots + x_n^kin mathbbZ[x_1,dots,x_n]$ is symmetric so The Theory implies it can be expressed as a polynomial in the elementary symmetric poynomials, i.e. there is an $s_k in mathbbZ[y_1,dots, y_n]$ (called a "Netwon Polynomial") such that $s_k(sigma_1, dots, sigma_n) = x_1^k + dots + x_n^k$. Then if $c_1(E),dots,c_n-1(E)$ vanish so will $sigma_1, dots sigma_n-1in H^*(S(X))$, so the recursion formula reduces to



                      $$ s_n = (-1)^n n sigma_n $$






                      share|cite|improve this answer











                      $endgroup$

















                        1












                        $begingroup$

                        I always find it helps to think about these things in terms of the Splitting Principle:




                        For every complex vector bundle $E to X$ of rank $n$ (where maybe $X$ has to be paracompact) there is a "splitting space" $S(X)$ and a continuous function $fcolon S(X) to X$ such that $f^*(E) cong oplus_i=1 ^n L_i$ where $L_i to S(X)$ is a complex line bundle and $f^*colon H^*(X) to H^*(S(X))$ is injective.




                        If we let $x_i = c_1(L_i)$ then $$f^*(c_k(E)) = sigma_k(x_1,dots,x_n)$$



                        where $sigma_kin mathbbZ[x_1,dots,x_n]$ is the $k$-th elementary symmetric polynomial. In particular $sigma_1 = x_1 + dots + x_n$ doesn't correspond to $c_n(E)$, it's $c_1(E)$.



                        The polynomial $x_1^k + dots + x_n^kin mathbbZ[x_1,dots,x_n]$ is symmetric so The Theory implies it can be expressed as a polynomial in the elementary symmetric poynomials, i.e. there is an $s_k in mathbbZ[y_1,dots, y_n]$ (called a "Netwon Polynomial") such that $s_k(sigma_1, dots, sigma_n) = x_1^k + dots + x_n^k$. Then if $c_1(E),dots,c_n-1(E)$ vanish so will $sigma_1, dots sigma_n-1in H^*(S(X))$, so the recursion formula reduces to



                        $$ s_n = (-1)^n n sigma_n $$






                        share|cite|improve this answer











                        $endgroup$















                          1












                          1








                          1





                          $begingroup$

                          I always find it helps to think about these things in terms of the Splitting Principle:




                          For every complex vector bundle $E to X$ of rank $n$ (where maybe $X$ has to be paracompact) there is a "splitting space" $S(X)$ and a continuous function $fcolon S(X) to X$ such that $f^*(E) cong oplus_i=1 ^n L_i$ where $L_i to S(X)$ is a complex line bundle and $f^*colon H^*(X) to H^*(S(X))$ is injective.




                          If we let $x_i = c_1(L_i)$ then $$f^*(c_k(E)) = sigma_k(x_1,dots,x_n)$$



                          where $sigma_kin mathbbZ[x_1,dots,x_n]$ is the $k$-th elementary symmetric polynomial. In particular $sigma_1 = x_1 + dots + x_n$ doesn't correspond to $c_n(E)$, it's $c_1(E)$.



                          The polynomial $x_1^k + dots + x_n^kin mathbbZ[x_1,dots,x_n]$ is symmetric so The Theory implies it can be expressed as a polynomial in the elementary symmetric poynomials, i.e. there is an $s_k in mathbbZ[y_1,dots, y_n]$ (called a "Netwon Polynomial") such that $s_k(sigma_1, dots, sigma_n) = x_1^k + dots + x_n^k$. Then if $c_1(E),dots,c_n-1(E)$ vanish so will $sigma_1, dots sigma_n-1in H^*(S(X))$, so the recursion formula reduces to



                          $$ s_n = (-1)^n n sigma_n $$






                          share|cite|improve this answer











                          $endgroup$



                          I always find it helps to think about these things in terms of the Splitting Principle:




                          For every complex vector bundle $E to X$ of rank $n$ (where maybe $X$ has to be paracompact) there is a "splitting space" $S(X)$ and a continuous function $fcolon S(X) to X$ such that $f^*(E) cong oplus_i=1 ^n L_i$ where $L_i to S(X)$ is a complex line bundle and $f^*colon H^*(X) to H^*(S(X))$ is injective.




                          If we let $x_i = c_1(L_i)$ then $$f^*(c_k(E)) = sigma_k(x_1,dots,x_n)$$



                          where $sigma_kin mathbbZ[x_1,dots,x_n]$ is the $k$-th elementary symmetric polynomial. In particular $sigma_1 = x_1 + dots + x_n$ doesn't correspond to $c_n(E)$, it's $c_1(E)$.



                          The polynomial $x_1^k + dots + x_n^kin mathbbZ[x_1,dots,x_n]$ is symmetric so The Theory implies it can be expressed as a polynomial in the elementary symmetric poynomials, i.e. there is an $s_k in mathbbZ[y_1,dots, y_n]$ (called a "Netwon Polynomial") such that $s_k(sigma_1, dots, sigma_n) = x_1^k + dots + x_n^k$. Then if $c_1(E),dots,c_n-1(E)$ vanish so will $sigma_1, dots sigma_n-1in H^*(S(X))$, so the recursion formula reduces to



                          $$ s_n = (-1)^n n sigma_n $$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Mar 24 at 6:01

























                          answered Mar 24 at 5:39









                          WilliamWilliam

                          3,2311228




                          3,2311228



























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                              Kathakali Contents Etymology and nomenclature History Repertoire Songs and musical instruments Traditional plays Styles: Sampradayam Training centers and awards Relationship to other dance forms See also Notes References External links Navigation menueThe Illustrated Encyclopedia of Hinduism: A-MSouth Asian Folklore: An EncyclopediaRoutledge International Encyclopedia of Women: Global Women's Issues and KnowledgeKathakali Dance-drama: Where Gods and Demons Come to PlayKathakali Dance-drama: Where Gods and Demons Come to PlayKathakali Dance-drama: Where Gods and Demons Come to Play10.1353/atj.2005.0004The Illustrated Encyclopedia of Hinduism: A-MEncyclopedia of HinduismKathakali Dance-drama: Where Gods and Demons Come to PlaySonic Liturgy: Ritual and Music in Hindu Tradition"The Mirror of Gesture"Kathakali Dance-drama: Where Gods and Demons Come to Play"Kathakali"Indian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceMedieval Indian Literature: An AnthologyThe Oxford Companion to Indian TheatreSouth Asian Folklore: An Encyclopedia : Afghanistan, Bangladesh, India, Nepal, Pakistan, Sri LankaThe Rise of Performance Studies: Rethinking Richard Schechner's Broad SpectrumIndian Theatre: Traditions of PerformanceModern Asian Theatre and Performance 1900-2000Critical Theory and PerformanceBetween Theater and AnthropologyKathakali603847011Indian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceIndian Theatre: Traditions of PerformanceBetween Theater and AnthropologyBetween Theater and AnthropologyNambeesan Smaraka AwardsArchivedThe Cambridge Guide to TheatreRoutledge International Encyclopedia of Women: Global Women's Issues and KnowledgeThe Garland Encyclopedia of World Music: South Asia : the Indian subcontinentThe Ethos of Noh: Actors and Their Art10.2307/1145740By Means of Performance: Intercultural Studies of Theatre and Ritual10.1017/s204912550000100xReconceiving the Renaissance: A Critical ReaderPerformance TheoryListening to Theatre: The Aural Dimension of Beijing Opera10.2307/1146013Kathakali: The Art of the Non-WorldlyOn KathakaliKathakali, the dance theatreThe Kathakali Complex: Performance & StructureKathakali Dance-Drama: Where Gods and Demons Come to Play10.1093/obo/9780195399318-0071Drama and Ritual of Early Hinduism"In the Shadow of Hollywood Orientalism: Authentic East Indian Dancing"10.1080/08949460490274013Sanskrit Play Production in Ancient IndiaIndian Music: History and StructureBharata, the Nāṭyaśāstra233639306Table of Contents2238067286469807Dance In Indian Painting10.2307/32047833204783Kathakali Dance-Theatre: A Visual Narrative of Sacred Indian MimeIndian Classical Dance: The Renaissance and BeyondKathakali: an indigenous art-form of Keralaeee

                              Urgehal History Discography Band members References External links Navigation menu"Mediateket: Urgehal""Interview with Enzifer of Urgehal, 2007""Urgehal - Interview"Urgehal"Urgehal Frontman Trondr Nefas Dies at 35"Urgehal9042691cb161873230(data)0000 0001 0669 4224no2016126817ee6ccef6-e558-44b6-b059-dbbb5b913b24145036459145036459