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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
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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









          $endgroup$




















            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






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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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                              How should I support this large drywall patch? Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?How do I cover large gaps in drywall?How do I keep drywall around a patch from crumbling?Can I glue a second layer of drywall?How to patch long strip on drywall?Large drywall patch: how to avoid bulging seams?Drywall Mesh Patch vs. Bulge? To remove or not to remove?How to fix this drywall job?Prep drywall before backsplashWhat's the best way to fix this horrible drywall patch job?Drywall patching using 3M Patch Plus Primer