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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
$begingroup$
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:
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
$endgroup$
add a comment |
$begingroup$
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:
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
$endgroup$
add a comment |
$begingroup$
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:
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
$endgroup$
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:
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
algebraic-topology vector-bundles topological-k-theory
edited Mar 24 at 5:02
Andrews
1,2812423
1,2812423
asked Mar 24 at 4:48
KarlPeterKarlPeter
6771416
6771416
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2 Answers
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$begingroup$
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 $$s_n(c_1,dots,c_n)=(-1)^n-1nc_n.$$
$endgroup$
add a comment |
$begingroup$
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
$$ s_n = (-1)^n n sigma_n $$
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
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 $$s_n(c_1,dots,c_n)=(-1)^n-1nc_n.$$
$endgroup$
add a comment |
$begingroup$
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 $$s_n(c_1,dots,c_n)=(-1)^n-1nc_n.$$
$endgroup$
add a comment |
$begingroup$
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 $$s_n(c_1,dots,c_n)=(-1)^n-1nc_n.$$
$endgroup$
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 $$s_n(c_1,dots,c_n)=(-1)^n-1nc_n.$$
answered Mar 24 at 5:36
Eric WofseyEric Wofsey
193k14220352
193k14220352
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$begingroup$
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
$$ s_n = (-1)^n n sigma_n $$
$endgroup$
add a comment |
$begingroup$
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
$$ s_n = (-1)^n n sigma_n $$
$endgroup$
add a comment |
$begingroup$
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
$$ s_n = (-1)^n n sigma_n $$
$endgroup$
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
$$ s_n = (-1)^n n sigma_n $$
edited Mar 24 at 6:01
answered Mar 24 at 5:39
WilliamWilliam
3,2311228
3,2311228
add a comment |
add a comment |
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