Homeomorphism between roots and coefficients of monic real polynomials
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Let us define an equivalence relation $sim$ on $mathbb C^n$ by $x sim y$ if and only if $x = (x_{sigma(1)}, dots, x_{sigma(n)}) = (y_1, dots,y_n)$ where $sigma in mathbb S^n$ belongs to the permutation group. We define the quotient space $mathbb C_{sym}^n = mathbb C^n/sim$. In Chapter $VI$ of Bhatia's Matrix Analysis, it states: if we define a map $f : mathbb C_{sym}^n to mathbb C^n$ by sending the roots of a monic $n^{th}$ polynomial to its coefficients, then $f$ is a homeomorphism.
Let $g = f^{-1}$. Now suppose I restrict the map $g$ on $mathbb R^n$. $g |_{mathbb R^n}: mathbb R^n to g(mathbb R^n)$. Then the image should be a subset in $mathbb C^n_{sym}$ that is invariant under complex conjugation, i.e., containing complex conjugate pairs. $g|_{mathbb R^n}$ is certainly a homeomorphism. Is there a place I can find this stated explicitly?
linear-algebra abstract-algebra general-topology polynomials reference-request
asked Nov 15 at 23:09
user9527
1,3891627
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up vote
2
down vote
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Let us define an equivalence relation $sim$ on $mathbb C^n$ by $x sim y$ if and only if $x = (x_{sigma(1)}, dots, x_{sigma(n)}) = (y_1, dots,y_n)$ where $sigma in mathbb S^n$ belongs to the permutation group. We define the quotient space $mathbb C_{sym}^n = mathbb C^n/sim$. In Chapter $VI$ of Bhatia's Matrix Analysis, it states: if we define a map $f : mathbb C_{sym}^n to mathbb C^n$ by sending the roots of a monic $n^{th}$ polynomial to its coefficients, then $f$ is a homeomorphism.
Let $g = f^{-1}$. Now suppose I restrict the map $g$ on $mathbb R^n$. $g |_{mathbb R^n}: mathbb R^n to g(mathbb R^n)$. Then the image should be a subset in $mathbb C^n_{sym}$ that is invariant under complex conjugation, i.e., containing complex conjugate pairs. $g|_{mathbb R^n}$ is certainly a homeomorphism. Is there a place I can find this stated explicitly?
linear-algebra abstract-algebra general-topology polynomials reference-request
asked Nov 15 at 23:09
user9527
1,3891627
The image is $(mathbb{C}^n_{sym})^{rho}$ the subset of $mathbb{C}^n_{sym}$ fixed by the complex conjugation $rho$, and $f : (mathbb{C}^n_{sym})^{rho} to mathbb{R}^n$ is an homeomorphism by the same argument as the one for $f : mathbb{C}^n_{sym} to mathbb{C}^n$
– reuns
Nov 16 at 0:30
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let us define an equivalence relation $sim$ on $mathbb C^n$ by $x sim y$ if and only if $x = (x_{sigma(1)}, dots, x_{sigma(n)}) = (y_1, dots,y_n)$ where $sigma in mathbb S^n$ belongs to the permutation group. We define the quotient space $mathbb C_{sym}^n = mathbb C^n/sim$. In Chapter $VI$ of Bhatia's Matrix Analysis, it states: if we define a map $f : mathbb C_{sym}^n to mathbb C^n$ by sending the roots of a monic $n^{th}$ polynomial to its coefficients, then $f$ is a homeomorphism.
Let $g = f^{-1}$. Now suppose I restrict the map $g$ on $mathbb R^n$. $g |_{mathbb R^n}: mathbb R^n to g(mathbb R^n)$. Then the image should be a subset in $mathbb C^n_{sym}$ that is invariant under complex conjugation, i.e., containing complex conjugate pairs. $g|_{mathbb R^n}$ is certainly a homeomorphism. Is there a place I can find this stated explicitly?
linear-algebra abstract-algebra general-topology polynomials reference-request
asked Nov 15 at 23:09
user9527
1,3891627
Let us define an equivalence relation $sim$ on $mathbb C^n$ by $x sim y$ if and only if $x = (x_{sigma(1)}, dots, x_{sigma(n)}) = (y_1, dots,y_n)$ where $sigma in mathbb S^n$ belongs to the permutation group. We define the quotient space $mathbb C_{sym}^n = mathbb C^n/sim$. In Chapter $VI$ of Bhatia's Matrix Analysis, it states: if we define a map $f : mathbb C_{sym}^n to mathbb C^n$ by sending the roots of a monic $n^{th}$ polynomial to its coefficients, then $f$ is a homeomorphism.
Let $g = f^{-1}$. Now suppose I restrict the map $g$ on $mathbb R^n$. $g |_{mathbb R^n}: mathbb R^n to g(mathbb R^n)$. Then the image should be a subset in $mathbb C^n_{sym}$ that is invariant under complex conjugation, i.e., containing complex conjugate pairs. $g|_{mathbb R^n}$ is certainly a homeomorphism. Is there a place I can find this stated explicitly?
linear-algebra abstract-algebra general-topology polynomials reference-request
linear-algebra abstract-algebra general-topology polynomials reference-request
asked Nov 15 at 23:09
user9527
1,3891627
asked Nov 15 at 23:09
user9527
1,3891627
edited Nov 15 at 23:18
asked Nov 15 at 23:09
user9527
1,3891627
asked Nov 15 at 23:09
user9527
1,3891627
asked Nov 15 at 23:09
user9527
1,3891627
1,3891627
The image is $(mathbb{C}^n_{sym})^{rho}$ the subset of $mathbb{C}^n_{sym}$ fixed by the complex conjugation $rho$, and $f : (mathbb{C}^n_{sym})^{rho} to mathbb{R}^n$ is an homeomorphism by the same argument as the one for $f : mathbb{C}^n_{sym} to mathbb{C}^n$
– reuns
Nov 16 at 0:30
add a comment |
The image is $(mathbb{C}^n_{sym})^{rho}$ the subset of $mathbb{C}^n_{sym}$ fixed by the complex conjugation $rho$, and $f : (mathbb{C}^n_{sym})^{rho} to mathbb{R}^n$ is an homeomorphism by the same argument as the one for $f : mathbb{C}^n_{sym} to mathbb{C}^n$
– reuns
Nov 16 at 0:30
The image is $(mathbb{C}^n_{sym})^{rho}$ the subset of $mathbb{C}^n_{sym}$ fixed by the complex conjugation $rho$, and $f : (mathbb{C}^n_{sym})^{rho} to mathbb{R}^n$ is an homeomorphism by the same argument as the one for $f : mathbb{C}^n_{sym} to mathbb{C}^n$
– reuns
Nov 16 at 0:30
The image is $(mathbb{C}^n_{sym})^{rho}$ the subset of $mathbb{C}^n_{sym}$ fixed by the complex conjugation $rho$, and $f : (mathbb{C}^n_{sym})^{rho} to mathbb{R}^n$ is an homeomorphism by the same argument as the one for $f : mathbb{C}^n_{sym} to mathbb{C}^n$
– reuns
Nov 16 at 0:30
add a comment |
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The image is $(mathbb{C}^n_{sym})^{rho}$ the subset of $mathbb{C}^n_{sym}$ fixed by the complex conjugation $rho$, and $f : (mathbb{C}^n_{sym})^{rho} to mathbb{R}^n$ is an homeomorphism by the same argument as the one for $f : mathbb{C}^n_{sym} to mathbb{C}^n$
– reuns
Nov 16 at 0:30