Covering space: preimage
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Let $A = {(x,y) in Bbb C^2: y^2 = f(x)}$ where $f$ is a polynomial of degree $d$ without repeated roots.
Let $f: A to Bbb C$ be defined by $f(x, y) = x$.
For large $R$, what is the preimage under $f$ of a circle centred at the origin of radius $R$ in $Bbb C$? Describe $A$ as a surface with boundary.
My thoughts on the question: for sufficiently large $R$, we have $f(Re^{itheta}) simeq R^de^{ditheta}$. The (local) inverses of $f$ are $q(x) = (x, pm sqrt{p(x)})$. For $d$ even, we can safely take the square root to get preimage similar to ${Re^{itheta}, R^{d/2}e^{d/2itheta}} cup {Re^{itheta}, -R^{d/2}e^{d/2itheta}}$. I would like to know what this looks like topologically.
For odd $d$ it seems like we need to make a branch cut to get the preimage. I don't know how that will affect the topology of the preimage or how it will affect the topology of $A$ as a whole.
Any hints are appreciated. I don't know differential or Riemannian geometry.
algebraic-topology covering-spaces riemann-sphere
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add a comment |
$begingroup$
Let $A = {(x,y) in Bbb C^2: y^2 = f(x)}$ where $f$ is a polynomial of degree $d$ without repeated roots.
Let $f: A to Bbb C$ be defined by $f(x, y) = x$.
For large $R$, what is the preimage under $f$ of a circle centred at the origin of radius $R$ in $Bbb C$? Describe $A$ as a surface with boundary.
My thoughts on the question: for sufficiently large $R$, we have $f(Re^{itheta}) simeq R^de^{ditheta}$. The (local) inverses of $f$ are $q(x) = (x, pm sqrt{p(x)})$. For $d$ even, we can safely take the square root to get preimage similar to ${Re^{itheta}, R^{d/2}e^{d/2itheta}} cup {Re^{itheta}, -R^{d/2}e^{d/2itheta}}$. I would like to know what this looks like topologically.
For odd $d$ it seems like we need to make a branch cut to get the preimage. I don't know how that will affect the topology of the preimage or how it will affect the topology of $A$ as a whole.
Any hints are appreciated. I don't know differential or Riemannian geometry.
algebraic-topology covering-spaces riemann-sphere
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$begingroup$
I'm sorry, but would you please go through this and fix all the symbolism errors in it? I keep trying to figure out what you really mean, but there are so many inconsistent uses of symbols that it is just too confusing to follow.
$endgroup$
– Paul Sinclair
Dec 13 '18 at 5:21
$begingroup$
@PaulSinclair Sorry about that! I think I've fixed it now.
$endgroup$
– user625807
Dec 13 '18 at 15:17
add a comment |
$begingroup$
Let $A = {(x,y) in Bbb C^2: y^2 = f(x)}$ where $f$ is a polynomial of degree $d$ without repeated roots.
Let $f: A to Bbb C$ be defined by $f(x, y) = x$.
For large $R$, what is the preimage under $f$ of a circle centred at the origin of radius $R$ in $Bbb C$? Describe $A$ as a surface with boundary.
My thoughts on the question: for sufficiently large $R$, we have $f(Re^{itheta}) simeq R^de^{ditheta}$. The (local) inverses of $f$ are $q(x) = (x, pm sqrt{p(x)})$. For $d$ even, we can safely take the square root to get preimage similar to ${Re^{itheta}, R^{d/2}e^{d/2itheta}} cup {Re^{itheta}, -R^{d/2}e^{d/2itheta}}$. I would like to know what this looks like topologically.
For odd $d$ it seems like we need to make a branch cut to get the preimage. I don't know how that will affect the topology of the preimage or how it will affect the topology of $A$ as a whole.
Any hints are appreciated. I don't know differential or Riemannian geometry.
algebraic-topology covering-spaces riemann-sphere
$endgroup$
Let $A = {(x,y) in Bbb C^2: y^2 = f(x)}$ where $f$ is a polynomial of degree $d$ without repeated roots.
Let $f: A to Bbb C$ be defined by $f(x, y) = x$.
For large $R$, what is the preimage under $f$ of a circle centred at the origin of radius $R$ in $Bbb C$? Describe $A$ as a surface with boundary.
My thoughts on the question: for sufficiently large $R$, we have $f(Re^{itheta}) simeq R^de^{ditheta}$. The (local) inverses of $f$ are $q(x) = (x, pm sqrt{p(x)})$. For $d$ even, we can safely take the square root to get preimage similar to ${Re^{itheta}, R^{d/2}e^{d/2itheta}} cup {Re^{itheta}, -R^{d/2}e^{d/2itheta}}$. I would like to know what this looks like topologically.
For odd $d$ it seems like we need to make a branch cut to get the preimage. I don't know how that will affect the topology of the preimage or how it will affect the topology of $A$ as a whole.
Any hints are appreciated. I don't know differential or Riemannian geometry.
algebraic-topology covering-spaces riemann-sphere
algebraic-topology covering-spaces riemann-sphere
edited Dec 13 '18 at 15:17
asked Dec 12 '18 at 19:16
user625807
$begingroup$
I'm sorry, but would you please go through this and fix all the symbolism errors in it? I keep trying to figure out what you really mean, but there are so many inconsistent uses of symbols that it is just too confusing to follow.
$endgroup$
– Paul Sinclair
Dec 13 '18 at 5:21
$begingroup$
@PaulSinclair Sorry about that! I think I've fixed it now.
$endgroup$
– user625807
Dec 13 '18 at 15:17
add a comment |
$begingroup$
I'm sorry, but would you please go through this and fix all the symbolism errors in it? I keep trying to figure out what you really mean, but there are so many inconsistent uses of symbols that it is just too confusing to follow.
$endgroup$
– Paul Sinclair
Dec 13 '18 at 5:21
$begingroup$
@PaulSinclair Sorry about that! I think I've fixed it now.
$endgroup$
– user625807
Dec 13 '18 at 15:17
$begingroup$
I'm sorry, but would you please go through this and fix all the symbolism errors in it? I keep trying to figure out what you really mean, but there are so many inconsistent uses of symbols that it is just too confusing to follow.
$endgroup$
– Paul Sinclair
Dec 13 '18 at 5:21
$begingroup$
I'm sorry, but would you please go through this and fix all the symbolism errors in it? I keep trying to figure out what you really mean, but there are so many inconsistent uses of symbols that it is just too confusing to follow.
$endgroup$
– Paul Sinclair
Dec 13 '18 at 5:21
$begingroup$
@PaulSinclair Sorry about that! I think I've fixed it now.
$endgroup$
– user625807
Dec 13 '18 at 15:17
$begingroup$
@PaulSinclair Sorry about that! I think I've fixed it now.
$endgroup$
– user625807
Dec 13 '18 at 15:17
add a comment |
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$begingroup$
I'm sorry, but would you please go through this and fix all the symbolism errors in it? I keep trying to figure out what you really mean, but there are so many inconsistent uses of symbols that it is just too confusing to follow.
$endgroup$
– Paul Sinclair
Dec 13 '18 at 5:21
$begingroup$
@PaulSinclair Sorry about that! I think I've fixed it now.
$endgroup$
– user625807
Dec 13 '18 at 15:17