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Which surfaces does a compact orientable surface with boundary cover?
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Suppose I have a bounded, orientable genus 5 surface with 4 boundary circles. Is there a way to determine what surfaces it covers?
First, I know that there is a covering map from the closed orientable surface of genus 5 to the closed non-orientable surface of genus 6. This covering map "corresponds" to the quotient maps that identifies anti-podal points. With this in mind, I think the the bounded, orientable genus 5 surface with 4 boundary circles covers a bounded non-orientable surface of genus 6 with 2 boundary circles.
Am I understanding this correctly? Can I take this further? Thanks for any responses!
algebraic-topology surfaces covering-spaces
edited 2 days ago
Michael Albanese
62.5k1598299
asked 2 days ago
MathUser_NotPrime
1,002112
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up vote
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Suppose I have a bounded, orientable genus 5 surface with 4 boundary circles. Is there a way to determine what surfaces it covers?
First, I know that there is a covering map from the closed orientable surface of genus 5 to the closed non-orientable surface of genus 6. This covering map "corresponds" to the quotient maps that identifies anti-podal points. With this in mind, I think the the bounded, orientable genus 5 surface with 4 boundary circles covers a bounded non-orientable surface of genus 6 with 2 boundary circles.
Am I understanding this correctly? Can I take this further? Thanks for any responses!
algebraic-topology surfaces covering-spaces
edited 2 days ago
Michael Albanese
62.5k1598299
asked 2 days ago
MathUser_NotPrime
1,002112
1
One obvious comment: if $S$ covers $S'$ Then $chi(S) = dchi(S')$ where $d>0$ is the degree of the cover. Also if $S$ has non- empty boundary $S'$ must have non-empty boundary. This gives a small number of possibilities to chrck, since for us $chi(S)=-12$.
– Nick L
2 days ago
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2
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up vote
2
down vote
favorite
Suppose I have a bounded, orientable genus 5 surface with 4 boundary circles. Is there a way to determine what surfaces it covers?
First, I know that there is a covering map from the closed orientable surface of genus 5 to the closed non-orientable surface of genus 6. This covering map "corresponds" to the quotient maps that identifies anti-podal points. With this in mind, I think the the bounded, orientable genus 5 surface with 4 boundary circles covers a bounded non-orientable surface of genus 6 with 2 boundary circles.
Am I understanding this correctly? Can I take this further? Thanks for any responses!
algebraic-topology surfaces covering-spaces
edited 2 days ago
Michael Albanese
62.5k1598299
asked 2 days ago
MathUser_NotPrime
1,002112
Suppose I have a bounded, orientable genus 5 surface with 4 boundary circles. Is there a way to determine what surfaces it covers?
First, I know that there is a covering map from the closed orientable surface of genus 5 to the closed non-orientable surface of genus 6. This covering map "corresponds" to the quotient maps that identifies anti-podal points. With this in mind, I think the the bounded, orientable genus 5 surface with 4 boundary circles covers a bounded non-orientable surface of genus 6 with 2 boundary circles.
Am I understanding this correctly? Can I take this further? Thanks for any responses!
algebraic-topology surfaces covering-spaces
algebraic-topology surfaces covering-spaces
edited 2 days ago
Michael Albanese
62.5k1598299
asked 2 days ago
MathUser_NotPrime
1,002112
edited 2 days ago
Michael Albanese
62.5k1598299
asked 2 days ago
MathUser_NotPrime
1,002112
edited 2 days ago
Michael Albanese
62.5k1598299
edited 2 days ago
Michael Albanese
62.5k1598299
edited 2 days ago
Michael Albanese
62.5k1598299
62.5k1598299
asked 2 days ago
MathUser_NotPrime
1,002112
asked 2 days ago
MathUser_NotPrime
1,002112
asked 2 days ago
MathUser_NotPrime
1,002112
1,002112
1
One obvious comment: if $S$ covers $S'$ Then $chi(S) = dchi(S')$ where $d>0$ is the degree of the cover. Also if $S$ has non- empty boundary $S'$ must have non-empty boundary. This gives a small number of possibilities to chrck, since for us $chi(S)=-12$.
– Nick L
2 days ago
add a comment |
1
One obvious comment: if $S$ covers $S'$ Then $chi(S) = dchi(S')$ where $d>0$ is the degree of the cover. Also if $S$ has non- empty boundary $S'$ must have non-empty boundary. This gives a small number of possibilities to chrck, since for us $chi(S)=-12$.
– Nick L
2 days ago
1
1
One obvious comment: if $S$ covers $S'$ Then $chi(S) = dchi(S')$ where $d>0$ is the degree of the cover. Also if $S$ has non- empty boundary $S'$ must have non-empty boundary. This gives a small number of possibilities to chrck, since for us $chi(S)=-12$.
– Nick L
2 days ago
One obvious comment: if $S$ covers $S'$ Then $chi(S) = dchi(S')$ where $d>0$ is the degree of the cover. Also if $S$ has non- empty boundary $S'$ must have non-empty boundary. This gives a small number of possibilities to chrck, since for us $chi(S)=-12$.
– Nick L
2 days ago
add a comment |
1 Answer
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oldest
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up vote
3
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accepted
In my answer below, I was implicitly assuming that the covering on the boundary was trivial. As Mike Miller points out below, this is not necessarily the case. See his comments for more details.
Denote the closed orientable surface of genus $g$ with $b$ boundary components by $Sigma_{g, b}$, and the closed non-orientable surface of genus $g$ with $b$ boundary components by $S_{g,b}$. Recall that $chi(Sigma_{g,b}) = 2 - 2g - b$ and $chi(S_{g,b}) = 2 - g - b$.
If $p : M to N$ is a covering map between manifolds with boundary, then it restricts to a covering map $p|_{partial M} : partial M to partial N$ of the same degree. So if $p : Sigma_{g,b} to Sigma_{g',b'}$ is a degree $k$ covering map, then $b = kb'$. Moreover,
begin{align*}
chi(Sigma_{g,b}) &= kchi(Sigma_{g',b'})\
2 - 2g - b &= k(2 - 2g' - b')\
2 - 2g - kb' &= k(2 - 2g' - b')\
2 - 2g &= k(2 - 2g')\
chi(Sigma_g) &= kchi(Sigma_{g'}).
end{align*}
The converse is also true. That is, if $chi(Sigma_g) = kchi(Sigma_{g'})$ and $b = kb'$, then there is a degree $k$ covering map $Sigma_{g,b} to Sigma_{g',b'}$. To see this, note that if $chi(Sigma_g) = kchi(Sigma_{g'})$, then there is a degree $k$ covering map $p : Sigma_g to Sigma_{g'}$; see this answer. If $D subset Sigma_{g'}$ is the interior of a closed disc in $Sigma_{g'}$, then $p^{-1}(D)$ is a disjoint union of the interiors of $k$ disjoint closed discs in $Sigma_g$. So if $D_1, dots, D_{b'}$ are the interiors of $b'$ disjoint closed discs in $Sigma_{g'}$, then $p^{-1}(Sigma_{g'}setminus(D_1cupdotscup D_{b'})$ is $Sigma_g$ with $kb' = b$ interiors of disjoint closed discs removed, i.e. $Sigma_{g,b}$. Therefore the restriction of $p$ to $Sigma_{g,b}$ is a degree $k$ covering $Sigma_{g,b} to Sigma_{g',b'}$.
Likewise, $Sigma_{g,b}$ is a $k$-sheeted covering of $S_{g'b'}$ if and only if $chi(Sigma_g) = kchi(S_g)$, $b = kb'$ and $k$ is even. Note that $k$ must be even as any orientable covering of a non-orientable manifold must factor through the orientation double cover.
Therefore, we have the following complete list of coverings:
$Sigma_{5,4} to Sigma_{5,4}$ of degree one,
$Sigma_{5,4} to Sigma_{3,2}$ of degree two,
$Sigma_{5,4} to S_{6,2}$ of degree two,
$Sigma_{5,4} to Sigma_{2,1}$ of degree four, and
$Sigma_{5,4} to S_{4,1}$ of degree four.
answered 2 days ago
Michael Albanese
62.5k1598299
I believe you've made an assumption here that the covering map is trivial on the boundary, which is not necessary; the condition on boundary components should be $b/k leq b' leq b$. I still believe your claim that (in addition to this constraint) the only condition is that $chi(Sigma_{g,b}) = k chi(Sigma_{g', b'})$, but I do not know a nice reference. I think there is some literature further trying to specify the data of $(g',b')$ and the $k$-partition of $b$ corresponding to the degrees of covers on the boundary, and IIRC all are realizable except possibly some with $g = 0$.
– Mike Miller
2 days ago
In particular, I think the following should also be on the list of possible covers: $Sigma_{2,4}, Sigma_{1,4}, Sigma_{1,3}, Sigma_{1,2}, Sigma_{0,4}, Sigma_{1,1}, Sigma_{0,3},$ and I think all should be realized. I think there is an evenness condition as you say on the number of boundary components and one should have similar results for covering $S_{g',b'}$. If you fill the boundary components in with discs, these covering maps correspond go branched covers with specified branching data, and yours correspond to those with trivial branch data (i.e., are actually covering maps).
– Mike Miller
2 days ago
You're right. Can't believe I missed that.
– Michael Albanese
yesterday
That subtlety has bitten me more than once.
– Mike Miller
yesterday
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
In my answer below, I was implicitly assuming that the covering on the boundary was trivial. As Mike Miller points out below, this is not necessarily the case. See his comments for more details.
Denote the closed orientable surface of genus $g$ with $b$ boundary components by $Sigma_{g, b}$, and the closed non-orientable surface of genus $g$ with $b$ boundary components by $S_{g,b}$. Recall that $chi(Sigma_{g,b}) = 2 - 2g - b$ and $chi(S_{g,b}) = 2 - g - b$.
If $p : M to N$ is a covering map between manifolds with boundary, then it restricts to a covering map $p|_{partial M} : partial M to partial N$ of the same degree. So if $p : Sigma_{g,b} to Sigma_{g',b'}$ is a degree $k$ covering map, then $b = kb'$. Moreover,
begin{align*}
chi(Sigma_{g,b}) &= kchi(Sigma_{g',b'})\
2 - 2g - b &= k(2 - 2g' - b')\
2 - 2g - kb' &= k(2 - 2g' - b')\
2 - 2g &= k(2 - 2g')\
chi(Sigma_g) &= kchi(Sigma_{g'}).
end{align*}
The converse is also true. That is, if $chi(Sigma_g) = kchi(Sigma_{g'})$ and $b = kb'$, then there is a degree $k$ covering map $Sigma_{g,b} to Sigma_{g',b'}$. To see this, note that if $chi(Sigma_g) = kchi(Sigma_{g'})$, then there is a degree $k$ covering map $p : Sigma_g to Sigma_{g'}$; see this answer. If $D subset Sigma_{g'}$ is the interior of a closed disc in $Sigma_{g'}$, then $p^{-1}(D)$ is a disjoint union of the interiors of $k$ disjoint closed discs in $Sigma_g$. So if $D_1, dots, D_{b'}$ are the interiors of $b'$ disjoint closed discs in $Sigma_{g'}$, then $p^{-1}(Sigma_{g'}setminus(D_1cupdotscup D_{b'})$ is $Sigma_g$ with $kb' = b$ interiors of disjoint closed discs removed, i.e. $Sigma_{g,b}$. Therefore the restriction of $p$ to $Sigma_{g,b}$ is a degree $k$ covering $Sigma_{g,b} to Sigma_{g',b'}$.
Likewise, $Sigma_{g,b}$ is a $k$-sheeted covering of $S_{g'b'}$ if and only if $chi(Sigma_g) = kchi(S_g)$, $b = kb'$ and $k$ is even. Note that $k$ must be even as any orientable covering of a non-orientable manifold must factor through the orientation double cover.
Therefore, we have the following complete list of coverings:
$Sigma_{5,4} to Sigma_{5,4}$ of degree one,
$Sigma_{5,4} to Sigma_{3,2}$ of degree two,
$Sigma_{5,4} to S_{6,2}$ of degree two,
$Sigma_{5,4} to Sigma_{2,1}$ of degree four, and
$Sigma_{5,4} to S_{4,1}$ of degree four.
answered 2 days ago
Michael Albanese
62.5k1598299
I believe you've made an assumption here that the covering map is trivial on the boundary, which is not necessary; the condition on boundary components should be $b/k leq b' leq b$. I still believe your claim that (in addition to this constraint) the only condition is that $chi(Sigma_{g,b}) = k chi(Sigma_{g', b'})$, but I do not know a nice reference. I think there is some literature further trying to specify the data of $(g',b')$ and the $k$-partition of $b$ corresponding to the degrees of covers on the boundary, and IIRC all are realizable except possibly some with $g = 0$.
– Mike Miller
2 days ago
In particular, I think the following should also be on the list of possible covers: $Sigma_{2,4}, Sigma_{1,4}, Sigma_{1,3}, Sigma_{1,2}, Sigma_{0,4}, Sigma_{1,1}, Sigma_{0,3},$ and I think all should be realized. I think there is an evenness condition as you say on the number of boundary components and one should have similar results for covering $S_{g',b'}$. If you fill the boundary components in with discs, these covering maps correspond go branched covers with specified branching data, and yours correspond to those with trivial branch data (i.e., are actually covering maps).
– Mike Miller
2 days ago
You're right. Can't believe I missed that.
– Michael Albanese
yesterday
That subtlety has bitten me more than once.
– Mike Miller
yesterday
add a comment |
up vote
3
down vote
accepted
In my answer below, I was implicitly assuming that the covering on the boundary was trivial. As Mike Miller points out below, this is not necessarily the case. See his comments for more details.
Denote the closed orientable surface of genus $g$ with $b$ boundary components by $Sigma_{g, b}$, and the closed non-orientable surface of genus $g$ with $b$ boundary components by $S_{g,b}$. Recall that $chi(Sigma_{g,b}) = 2 - 2g - b$ and $chi(S_{g,b}) = 2 - g - b$.
If $p : M to N$ is a covering map between manifolds with boundary, then it restricts to a covering map $p|_{partial M} : partial M to partial N$ of the same degree. So if $p : Sigma_{g,b} to Sigma_{g',b'}$ is a degree $k$ covering map, then $b = kb'$. Moreover,
begin{align*}
chi(Sigma_{g,b}) &= kchi(Sigma_{g',b'})\
2 - 2g - b &= k(2 - 2g' - b')\
2 - 2g - kb' &= k(2 - 2g' - b')\
2 - 2g &= k(2 - 2g')\
chi(Sigma_g) &= kchi(Sigma_{g'}).
end{align*}
The converse is also true. That is, if $chi(Sigma_g) = kchi(Sigma_{g'})$ and $b = kb'$, then there is a degree $k$ covering map $Sigma_{g,b} to Sigma_{g',b'}$. To see this, note that if $chi(Sigma_g) = kchi(Sigma_{g'})$, then there is a degree $k$ covering map $p : Sigma_g to Sigma_{g'}$; see this answer. If $D subset Sigma_{g'}$ is the interior of a closed disc in $Sigma_{g'}$, then $p^{-1}(D)$ is a disjoint union of the interiors of $k$ disjoint closed discs in $Sigma_g$. So if $D_1, dots, D_{b'}$ are the interiors of $b'$ disjoint closed discs in $Sigma_{g'}$, then $p^{-1}(Sigma_{g'}setminus(D_1cupdotscup D_{b'})$ is $Sigma_g$ with $kb' = b$ interiors of disjoint closed discs removed, i.e. $Sigma_{g,b}$. Therefore the restriction of $p$ to $Sigma_{g,b}$ is a degree $k$ covering $Sigma_{g,b} to Sigma_{g',b'}$.
Likewise, $Sigma_{g,b}$ is a $k$-sheeted covering of $S_{g'b'}$ if and only if $chi(Sigma_g) = kchi(S_g)$, $b = kb'$ and $k$ is even. Note that $k$ must be even as any orientable covering of a non-orientable manifold must factor through the orientation double cover.
Therefore, we have the following complete list of coverings:
$Sigma_{5,4} to Sigma_{5,4}$ of degree one,
$Sigma_{5,4} to Sigma_{3,2}$ of degree two,
$Sigma_{5,4} to S_{6,2}$ of degree two,
$Sigma_{5,4} to Sigma_{2,1}$ of degree four, and
$Sigma_{5,4} to S_{4,1}$ of degree four.
answered 2 days ago
Michael Albanese
62.5k1598299
I believe you've made an assumption here that the covering map is trivial on the boundary, which is not necessary; the condition on boundary components should be $b/k leq b' leq b$. I still believe your claim that (in addition to this constraint) the only condition is that $chi(Sigma_{g,b}) = k chi(Sigma_{g', b'})$, but I do not know a nice reference. I think there is some literature further trying to specify the data of $(g',b')$ and the $k$-partition of $b$ corresponding to the degrees of covers on the boundary, and IIRC all are realizable except possibly some with $g = 0$.
– Mike Miller
2 days ago
In particular, I think the following should also be on the list of possible covers: $Sigma_{2,4}, Sigma_{1,4}, Sigma_{1,3}, Sigma_{1,2}, Sigma_{0,4}, Sigma_{1,1}, Sigma_{0,3},$ and I think all should be realized. I think there is an evenness condition as you say on the number of boundary components and one should have similar results for covering $S_{g',b'}$. If you fill the boundary components in with discs, these covering maps correspond go branched covers with specified branching data, and yours correspond to those with trivial branch data (i.e., are actually covering maps).
– Mike Miller
2 days ago
You're right. Can't believe I missed that.
– Michael Albanese
yesterday
That subtlety has bitten me more than once.
– Mike Miller
yesterday
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
In my answer below, I was implicitly assuming that the covering on the boundary was trivial. As Mike Miller points out below, this is not necessarily the case. See his comments for more details.
Denote the closed orientable surface of genus $g$ with $b$ boundary components by $Sigma_{g, b}$, and the closed non-orientable surface of genus $g$ with $b$ boundary components by $S_{g,b}$. Recall that $chi(Sigma_{g,b}) = 2 - 2g - b$ and $chi(S_{g,b}) = 2 - g - b$.
If $p : M to N$ is a covering map between manifolds with boundary, then it restricts to a covering map $p|_{partial M} : partial M to partial N$ of the same degree. So if $p : Sigma_{g,b} to Sigma_{g',b'}$ is a degree $k$ covering map, then $b = kb'$. Moreover,
begin{align*}
chi(Sigma_{g,b}) &= kchi(Sigma_{g',b'})\
2 - 2g - b &= k(2 - 2g' - b')\
2 - 2g - kb' &= k(2 - 2g' - b')\
2 - 2g &= k(2 - 2g')\
chi(Sigma_g) &= kchi(Sigma_{g'}).
end{align*}
The converse is also true. That is, if $chi(Sigma_g) = kchi(Sigma_{g'})$ and $b = kb'$, then there is a degree $k$ covering map $Sigma_{g,b} to Sigma_{g',b'}$. To see this, note that if $chi(Sigma_g) = kchi(Sigma_{g'})$, then there is a degree $k$ covering map $p : Sigma_g to Sigma_{g'}$; see this answer. If $D subset Sigma_{g'}$ is the interior of a closed disc in $Sigma_{g'}$, then $p^{-1}(D)$ is a disjoint union of the interiors of $k$ disjoint closed discs in $Sigma_g$. So if $D_1, dots, D_{b'}$ are the interiors of $b'$ disjoint closed discs in $Sigma_{g'}$, then $p^{-1}(Sigma_{g'}setminus(D_1cupdotscup D_{b'})$ is $Sigma_g$ with $kb' = b$ interiors of disjoint closed discs removed, i.e. $Sigma_{g,b}$. Therefore the restriction of $p$ to $Sigma_{g,b}$ is a degree $k$ covering $Sigma_{g,b} to Sigma_{g',b'}$.
Likewise, $Sigma_{g,b}$ is a $k$-sheeted covering of $S_{g'b'}$ if and only if $chi(Sigma_g) = kchi(S_g)$, $b = kb'$ and $k$ is even. Note that $k$ must be even as any orientable covering of a non-orientable manifold must factor through the orientation double cover.
Therefore, we have the following complete list of coverings:
$Sigma_{5,4} to Sigma_{5,4}$ of degree one,
$Sigma_{5,4} to Sigma_{3,2}$ of degree two,
$Sigma_{5,4} to S_{6,2}$ of degree two,
$Sigma_{5,4} to Sigma_{2,1}$ of degree four, and
$Sigma_{5,4} to S_{4,1}$ of degree four.
answered 2 days ago
Michael Albanese
62.5k1598299
In my answer below, I was implicitly assuming that the covering on the boundary was trivial. As Mike Miller points out below, this is not necessarily the case. See his comments for more details.
Denote the closed orientable surface of genus $g$ with $b$ boundary components by $Sigma_{g, b}$, and the closed non-orientable surface of genus $g$ with $b$ boundary components by $S_{g,b}$. Recall that $chi(Sigma_{g,b}) = 2 - 2g - b$ and $chi(S_{g,b}) = 2 - g - b$.
If $p : M to N$ is a covering map between manifolds with boundary, then it restricts to a covering map $p|_{partial M} : partial M to partial N$ of the same degree. So if $p : Sigma_{g,b} to Sigma_{g',b'}$ is a degree $k$ covering map, then $b = kb'$. Moreover,
begin{align*}
chi(Sigma_{g,b}) &= kchi(Sigma_{g',b'})\
2 - 2g - b &= k(2 - 2g' - b')\
2 - 2g - kb' &= k(2 - 2g' - b')\
2 - 2g &= k(2 - 2g')\
chi(Sigma_g) &= kchi(Sigma_{g'}).
end{align*}
The converse is also true. That is, if $chi(Sigma_g) = kchi(Sigma_{g'})$ and $b = kb'$, then there is a degree $k$ covering map $Sigma_{g,b} to Sigma_{g',b'}$. To see this, note that if $chi(Sigma_g) = kchi(Sigma_{g'})$, then there is a degree $k$ covering map $p : Sigma_g to Sigma_{g'}$; see this answer. If $D subset Sigma_{g'}$ is the interior of a closed disc in $Sigma_{g'}$, then $p^{-1}(D)$ is a disjoint union of the interiors of $k$ disjoint closed discs in $Sigma_g$. So if $D_1, dots, D_{b'}$ are the interiors of $b'$ disjoint closed discs in $Sigma_{g'}$, then $p^{-1}(Sigma_{g'}setminus(D_1cupdotscup D_{b'})$ is $Sigma_g$ with $kb' = b$ interiors of disjoint closed discs removed, i.e. $Sigma_{g,b}$. Therefore the restriction of $p$ to $Sigma_{g,b}$ is a degree $k$ covering $Sigma_{g,b} to Sigma_{g',b'}$.
Likewise, $Sigma_{g,b}$ is a $k$-sheeted covering of $S_{g'b'}$ if and only if $chi(Sigma_g) = kchi(S_g)$, $b = kb'$ and $k$ is even. Note that $k$ must be even as any orientable covering of a non-orientable manifold must factor through the orientation double cover.
Therefore, we have the following complete list of coverings:
$Sigma_{5,4} to Sigma_{5,4}$ of degree one,
$Sigma_{5,4} to Sigma_{3,2}$ of degree two,
$Sigma_{5,4} to S_{6,2}$ of degree two,
$Sigma_{5,4} to Sigma_{2,1}$ of degree four, and
$Sigma_{5,4} to S_{4,1}$ of degree four.
answered 2 days ago
Michael Albanese
62.5k1598299
edited yesterday
answered 2 days ago
Michael Albanese
62.5k1598299
answered 2 days ago
Michael Albanese
62.5k1598299
answered 2 days ago
Michael Albanese
62.5k1598299
62.5k1598299
I believe you've made an assumption here that the covering map is trivial on the boundary, which is not necessary; the condition on boundary components should be $b/k leq b' leq b$. I still believe your claim that (in addition to this constraint) the only condition is that $chi(Sigma_{g,b}) = k chi(Sigma_{g', b'})$, but I do not know a nice reference. I think there is some literature further trying to specify the data of $(g',b')$ and the $k$-partition of $b$ corresponding to the degrees of covers on the boundary, and IIRC all are realizable except possibly some with $g = 0$.
– Mike Miller
2 days ago
In particular, I think the following should also be on the list of possible covers: $Sigma_{2,4}, Sigma_{1,4}, Sigma_{1,3}, Sigma_{1,2}, Sigma_{0,4}, Sigma_{1,1}, Sigma_{0,3},$ and I think all should be realized. I think there is an evenness condition as you say on the number of boundary components and one should have similar results for covering $S_{g',b'}$. If you fill the boundary components in with discs, these covering maps correspond go branched covers with specified branching data, and yours correspond to those with trivial branch data (i.e., are actually covering maps).
– Mike Miller
2 days ago
You're right. Can't believe I missed that.
– Michael Albanese
yesterday
That subtlety has bitten me more than once.
– Mike Miller
yesterday
add a comment |
I believe you've made an assumption here that the covering map is trivial on the boundary, which is not necessary; the condition on boundary components should be $b/k leq b' leq b$. I still believe your claim that (in addition to this constraint) the only condition is that $chi(Sigma_{g,b}) = k chi(Sigma_{g', b'})$, but I do not know a nice reference. I think there is some literature further trying to specify the data of $(g',b')$ and the $k$-partition of $b$ corresponding to the degrees of covers on the boundary, and IIRC all are realizable except possibly some with $g = 0$.
– Mike Miller
2 days ago
In particular, I think the following should also be on the list of possible covers: $Sigma_{2,4}, Sigma_{1,4}, Sigma_{1,3}, Sigma_{1,2}, Sigma_{0,4}, Sigma_{1,1}, Sigma_{0,3},$ and I think all should be realized. I think there is an evenness condition as you say on the number of boundary components and one should have similar results for covering $S_{g',b'}$. If you fill the boundary components in with discs, these covering maps correspond go branched covers with specified branching data, and yours correspond to those with trivial branch data (i.e., are actually covering maps).
– Mike Miller
2 days ago
You're right. Can't believe I missed that.
– Michael Albanese
yesterday
That subtlety has bitten me more than once.
– Mike Miller
yesterday
I believe you've made an assumption here that the covering map is trivial on the boundary, which is not necessary; the condition on boundary components should be $b/k leq b' leq b$. I still believe your claim that (in addition to this constraint) the only condition is that $chi(Sigma_{g,b}) = k chi(Sigma_{g', b'})$, but I do not know a nice reference. I think there is some literature further trying to specify the data of $(g',b')$ and the $k$-partition of $b$ corresponding to the degrees of covers on the boundary, and IIRC all are realizable except possibly some with $g = 0$.
– Mike Miller
2 days ago
I believe you've made an assumption here that the covering map is trivial on the boundary, which is not necessary; the condition on boundary components should be $b/k leq b' leq b$. I still believe your claim that (in addition to this constraint) the only condition is that $chi(Sigma_{g,b}) = k chi(Sigma_{g', b'})$, but I do not know a nice reference. I think there is some literature further trying to specify the data of $(g',b')$ and the $k$-partition of $b$ corresponding to the degrees of covers on the boundary, and IIRC all are realizable except possibly some with $g = 0$.
– Mike Miller
2 days ago
In particular, I think the following should also be on the list of possible covers: $Sigma_{2,4}, Sigma_{1,4}, Sigma_{1,3}, Sigma_{1,2}, Sigma_{0,4}, Sigma_{1,1}, Sigma_{0,3},$ and I think all should be realized. I think there is an evenness condition as you say on the number of boundary components and one should have similar results for covering $S_{g',b'}$. If you fill the boundary components in with discs, these covering maps correspond go branched covers with specified branching data, and yours correspond to those with trivial branch data (i.e., are actually covering maps).
– Mike Miller
2 days ago
In particular, I think the following should also be on the list of possible covers: $Sigma_{2,4}, Sigma_{1,4}, Sigma_{1,3}, Sigma_{1,2}, Sigma_{0,4}, Sigma_{1,1}, Sigma_{0,3},$ and I think all should be realized. I think there is an evenness condition as you say on the number of boundary components and one should have similar results for covering $S_{g',b'}$. If you fill the boundary components in with discs, these covering maps correspond go branched covers with specified branching data, and yours correspond to those with trivial branch data (i.e., are actually covering maps).
– Mike Miller
2 days ago
You're right. Can't believe I missed that.
– Michael Albanese
yesterday
You're right. Can't believe I missed that.
– Michael Albanese
yesterday
That subtlety has bitten me more than once.
– Mike Miller
yesterday
That subtlety has bitten me more than once.
– Mike Miller
yesterday
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1
One obvious comment: if $S$ covers $S'$ Then $chi(S) = dchi(S')$ where $d>0$ is the degree of the cover. Also if $S$ has non- empty boundary $S'$ must have non-empty boundary. This gives a small number of possibilities to chrck, since for us $chi(S)=-12$.
– Nick L
2 days ago