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Ramified covering of a torus [closed]
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Let $Mto T$ be a double covering of a torus $T=S^1times S^1$ ramified over $n$ disjoint circles. Then what is the Euler characteristics of $M$?
algebraic-topology differential-topology geometric-topology
asked Jul 22 at 18:55
Alfred
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closed as off-topic by Lee Mosher, Isaac Browne, Gibbs, Claude Leibovici, Tyrone Jul 24 at 10:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Isaac Browne, Gibbs, Claude Leibovici, Tyrone
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Let $Mto T$ be a double covering of a torus $T=S^1times S^1$ ramified over $n$ disjoint circles. Then what is the Euler characteristics of $M$?
algebraic-topology differential-topology geometric-topology
asked Jul 22 at 18:55
Alfred
1725
closed as off-topic by Lee Mosher, Isaac Browne, Gibbs, Claude Leibovici, Tyrone Jul 24 at 10:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Isaac Browne, Gibbs, Claude Leibovici, Tyrone
I know what it means for a map between 2-dimensional manifolds to be a ramified covering over a collection of points. But what does it mean to be a ramified covering over a collection of circles?
– Lee Mosher
Jul 23 at 14:15
@LeeMosher : probably that the locus $ t in T : #f^-1(t) = 1$ is a circle. For example, the projection of the 2-sphere on the disk is a ramified cover branched over a circle.
– Nicolas Hemelsoet
Jul 23 at 22:21
That would make sense for a circle which is a boundary component of the image surface. However, in this question the image is a surface with empty boundary. So the question is quite unclear.
– Lee Mosher
Jul 24 at 1:29
@LeeMosher ah of course you are right. I feel silly I answered without having noticed this, I'll modify my answer. Sorry for the silly comment.
– Nicolas Hemelsoet
Jul 24 at 9:38
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $Mto T$ be a double covering of a torus $T=S^1times S^1$ ramified over $n$ disjoint circles. Then what is the Euler characteristics of $M$?
algebraic-topology differential-topology geometric-topology
asked Jul 22 at 18:55
Alfred
1725
Let $Mto T$ be a double covering of a torus $T=S^1times S^1$ ramified over $n$ disjoint circles. Then what is the Euler characteristics of $M$?
algebraic-topology differential-topology geometric-topology
asked Jul 22 at 18:55
Alfred
1725
asked Jul 22 at 18:55
Alfred
1725
asked Jul 22 at 18:55
Alfred
1725
asked Jul 22 at 18:55
Alfred
1725
1725
closed as off-topic by Lee Mosher, Isaac Browne, Gibbs, Claude Leibovici, Tyrone Jul 24 at 10:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Isaac Browne, Gibbs, Claude Leibovici, Tyrone
closed as off-topic by Lee Mosher, Isaac Browne, Gibbs, Claude Leibovici, Tyrone Jul 24 at 10:56
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Isaac Browne, Gibbs, Claude Leibovici, Tyrone
I know what it means for a map between 2-dimensional manifolds to be a ramified covering over a collection of points. But what does it mean to be a ramified covering over a collection of circles?
– Lee Mosher
Jul 23 at 14:15
@LeeMosher : probably that the locus $ t in T : #f^-1(t) = 1$ is a circle. For example, the projection of the 2-sphere on the disk is a ramified cover branched over a circle.
– Nicolas Hemelsoet
Jul 23 at 22:21
That would make sense for a circle which is a boundary component of the image surface. However, in this question the image is a surface with empty boundary. So the question is quite unclear.
– Lee Mosher
Jul 24 at 1:29
@LeeMosher ah of course you are right. I feel silly I answered without having noticed this, I'll modify my answer. Sorry for the silly comment.
– Nicolas Hemelsoet
Jul 24 at 9:38
add a comment |Â
I know what it means for a map between 2-dimensional manifolds to be a ramified covering over a collection of points. But what does it mean to be a ramified covering over a collection of circles?
– Lee Mosher
Jul 23 at 14:15
@LeeMosher : probably that the locus $ t in T : #f^-1(t) = 1$ is a circle. For example, the projection of the 2-sphere on the disk is a ramified cover branched over a circle.
– Nicolas Hemelsoet
Jul 23 at 22:21
That would make sense for a circle which is a boundary component of the image surface. However, in this question the image is a surface with empty boundary. So the question is quite unclear.
– Lee Mosher
Jul 24 at 1:29
@LeeMosher ah of course you are right. I feel silly I answered without having noticed this, I'll modify my answer. Sorry for the silly comment.
– Nicolas Hemelsoet
Jul 24 at 9:38
I know what it means for a map between 2-dimensional manifolds to be a ramified covering over a collection of points. But what does it mean to be a ramified covering over a collection of circles?
– Lee Mosher
Jul 23 at 14:15
I know what it means for a map between 2-dimensional manifolds to be a ramified covering over a collection of points. But what does it mean to be a ramified covering over a collection of circles?
– Lee Mosher
Jul 23 at 14:15
@LeeMosher : probably that the locus $ t in T : #f^-1(t) = 1$ is a circle. For example, the projection of the 2-sphere on the disk is a ramified cover branched over a circle.
– Nicolas Hemelsoet
Jul 23 at 22:21
@LeeMosher : probably that the locus $ t in T : #f^-1(t) = 1$ is a circle. For example, the projection of the 2-sphere on the disk is a ramified cover branched over a circle.
– Nicolas Hemelsoet
Jul 23 at 22:21
That would make sense for a circle which is a boundary component of the image surface. However, in this question the image is a surface with empty boundary. So the question is quite unclear.
– Lee Mosher
Jul 24 at 1:29
That would make sense for a circle which is a boundary component of the image surface. However, in this question the image is a surface with empty boundary. So the question is quite unclear.
– Lee Mosher
Jul 24 at 1:29
@LeeMosher ah of course you are right. I feel silly I answered without having noticed this, I'll modify my answer. Sorry for the silly comment.
– Nicolas Hemelsoet
Jul 24 at 9:38
@LeeMosher ah of course you are right. I feel silly I answered without having noticed this, I'll modify my answer. Sorry for the silly comment.
– Nicolas Hemelsoet
Jul 24 at 9:38
add a comment |Â
1 Answer
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Hint : if $chi(A) = 0$ and $p :X to Y$ is a $2$-fold covering with branch locus $A$ then $chi(X) = 2 chi(Y)$. (To prove it, just triangulate appropriately everything. )
Remark : As mentioned by Lee Mosher in the comments, this can only happens if $n=0$. Indeed, if $f : M to N$ is a covering between surfaces ramified say along a circle $C subset N$, and $U$ is a little neighborhood of some $c in C$, $f^-1(U) cong (x,y,z) : xy = 0$ in $Bbb R^3$ and in particular $M$ is not a manifold. The only possibility for $n neq 0$ is when $N$ has boundary and the ramification locus is contained in the boundary $partial N$.
answered Jul 22 at 19:12
Nicolas Hemelsoet
4,855316
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Hint : if $chi(A) = 0$ and $p :X to Y$ is a $2$-fold covering with branch locus $A$ then $chi(X) = 2 chi(Y)$. (To prove it, just triangulate appropriately everything. )
Remark : As mentioned by Lee Mosher in the comments, this can only happens if $n=0$. Indeed, if $f : M to N$ is a covering between surfaces ramified say along a circle $C subset N$, and $U$ is a little neighborhood of some $c in C$, $f^-1(U) cong (x,y,z) : xy = 0$ in $Bbb R^3$ and in particular $M$ is not a manifold. The only possibility for $n neq 0$ is when $N$ has boundary and the ramification locus is contained in the boundary $partial N$.
answered Jul 22 at 19:12
Nicolas Hemelsoet
4,855316
add a comment |Â
up vote
2
down vote
Hint : if $chi(A) = 0$ and $p :X to Y$ is a $2$-fold covering with branch locus $A$ then $chi(X) = 2 chi(Y)$. (To prove it, just triangulate appropriately everything. )
Remark : As mentioned by Lee Mosher in the comments, this can only happens if $n=0$. Indeed, if $f : M to N$ is a covering between surfaces ramified say along a circle $C subset N$, and $U$ is a little neighborhood of some $c in C$, $f^-1(U) cong (x,y,z) : xy = 0$ in $Bbb R^3$ and in particular $M$ is not a manifold. The only possibility for $n neq 0$ is when $N$ has boundary and the ramification locus is contained in the boundary $partial N$.
answered Jul 22 at 19:12
Nicolas Hemelsoet
4,855316
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Hint : if $chi(A) = 0$ and $p :X to Y$ is a $2$-fold covering with branch locus $A$ then $chi(X) = 2 chi(Y)$. (To prove it, just triangulate appropriately everything. )
Remark : As mentioned by Lee Mosher in the comments, this can only happens if $n=0$. Indeed, if $f : M to N$ is a covering between surfaces ramified say along a circle $C subset N$, and $U$ is a little neighborhood of some $c in C$, $f^-1(U) cong (x,y,z) : xy = 0$ in $Bbb R^3$ and in particular $M$ is not a manifold. The only possibility for $n neq 0$ is when $N$ has boundary and the ramification locus is contained in the boundary $partial N$.
answered Jul 22 at 19:12
Nicolas Hemelsoet
4,855316
Hint : if $chi(A) = 0$ and $p :X to Y$ is a $2$-fold covering with branch locus $A$ then $chi(X) = 2 chi(Y)$. (To prove it, just triangulate appropriately everything. )
Remark : As mentioned by Lee Mosher in the comments, this can only happens if $n=0$. Indeed, if $f : M to N$ is a covering between surfaces ramified say along a circle $C subset N$, and $U$ is a little neighborhood of some $c in C$, $f^-1(U) cong (x,y,z) : xy = 0$ in $Bbb R^3$ and in particular $M$ is not a manifold. The only possibility for $n neq 0$ is when $N$ has boundary and the ramification locus is contained in the boundary $partial N$.
answered Jul 22 at 19:12
Nicolas Hemelsoet
4,855316
edited Jul 24 at 9:42
answered Jul 22 at 19:12
Nicolas Hemelsoet
4,855316
answered Jul 22 at 19:12
Nicolas Hemelsoet
4,855316
answered Jul 22 at 19:12
Nicolas Hemelsoet
4,855316
4,855316
add a comment |Â
add a comment |Â
I know what it means for a map between 2-dimensional manifolds to be a ramified covering over a collection of points. But what does it mean to be a ramified covering over a collection of circles?
– Lee Mosher
Jul 23 at 14:15
@LeeMosher : probably that the locus $ t in T : #f^-1(t) = 1$ is a circle. For example, the projection of the 2-sphere on the disk is a ramified cover branched over a circle.
– Nicolas Hemelsoet
Jul 23 at 22:21
That would make sense for a circle which is a boundary component of the image surface. However, in this question the image is a surface with empty boundary. So the question is quite unclear.
– Lee Mosher
Jul 24 at 1:29
@LeeMosher ah of course you are right. I feel silly I answered without having noticed this, I'll modify my answer. Sorry for the silly comment.
– Nicolas Hemelsoet
Jul 24 at 9:38