Euler characteristic of a Y-shaped pipe?
$begingroup$
I'm familiar with the idea in topology that shapes that can be continuously deformed into one another are considered "equivalent". I read about the Euler Characteristic as being Vertices-Edges+Faces. Thinking of this number as related to the genus of an object, and the genus as the number of holes it has (which is probably where I'm going wrong), I began wondering what the Euler number or genus of a Y-shaped pipe (see below) would be, having three openings that all converge together.
Y-shaped pipe:
Thanks in advance for any help!
general-topology algebraic-topology
edited Dec 24 '18 at 7:26
Glorfindel
3,42981830
asked Feb 13 '14 at 18:41
aleph_aleph_nullaleph_aleph_null
19028
$endgroup$
add a comment |
$begingroup$
I'm familiar with the idea in topology that shapes that can be continuously deformed into one another are considered "equivalent". I read about the Euler Characteristic as being Vertices-Edges+Faces. Thinking of this number as related to the genus of an object, and the genus as the number of holes it has (which is probably where I'm going wrong), I began wondering what the Euler number or genus of a Y-shaped pipe (see below) would be, having three openings that all converge together.
Y-shaped pipe:
Thanks in advance for any help!
general-topology algebraic-topology
edited Dec 24 '18 at 7:26
Glorfindel
3,42981830
asked Feb 13 '14 at 18:41
aleph_aleph_nullaleph_aleph_null
19028
$endgroup$
1
$begingroup$
should be $-1$. Note that the pipe is homeomorphic to a sphere with three holes, which is homeomorphic to a 2-disk with two holes, and this deformation retracts to a wedge of two circles. And this wedge has one vertex and two edges.
$endgroup$
– Stefan Hamcke
Feb 13 '14 at 18:57
$begingroup$
To see a triangulation for your space, take 3 cubes with 2 opposite faces removed from each, and 2 equilateral triangles (with side length the same as your cube). It should be easy to figure out how to glue those together to get your space.
$endgroup$
– PVAL-inactive
Feb 13 '14 at 19:17
add a comment |
$begingroup$
I'm familiar with the idea in topology that shapes that can be continuously deformed into one another are considered "equivalent". I read about the Euler Characteristic as being Vertices-Edges+Faces. Thinking of this number as related to the genus of an object, and the genus as the number of holes it has (which is probably where I'm going wrong), I began wondering what the Euler number or genus of a Y-shaped pipe (see below) would be, having three openings that all converge together.
Y-shaped pipe:
Thanks in advance for any help!
general-topology algebraic-topology
edited Dec 24 '18 at 7:26
Glorfindel
3,42981830
asked Feb 13 '14 at 18:41
aleph_aleph_nullaleph_aleph_null
19028
$endgroup$
I'm familiar with the idea in topology that shapes that can be continuously deformed into one another are considered "equivalent". I read about the Euler Characteristic as being Vertices-Edges+Faces. Thinking of this number as related to the genus of an object, and the genus as the number of holes it has (which is probably where I'm going wrong), I began wondering what the Euler number or genus of a Y-shaped pipe (see below) would be, having three openings that all converge together.
Y-shaped pipe:
Thanks in advance for any help!
general-topology algebraic-topology
general-topology algebraic-topology
edited Dec 24 '18 at 7:26
Glorfindel
3,42981830
asked Feb 13 '14 at 18:41
aleph_aleph_nullaleph_aleph_null
19028
edited Dec 24 '18 at 7:26
Glorfindel
3,42981830
asked Feb 13 '14 at 18:41
aleph_aleph_nullaleph_aleph_null
19028
edited Dec 24 '18 at 7:26
Glorfindel
3,42981830
edited Dec 24 '18 at 7:26
Glorfindel
3,42981830
edited Dec 24 '18 at 7:26
Glorfindel
3,42981830
3,42981830
asked Feb 13 '14 at 18:41
aleph_aleph_nullaleph_aleph_null
19028
asked Feb 13 '14 at 18:41
aleph_aleph_nullaleph_aleph_null
19028
asked Feb 13 '14 at 18:41
aleph_aleph_nullaleph_aleph_null
19028
19028
1
$begingroup$
should be $-1$. Note that the pipe is homeomorphic to a sphere with three holes, which is homeomorphic to a 2-disk with two holes, and this deformation retracts to a wedge of two circles. And this wedge has one vertex and two edges.
$endgroup$
– Stefan Hamcke
Feb 13 '14 at 18:57
$begingroup$
To see a triangulation for your space, take 3 cubes with 2 opposite faces removed from each, and 2 equilateral triangles (with side length the same as your cube). It should be easy to figure out how to glue those together to get your space.
$endgroup$
– PVAL-inactive
Feb 13 '14 at 19:17
add a comment |
1
$begingroup$
should be $-1$. Note that the pipe is homeomorphic to a sphere with three holes, which is homeomorphic to a 2-disk with two holes, and this deformation retracts to a wedge of two circles. And this wedge has one vertex and two edges.
$endgroup$
– Stefan Hamcke
Feb 13 '14 at 18:57
$begingroup$
To see a triangulation for your space, take 3 cubes with 2 opposite faces removed from each, and 2 equilateral triangles (with side length the same as your cube). It should be easy to figure out how to glue those together to get your space.
$endgroup$
– PVAL-inactive
Feb 13 '14 at 19:17
1
1
$begingroup$
should be $-1$. Note that the pipe is homeomorphic to a sphere with three holes, which is homeomorphic to a 2-disk with two holes, and this deformation retracts to a wedge of two circles. And this wedge has one vertex and two edges.
$endgroup$
– Stefan Hamcke
Feb 13 '14 at 18:57
$begingroup$
should be $-1$. Note that the pipe is homeomorphic to a sphere with three holes, which is homeomorphic to a 2-disk with two holes, and this deformation retracts to a wedge of two circles. And this wedge has one vertex and two edges.
$endgroup$
– Stefan Hamcke
Feb 13 '14 at 18:57
$begingroup$
To see a triangulation for your space, take 3 cubes with 2 opposite faces removed from each, and 2 equilateral triangles (with side length the same as your cube). It should be easy to figure out how to glue those together to get your space.
$endgroup$
– PVAL-inactive
Feb 13 '14 at 19:17
$begingroup$
To see a triangulation for your space, take 3 cubes with 2 opposite faces removed from each, and 2 equilateral triangles (with side length the same as your cube). It should be easy to figure out how to glue those together to get your space.
$endgroup$
– PVAL-inactive
Feb 13 '14 at 19:17
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
This can be answered with some knot theory. The openings can be thought of as an unlink on three components. This just means three circles that are not "linked" together. This Y shaped pipe can be deformed to a sphere with three holes missing from it. Just think about the pipe as if it were a weird ballon. If we would fill these holes, we get a sphere, which means that it has genus zero. Then, we know that the Euler characteristic of sphere is two. (For a genus $g$ "shape," the Euler characteristic is $2-2g$. )
But we added those disks, which each count as a face. So, to get back to our Y shaped pipe, we need to subtract those disks (faces), which means we have $2-3=-1$.
Or, we could have used this: From Rolfsen's Knots and Links, we have a relation which says $$ g(M)=1-frac{chi(M)+b}{2} $$ where $g(M)$ is the genus of a manifold $M$ (our Y shaped pipe), $chi(M)$ is the Euler characteristic, and $b$ is the number of boundary components, which means the holes. We know the genus, and the number of holes, all you have to do is solve for $chi(M)$.
answered Feb 13 '14 at 19:04
N. OwadN. Owad
4,66521233
$endgroup$
$begingroup$
How have you used any knot theory here?
$endgroup$
– Dan Rust
Feb 13 '14 at 22:48
$begingroup$
@DanielRust It is more the way you view the problem. If you think of the boundary as the unlink, then there is a unique minimal genus orientable surface. This is at least how I approached the problem.
$endgroup$
– N. Owad
Feb 14 '14 at 16:16
add a comment |
$begingroup$
The easy way to find the Euler characteristic is to note that we can deformation retract the pair of pants (that's the name topologists normally give this space) onto a subset which is homeomorphic to a circle with an interval glued at one end to the north pole or the circle and the other end glued to the south pole of the circle. This in turn is homotopy equivalent to a wedge of two circles by shrinking the extra interval to a point. This space has one vertex and two edges so we get $chi=1-2=-1$. As Euler characteristic is a homotopy invariant, it follows that this is also the Euler characteristic of the pair of pants.
It's actually pretty easy to put a cellular decomposition straight onto the pair of pants without worrying about homotopy equivalences. Note that the pair of pants is homeomorphic to a closed disk with two smaller open disks removed from its interior. We can decompose this into cells by placing an edge between the boundaries of the removed disks and also an edge from each boundary of the removed disks to the boundary of the large disk. This gives us a cellular decomposition consisting of $6$ vertices on the boundary, $3$ edges in the interior, $6$ edges on the boundary, and $2$ faces in the interior. This gives $chi=6-(3+6)+2=-1$ as expected.
answered Feb 13 '14 at 22:47
Dan RustDan Rust
22.9k114984
$endgroup$
add a comment |
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$begingroup$
This can be answered with some knot theory. The openings can be thought of as an unlink on three components. This just means three circles that are not "linked" together. This Y shaped pipe can be deformed to a sphere with three holes missing from it. Just think about the pipe as if it were a weird ballon. If we would fill these holes, we get a sphere, which means that it has genus zero. Then, we know that the Euler characteristic of sphere is two. (For a genus $g$ "shape," the Euler characteristic is $2-2g$. )
But we added those disks, which each count as a face. So, to get back to our Y shaped pipe, we need to subtract those disks (faces), which means we have $2-3=-1$.
Or, we could have used this: From Rolfsen's Knots and Links, we have a relation which says $$ g(M)=1-frac{chi(M)+b}{2} $$ where $g(M)$ is the genus of a manifold $M$ (our Y shaped pipe), $chi(M)$ is the Euler characteristic, and $b$ is the number of boundary components, which means the holes. We know the genus, and the number of holes, all you have to do is solve for $chi(M)$.
answered Feb 13 '14 at 19:04
N. OwadN. Owad
4,66521233
$endgroup$
$begingroup$
How have you used any knot theory here?
$endgroup$
– Dan Rust
Feb 13 '14 at 22:48
$begingroup$
@DanielRust It is more the way you view the problem. If you think of the boundary as the unlink, then there is a unique minimal genus orientable surface. This is at least how I approached the problem.
$endgroup$
– N. Owad
Feb 14 '14 at 16:16
add a comment |
$begingroup$
This can be answered with some knot theory. The openings can be thought of as an unlink on three components. This just means three circles that are not "linked" together. This Y shaped pipe can be deformed to a sphere with three holes missing from it. Just think about the pipe as if it were a weird ballon. If we would fill these holes, we get a sphere, which means that it has genus zero. Then, we know that the Euler characteristic of sphere is two. (For a genus $g$ "shape," the Euler characteristic is $2-2g$. )
But we added those disks, which each count as a face. So, to get back to our Y shaped pipe, we need to subtract those disks (faces), which means we have $2-3=-1$.
Or, we could have used this: From Rolfsen's Knots and Links, we have a relation which says $$ g(M)=1-frac{chi(M)+b}{2} $$ where $g(M)$ is the genus of a manifold $M$ (our Y shaped pipe), $chi(M)$ is the Euler characteristic, and $b$ is the number of boundary components, which means the holes. We know the genus, and the number of holes, all you have to do is solve for $chi(M)$.
answered Feb 13 '14 at 19:04
N. OwadN. Owad
4,66521233
$endgroup$
$begingroup$
How have you used any knot theory here?
$endgroup$
– Dan Rust
Feb 13 '14 at 22:48
$begingroup$
@DanielRust It is more the way you view the problem. If you think of the boundary as the unlink, then there is a unique minimal genus orientable surface. This is at least how I approached the problem.
$endgroup$
– N. Owad
Feb 14 '14 at 16:16
add a comment |
$begingroup$
This can be answered with some knot theory. The openings can be thought of as an unlink on three components. This just means three circles that are not "linked" together. This Y shaped pipe can be deformed to a sphere with three holes missing from it. Just think about the pipe as if it were a weird ballon. If we would fill these holes, we get a sphere, which means that it has genus zero. Then, we know that the Euler characteristic of sphere is two. (For a genus $g$ "shape," the Euler characteristic is $2-2g$. )
But we added those disks, which each count as a face. So, to get back to our Y shaped pipe, we need to subtract those disks (faces), which means we have $2-3=-1$.
Or, we could have used this: From Rolfsen's Knots and Links, we have a relation which says $$ g(M)=1-frac{chi(M)+b}{2} $$ where $g(M)$ is the genus of a manifold $M$ (our Y shaped pipe), $chi(M)$ is the Euler characteristic, and $b$ is the number of boundary components, which means the holes. We know the genus, and the number of holes, all you have to do is solve for $chi(M)$.
answered Feb 13 '14 at 19:04
N. OwadN. Owad
4,66521233
$endgroup$
This can be answered with some knot theory. The openings can be thought of as an unlink on three components. This just means three circles that are not "linked" together. This Y shaped pipe can be deformed to a sphere with three holes missing from it. Just think about the pipe as if it were a weird ballon. If we would fill these holes, we get a sphere, which means that it has genus zero. Then, we know that the Euler characteristic of sphere is two. (For a genus $g$ "shape," the Euler characteristic is $2-2g$. )
But we added those disks, which each count as a face. So, to get back to our Y shaped pipe, we need to subtract those disks (faces), which means we have $2-3=-1$.
Or, we could have used this: From Rolfsen's Knots and Links, we have a relation which says $$ g(M)=1-frac{chi(M)+b}{2} $$ where $g(M)$ is the genus of a manifold $M$ (our Y shaped pipe), $chi(M)$ is the Euler characteristic, and $b$ is the number of boundary components, which means the holes. We know the genus, and the number of holes, all you have to do is solve for $chi(M)$.
answered Feb 13 '14 at 19:04
N. OwadN. Owad
4,66521233
edited Feb 13 '14 at 19:15
answered Feb 13 '14 at 19:04
N. OwadN. Owad
4,66521233
answered Feb 13 '14 at 19:04
N. OwadN. Owad
4,66521233
answered Feb 13 '14 at 19:04
N. OwadN. Owad
4,66521233
4,66521233
$begingroup$
How have you used any knot theory here?
$endgroup$
– Dan Rust
Feb 13 '14 at 22:48
$begingroup$
@DanielRust It is more the way you view the problem. If you think of the boundary as the unlink, then there is a unique minimal genus orientable surface. This is at least how I approached the problem.
$endgroup$
– N. Owad
Feb 14 '14 at 16:16
add a comment |
$begingroup$
How have you used any knot theory here?
$endgroup$
– Dan Rust
Feb 13 '14 at 22:48
$begingroup$
@DanielRust It is more the way you view the problem. If you think of the boundary as the unlink, then there is a unique minimal genus orientable surface. This is at least how I approached the problem.
$endgroup$
– N. Owad
Feb 14 '14 at 16:16
$begingroup$
How have you used any knot theory here?
$endgroup$
– Dan Rust
Feb 13 '14 at 22:48
$begingroup$
How have you used any knot theory here?
$endgroup$
– Dan Rust
Feb 13 '14 at 22:48
$begingroup$
@DanielRust It is more the way you view the problem. If you think of the boundary as the unlink, then there is a unique minimal genus orientable surface. This is at least how I approached the problem.
$endgroup$
– N. Owad
Feb 14 '14 at 16:16
$begingroup$
@DanielRust It is more the way you view the problem. If you think of the boundary as the unlink, then there is a unique minimal genus orientable surface. This is at least how I approached the problem.
$endgroup$
– N. Owad
Feb 14 '14 at 16:16
add a comment |
$begingroup$
The easy way to find the Euler characteristic is to note that we can deformation retract the pair of pants (that's the name topologists normally give this space) onto a subset which is homeomorphic to a circle with an interval glued at one end to the north pole or the circle and the other end glued to the south pole of the circle. This in turn is homotopy equivalent to a wedge of two circles by shrinking the extra interval to a point. This space has one vertex and two edges so we get $chi=1-2=-1$. As Euler characteristic is a homotopy invariant, it follows that this is also the Euler characteristic of the pair of pants.
It's actually pretty easy to put a cellular decomposition straight onto the pair of pants without worrying about homotopy equivalences. Note that the pair of pants is homeomorphic to a closed disk with two smaller open disks removed from its interior. We can decompose this into cells by placing an edge between the boundaries of the removed disks and also an edge from each boundary of the removed disks to the boundary of the large disk. This gives us a cellular decomposition consisting of $6$ vertices on the boundary, $3$ edges in the interior, $6$ edges on the boundary, and $2$ faces in the interior. This gives $chi=6-(3+6)+2=-1$ as expected.
answered Feb 13 '14 at 22:47
Dan RustDan Rust
22.9k114984
$endgroup$
add a comment |
$begingroup$
The easy way to find the Euler characteristic is to note that we can deformation retract the pair of pants (that's the name topologists normally give this space) onto a subset which is homeomorphic to a circle with an interval glued at one end to the north pole or the circle and the other end glued to the south pole of the circle. This in turn is homotopy equivalent to a wedge of two circles by shrinking the extra interval to a point. This space has one vertex and two edges so we get $chi=1-2=-1$. As Euler characteristic is a homotopy invariant, it follows that this is also the Euler characteristic of the pair of pants.
It's actually pretty easy to put a cellular decomposition straight onto the pair of pants without worrying about homotopy equivalences. Note that the pair of pants is homeomorphic to a closed disk with two smaller open disks removed from its interior. We can decompose this into cells by placing an edge between the boundaries of the removed disks and also an edge from each boundary of the removed disks to the boundary of the large disk. This gives us a cellular decomposition consisting of $6$ vertices on the boundary, $3$ edges in the interior, $6$ edges on the boundary, and $2$ faces in the interior. This gives $chi=6-(3+6)+2=-1$ as expected.
answered Feb 13 '14 at 22:47
Dan RustDan Rust
22.9k114984
$endgroup$
add a comment |
$begingroup$
The easy way to find the Euler characteristic is to note that we can deformation retract the pair of pants (that's the name topologists normally give this space) onto a subset which is homeomorphic to a circle with an interval glued at one end to the north pole or the circle and the other end glued to the south pole of the circle. This in turn is homotopy equivalent to a wedge of two circles by shrinking the extra interval to a point. This space has one vertex and two edges so we get $chi=1-2=-1$. As Euler characteristic is a homotopy invariant, it follows that this is also the Euler characteristic of the pair of pants.
It's actually pretty easy to put a cellular decomposition straight onto the pair of pants without worrying about homotopy equivalences. Note that the pair of pants is homeomorphic to a closed disk with two smaller open disks removed from its interior. We can decompose this into cells by placing an edge between the boundaries of the removed disks and also an edge from each boundary of the removed disks to the boundary of the large disk. This gives us a cellular decomposition consisting of $6$ vertices on the boundary, $3$ edges in the interior, $6$ edges on the boundary, and $2$ faces in the interior. This gives $chi=6-(3+6)+2=-1$ as expected.
answered Feb 13 '14 at 22:47
Dan RustDan Rust
22.9k114984
$endgroup$
The easy way to find the Euler characteristic is to note that we can deformation retract the pair of pants (that's the name topologists normally give this space) onto a subset which is homeomorphic to a circle with an interval glued at one end to the north pole or the circle and the other end glued to the south pole of the circle. This in turn is homotopy equivalent to a wedge of two circles by shrinking the extra interval to a point. This space has one vertex and two edges so we get $chi=1-2=-1$. As Euler characteristic is a homotopy invariant, it follows that this is also the Euler characteristic of the pair of pants.
It's actually pretty easy to put a cellular decomposition straight onto the pair of pants without worrying about homotopy equivalences. Note that the pair of pants is homeomorphic to a closed disk with two smaller open disks removed from its interior. We can decompose this into cells by placing an edge between the boundaries of the removed disks and also an edge from each boundary of the removed disks to the boundary of the large disk. This gives us a cellular decomposition consisting of $6$ vertices on the boundary, $3$ edges in the interior, $6$ edges on the boundary, and $2$ faces in the interior. This gives $chi=6-(3+6)+2=-1$ as expected.
answered Feb 13 '14 at 22:47
Dan RustDan Rust
22.9k114984
answered Feb 13 '14 at 22:47
Dan RustDan Rust
22.9k114984
answered Feb 13 '14 at 22:47
Dan RustDan Rust
22.9k114984
answered Feb 13 '14 at 22:47
Dan RustDan Rust
22.9k114984
22.9k114984
add a comment |
add a comment |
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$begingroup$
should be $-1$. Note that the pipe is homeomorphic to a sphere with three holes, which is homeomorphic to a 2-disk with two holes, and this deformation retracts to a wedge of two circles. And this wedge has one vertex and two edges.
$endgroup$
– Stefan Hamcke
Feb 13 '14 at 18:57
$begingroup$
To see a triangulation for your space, take 3 cubes with 2 opposite faces removed from each, and 2 equilateral triangles (with side length the same as your cube). It should be easy to figure out how to glue those together to get your space.
$endgroup$
– PVAL-inactive
Feb 13 '14 at 19:17