It's confusing to calculate Euler characteristic of this surface
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This pic below is an exploded view of a cone.
I'm trying to calculate the Euler characteristic of the surface made from the fragment $M$, i.e.,
At first I thought the Euler characteristic is 0, but the one who made this question says it is actually 1.
And this is actually a part of an exercise to find the total geodesic curvature of $partial M$. I tried to figure it out using the Gauss-Bonnet theorem.
I'm sorry that the question is not clear.
$M$ is the region between top circle of cone and closed geodesic meeting the top circle of cone at one point.
differential-geometry
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show 16 more comments
up vote
2
down vote
favorite
This pic below is an exploded view of a cone.
I'm trying to calculate the Euler characteristic of the surface made from the fragment $M$, i.e.,
At first I thought the Euler characteristic is 0, but the one who made this question says it is actually 1.
And this is actually a part of an exercise to find the total geodesic curvature of $partial M$. I tried to figure it out using the Gauss-Bonnet theorem.
I'm sorry that the question is not clear.
$M$ is the region between top circle of cone and closed geodesic meeting the top circle of cone at one point.
differential-geometry
1
If we agree that that's what $M$ is, then the Euler characteristic is indeed $0$.
â Ted Shifrin
Sep 6 at 4:31
1
@민ê´Âê±´ I see what the problem is. If one incorrectly computes exterior angles for $partial M$, one is led to the conclusion that $chi(M)=1$. But $chi(M)=0$ is absolutely correct. You cannot compute exterior angles from this picture in the plane.
â Ted Shifrin
Sep 6 at 4:51
1
@민ê´Âê±´ Hello! I've edited your question (and added a picture), since I think it may help clarify what you want. If my edit does not clarify what you intended to mean, feel free to roll it back. In the future, please try to present the objects which come from your context in a way that makes it immediately clear what is being talked about.
â Aloizio Macedoâ¦
Sep 6 at 5:43
1
@AloizioMacedo thanks actually it was first time for me to upload a question and it was somewhat difficult to make a clear question. I'm gonna try to do better in the future.
â ë¯¼ê´Âê±´
Sep 6 at 7:16
2
I think the closing was overly quick, @JohnMa and Aloizio, given that some of us were engaged in the question. It still turns out to be a fascinating issue. Unless I'm being stupid, if I think of $partial M$ as a piecewise-smooth curve in one way, I get the wrong answer when I apply Gauss-Bonnet; if I think it the other way, I get the right answer. I'm a bit puzzled.
â Ted Shifrin
Sep 6 at 16:49
 |Â
show 16 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This pic below is an exploded view of a cone.
I'm trying to calculate the Euler characteristic of the surface made from the fragment $M$, i.e.,
At first I thought the Euler characteristic is 0, but the one who made this question says it is actually 1.
And this is actually a part of an exercise to find the total geodesic curvature of $partial M$. I tried to figure it out using the Gauss-Bonnet theorem.
I'm sorry that the question is not clear.
$M$ is the region between top circle of cone and closed geodesic meeting the top circle of cone at one point.
differential-geometry
This pic below is an exploded view of a cone.
I'm trying to calculate the Euler characteristic of the surface made from the fragment $M$, i.e.,
At first I thought the Euler characteristic is 0, but the one who made this question says it is actually 1.
And this is actually a part of an exercise to find the total geodesic curvature of $partial M$. I tried to figure it out using the Gauss-Bonnet theorem.
I'm sorry that the question is not clear.
$M$ is the region between top circle of cone and closed geodesic meeting the top circle of cone at one point.
differential-geometry
differential-geometry
edited Sep 6 at 5:32
Aloizio Macedoâ¦
22.8k23483
22.8k23483
asked Sep 5 at 8:07
민ê´Âê±´
163
163
1
If we agree that that's what $M$ is, then the Euler characteristic is indeed $0$.
â Ted Shifrin
Sep 6 at 4:31
1
@민ê´Âê±´ I see what the problem is. If one incorrectly computes exterior angles for $partial M$, one is led to the conclusion that $chi(M)=1$. But $chi(M)=0$ is absolutely correct. You cannot compute exterior angles from this picture in the plane.
â Ted Shifrin
Sep 6 at 4:51
1
@민ê´Âê±´ Hello! I've edited your question (and added a picture), since I think it may help clarify what you want. If my edit does not clarify what you intended to mean, feel free to roll it back. In the future, please try to present the objects which come from your context in a way that makes it immediately clear what is being talked about.
â Aloizio Macedoâ¦
Sep 6 at 5:43
1
@AloizioMacedo thanks actually it was first time for me to upload a question and it was somewhat difficult to make a clear question. I'm gonna try to do better in the future.
â ë¯¼ê´Âê±´
Sep 6 at 7:16
2
I think the closing was overly quick, @JohnMa and Aloizio, given that some of us were engaged in the question. It still turns out to be a fascinating issue. Unless I'm being stupid, if I think of $partial M$ as a piecewise-smooth curve in one way, I get the wrong answer when I apply Gauss-Bonnet; if I think it the other way, I get the right answer. I'm a bit puzzled.
â Ted Shifrin
Sep 6 at 16:49
 |Â
show 16 more comments
1
If we agree that that's what $M$ is, then the Euler characteristic is indeed $0$.
â Ted Shifrin
Sep 6 at 4:31
1
@민ê´Âê±´ I see what the problem is. If one incorrectly computes exterior angles for $partial M$, one is led to the conclusion that $chi(M)=1$. But $chi(M)=0$ is absolutely correct. You cannot compute exterior angles from this picture in the plane.
â Ted Shifrin
Sep 6 at 4:51
1
@민ê´Âê±´ Hello! I've edited your question (and added a picture), since I think it may help clarify what you want. If my edit does not clarify what you intended to mean, feel free to roll it back. In the future, please try to present the objects which come from your context in a way that makes it immediately clear what is being talked about.
â Aloizio Macedoâ¦
Sep 6 at 5:43
1
@AloizioMacedo thanks actually it was first time for me to upload a question and it was somewhat difficult to make a clear question. I'm gonna try to do better in the future.
â ë¯¼ê´Âê±´
Sep 6 at 7:16
2
I think the closing was overly quick, @JohnMa and Aloizio, given that some of us were engaged in the question. It still turns out to be a fascinating issue. Unless I'm being stupid, if I think of $partial M$ as a piecewise-smooth curve in one way, I get the wrong answer when I apply Gauss-Bonnet; if I think it the other way, I get the right answer. I'm a bit puzzled.
â Ted Shifrin
Sep 6 at 16:49
1
1
If we agree that that's what $M$ is, then the Euler characteristic is indeed $0$.
â Ted Shifrin
Sep 6 at 4:31
If we agree that that's what $M$ is, then the Euler characteristic is indeed $0$.
â Ted Shifrin
Sep 6 at 4:31
1
1
@민ê´Âê±´ I see what the problem is. If one incorrectly computes exterior angles for $partial M$, one is led to the conclusion that $chi(M)=1$. But $chi(M)=0$ is absolutely correct. You cannot compute exterior angles from this picture in the plane.
â Ted Shifrin
Sep 6 at 4:51
@민ê´Âê±´ I see what the problem is. If one incorrectly computes exterior angles for $partial M$, one is led to the conclusion that $chi(M)=1$. But $chi(M)=0$ is absolutely correct. You cannot compute exterior angles from this picture in the plane.
â Ted Shifrin
Sep 6 at 4:51
1
1
@민ê´Âê±´ Hello! I've edited your question (and added a picture), since I think it may help clarify what you want. If my edit does not clarify what you intended to mean, feel free to roll it back. In the future, please try to present the objects which come from your context in a way that makes it immediately clear what is being talked about.
â Aloizio Macedoâ¦
Sep 6 at 5:43
@민ê´Âê±´ Hello! I've edited your question (and added a picture), since I think it may help clarify what you want. If my edit does not clarify what you intended to mean, feel free to roll it back. In the future, please try to present the objects which come from your context in a way that makes it immediately clear what is being talked about.
â Aloizio Macedoâ¦
Sep 6 at 5:43
1
1
@AloizioMacedo thanks actually it was first time for me to upload a question and it was somewhat difficult to make a clear question. I'm gonna try to do better in the future.
â ë¯¼ê´Âê±´
Sep 6 at 7:16
@AloizioMacedo thanks actually it was first time for me to upload a question and it was somewhat difficult to make a clear question. I'm gonna try to do better in the future.
â ë¯¼ê´Âê±´
Sep 6 at 7:16
2
2
I think the closing was overly quick, @JohnMa and Aloizio, given that some of us were engaged in the question. It still turns out to be a fascinating issue. Unless I'm being stupid, if I think of $partial M$ as a piecewise-smooth curve in one way, I get the wrong answer when I apply Gauss-Bonnet; if I think it the other way, I get the right answer. I'm a bit puzzled.
â Ted Shifrin
Sep 6 at 16:49
I think the closing was overly quick, @JohnMa and Aloizio, given that some of us were engaged in the question. It still turns out to be a fascinating issue. Unless I'm being stupid, if I think of $partial M$ as a piecewise-smooth curve in one way, I get the wrong answer when I apply Gauss-Bonnet; if I think it the other way, I get the right answer. I'm a bit puzzled.
â Ted Shifrin
Sep 6 at 16:49
 |Â
show 16 more comments
3 Answers
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0
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If I'm reading that picture correctly, the Euler characteristic shouldn't exist.
The subset $M$ of the cone looks like it started as a topological disk facing left, and is curving to the right to connect back with itself, and will become a topological cylinder (a sphere with 2 disks removed).
A disk has $chi=1$ , and a cylinder has $chi=0$ . The surface $M$ is between those.
Nevertheless, Euler's formula can be applied to a polygonalization of $M$, formed as a square pyramid with 2 opposite triangles removed; the peak of the pyramid is the singular point.
The pyramid has $5$ vertices, $8$ edges, and $3$ faces. Thus $chi=5-8+3=0$.
add a comment |Â
up vote
-2
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We have Gauss Bonnet theorem
$$ int K dA + int kappa_g ,ds= 2 pi chi $$
Solid angle $int K dA= 0$ on the flat development of cone.
Rotation in the plane of development (across cone base and geodesic) consists of summing three exterior angles counter-clockwise $Sigma psi_i= 2 pi $ around the contour for the line integral which can be readily found as:
$$ Sigma psi_i = int kappa_geodesic, ds = (pi- beta) + 2 beta + (pi- beta) = 2 pi$$
The angles are invariant in isometric mappings from 3D cone to a cone development and so it is a good way to compute these angles as above.
Plug these angles into GB theorem and we have
$$ 0+ 2 pi = 2 pi chirightarrow chi=1. $$
It can be noted that the geodesic arc is not an arc of a circle but a segment of a generalized sine-curve.
1
Your computation is wrong, as $chi = 0$, as I've explained in comments.
â Ted Shifrin
Sep 9 at 4:20
ItâÂÂs wrong again. Please integrate total geodesic curvuture with its defnition and see what happens.
â ë¯¼ê´Âê±´
Sep 9 at 11:33
add a comment |Â
up vote
-3
down vote
Gauss- Bonnet theorem establishes the connection/identity between isometric (left hand side of equation) and Euler topological constant (right hand side ). In short
$ int K dA + int kappa_g ,ds= 2 pi chi$
where for compact surfaces the first term is total/integral curvature or solid angle in steridians, second term is rotation in tangent plane measured in radians which together elegantly sum up to Euler charactristic $2 pi chi.$ Sudden jumps with geodesics can be accommodated/interpreted as external angles sum $Sigma psi_i$ around the contour for the line integral.
For a Torus with cancelling geodesic sections:
$0+ 0=2 pi chi rightarrow , chi=0 tag1$
For a hemisphere bounded by an equator:
$2 pi +0 = 2 pi chi rightarrow , chi=1 tag2 $
For a closed convex loop on a developable surface (Gauss curvature $K=0,$ for cones, cylinders/developable helicoids) for either continuous ( like a circle in a flat plane or non-intersecting continuous loop on a curved surface) curves or discontinuous sloped curves (segment of circle $M$ like the one you sketched of developed cone patch):
$0+ 2 pi=2 pi chi rightarrow , chi=1 tag3 $
The matter is thus established by isometry/topological considerations.
EDIT1:
I am in agreement with the question setter. I would suppose he wanted the student to recognize a group of such isometric/homeomorphic equivalents.
This doesn't really address the question.
â Ted Shifrin
Sep 7 at 0:52
Sorry but I think you didn't get what the question is.
â ë¯¼ê´Âê±´
Sep 7 at 2:02
You're in agreement with the question setter? That person is totally wrong. As i said earlier, the puzzle comes with applying Gauss-Bonnet, calculating total geodesic curvature, and exterior angles. You should try that instead of writing irrelevant prose.
â Ted Shifrin
Sep 7 at 2:15
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
If I'm reading that picture correctly, the Euler characteristic shouldn't exist.
The subset $M$ of the cone looks like it started as a topological disk facing left, and is curving to the right to connect back with itself, and will become a topological cylinder (a sphere with 2 disks removed).
A disk has $chi=1$ , and a cylinder has $chi=0$ . The surface $M$ is between those.
Nevertheless, Euler's formula can be applied to a polygonalization of $M$, formed as a square pyramid with 2 opposite triangles removed; the peak of the pyramid is the singular point.
The pyramid has $5$ vertices, $8$ edges, and $3$ faces. Thus $chi=5-8+3=0$.
add a comment |Â
up vote
0
down vote
If I'm reading that picture correctly, the Euler characteristic shouldn't exist.
The subset $M$ of the cone looks like it started as a topological disk facing left, and is curving to the right to connect back with itself, and will become a topological cylinder (a sphere with 2 disks removed).
A disk has $chi=1$ , and a cylinder has $chi=0$ . The surface $M$ is between those.
Nevertheless, Euler's formula can be applied to a polygonalization of $M$, formed as a square pyramid with 2 opposite triangles removed; the peak of the pyramid is the singular point.
The pyramid has $5$ vertices, $8$ edges, and $3$ faces. Thus $chi=5-8+3=0$.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
If I'm reading that picture correctly, the Euler characteristic shouldn't exist.
The subset $M$ of the cone looks like it started as a topological disk facing left, and is curving to the right to connect back with itself, and will become a topological cylinder (a sphere with 2 disks removed).
A disk has $chi=1$ , and a cylinder has $chi=0$ . The surface $M$ is between those.
Nevertheless, Euler's formula can be applied to a polygonalization of $M$, formed as a square pyramid with 2 opposite triangles removed; the peak of the pyramid is the singular point.
The pyramid has $5$ vertices, $8$ edges, and $3$ faces. Thus $chi=5-8+3=0$.
If I'm reading that picture correctly, the Euler characteristic shouldn't exist.
The subset $M$ of the cone looks like it started as a topological disk facing left, and is curving to the right to connect back with itself, and will become a topological cylinder (a sphere with 2 disks removed).
A disk has $chi=1$ , and a cylinder has $chi=0$ . The surface $M$ is between those.
Nevertheless, Euler's formula can be applied to a polygonalization of $M$, formed as a square pyramid with 2 opposite triangles removed; the peak of the pyramid is the singular point.
The pyramid has $5$ vertices, $8$ edges, and $3$ faces. Thus $chi=5-8+3=0$.
answered yesterday
mr_e_man
1,114422
1,114422
add a comment |Â
add a comment |Â
up vote
-2
down vote
We have Gauss Bonnet theorem
$$ int K dA + int kappa_g ,ds= 2 pi chi $$
Solid angle $int K dA= 0$ on the flat development of cone.
Rotation in the plane of development (across cone base and geodesic) consists of summing three exterior angles counter-clockwise $Sigma psi_i= 2 pi $ around the contour for the line integral which can be readily found as:
$$ Sigma psi_i = int kappa_geodesic, ds = (pi- beta) + 2 beta + (pi- beta) = 2 pi$$
The angles are invariant in isometric mappings from 3D cone to a cone development and so it is a good way to compute these angles as above.
Plug these angles into GB theorem and we have
$$ 0+ 2 pi = 2 pi chirightarrow chi=1. $$
It can be noted that the geodesic arc is not an arc of a circle but a segment of a generalized sine-curve.
1
Your computation is wrong, as $chi = 0$, as I've explained in comments.
â Ted Shifrin
Sep 9 at 4:20
ItâÂÂs wrong again. Please integrate total geodesic curvuture with its defnition and see what happens.
â ë¯¼ê´Âê±´
Sep 9 at 11:33
add a comment |Â
up vote
-2
down vote
We have Gauss Bonnet theorem
$$ int K dA + int kappa_g ,ds= 2 pi chi $$
Solid angle $int K dA= 0$ on the flat development of cone.
Rotation in the plane of development (across cone base and geodesic) consists of summing three exterior angles counter-clockwise $Sigma psi_i= 2 pi $ around the contour for the line integral which can be readily found as:
$$ Sigma psi_i = int kappa_geodesic, ds = (pi- beta) + 2 beta + (pi- beta) = 2 pi$$
The angles are invariant in isometric mappings from 3D cone to a cone development and so it is a good way to compute these angles as above.
Plug these angles into GB theorem and we have
$$ 0+ 2 pi = 2 pi chirightarrow chi=1. $$
It can be noted that the geodesic arc is not an arc of a circle but a segment of a generalized sine-curve.
1
Your computation is wrong, as $chi = 0$, as I've explained in comments.
â Ted Shifrin
Sep 9 at 4:20
ItâÂÂs wrong again. Please integrate total geodesic curvuture with its defnition and see what happens.
â ë¯¼ê´Âê±´
Sep 9 at 11:33
add a comment |Â
up vote
-2
down vote
up vote
-2
down vote
We have Gauss Bonnet theorem
$$ int K dA + int kappa_g ,ds= 2 pi chi $$
Solid angle $int K dA= 0$ on the flat development of cone.
Rotation in the plane of development (across cone base and geodesic) consists of summing three exterior angles counter-clockwise $Sigma psi_i= 2 pi $ around the contour for the line integral which can be readily found as:
$$ Sigma psi_i = int kappa_geodesic, ds = (pi- beta) + 2 beta + (pi- beta) = 2 pi$$
The angles are invariant in isometric mappings from 3D cone to a cone development and so it is a good way to compute these angles as above.
Plug these angles into GB theorem and we have
$$ 0+ 2 pi = 2 pi chirightarrow chi=1. $$
It can be noted that the geodesic arc is not an arc of a circle but a segment of a generalized sine-curve.
We have Gauss Bonnet theorem
$$ int K dA + int kappa_g ,ds= 2 pi chi $$
Solid angle $int K dA= 0$ on the flat development of cone.
Rotation in the plane of development (across cone base and geodesic) consists of summing three exterior angles counter-clockwise $Sigma psi_i= 2 pi $ around the contour for the line integral which can be readily found as:
$$ Sigma psi_i = int kappa_geodesic, ds = (pi- beta) + 2 beta + (pi- beta) = 2 pi$$
The angles are invariant in isometric mappings from 3D cone to a cone development and so it is a good way to compute these angles as above.
Plug these angles into GB theorem and we have
$$ 0+ 2 pi = 2 pi chirightarrow chi=1. $$
It can be noted that the geodesic arc is not an arc of a circle but a segment of a generalized sine-curve.
edited Sep 9 at 0:05
answered Sep 8 at 21:02
Narasimham
20.3k52057
20.3k52057
1
Your computation is wrong, as $chi = 0$, as I've explained in comments.
â Ted Shifrin
Sep 9 at 4:20
ItâÂÂs wrong again. Please integrate total geodesic curvuture with its defnition and see what happens.
â ë¯¼ê´Âê±´
Sep 9 at 11:33
add a comment |Â
1
Your computation is wrong, as $chi = 0$, as I've explained in comments.
â Ted Shifrin
Sep 9 at 4:20
ItâÂÂs wrong again. Please integrate total geodesic curvuture with its defnition and see what happens.
â ë¯¼ê´Âê±´
Sep 9 at 11:33
1
1
Your computation is wrong, as $chi = 0$, as I've explained in comments.
â Ted Shifrin
Sep 9 at 4:20
Your computation is wrong, as $chi = 0$, as I've explained in comments.
â Ted Shifrin
Sep 9 at 4:20
ItâÂÂs wrong again. Please integrate total geodesic curvuture with its defnition and see what happens.
â ë¯¼ê´Âê±´
Sep 9 at 11:33
ItâÂÂs wrong again. Please integrate total geodesic curvuture with its defnition and see what happens.
â ë¯¼ê´Âê±´
Sep 9 at 11:33
add a comment |Â
up vote
-3
down vote
Gauss- Bonnet theorem establishes the connection/identity between isometric (left hand side of equation) and Euler topological constant (right hand side ). In short
$ int K dA + int kappa_g ,ds= 2 pi chi$
where for compact surfaces the first term is total/integral curvature or solid angle in steridians, second term is rotation in tangent plane measured in radians which together elegantly sum up to Euler charactristic $2 pi chi.$ Sudden jumps with geodesics can be accommodated/interpreted as external angles sum $Sigma psi_i$ around the contour for the line integral.
For a Torus with cancelling geodesic sections:
$0+ 0=2 pi chi rightarrow , chi=0 tag1$
For a hemisphere bounded by an equator:
$2 pi +0 = 2 pi chi rightarrow , chi=1 tag2 $
For a closed convex loop on a developable surface (Gauss curvature $K=0,$ for cones, cylinders/developable helicoids) for either continuous ( like a circle in a flat plane or non-intersecting continuous loop on a curved surface) curves or discontinuous sloped curves (segment of circle $M$ like the one you sketched of developed cone patch):
$0+ 2 pi=2 pi chi rightarrow , chi=1 tag3 $
The matter is thus established by isometry/topological considerations.
EDIT1:
I am in agreement with the question setter. I would suppose he wanted the student to recognize a group of such isometric/homeomorphic equivalents.
This doesn't really address the question.
â Ted Shifrin
Sep 7 at 0:52
Sorry but I think you didn't get what the question is.
â ë¯¼ê´Âê±´
Sep 7 at 2:02
You're in agreement with the question setter? That person is totally wrong. As i said earlier, the puzzle comes with applying Gauss-Bonnet, calculating total geodesic curvature, and exterior angles. You should try that instead of writing irrelevant prose.
â Ted Shifrin
Sep 7 at 2:15
add a comment |Â
up vote
-3
down vote
Gauss- Bonnet theorem establishes the connection/identity between isometric (left hand side of equation) and Euler topological constant (right hand side ). In short
$ int K dA + int kappa_g ,ds= 2 pi chi$
where for compact surfaces the first term is total/integral curvature or solid angle in steridians, second term is rotation in tangent plane measured in radians which together elegantly sum up to Euler charactristic $2 pi chi.$ Sudden jumps with geodesics can be accommodated/interpreted as external angles sum $Sigma psi_i$ around the contour for the line integral.
For a Torus with cancelling geodesic sections:
$0+ 0=2 pi chi rightarrow , chi=0 tag1$
For a hemisphere bounded by an equator:
$2 pi +0 = 2 pi chi rightarrow , chi=1 tag2 $
For a closed convex loop on a developable surface (Gauss curvature $K=0,$ for cones, cylinders/developable helicoids) for either continuous ( like a circle in a flat plane or non-intersecting continuous loop on a curved surface) curves or discontinuous sloped curves (segment of circle $M$ like the one you sketched of developed cone patch):
$0+ 2 pi=2 pi chi rightarrow , chi=1 tag3 $
The matter is thus established by isometry/topological considerations.
EDIT1:
I am in agreement with the question setter. I would suppose he wanted the student to recognize a group of such isometric/homeomorphic equivalents.
This doesn't really address the question.
â Ted Shifrin
Sep 7 at 0:52
Sorry but I think you didn't get what the question is.
â ë¯¼ê´Âê±´
Sep 7 at 2:02
You're in agreement with the question setter? That person is totally wrong. As i said earlier, the puzzle comes with applying Gauss-Bonnet, calculating total geodesic curvature, and exterior angles. You should try that instead of writing irrelevant prose.
â Ted Shifrin
Sep 7 at 2:15
add a comment |Â
up vote
-3
down vote
up vote
-3
down vote
Gauss- Bonnet theorem establishes the connection/identity between isometric (left hand side of equation) and Euler topological constant (right hand side ). In short
$ int K dA + int kappa_g ,ds= 2 pi chi$
where for compact surfaces the first term is total/integral curvature or solid angle in steridians, second term is rotation in tangent plane measured in radians which together elegantly sum up to Euler charactristic $2 pi chi.$ Sudden jumps with geodesics can be accommodated/interpreted as external angles sum $Sigma psi_i$ around the contour for the line integral.
For a Torus with cancelling geodesic sections:
$0+ 0=2 pi chi rightarrow , chi=0 tag1$
For a hemisphere bounded by an equator:
$2 pi +0 = 2 pi chi rightarrow , chi=1 tag2 $
For a closed convex loop on a developable surface (Gauss curvature $K=0,$ for cones, cylinders/developable helicoids) for either continuous ( like a circle in a flat plane or non-intersecting continuous loop on a curved surface) curves or discontinuous sloped curves (segment of circle $M$ like the one you sketched of developed cone patch):
$0+ 2 pi=2 pi chi rightarrow , chi=1 tag3 $
The matter is thus established by isometry/topological considerations.
EDIT1:
I am in agreement with the question setter. I would suppose he wanted the student to recognize a group of such isometric/homeomorphic equivalents.
Gauss- Bonnet theorem establishes the connection/identity between isometric (left hand side of equation) and Euler topological constant (right hand side ). In short
$ int K dA + int kappa_g ,ds= 2 pi chi$
where for compact surfaces the first term is total/integral curvature or solid angle in steridians, second term is rotation in tangent plane measured in radians which together elegantly sum up to Euler charactristic $2 pi chi.$ Sudden jumps with geodesics can be accommodated/interpreted as external angles sum $Sigma psi_i$ around the contour for the line integral.
For a Torus with cancelling geodesic sections:
$0+ 0=2 pi chi rightarrow , chi=0 tag1$
For a hemisphere bounded by an equator:
$2 pi +0 = 2 pi chi rightarrow , chi=1 tag2 $
For a closed convex loop on a developable surface (Gauss curvature $K=0,$ for cones, cylinders/developable helicoids) for either continuous ( like a circle in a flat plane or non-intersecting continuous loop on a curved surface) curves or discontinuous sloped curves (segment of circle $M$ like the one you sketched of developed cone patch):
$0+ 2 pi=2 pi chi rightarrow , chi=1 tag3 $
The matter is thus established by isometry/topological considerations.
EDIT1:
I am in agreement with the question setter. I would suppose he wanted the student to recognize a group of such isometric/homeomorphic equivalents.
edited Sep 6 at 23:44
answered Sep 6 at 20:59
Narasimham
20.3k52057
20.3k52057
This doesn't really address the question.
â Ted Shifrin
Sep 7 at 0:52
Sorry but I think you didn't get what the question is.
â ë¯¼ê´Âê±´
Sep 7 at 2:02
You're in agreement with the question setter? That person is totally wrong. As i said earlier, the puzzle comes with applying Gauss-Bonnet, calculating total geodesic curvature, and exterior angles. You should try that instead of writing irrelevant prose.
â Ted Shifrin
Sep 7 at 2:15
add a comment |Â
This doesn't really address the question.
â Ted Shifrin
Sep 7 at 0:52
Sorry but I think you didn't get what the question is.
â ë¯¼ê´Âê±´
Sep 7 at 2:02
You're in agreement with the question setter? That person is totally wrong. As i said earlier, the puzzle comes with applying Gauss-Bonnet, calculating total geodesic curvature, and exterior angles. You should try that instead of writing irrelevant prose.
â Ted Shifrin
Sep 7 at 2:15
This doesn't really address the question.
â Ted Shifrin
Sep 7 at 0:52
This doesn't really address the question.
â Ted Shifrin
Sep 7 at 0:52
Sorry but I think you didn't get what the question is.
â ë¯¼ê´Âê±´
Sep 7 at 2:02
Sorry but I think you didn't get what the question is.
â ë¯¼ê´Âê±´
Sep 7 at 2:02
You're in agreement with the question setter? That person is totally wrong. As i said earlier, the puzzle comes with applying Gauss-Bonnet, calculating total geodesic curvature, and exterior angles. You should try that instead of writing irrelevant prose.
â Ted Shifrin
Sep 7 at 2:15
You're in agreement with the question setter? That person is totally wrong. As i said earlier, the puzzle comes with applying Gauss-Bonnet, calculating total geodesic curvature, and exterior angles. You should try that instead of writing irrelevant prose.
â Ted Shifrin
Sep 7 at 2:15
add a comment |Â
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1
If we agree that that's what $M$ is, then the Euler characteristic is indeed $0$.
â Ted Shifrin
Sep 6 at 4:31
1
@민ê´Âê±´ I see what the problem is. If one incorrectly computes exterior angles for $partial M$, one is led to the conclusion that $chi(M)=1$. But $chi(M)=0$ is absolutely correct. You cannot compute exterior angles from this picture in the plane.
â Ted Shifrin
Sep 6 at 4:51
1
@민ê´Âê±´ Hello! I've edited your question (and added a picture), since I think it may help clarify what you want. If my edit does not clarify what you intended to mean, feel free to roll it back. In the future, please try to present the objects which come from your context in a way that makes it immediately clear what is being talked about.
â Aloizio Macedoâ¦
Sep 6 at 5:43
1
@AloizioMacedo thanks actually it was first time for me to upload a question and it was somewhat difficult to make a clear question. I'm gonna try to do better in the future.
â ë¯¼ê´Âê±´
Sep 6 at 7:16
2
I think the closing was overly quick, @JohnMa and Aloizio, given that some of us were engaged in the question. It still turns out to be a fascinating issue. Unless I'm being stupid, if I think of $partial M$ as a piecewise-smooth curve in one way, I get the wrong answer when I apply Gauss-Bonnet; if I think it the other way, I get the right answer. I'm a bit puzzled.
â Ted Shifrin
Sep 6 at 16:49