triangulation of a circle and the way to solve a problem
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Consider the circle $S^1$
with multiplication given by the complex numbers.
Prove that the map $f(x) = x
^n$
, $n$ a positive integer, has degree $n$. What is the
degree of the map $g(x) = 1/x$.
This is exercise 25 of Croom's book "basic concepts in algebraic topology". I have not solved it yet.
My approach is that I triangulate the circle as follows: denote $a=(1,0),b=(-1,0),c=(0,1), d=(0,-1)$. The complex contains the 1-simplexes $(ab),(ac),(ad),(bc),(bd)$ and their faces. However, with this triangulation, I computed $H_1(S^1)=mathbbZ^2$, which is wrong.
So could u explain why I was wrong, and how to solve the original problem? Thank you
algebraic-topology homology-cohomology simplex triangulation
asked Sep 6 at 11:46
chiÌ trung châu
1,2591725
add a comment |Â
up vote
0
down vote
favorite
Consider the circle $S^1$
with multiplication given by the complex numbers.
Prove that the map $f(x) = x
^n$
, $n$ a positive integer, has degree $n$. What is the
degree of the map $g(x) = 1/x$.
This is exercise 25 of Croom's book "basic concepts in algebraic topology". I have not solved it yet.
My approach is that I triangulate the circle as follows: denote $a=(1,0),b=(-1,0),c=(0,1), d=(0,-1)$. The complex contains the 1-simplexes $(ab),(ac),(ad),(bc),(bd)$ and their faces. However, with this triangulation, I computed $H_1(S^1)=mathbbZ^2$, which is wrong.
So could u explain why I was wrong, and how to solve the original problem? Thank you
algebraic-topology homology-cohomology simplex triangulation
asked Sep 6 at 11:46
chiÌ trung châu
1,2591725
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Consider the circle $S^1$
with multiplication given by the complex numbers.
Prove that the map $f(x) = x
^n$
, $n$ a positive integer, has degree $n$. What is the
degree of the map $g(x) = 1/x$.
This is exercise 25 of Croom's book "basic concepts in algebraic topology". I have not solved it yet.
My approach is that I triangulate the circle as follows: denote $a=(1,0),b=(-1,0),c=(0,1), d=(0,-1)$. The complex contains the 1-simplexes $(ab),(ac),(ad),(bc),(bd)$ and their faces. However, with this triangulation, I computed $H_1(S^1)=mathbbZ^2$, which is wrong.
So could u explain why I was wrong, and how to solve the original problem? Thank you
algebraic-topology homology-cohomology simplex triangulation
asked Sep 6 at 11:46
chiÌ trung châu
1,2591725
Consider the circle $S^1$
with multiplication given by the complex numbers.
Prove that the map $f(x) = x
^n$
, $n$ a positive integer, has degree $n$. What is the
degree of the map $g(x) = 1/x$.
This is exercise 25 of Croom's book "basic concepts in algebraic topology". I have not solved it yet.
My approach is that I triangulate the circle as follows: denote $a=(1,0),b=(-1,0),c=(0,1), d=(0,-1)$. The complex contains the 1-simplexes $(ab),(ac),(ad),(bc),(bd)$ and their faces. However, with this triangulation, I computed $H_1(S^1)=mathbbZ^2$, which is wrong.
So could u explain why I was wrong, and how to solve the original problem? Thank you
algebraic-topology homology-cohomology simplex triangulation
algebraic-topology homology-cohomology simplex triangulation
asked Sep 6 at 11:46
chiÌ trung châu
1,2591725
asked Sep 6 at 11:46
chiÌ trung châu
1,2591725
asked Sep 6 at 11:46
chiÌ trung châu
1,2591725
asked Sep 6 at 11:46
chiÌ trung châu
1,2591725
asked Sep 6 at 11:46
chiÌ trung châu
1,2591725
1,2591725
add a comment |Â
add a comment |Â
1 Answer
1
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1
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The first place you went wrong was this:
The 1-simplexes should be $ac$, $cb$, $bd$, and $da$. The edge $ab$ is a diameter of the circle.
With that in hand, perhaps you can complete the problem yourself. One way to solve it is indeed to triangulate, although you'll probably want quite a lot more vertices to make that easier.
answered Sep 6 at 11:50
John Hughes
60k23987
thank u. Can u explain why u think it would make the problem easier if we have more vertices? Personally, I think these 4 would do it since $x^n$ and $1/x$ map these four to these four, hence simplicial maps.
– chiÌ trung châu
Sep 6 at 11:55
Your complex was a circle with one diameter included, or, drawn with straight lines, it's a pair of triangles that share one edge. The first homology of this object is, in face $Bbb Z oplus Bbb Z$.
– John Hughes
Sep 6 at 11:59
tks. I just realize that and delete the question :)). What about my first ques?
– chiÌ trung châu
Sep 6 at 12:00
As for your map: even for $n = 2$, it's not simplicial, for the images of $a$ and $c$ end up being $a$ and $b$, and while $ac$ is an edge, $ab$ is not. You'll need to triangulate the domain and codomain with different numbers of vertices to make this all work out (if you're approaching it the way you seem to be).
– John Hughes
Sep 6 at 12:01
wow, seems like much more work than I thought. Thank u
– chiÌ trung châu
Sep 6 at 12:09
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The first place you went wrong was this:
The 1-simplexes should be $ac$, $cb$, $bd$, and $da$. The edge $ab$ is a diameter of the circle.
With that in hand, perhaps you can complete the problem yourself. One way to solve it is indeed to triangulate, although you'll probably want quite a lot more vertices to make that easier.
answered Sep 6 at 11:50
John Hughes
60k23987
thank u. Can u explain why u think it would make the problem easier if we have more vertices? Personally, I think these 4 would do it since $x^n$ and $1/x$ map these four to these four, hence simplicial maps.
– chiÌ trung châu
Sep 6 at 11:55
Your complex was a circle with one diameter included, or, drawn with straight lines, it's a pair of triangles that share one edge. The first homology of this object is, in face $Bbb Z oplus Bbb Z$.
– John Hughes
Sep 6 at 11:59
tks. I just realize that and delete the question :)). What about my first ques?
– chiÌ trung châu
Sep 6 at 12:00
As for your map: even for $n = 2$, it's not simplicial, for the images of $a$ and $c$ end up being $a$ and $b$, and while $ac$ is an edge, $ab$ is not. You'll need to triangulate the domain and codomain with different numbers of vertices to make this all work out (if you're approaching it the way you seem to be).
– John Hughes
Sep 6 at 12:01
wow, seems like much more work than I thought. Thank u
– chiÌ trung châu
Sep 6 at 12:09
add a comment |Â
up vote
1
down vote
accepted
The first place you went wrong was this:
The 1-simplexes should be $ac$, $cb$, $bd$, and $da$. The edge $ab$ is a diameter of the circle.
With that in hand, perhaps you can complete the problem yourself. One way to solve it is indeed to triangulate, although you'll probably want quite a lot more vertices to make that easier.
answered Sep 6 at 11:50
John Hughes
60k23987
thank u. Can u explain why u think it would make the problem easier if we have more vertices? Personally, I think these 4 would do it since $x^n$ and $1/x$ map these four to these four, hence simplicial maps.
– chiÌ trung châu
Sep 6 at 11:55
Your complex was a circle with one diameter included, or, drawn with straight lines, it's a pair of triangles that share one edge. The first homology of this object is, in face $Bbb Z oplus Bbb Z$.
– John Hughes
Sep 6 at 11:59
tks. I just realize that and delete the question :)). What about my first ques?
– chiÌ trung châu
Sep 6 at 12:00
As for your map: even for $n = 2$, it's not simplicial, for the images of $a$ and $c$ end up being $a$ and $b$, and while $ac$ is an edge, $ab$ is not. You'll need to triangulate the domain and codomain with different numbers of vertices to make this all work out (if you're approaching it the way you seem to be).
– John Hughes
Sep 6 at 12:01
wow, seems like much more work than I thought. Thank u
– chiÌ trung châu
Sep 6 at 12:09
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The first place you went wrong was this:
The 1-simplexes should be $ac$, $cb$, $bd$, and $da$. The edge $ab$ is a diameter of the circle.
With that in hand, perhaps you can complete the problem yourself. One way to solve it is indeed to triangulate, although you'll probably want quite a lot more vertices to make that easier.
answered Sep 6 at 11:50
John Hughes
60k23987
The first place you went wrong was this:
The 1-simplexes should be $ac$, $cb$, $bd$, and $da$. The edge $ab$ is a diameter of the circle.
With that in hand, perhaps you can complete the problem yourself. One way to solve it is indeed to triangulate, although you'll probably want quite a lot more vertices to make that easier.
answered Sep 6 at 11:50
John Hughes
60k23987
answered Sep 6 at 11:50
John Hughes
60k23987
answered Sep 6 at 11:50
John Hughes
60k23987
answered Sep 6 at 11:50
John Hughes
60k23987
60k23987
thank u. Can u explain why u think it would make the problem easier if we have more vertices? Personally, I think these 4 would do it since $x^n$ and $1/x$ map these four to these four, hence simplicial maps.
– chiÌ trung châu
Sep 6 at 11:55
Your complex was a circle with one diameter included, or, drawn with straight lines, it's a pair of triangles that share one edge. The first homology of this object is, in face $Bbb Z oplus Bbb Z$.
– John Hughes
Sep 6 at 11:59
tks. I just realize that and delete the question :)). What about my first ques?
– chiÌ trung châu
Sep 6 at 12:00
As for your map: even for $n = 2$, it's not simplicial, for the images of $a$ and $c$ end up being $a$ and $b$, and while $ac$ is an edge, $ab$ is not. You'll need to triangulate the domain and codomain with different numbers of vertices to make this all work out (if you're approaching it the way you seem to be).
– John Hughes
Sep 6 at 12:01
wow, seems like much more work than I thought. Thank u
– chiÌ trung châu
Sep 6 at 12:09
add a comment |Â
thank u. Can u explain why u think it would make the problem easier if we have more vertices? Personally, I think these 4 would do it since $x^n$ and $1/x$ map these four to these four, hence simplicial maps.
– chiÌ trung châu
Sep 6 at 11:55
Your complex was a circle with one diameter included, or, drawn with straight lines, it's a pair of triangles that share one edge. The first homology of this object is, in face $Bbb Z oplus Bbb Z$.
– John Hughes
Sep 6 at 11:59
tks. I just realize that and delete the question :)). What about my first ques?
– chiÌ trung châu
Sep 6 at 12:00
As for your map: even for $n = 2$, it's not simplicial, for the images of $a$ and $c$ end up being $a$ and $b$, and while $ac$ is an edge, $ab$ is not. You'll need to triangulate the domain and codomain with different numbers of vertices to make this all work out (if you're approaching it the way you seem to be).
– John Hughes
Sep 6 at 12:01
wow, seems like much more work than I thought. Thank u
– chiÌ trung châu
Sep 6 at 12:09
thank u. Can u explain why u think it would make the problem easier if we have more vertices? Personally, I think these 4 would do it since $x^n$ and $1/x$ map these four to these four, hence simplicial maps.
– chiÌ trung châu
Sep 6 at 11:55
thank u. Can u explain why u think it would make the problem easier if we have more vertices? Personally, I think these 4 would do it since $x^n$ and $1/x$ map these four to these four, hence simplicial maps.
– chiÌ trung châu
Sep 6 at 11:55
Your complex was a circle with one diameter included, or, drawn with straight lines, it's a pair of triangles that share one edge. The first homology of this object is, in face $Bbb Z oplus Bbb Z$.
– John Hughes
Sep 6 at 11:59
Your complex was a circle with one diameter included, or, drawn with straight lines, it's a pair of triangles that share one edge. The first homology of this object is, in face $Bbb Z oplus Bbb Z$.
– John Hughes
Sep 6 at 11:59
tks. I just realize that and delete the question :)). What about my first ques?
– chiÌ trung châu
Sep 6 at 12:00
tks. I just realize that and delete the question :)). What about my first ques?
– chiÌ trung châu
Sep 6 at 12:00
As for your map: even for $n = 2$, it's not simplicial, for the images of $a$ and $c$ end up being $a$ and $b$, and while $ac$ is an edge, $ab$ is not. You'll need to triangulate the domain and codomain with different numbers of vertices to make this all work out (if you're approaching it the way you seem to be).
– John Hughes
Sep 6 at 12:01
As for your map: even for $n = 2$, it's not simplicial, for the images of $a$ and $c$ end up being $a$ and $b$, and while $ac$ is an edge, $ab$ is not. You'll need to triangulate the domain and codomain with different numbers of vertices to make this all work out (if you're approaching it the way you seem to be).
– John Hughes
Sep 6 at 12:01
wow, seems like much more work than I thought. Thank u
– chiÌ trung châu
Sep 6 at 12:09
wow, seems like much more work than I thought. Thank u
– chiÌ trung châu
Sep 6 at 12:09
add a comment |Â
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