Find the fundamental group of $Bbb C^2 setminus {(x,y):xy=0 }$. [closed]
What is the fundamental group of $Bbb C^2 setminus {(x,y):xy=0 }$?
How do I proceed? Please help me in this regard.
Thank you very much.
algebraic-topology fundamental-groups
asked Nov 28 '18 at 12:08
Dbchatto67
52215
closed as off-topic by Carl Mummert, Alexander Gruber♦ Nov 30 '18 at 3:33
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." – Carl Mummert, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
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show 3 more comments
What is the fundamental group of $Bbb C^2 setminus {(x,y):xy=0 }$?
How do I proceed? Please help me in this regard.
Thank you very much.
algebraic-topology fundamental-groups
asked Nov 28 '18 at 12:08
Dbchatto67
52215
closed as off-topic by Carl Mummert, Alexander Gruber♦ Nov 30 '18 at 3:33
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." – Carl Mummert, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
Hint: $xy=0$ is equivalent to $x=0$ or $y=0$. Both describe real hyperplanes in $mathbb R^4congmathbb C^2$.
– Christoph
Nov 28 '18 at 12:19
Yeah that is correct. How do I somehow invoke here the concept of deformation retract or homotopy equivalence? Can I use Van Kampen's theorem?
– Dbchatto67
Nov 28 '18 at 12:26
Actually that is same as saying what is the fundamental group of $Bbb R^4$ without two intersecting hyperplanes which sounds much like $Bbb R^3$ without two intersecting planes upon taking projection. Right?
– Dbchatto67
Nov 28 '18 at 12:31
The description of the set you study is ambiguous: what are $x,y$ ? Complex or real numbers?
– Georges Elencwajg
Nov 28 '18 at 12:33
I have found it from an online source. This is not the question which I made myself.
– Dbchatto67
Nov 28 '18 at 12:35
|
show 3 more comments
What is the fundamental group of $Bbb C^2 setminus {(x,y):xy=0 }$?
How do I proceed? Please help me in this regard.
Thank you very much.
algebraic-topology fundamental-groups
asked Nov 28 '18 at 12:08
Dbchatto67
52215
What is the fundamental group of $Bbb C^2 setminus {(x,y):xy=0 }$?
How do I proceed? Please help me in this regard.
Thank you very much.
algebraic-topology fundamental-groups
algebraic-topology fundamental-groups
asked Nov 28 '18 at 12:08
Dbchatto67
52215
asked Nov 28 '18 at 12:08
Dbchatto67
52215
asked Nov 28 '18 at 12:08
Dbchatto67
52215
asked Nov 28 '18 at 12:08
Dbchatto67
52215
asked Nov 28 '18 at 12:08
Dbchatto67
52215
52215
closed as off-topic by Carl Mummert, Alexander Gruber♦ Nov 30 '18 at 3:33
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." – Carl Mummert, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Carl Mummert, Alexander Gruber♦ Nov 30 '18 at 3:33
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." – Carl Mummert, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
Hint: $xy=0$ is equivalent to $x=0$ or $y=0$. Both describe real hyperplanes in $mathbb R^4congmathbb C^2$.
– Christoph
Nov 28 '18 at 12:19
Yeah that is correct. How do I somehow invoke here the concept of deformation retract or homotopy equivalence? Can I use Van Kampen's theorem?
– Dbchatto67
Nov 28 '18 at 12:26
Actually that is same as saying what is the fundamental group of $Bbb R^4$ without two intersecting hyperplanes which sounds much like $Bbb R^3$ without two intersecting planes upon taking projection. Right?
– Dbchatto67
Nov 28 '18 at 12:31
The description of the set you study is ambiguous: what are $x,y$ ? Complex or real numbers?
– Georges Elencwajg
Nov 28 '18 at 12:33
I have found it from an online source. This is not the question which I made myself.
– Dbchatto67
Nov 28 '18 at 12:35
|
show 3 more comments
Hint: $xy=0$ is equivalent to $x=0$ or $y=0$. Both describe real hyperplanes in $mathbb R^4congmathbb C^2$.
– Christoph
Nov 28 '18 at 12:19
Yeah that is correct. How do I somehow invoke here the concept of deformation retract or homotopy equivalence? Can I use Van Kampen's theorem?
– Dbchatto67
Nov 28 '18 at 12:26
Actually that is same as saying what is the fundamental group of $Bbb R^4$ without two intersecting hyperplanes which sounds much like $Bbb R^3$ without two intersecting planes upon taking projection. Right?
– Dbchatto67
Nov 28 '18 at 12:31
The description of the set you study is ambiguous: what are $x,y$ ? Complex or real numbers?
– Georges Elencwajg
Nov 28 '18 at 12:33
I have found it from an online source. This is not the question which I made myself.
– Dbchatto67
Nov 28 '18 at 12:35
Hint: $xy=0$ is equivalent to $x=0$ or $y=0$. Both describe real hyperplanes in $mathbb R^4congmathbb C^2$.
– Christoph
Nov 28 '18 at 12:19
Hint: $xy=0$ is equivalent to $x=0$ or $y=0$. Both describe real hyperplanes in $mathbb R^4congmathbb C^2$.
– Christoph
Nov 28 '18 at 12:19
Yeah that is correct. How do I somehow invoke here the concept of deformation retract or homotopy equivalence? Can I use Van Kampen's theorem?
– Dbchatto67
Nov 28 '18 at 12:26
Yeah that is correct. How do I somehow invoke here the concept of deformation retract or homotopy equivalence? Can I use Van Kampen's theorem?
– Dbchatto67
Nov 28 '18 at 12:26
Actually that is same as saying what is the fundamental group of $Bbb R^4$ without two intersecting hyperplanes which sounds much like $Bbb R^3$ without two intersecting planes upon taking projection. Right?
– Dbchatto67
Nov 28 '18 at 12:31
Actually that is same as saying what is the fundamental group of $Bbb R^4$ without two intersecting hyperplanes which sounds much like $Bbb R^3$ without two intersecting planes upon taking projection. Right?
– Dbchatto67
Nov 28 '18 at 12:31
The description of the set you study is ambiguous: what are $x,y$ ? Complex or real numbers?
– Georges Elencwajg
Nov 28 '18 at 12:33
The description of the set you study is ambiguous: what are $x,y$ ? Complex or real numbers?
– Georges Elencwajg
Nov 28 '18 at 12:33
I have found it from an online source. This is not the question which I made myself.
– Dbchatto67
Nov 28 '18 at 12:35
I have found it from an online source. This is not the question which I made myself.
– Dbchatto67
Nov 28 '18 at 12:35
|
show 3 more comments
2 Answers
2
active
oldest
votes
Your space $X$ is exactly equal to $mathbb C^ast times mathbb C ^ast$ so that $$pi_1(X)= pi_1(mathbb C^ast times mathbb C ^ast)=pi_1(mathbb C^ast )times pi_1(mathbb C^ast )=mathbb Ztimes mathbb Z$$
answered Nov 28 '18 at 13:04
Georges Elencwajg
118k7180329
Much simpler, upvoted. Maybe I was drawn into $ mathbb{R}^4 $ by the first comment.
– hellHound
Nov 28 '18 at 13:20
1
Dear @hellHound: congratulations for your sense of fair-play, which is more valuable in my eyes than any mathematical calculation!
– Georges Elencwajg
Nov 28 '18 at 14:00
Dear @GeorgesElencwajg, sincere thanks for your kind words. As a student, the learning experience I take from your solution is to always think of the simplest concepts first while tackling any problem.
– hellHound
Nov 28 '18 at 16:05
add a comment |
Sketch, but almost a complete solution: First observe that topologically, the space is $ mathbb{R}^4 setminus (H_1 cup H_2) $ where $$ H_1 = {(x_1, x_2, x_3, x_4) | x_1=x_2=0 } , H_2 = {(x_1, x_2, x_3, x_4) | x_3=x_4=0 } $$ are two dimensional subspaces. This space clearly deformation retracts to $ S^3 $ minus two copies of $ S^1 $, which are given by the equations $ C_1:x_3^2 + x_4^2 = 1 $ and $ C_2:x_1^2 + x_2^2 = 1 $. Now use stereographic projection from the north pole $ (0,0,0,1) $ to see that $ C_2 $ goes to the same circle in $ mathbb{R}^3 $ while $ C_1 $ maps to the $ z$-axis of $ mathbb{R}^3 $. So finally, you need to compute the fundamental group of $ mathbb{R}^3 setminus ({ x^2 + y^2 = 1 } cup { (0,0,z) | z in mathbb{R} }) $. This space deformation retracts to a torus, whose fundamental group is $ mathbb{Z} oplus mathbb{Z} $.
answered Nov 28 '18 at 12:33
hellHound
48328
Why do you take two dimensional subspaces? Here the hyperplanes have dimension one less than the dimension of $Bbb R^4$ and hence their dimension are $3$. Am I right?
– Dbchatto67
Nov 28 '18 at 12:40
1
No. What is the dimension of the subspace $ H_1 $? It is $ 2 $, the $ x_3, x_4 $ are free variables. As for why do I take two dimensional subspaces, isn't that is what's asked in your problem? A point $ { z_1, z_2 } in mathbb{C}^2 $ where $ z_1 =0 $ is given by a point in $ H_1 $ in $ mathbb{R}^4 $.
– hellHound
Nov 28 '18 at 12:42
How do I visualise the stereographic projection in $4$-dimension?
– Dbchatto67
Nov 28 '18 at 12:49
The same way as you do in three (or two) dimensions. The formulae are identical.
– hellHound
Nov 28 '18 at 12:51
1
I found that $C_1$ maps to the $z$-axis under the formula $(x,y,z)= left (frac {x_1} {1-x_4} , frac {x_2} {1-x_4} , frac {x_3} {1-x_4} right)$ by substituting $x_1 = x_2 = 0$ and $C_2$ maps to a circle $x^2 + y^2 = 1$ by substituting $x_3=x_4=0$. Am I right now?
– Dbchatto67
Nov 28 '18 at 13:33
|
show 4 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Your space $X$ is exactly equal to $mathbb C^ast times mathbb C ^ast$ so that $$pi_1(X)= pi_1(mathbb C^ast times mathbb C ^ast)=pi_1(mathbb C^ast )times pi_1(mathbb C^ast )=mathbb Ztimes mathbb Z$$
answered Nov 28 '18 at 13:04
Georges Elencwajg
118k7180329
Much simpler, upvoted. Maybe I was drawn into $ mathbb{R}^4 $ by the first comment.
– hellHound
Nov 28 '18 at 13:20
1
Dear @hellHound: congratulations for your sense of fair-play, which is more valuable in my eyes than any mathematical calculation!
– Georges Elencwajg
Nov 28 '18 at 14:00
Dear @GeorgesElencwajg, sincere thanks for your kind words. As a student, the learning experience I take from your solution is to always think of the simplest concepts first while tackling any problem.
– hellHound
Nov 28 '18 at 16:05
add a comment |
Your space $X$ is exactly equal to $mathbb C^ast times mathbb C ^ast$ so that $$pi_1(X)= pi_1(mathbb C^ast times mathbb C ^ast)=pi_1(mathbb C^ast )times pi_1(mathbb C^ast )=mathbb Ztimes mathbb Z$$
answered Nov 28 '18 at 13:04
Georges Elencwajg
118k7180329
Much simpler, upvoted. Maybe I was drawn into $ mathbb{R}^4 $ by the first comment.
– hellHound
Nov 28 '18 at 13:20
1
Dear @hellHound: congratulations for your sense of fair-play, which is more valuable in my eyes than any mathematical calculation!
– Georges Elencwajg
Nov 28 '18 at 14:00
Dear @GeorgesElencwajg, sincere thanks for your kind words. As a student, the learning experience I take from your solution is to always think of the simplest concepts first while tackling any problem.
– hellHound
Nov 28 '18 at 16:05
add a comment |
Your space $X$ is exactly equal to $mathbb C^ast times mathbb C ^ast$ so that $$pi_1(X)= pi_1(mathbb C^ast times mathbb C ^ast)=pi_1(mathbb C^ast )times pi_1(mathbb C^ast )=mathbb Ztimes mathbb Z$$
answered Nov 28 '18 at 13:04
Georges Elencwajg
118k7180329
Your space $X$ is exactly equal to $mathbb C^ast times mathbb C ^ast$ so that $$pi_1(X)= pi_1(mathbb C^ast times mathbb C ^ast)=pi_1(mathbb C^ast )times pi_1(mathbb C^ast )=mathbb Ztimes mathbb Z$$
answered Nov 28 '18 at 13:04
Georges Elencwajg
118k7180329
answered Nov 28 '18 at 13:04
Georges Elencwajg
118k7180329
answered Nov 28 '18 at 13:04
Georges Elencwajg
118k7180329
answered Nov 28 '18 at 13:04
Georges Elencwajg
118k7180329
118k7180329
Much simpler, upvoted. Maybe I was drawn into $ mathbb{R}^4 $ by the first comment.
– hellHound
Nov 28 '18 at 13:20
1
Dear @hellHound: congratulations for your sense of fair-play, which is more valuable in my eyes than any mathematical calculation!
– Georges Elencwajg
Nov 28 '18 at 14:00
Dear @GeorgesElencwajg, sincere thanks for your kind words. As a student, the learning experience I take from your solution is to always think of the simplest concepts first while tackling any problem.
– hellHound
Nov 28 '18 at 16:05
add a comment |
Much simpler, upvoted. Maybe I was drawn into $ mathbb{R}^4 $ by the first comment.
– hellHound
Nov 28 '18 at 13:20
1
Dear @hellHound: congratulations for your sense of fair-play, which is more valuable in my eyes than any mathematical calculation!
– Georges Elencwajg
Nov 28 '18 at 14:00
Dear @GeorgesElencwajg, sincere thanks for your kind words. As a student, the learning experience I take from your solution is to always think of the simplest concepts first while tackling any problem.
– hellHound
Nov 28 '18 at 16:05
Much simpler, upvoted. Maybe I was drawn into $ mathbb{R}^4 $ by the first comment.
– hellHound
Nov 28 '18 at 13:20
Much simpler, upvoted. Maybe I was drawn into $ mathbb{R}^4 $ by the first comment.
– hellHound
Nov 28 '18 at 13:20
1
1
Dear @hellHound: congratulations for your sense of fair-play, which is more valuable in my eyes than any mathematical calculation!
– Georges Elencwajg
Nov 28 '18 at 14:00
Dear @hellHound: congratulations for your sense of fair-play, which is more valuable in my eyes than any mathematical calculation!
– Georges Elencwajg
Nov 28 '18 at 14:00
Dear @GeorgesElencwajg, sincere thanks for your kind words. As a student, the learning experience I take from your solution is to always think of the simplest concepts first while tackling any problem.
– hellHound
Nov 28 '18 at 16:05
Dear @GeorgesElencwajg, sincere thanks for your kind words. As a student, the learning experience I take from your solution is to always think of the simplest concepts first while tackling any problem.
– hellHound
Nov 28 '18 at 16:05
add a comment |
Sketch, but almost a complete solution: First observe that topologically, the space is $ mathbb{R}^4 setminus (H_1 cup H_2) $ where $$ H_1 = {(x_1, x_2, x_3, x_4) | x_1=x_2=0 } , H_2 = {(x_1, x_2, x_3, x_4) | x_3=x_4=0 } $$ are two dimensional subspaces. This space clearly deformation retracts to $ S^3 $ minus two copies of $ S^1 $, which are given by the equations $ C_1:x_3^2 + x_4^2 = 1 $ and $ C_2:x_1^2 + x_2^2 = 1 $. Now use stereographic projection from the north pole $ (0,0,0,1) $ to see that $ C_2 $ goes to the same circle in $ mathbb{R}^3 $ while $ C_1 $ maps to the $ z$-axis of $ mathbb{R}^3 $. So finally, you need to compute the fundamental group of $ mathbb{R}^3 setminus ({ x^2 + y^2 = 1 } cup { (0,0,z) | z in mathbb{R} }) $. This space deformation retracts to a torus, whose fundamental group is $ mathbb{Z} oplus mathbb{Z} $.
answered Nov 28 '18 at 12:33
hellHound
48328
Why do you take two dimensional subspaces? Here the hyperplanes have dimension one less than the dimension of $Bbb R^4$ and hence their dimension are $3$. Am I right?
– Dbchatto67
Nov 28 '18 at 12:40
1
No. What is the dimension of the subspace $ H_1 $? It is $ 2 $, the $ x_3, x_4 $ are free variables. As for why do I take two dimensional subspaces, isn't that is what's asked in your problem? A point $ { z_1, z_2 } in mathbb{C}^2 $ where $ z_1 =0 $ is given by a point in $ H_1 $ in $ mathbb{R}^4 $.
– hellHound
Nov 28 '18 at 12:42
How do I visualise the stereographic projection in $4$-dimension?
– Dbchatto67
Nov 28 '18 at 12:49
The same way as you do in three (or two) dimensions. The formulae are identical.
– hellHound
Nov 28 '18 at 12:51
1
I found that $C_1$ maps to the $z$-axis under the formula $(x,y,z)= left (frac {x_1} {1-x_4} , frac {x_2} {1-x_4} , frac {x_3} {1-x_4} right)$ by substituting $x_1 = x_2 = 0$ and $C_2$ maps to a circle $x^2 + y^2 = 1$ by substituting $x_3=x_4=0$. Am I right now?
– Dbchatto67
Nov 28 '18 at 13:33
|
show 4 more comments
Sketch, but almost a complete solution: First observe that topologically, the space is $ mathbb{R}^4 setminus (H_1 cup H_2) $ where $$ H_1 = {(x_1, x_2, x_3, x_4) | x_1=x_2=0 } , H_2 = {(x_1, x_2, x_3, x_4) | x_3=x_4=0 } $$ are two dimensional subspaces. This space clearly deformation retracts to $ S^3 $ minus two copies of $ S^1 $, which are given by the equations $ C_1:x_3^2 + x_4^2 = 1 $ and $ C_2:x_1^2 + x_2^2 = 1 $. Now use stereographic projection from the north pole $ (0,0,0,1) $ to see that $ C_2 $ goes to the same circle in $ mathbb{R}^3 $ while $ C_1 $ maps to the $ z$-axis of $ mathbb{R}^3 $. So finally, you need to compute the fundamental group of $ mathbb{R}^3 setminus ({ x^2 + y^2 = 1 } cup { (0,0,z) | z in mathbb{R} }) $. This space deformation retracts to a torus, whose fundamental group is $ mathbb{Z} oplus mathbb{Z} $.
answered Nov 28 '18 at 12:33
hellHound
48328
Why do you take two dimensional subspaces? Here the hyperplanes have dimension one less than the dimension of $Bbb R^4$ and hence their dimension are $3$. Am I right?
– Dbchatto67
Nov 28 '18 at 12:40
1
No. What is the dimension of the subspace $ H_1 $? It is $ 2 $, the $ x_3, x_4 $ are free variables. As for why do I take two dimensional subspaces, isn't that is what's asked in your problem? A point $ { z_1, z_2 } in mathbb{C}^2 $ where $ z_1 =0 $ is given by a point in $ H_1 $ in $ mathbb{R}^4 $.
– hellHound
Nov 28 '18 at 12:42
How do I visualise the stereographic projection in $4$-dimension?
– Dbchatto67
Nov 28 '18 at 12:49
The same way as you do in three (or two) dimensions. The formulae are identical.
– hellHound
Nov 28 '18 at 12:51
1
I found that $C_1$ maps to the $z$-axis under the formula $(x,y,z)= left (frac {x_1} {1-x_4} , frac {x_2} {1-x_4} , frac {x_3} {1-x_4} right)$ by substituting $x_1 = x_2 = 0$ and $C_2$ maps to a circle $x^2 + y^2 = 1$ by substituting $x_3=x_4=0$. Am I right now?
– Dbchatto67
Nov 28 '18 at 13:33
|
show 4 more comments
Sketch, but almost a complete solution: First observe that topologically, the space is $ mathbb{R}^4 setminus (H_1 cup H_2) $ where $$ H_1 = {(x_1, x_2, x_3, x_4) | x_1=x_2=0 } , H_2 = {(x_1, x_2, x_3, x_4) | x_3=x_4=0 } $$ are two dimensional subspaces. This space clearly deformation retracts to $ S^3 $ minus two copies of $ S^1 $, which are given by the equations $ C_1:x_3^2 + x_4^2 = 1 $ and $ C_2:x_1^2 + x_2^2 = 1 $. Now use stereographic projection from the north pole $ (0,0,0,1) $ to see that $ C_2 $ goes to the same circle in $ mathbb{R}^3 $ while $ C_1 $ maps to the $ z$-axis of $ mathbb{R}^3 $. So finally, you need to compute the fundamental group of $ mathbb{R}^3 setminus ({ x^2 + y^2 = 1 } cup { (0,0,z) | z in mathbb{R} }) $. This space deformation retracts to a torus, whose fundamental group is $ mathbb{Z} oplus mathbb{Z} $.
answered Nov 28 '18 at 12:33
hellHound
48328
Sketch, but almost a complete solution: First observe that topologically, the space is $ mathbb{R}^4 setminus (H_1 cup H_2) $ where $$ H_1 = {(x_1, x_2, x_3, x_4) | x_1=x_2=0 } , H_2 = {(x_1, x_2, x_3, x_4) | x_3=x_4=0 } $$ are two dimensional subspaces. This space clearly deformation retracts to $ S^3 $ minus two copies of $ S^1 $, which are given by the equations $ C_1:x_3^2 + x_4^2 = 1 $ and $ C_2:x_1^2 + x_2^2 = 1 $. Now use stereographic projection from the north pole $ (0,0,0,1) $ to see that $ C_2 $ goes to the same circle in $ mathbb{R}^3 $ while $ C_1 $ maps to the $ z$-axis of $ mathbb{R}^3 $. So finally, you need to compute the fundamental group of $ mathbb{R}^3 setminus ({ x^2 + y^2 = 1 } cup { (0,0,z) | z in mathbb{R} }) $. This space deformation retracts to a torus, whose fundamental group is $ mathbb{Z} oplus mathbb{Z} $.
answered Nov 28 '18 at 12:33
hellHound
48328
edited Nov 28 '18 at 15:53
answered Nov 28 '18 at 12:33
hellHound
48328
answered Nov 28 '18 at 12:33
hellHound
48328
answered Nov 28 '18 at 12:33
hellHound
48328
48328
Why do you take two dimensional subspaces? Here the hyperplanes have dimension one less than the dimension of $Bbb R^4$ and hence their dimension are $3$. Am I right?
– Dbchatto67
Nov 28 '18 at 12:40
1
No. What is the dimension of the subspace $ H_1 $? It is $ 2 $, the $ x_3, x_4 $ are free variables. As for why do I take two dimensional subspaces, isn't that is what's asked in your problem? A point $ { z_1, z_2 } in mathbb{C}^2 $ where $ z_1 =0 $ is given by a point in $ H_1 $ in $ mathbb{R}^4 $.
– hellHound
Nov 28 '18 at 12:42
How do I visualise the stereographic projection in $4$-dimension?
– Dbchatto67
Nov 28 '18 at 12:49
The same way as you do in three (or two) dimensions. The formulae are identical.
– hellHound
Nov 28 '18 at 12:51
1
I found that $C_1$ maps to the $z$-axis under the formula $(x,y,z)= left (frac {x_1} {1-x_4} , frac {x_2} {1-x_4} , frac {x_3} {1-x_4} right)$ by substituting $x_1 = x_2 = 0$ and $C_2$ maps to a circle $x^2 + y^2 = 1$ by substituting $x_3=x_4=0$. Am I right now?
– Dbchatto67
Nov 28 '18 at 13:33
|
show 4 more comments
Why do you take two dimensional subspaces? Here the hyperplanes have dimension one less than the dimension of $Bbb R^4$ and hence their dimension are $3$. Am I right?
– Dbchatto67
Nov 28 '18 at 12:40
1
No. What is the dimension of the subspace $ H_1 $? It is $ 2 $, the $ x_3, x_4 $ are free variables. As for why do I take two dimensional subspaces, isn't that is what's asked in your problem? A point $ { z_1, z_2 } in mathbb{C}^2 $ where $ z_1 =0 $ is given by a point in $ H_1 $ in $ mathbb{R}^4 $.
– hellHound
Nov 28 '18 at 12:42
How do I visualise the stereographic projection in $4$-dimension?
– Dbchatto67
Nov 28 '18 at 12:49
The same way as you do in three (or two) dimensions. The formulae are identical.
– hellHound
Nov 28 '18 at 12:51
1
I found that $C_1$ maps to the $z$-axis under the formula $(x,y,z)= left (frac {x_1} {1-x_4} , frac {x_2} {1-x_4} , frac {x_3} {1-x_4} right)$ by substituting $x_1 = x_2 = 0$ and $C_2$ maps to a circle $x^2 + y^2 = 1$ by substituting $x_3=x_4=0$. Am I right now?
– Dbchatto67
Nov 28 '18 at 13:33
Why do you take two dimensional subspaces? Here the hyperplanes have dimension one less than the dimension of $Bbb R^4$ and hence their dimension are $3$. Am I right?
– Dbchatto67
Nov 28 '18 at 12:40
Why do you take two dimensional subspaces? Here the hyperplanes have dimension one less than the dimension of $Bbb R^4$ and hence their dimension are $3$. Am I right?
– Dbchatto67
Nov 28 '18 at 12:40
1
1
No. What is the dimension of the subspace $ H_1 $? It is $ 2 $, the $ x_3, x_4 $ are free variables. As for why do I take two dimensional subspaces, isn't that is what's asked in your problem? A point $ { z_1, z_2 } in mathbb{C}^2 $ where $ z_1 =0 $ is given by a point in $ H_1 $ in $ mathbb{R}^4 $.
– hellHound
Nov 28 '18 at 12:42
No. What is the dimension of the subspace $ H_1 $? It is $ 2 $, the $ x_3, x_4 $ are free variables. As for why do I take two dimensional subspaces, isn't that is what's asked in your problem? A point $ { z_1, z_2 } in mathbb{C}^2 $ where $ z_1 =0 $ is given by a point in $ H_1 $ in $ mathbb{R}^4 $.
– hellHound
Nov 28 '18 at 12:42
How do I visualise the stereographic projection in $4$-dimension?
– Dbchatto67
Nov 28 '18 at 12:49
How do I visualise the stereographic projection in $4$-dimension?
– Dbchatto67
Nov 28 '18 at 12:49
The same way as you do in three (or two) dimensions. The formulae are identical.
– hellHound
Nov 28 '18 at 12:51
The same way as you do in three (or two) dimensions. The formulae are identical.
– hellHound
Nov 28 '18 at 12:51
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I found that $C_1$ maps to the $z$-axis under the formula $(x,y,z)= left (frac {x_1} {1-x_4} , frac {x_2} {1-x_4} , frac {x_3} {1-x_4} right)$ by substituting $x_1 = x_2 = 0$ and $C_2$ maps to a circle $x^2 + y^2 = 1$ by substituting $x_3=x_4=0$. Am I right now?
– Dbchatto67
Nov 28 '18 at 13:33
I found that $C_1$ maps to the $z$-axis under the formula $(x,y,z)= left (frac {x_1} {1-x_4} , frac {x_2} {1-x_4} , frac {x_3} {1-x_4} right)$ by substituting $x_1 = x_2 = 0$ and $C_2$ maps to a circle $x^2 + y^2 = 1$ by substituting $x_3=x_4=0$. Am I right now?
– Dbchatto67
Nov 28 '18 at 13:33
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show 4 more comments
Hint: $xy=0$ is equivalent to $x=0$ or $y=0$. Both describe real hyperplanes in $mathbb R^4congmathbb C^2$.
– Christoph
Nov 28 '18 at 12:19
Yeah that is correct. How do I somehow invoke here the concept of deformation retract or homotopy equivalence? Can I use Van Kampen's theorem?
– Dbchatto67
Nov 28 '18 at 12:26
Actually that is same as saying what is the fundamental group of $Bbb R^4$ without two intersecting hyperplanes which sounds much like $Bbb R^3$ without two intersecting planes upon taking projection. Right?
– Dbchatto67
Nov 28 '18 at 12:31
The description of the set you study is ambiguous: what are $x,y$ ? Complex or real numbers?
– Georges Elencwajg
Nov 28 '18 at 12:33
I have found it from an online source. This is not the question which I made myself.
– Dbchatto67
Nov 28 '18 at 12:35