$mathbb CP^1 approx S^2$ proof checkFlag manifold to Complex Projective lineDoes the completeness of a normed vector space only depend on its topology?existence continuous function $f:Xto [0,1]$ such that $f(x_1)=0 , f(x_2)=frac12, f(x_3)=1$Verification of a proof regarding the connected sum of two surfacesProb. 2, Sec. 20 in Munkres' TOPOLOGY, 2nd ed: The dictionary order topology on $mathbbR times mathbbR$ is metrizable.Easy proof of $A(x)$ is bounded under some condition.Prob. 5, Sec. 20 in Munkres' TOPOLOGY, 2nd ed: What is the closure of $mathbbR^infty$ in $mathbbR^omega$ in the uniform topology?Following a proof of “every semimetric space is normal and completely regular”Continuity of a linear functional on various spaces.Every two norms on a finite dimensional (real or complex) vector space $V$ are equivalent.show that $U(x,epsilon)$ is not even open in the uniform topology.
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$mathbb CP^1 approx S^2$ proof check
Flag manifold to Complex Projective lineDoes the completeness of a normed vector space only depend on its topology?existence continuous function $f:Xto [0,1]$ such that $f(x_1)=0 , f(x_2)=frac12, f(x_3)=1$Verification of a proof regarding the connected sum of two surfacesProb. 2, Sec. 20 in Munkres' TOPOLOGY, 2nd ed: The dictionary order topology on $mathbbR times mathbbR$ is metrizable.Easy proof of $A(x)$ is bounded under some condition.Prob. 5, Sec. 20 in Munkres' TOPOLOGY, 2nd ed: What is the closure of $mathbbR^infty$ in $mathbbR^omega$ in the uniform topology?Following a proof of “every semimetric space is normal and completely regular”Continuity of a linear functional on various spaces.Every two norms on a finite dimensional (real or complex) vector space $V$ are equivalent.show that $U(x,epsilon)$ is not even open in the uniform topology.
$begingroup$
I wanted to give a whole proof of this fact as I was not able to find a detailed one myself. I have the feeling that such a proof has been asked quite frequently by several users and I hope this may help other students as me who are touching on this topic for the first time.
For $mathbb CP^1$ we take the definition $(mathbb C^2-0)/(ztildelambda z)$ for any $lambda$ non zero complex number.
Define $$F: mathbb R^4 to S^2/tilde_antipodal, space (x_1,x_2,x_3,x_4) mapsto [frac1Vert(x_1,x_2,x_3)Vert_2(x_1,x_2,x_3)] $$
Observe that $$F(lambda x)=[fraclambdavert lambda vert Vert(x_1,x_2,x_3)Vert_2(x_1,x_2,x_3)]=stackrel+-frac1Vert(x_1,x_2,x_3)Vert_2(x_1,x_2,x_3)=[frac1Vert(x_1,x_2,x_3)Vert_2(x_1,x_2,x_3)]=F(x)$$
hence $F$ descend to $tildeF: mathbb R P^3 to S^2/tilde_antipodal$ and is continuous since $F$ is continuous. $tildeF$ is obviously surjective and also injective since $tildeF([x])=tildeF([y]) implies x spacetilde_antipodalspace y$ and therefore $x=-y implies [x]=[y]$. The domain is compact and the target is Hausdorff so we obtain $$mathbb R P^3 approx S^2 / tilde_antipodal$$By some similiar argument one can show that $S^2 approx S^2/tilde_antipodal$ and hence $mathbb R P^3 approx S^2$ and so we conclude $$mathbb C P^1 approx S^2$$
Question:
Is this approach of viewing $mathbb C P^1$ as $mathbb R P^3$ right? Is the argument right, are there any flaws?
general-topology geometry algebraic-topology proof-verification alternative-proof
asked Feb 2 '16 at 12:52
noctusraidnoctusraid
8691727
$endgroup$
|
show 4 more comments
$begingroup$
I wanted to give a whole proof of this fact as I was not able to find a detailed one myself. I have the feeling that such a proof has been asked quite frequently by several users and I hope this may help other students as me who are touching on this topic for the first time.
For $mathbb CP^1$ we take the definition $(mathbb C^2-0)/(ztildelambda z)$ for any $lambda$ non zero complex number.
Define $$F: mathbb R^4 to S^2/tilde_antipodal, space (x_1,x_2,x_3,x_4) mapsto [frac1Vert(x_1,x_2,x_3)Vert_2(x_1,x_2,x_3)] $$
Observe that $$F(lambda x)=[fraclambdavert lambda vert Vert(x_1,x_2,x_3)Vert_2(x_1,x_2,x_3)]=stackrel+-frac1Vert(x_1,x_2,x_3)Vert_2(x_1,x_2,x_3)=[frac1Vert(x_1,x_2,x_3)Vert_2(x_1,x_2,x_3)]=F(x)$$
hence $F$ descend to $tildeF: mathbb R P^3 to S^2/tilde_antipodal$ and is continuous since $F$ is continuous. $tildeF$ is obviously surjective and also injective since $tildeF([x])=tildeF([y]) implies x spacetilde_antipodalspace y$ and therefore $x=-y implies [x]=[y]$. The domain is compact and the target is Hausdorff so we obtain $$mathbb R P^3 approx S^2 / tilde_antipodal$$By some similiar argument one can show that $S^2 approx S^2/tilde_antipodal$ and hence $mathbb R P^3 approx S^2$ and so we conclude $$mathbb C P^1 approx S^2$$
Question:
Is this approach of viewing $mathbb C P^1$ as $mathbb R P^3$ right? Is the argument right, are there any flaws?
general-topology geometry algebraic-topology proof-verification alternative-proof
asked Feb 2 '16 at 12:52
noctusraidnoctusraid
8691727
$endgroup$
1
$begingroup$
There is definitely a problem with this: $mathbbRP^2$ is not homeomorphic to $S^2$.
$endgroup$
– Simon Rose
Feb 2 '16 at 13:23
$begingroup$
I think something wrong is going on here. RP2 is not homeomorphic to the sphere. Is not even orientable
$endgroup$
– Riccardo
Feb 2 '16 at 13:25
$begingroup$
Oh wait I meant $mathbb R P^3$.
$endgroup$
– noctusraid
Feb 2 '16 at 13:25
2
$begingroup$
$mathbbRP^3$ is three dimensional. $mathbbCP^1$ is two-dimensional.
$endgroup$
– Cheerful Parsnip
Feb 2 '16 at 13:29
1
$begingroup$
there is something wrong here as well. I mean $mathbbRP^3$ has fundamental group $mathbbZ/2mathbbZ$, but $S^2$ is simply connected
$endgroup$
– Riccardo
Feb 2 '16 at 13:31
|
show 4 more comments
$begingroup$
I wanted to give a whole proof of this fact as I was not able to find a detailed one myself. I have the feeling that such a proof has been asked quite frequently by several users and I hope this may help other students as me who are touching on this topic for the first time.
For $mathbb CP^1$ we take the definition $(mathbb C^2-0)/(ztildelambda z)$ for any $lambda$ non zero complex number.
Define $$F: mathbb R^4 to S^2/tilde_antipodal, space (x_1,x_2,x_3,x_4) mapsto [frac1Vert(x_1,x_2,x_3)Vert_2(x_1,x_2,x_3)] $$
Observe that $$F(lambda x)=[fraclambdavert lambda vert Vert(x_1,x_2,x_3)Vert_2(x_1,x_2,x_3)]=stackrel+-frac1Vert(x_1,x_2,x_3)Vert_2(x_1,x_2,x_3)=[frac1Vert(x_1,x_2,x_3)Vert_2(x_1,x_2,x_3)]=F(x)$$
hence $F$ descend to $tildeF: mathbb R P^3 to S^2/tilde_antipodal$ and is continuous since $F$ is continuous. $tildeF$ is obviously surjective and also injective since $tildeF([x])=tildeF([y]) implies x spacetilde_antipodalspace y$ and therefore $x=-y implies [x]=[y]$. The domain is compact and the target is Hausdorff so we obtain $$mathbb R P^3 approx S^2 / tilde_antipodal$$By some similiar argument one can show that $S^2 approx S^2/tilde_antipodal$ and hence $mathbb R P^3 approx S^2$ and so we conclude $$mathbb C P^1 approx S^2$$
Question:
Is this approach of viewing $mathbb C P^1$ as $mathbb R P^3$ right? Is the argument right, are there any flaws?
general-topology geometry algebraic-topology proof-verification alternative-proof
asked Feb 2 '16 at 12:52
noctusraidnoctusraid
8691727
$endgroup$
I wanted to give a whole proof of this fact as I was not able to find a detailed one myself. I have the feeling that such a proof has been asked quite frequently by several users and I hope this may help other students as me who are touching on this topic for the first time.
For $mathbb CP^1$ we take the definition $(mathbb C^2-0)/(ztildelambda z)$ for any $lambda$ non zero complex number.
Define $$F: mathbb R^4 to S^2/tilde_antipodal, space (x_1,x_2,x_3,x_4) mapsto [frac1Vert(x_1,x_2,x_3)Vert_2(x_1,x_2,x_3)] $$
Observe that $$F(lambda x)=[fraclambdavert lambda vert Vert(x_1,x_2,x_3)Vert_2(x_1,x_2,x_3)]=stackrel+-frac1Vert(x_1,x_2,x_3)Vert_2(x_1,x_2,x_3)=[frac1Vert(x_1,x_2,x_3)Vert_2(x_1,x_2,x_3)]=F(x)$$
hence $F$ descend to $tildeF: mathbb R P^3 to S^2/tilde_antipodal$ and is continuous since $F$ is continuous. $tildeF$ is obviously surjective and also injective since $tildeF([x])=tildeF([y]) implies x spacetilde_antipodalspace y$ and therefore $x=-y implies [x]=[y]$. The domain is compact and the target is Hausdorff so we obtain $$mathbb R P^3 approx S^2 / tilde_antipodal$$By some similiar argument one can show that $S^2 approx S^2/tilde_antipodal$ and hence $mathbb R P^3 approx S^2$ and so we conclude $$mathbb C P^1 approx S^2$$
Question:
Is this approach of viewing $mathbb C P^1$ as $mathbb R P^3$ right? Is the argument right, are there any flaws?
general-topology geometry algebraic-topology proof-verification alternative-proof
general-topology geometry algebraic-topology proof-verification alternative-proof
asked Feb 2 '16 at 12:52
noctusraidnoctusraid
8691727
asked Feb 2 '16 at 12:52
noctusraidnoctusraid
8691727
edited Feb 2 '16 at 13:27
noctusraid
asked Feb 2 '16 at 12:52
noctusraidnoctusraid
8691727
asked Feb 2 '16 at 12:52
noctusraidnoctusraid
8691727
asked Feb 2 '16 at 12:52
noctusraidnoctusraid
8691727
8691727
1
$begingroup$
There is definitely a problem with this: $mathbbRP^2$ is not homeomorphic to $S^2$.
$endgroup$
– Simon Rose
Feb 2 '16 at 13:23
$begingroup$
I think something wrong is going on here. RP2 is not homeomorphic to the sphere. Is not even orientable
$endgroup$
– Riccardo
Feb 2 '16 at 13:25
$begingroup$
Oh wait I meant $mathbb R P^3$.
$endgroup$
– noctusraid
Feb 2 '16 at 13:25
2
$begingroup$
$mathbbRP^3$ is three dimensional. $mathbbCP^1$ is two-dimensional.
$endgroup$
– Cheerful Parsnip
Feb 2 '16 at 13:29
1
$begingroup$
there is something wrong here as well. I mean $mathbbRP^3$ has fundamental group $mathbbZ/2mathbbZ$, but $S^2$ is simply connected
$endgroup$
– Riccardo
Feb 2 '16 at 13:31
|
show 4 more comments
1
$begingroup$
There is definitely a problem with this: $mathbbRP^2$ is not homeomorphic to $S^2$.
$endgroup$
– Simon Rose
Feb 2 '16 at 13:23
$begingroup$
I think something wrong is going on here. RP2 is not homeomorphic to the sphere. Is not even orientable
$endgroup$
– Riccardo
Feb 2 '16 at 13:25
$begingroup$
Oh wait I meant $mathbb R P^3$.
$endgroup$
– noctusraid
Feb 2 '16 at 13:25
2
$begingroup$
$mathbbRP^3$ is three dimensional. $mathbbCP^1$ is two-dimensional.
$endgroup$
– Cheerful Parsnip
Feb 2 '16 at 13:29
1
$begingroup$
there is something wrong here as well. I mean $mathbbRP^3$ has fundamental group $mathbbZ/2mathbbZ$, but $S^2$ is simply connected
$endgroup$
– Riccardo
Feb 2 '16 at 13:31
1
1
$begingroup$
There is definitely a problem with this: $mathbbRP^2$ is not homeomorphic to $S^2$.
$endgroup$
– Simon Rose
Feb 2 '16 at 13:23
$begingroup$
There is definitely a problem with this: $mathbbRP^2$ is not homeomorphic to $S^2$.
$endgroup$
– Simon Rose
Feb 2 '16 at 13:23
$begingroup$
I think something wrong is going on here. RP2 is not homeomorphic to the sphere. Is not even orientable
$endgroup$
– Riccardo
Feb 2 '16 at 13:25
$begingroup$
I think something wrong is going on here. RP2 is not homeomorphic to the sphere. Is not even orientable
$endgroup$
– Riccardo
Feb 2 '16 at 13:25
$begingroup$
Oh wait I meant $mathbb R P^3$.
$endgroup$
– noctusraid
Feb 2 '16 at 13:25
$begingroup$
Oh wait I meant $mathbb R P^3$.
$endgroup$
– noctusraid
Feb 2 '16 at 13:25
2
2
$begingroup$
$mathbbRP^3$ is three dimensional. $mathbbCP^1$ is two-dimensional.
$endgroup$
– Cheerful Parsnip
Feb 2 '16 at 13:29
$begingroup$
$mathbbRP^3$ is three dimensional. $mathbbCP^1$ is two-dimensional.
$endgroup$
– Cheerful Parsnip
Feb 2 '16 at 13:29
1
1
$begingroup$
there is something wrong here as well. I mean $mathbbRP^3$ has fundamental group $mathbbZ/2mathbbZ$, but $S^2$ is simply connected
$endgroup$
– Riccardo
Feb 2 '16 at 13:31
$begingroup$
there is something wrong here as well. I mean $mathbbRP^3$ has fundamental group $mathbbZ/2mathbbZ$, but $S^2$ is simply connected
$endgroup$
– Riccardo
Feb 2 '16 at 13:31
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Personally, the way I think of this is to look at the following decomposition of $mathbbC^2$ into two sets, and see how they fit together into the quotient.
Let $U = (x, y) in mathbbC^2 mid y neq 0$ and let $P = (x, 0) in mathbbC^2$. Note that these are disjoint and their union is all of $mathbbC^2$.
When we quotient $P$ by the action of $mathbbC^times$, we get a single point (since $P$ can be easily identified with $mathbbC^times$ in a way that preserves the action). When we quotient $U$ by the action of $mathbbC^times$, one can show that it is homeomorphic to $mathbbC cong mathbbR^2$. So the resulting quotient can be seen as the union of the plane with one point.
Can you see where to go from there?
answered Feb 2 '16 at 13:27
Simon RoseSimon Rose
5,6601014
$endgroup$
$begingroup$
Are you heading towards the stereographic projection?
$endgroup$
– noctusraid
Feb 2 '16 at 13:28
$begingroup$
Basically, yes.
$endgroup$
– Simon Rose
Feb 2 '16 at 13:29
$begingroup$
Well we can extend the stereographic projection $p: S^2-0 to mathbb R^2$ to $tildep: S^2 to mathbb R^2 cup infty$ homeomorphically.
$endgroup$
– noctusraid
Feb 2 '16 at 13:37
add a comment |
$begingroup$
Considering the CW complex structure of $Bbb C P^1$ , note that $Bbb C P^1$ is obtained from $Bbb C P^0$ by attaching a $2-cell$ but $Bbb C P^0 cong *$ , thus $$Bbb C P^1 cong D^2 / partial D^2 cong S^2 $$
answered Mar 20 at 16:41
reflexivereflexive
1,183625
$endgroup$
add a comment |
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2 Answers
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votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Personally, the way I think of this is to look at the following decomposition of $mathbbC^2$ into two sets, and see how they fit together into the quotient.
Let $U = (x, y) in mathbbC^2 mid y neq 0$ and let $P = (x, 0) in mathbbC^2$. Note that these are disjoint and their union is all of $mathbbC^2$.
When we quotient $P$ by the action of $mathbbC^times$, we get a single point (since $P$ can be easily identified with $mathbbC^times$ in a way that preserves the action). When we quotient $U$ by the action of $mathbbC^times$, one can show that it is homeomorphic to $mathbbC cong mathbbR^2$. So the resulting quotient can be seen as the union of the plane with one point.
Can you see where to go from there?
answered Feb 2 '16 at 13:27
Simon RoseSimon Rose
5,6601014
$endgroup$
$begingroup$
Are you heading towards the stereographic projection?
$endgroup$
– noctusraid
Feb 2 '16 at 13:28
$begingroup$
Basically, yes.
$endgroup$
– Simon Rose
Feb 2 '16 at 13:29
$begingroup$
Well we can extend the stereographic projection $p: S^2-0 to mathbb R^2$ to $tildep: S^2 to mathbb R^2 cup infty$ homeomorphically.
$endgroup$
– noctusraid
Feb 2 '16 at 13:37
add a comment |
$begingroup$
Personally, the way I think of this is to look at the following decomposition of $mathbbC^2$ into two sets, and see how they fit together into the quotient.
Let $U = (x, y) in mathbbC^2 mid y neq 0$ and let $P = (x, 0) in mathbbC^2$. Note that these are disjoint and their union is all of $mathbbC^2$.
When we quotient $P$ by the action of $mathbbC^times$, we get a single point (since $P$ can be easily identified with $mathbbC^times$ in a way that preserves the action). When we quotient $U$ by the action of $mathbbC^times$, one can show that it is homeomorphic to $mathbbC cong mathbbR^2$. So the resulting quotient can be seen as the union of the plane with one point.
Can you see where to go from there?
answered Feb 2 '16 at 13:27
Simon RoseSimon Rose
5,6601014
$endgroup$
$begingroup$
Are you heading towards the stereographic projection?
$endgroup$
– noctusraid
Feb 2 '16 at 13:28
$begingroup$
Basically, yes.
$endgroup$
– Simon Rose
Feb 2 '16 at 13:29
$begingroup$
Well we can extend the stereographic projection $p: S^2-0 to mathbb R^2$ to $tildep: S^2 to mathbb R^2 cup infty$ homeomorphically.
$endgroup$
– noctusraid
Feb 2 '16 at 13:37
add a comment |
$begingroup$
Personally, the way I think of this is to look at the following decomposition of $mathbbC^2$ into two sets, and see how they fit together into the quotient.
Let $U = (x, y) in mathbbC^2 mid y neq 0$ and let $P = (x, 0) in mathbbC^2$. Note that these are disjoint and their union is all of $mathbbC^2$.
When we quotient $P$ by the action of $mathbbC^times$, we get a single point (since $P$ can be easily identified with $mathbbC^times$ in a way that preserves the action). When we quotient $U$ by the action of $mathbbC^times$, one can show that it is homeomorphic to $mathbbC cong mathbbR^2$. So the resulting quotient can be seen as the union of the plane with one point.
Can you see where to go from there?
answered Feb 2 '16 at 13:27
Simon RoseSimon Rose
5,6601014
$endgroup$
Personally, the way I think of this is to look at the following decomposition of $mathbbC^2$ into two sets, and see how they fit together into the quotient.
Let $U = (x, y) in mathbbC^2 mid y neq 0$ and let $P = (x, 0) in mathbbC^2$. Note that these are disjoint and their union is all of $mathbbC^2$.
When we quotient $P$ by the action of $mathbbC^times$, we get a single point (since $P$ can be easily identified with $mathbbC^times$ in a way that preserves the action). When we quotient $U$ by the action of $mathbbC^times$, one can show that it is homeomorphic to $mathbbC cong mathbbR^2$. So the resulting quotient can be seen as the union of the plane with one point.
Can you see where to go from there?
answered Feb 2 '16 at 13:27
Simon RoseSimon Rose
5,6601014
answered Feb 2 '16 at 13:27
Simon RoseSimon Rose
5,6601014
answered Feb 2 '16 at 13:27
Simon RoseSimon Rose
5,6601014
answered Feb 2 '16 at 13:27
Simon RoseSimon Rose
5,6601014
5,6601014
$begingroup$
Are you heading towards the stereographic projection?
$endgroup$
– noctusraid
Feb 2 '16 at 13:28
$begingroup$
Basically, yes.
$endgroup$
– Simon Rose
Feb 2 '16 at 13:29
$begingroup$
Well we can extend the stereographic projection $p: S^2-0 to mathbb R^2$ to $tildep: S^2 to mathbb R^2 cup infty$ homeomorphically.
$endgroup$
– noctusraid
Feb 2 '16 at 13:37
add a comment |
$begingroup$
Are you heading towards the stereographic projection?
$endgroup$
– noctusraid
Feb 2 '16 at 13:28
$begingroup$
Basically, yes.
$endgroup$
– Simon Rose
Feb 2 '16 at 13:29
$begingroup$
Well we can extend the stereographic projection $p: S^2-0 to mathbb R^2$ to $tildep: S^2 to mathbb R^2 cup infty$ homeomorphically.
$endgroup$
– noctusraid
Feb 2 '16 at 13:37
$begingroup$
Are you heading towards the stereographic projection?
$endgroup$
– noctusraid
Feb 2 '16 at 13:28
$begingroup$
Are you heading towards the stereographic projection?
$endgroup$
– noctusraid
Feb 2 '16 at 13:28
$begingroup$
Basically, yes.
$endgroup$
– Simon Rose
Feb 2 '16 at 13:29
$begingroup$
Basically, yes.
$endgroup$
– Simon Rose
Feb 2 '16 at 13:29
$begingroup$
Well we can extend the stereographic projection $p: S^2-0 to mathbb R^2$ to $tildep: S^2 to mathbb R^2 cup infty$ homeomorphically.
$endgroup$
– noctusraid
Feb 2 '16 at 13:37
$begingroup$
Well we can extend the stereographic projection $p: S^2-0 to mathbb R^2$ to $tildep: S^2 to mathbb R^2 cup infty$ homeomorphically.
$endgroup$
– noctusraid
Feb 2 '16 at 13:37
add a comment |
$begingroup$
Considering the CW complex structure of $Bbb C P^1$ , note that $Bbb C P^1$ is obtained from $Bbb C P^0$ by attaching a $2-cell$ but $Bbb C P^0 cong *$ , thus $$Bbb C P^1 cong D^2 / partial D^2 cong S^2 $$
answered Mar 20 at 16:41
reflexivereflexive
1,183625
$endgroup$
add a comment |
$begingroup$
Considering the CW complex structure of $Bbb C P^1$ , note that $Bbb C P^1$ is obtained from $Bbb C P^0$ by attaching a $2-cell$ but $Bbb C P^0 cong *$ , thus $$Bbb C P^1 cong D^2 / partial D^2 cong S^2 $$
answered Mar 20 at 16:41
reflexivereflexive
1,183625
$endgroup$
add a comment |
$begingroup$
Considering the CW complex structure of $Bbb C P^1$ , note that $Bbb C P^1$ is obtained from $Bbb C P^0$ by attaching a $2-cell$ but $Bbb C P^0 cong *$ , thus $$Bbb C P^1 cong D^2 / partial D^2 cong S^2 $$
answered Mar 20 at 16:41
reflexivereflexive
1,183625
$endgroup$
Considering the CW complex structure of $Bbb C P^1$ , note that $Bbb C P^1$ is obtained from $Bbb C P^0$ by attaching a $2-cell$ but $Bbb C P^0 cong *$ , thus $$Bbb C P^1 cong D^2 / partial D^2 cong S^2 $$
answered Mar 20 at 16:41
reflexivereflexive
1,183625
answered Mar 20 at 16:41
reflexivereflexive
1,183625
answered Mar 20 at 16:41
reflexivereflexive
1,183625
answered Mar 20 at 16:41
reflexivereflexive
1,183625
1,183625
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1
$begingroup$
There is definitely a problem with this: $mathbbRP^2$ is not homeomorphic to $S^2$.
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– Simon Rose
Feb 2 '16 at 13:23
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I think something wrong is going on here. RP2 is not homeomorphic to the sphere. Is not even orientable
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– Riccardo
Feb 2 '16 at 13:25
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Oh wait I meant $mathbb R P^3$.
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– noctusraid
Feb 2 '16 at 13:25
2
$begingroup$
$mathbbRP^3$ is three dimensional. $mathbbCP^1$ is two-dimensional.
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– Cheerful Parsnip
Feb 2 '16 at 13:29
1
$begingroup$
there is something wrong here as well. I mean $mathbbRP^3$ has fundamental group $mathbbZ/2mathbbZ$, but $S^2$ is simply connected
$endgroup$
– Riccardo
Feb 2 '16 at 13:31