Mixing
Homogeneous space and quotient space for complex projective spaces
Clash Royale CLAN TAG#URR8PPP
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1
down vote
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We know that
$$
O(n+1)/O(n) simeq SO(n+1)/SO(n) simeq S^n,
$$
based on the result of homogeneous space.
Also
$$
PO(n+1)/PO(n) simeq P^n,
$$
$P^n$ is the projective space.
These are in some sense spheres.
If we embed the complex projective space $mathbbCP^n$ into $mathbbCP^n+1$, we may be able to define the quotient space
$$
mathbbCP^n+1/mathbbCP^n simeq ?
$$
Do we have simpler expressions for the above "manifolds" or "quotient space"?
Homotopy group $pi_i(mathbbCP^n+1/mathbbCP^n)=?$
Attempt: Note that $mathbbCP^n=S^2n+1/U(1)$, so for
$mathbbCP^1simeq S^2$ and $mathbbCP^0simeq 0$,
so
$$pi_i(mathbbCP^1/mathbbCP^0)=pi_i(S^2/0)=pi_i(S^2),$$
which homotopy group is known.
differential-topology lie-groups projective-space quotient-spaces homogeneous-spaces
asked Jul 31 at 3:30
wonderich
1,60011226
add a comment |Â
up vote
1
down vote
favorite
We know that
$$
O(n+1)/O(n) simeq SO(n+1)/SO(n) simeq S^n,
$$
based on the result of homogeneous space.
Also
$$
PO(n+1)/PO(n) simeq P^n,
$$
$P^n$ is the projective space.
These are in some sense spheres.
If we embed the complex projective space $mathbbCP^n$ into $mathbbCP^n+1$, we may be able to define the quotient space
$$
mathbbCP^n+1/mathbbCP^n simeq ?
$$
Do we have simpler expressions for the above "manifolds" or "quotient space"?
Homotopy group $pi_i(mathbbCP^n+1/mathbbCP^n)=?$
Attempt: Note that $mathbbCP^n=S^2n+1/U(1)$, so for
$mathbbCP^1simeq S^2$ and $mathbbCP^0simeq 0$,
so
$$pi_i(mathbbCP^1/mathbbCP^0)=pi_i(S^2/0)=pi_i(S^2),$$
which homotopy group is known.
differential-topology lie-groups projective-space quotient-spaces homogeneous-spaces
asked Jul 31 at 3:30
wonderich
1,60011226
The notation $mathbbCP^n+1/mathbbCP^n$ is typically understood to be the quotient space where we identify $mathbbCP^n$ to a point. (But not other identifications are made). Is this what you meant? If so, for the usual embedding of $mathbbCP^n$ into $mathbbCP^n+1$, $mathbbCP^n+1/mathbbCP^n$ is homeomorphic to $S^2n+2$. That said, as a general rule, for $Nsubseteq M$ closed manifolds, there is no reason for $M/N$ to be homotopy equivalent to a manifold.
– Jason DeVito
Aug 1 at 0:43
thanks - if this is correct - this counts as an answer.
– wonderich
Aug 1 at 14:24
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
We know that
$$
O(n+1)/O(n) simeq SO(n+1)/SO(n) simeq S^n,
$$
based on the result of homogeneous space.
Also
$$
PO(n+1)/PO(n) simeq P^n,
$$
$P^n$ is the projective space.
These are in some sense spheres.
If we embed the complex projective space $mathbbCP^n$ into $mathbbCP^n+1$, we may be able to define the quotient space
$$
mathbbCP^n+1/mathbbCP^n simeq ?
$$
Do we have simpler expressions for the above "manifolds" or "quotient space"?
Homotopy group $pi_i(mathbbCP^n+1/mathbbCP^n)=?$
Attempt: Note that $mathbbCP^n=S^2n+1/U(1)$, so for
$mathbbCP^1simeq S^2$ and $mathbbCP^0simeq 0$,
so
$$pi_i(mathbbCP^1/mathbbCP^0)=pi_i(S^2/0)=pi_i(S^2),$$
which homotopy group is known.
differential-topology lie-groups projective-space quotient-spaces homogeneous-spaces
asked Jul 31 at 3:30
wonderich
1,60011226
We know that
$$
O(n+1)/O(n) simeq SO(n+1)/SO(n) simeq S^n,
$$
based on the result of homogeneous space.
Also
$$
PO(n+1)/PO(n) simeq P^n,
$$
$P^n$ is the projective space.
These are in some sense spheres.
If we embed the complex projective space $mathbbCP^n$ into $mathbbCP^n+1$, we may be able to define the quotient space
$$
mathbbCP^n+1/mathbbCP^n simeq ?
$$
Do we have simpler expressions for the above "manifolds" or "quotient space"?
Homotopy group $pi_i(mathbbCP^n+1/mathbbCP^n)=?$
Attempt: Note that $mathbbCP^n=S^2n+1/U(1)$, so for
$mathbbCP^1simeq S^2$ and $mathbbCP^0simeq 0$,
so
$$pi_i(mathbbCP^1/mathbbCP^0)=pi_i(S^2/0)=pi_i(S^2),$$
which homotopy group is known.
differential-topology lie-groups projective-space quotient-spaces homogeneous-spaces
asked Jul 31 at 3:30
wonderich
1,60011226
edited Aug 1 at 18:08
asked Jul 31 at 3:30
wonderich
1,60011226
asked Jul 31 at 3:30
wonderich
1,60011226
asked Jul 31 at 3:30
wonderich
1,60011226
1,60011226
The notation $mathbbCP^n+1/mathbbCP^n$ is typically understood to be the quotient space where we identify $mathbbCP^n$ to a point. (But not other identifications are made). Is this what you meant? If so, for the usual embedding of $mathbbCP^n$ into $mathbbCP^n+1$, $mathbbCP^n+1/mathbbCP^n$ is homeomorphic to $S^2n+2$. That said, as a general rule, for $Nsubseteq M$ closed manifolds, there is no reason for $M/N$ to be homotopy equivalent to a manifold.
– Jason DeVito
Aug 1 at 0:43
thanks - if this is correct - this counts as an answer.
– wonderich
Aug 1 at 14:24
add a comment |Â
The notation $mathbbCP^n+1/mathbbCP^n$ is typically understood to be the quotient space where we identify $mathbbCP^n$ to a point. (But not other identifications are made). Is this what you meant? If so, for the usual embedding of $mathbbCP^n$ into $mathbbCP^n+1$, $mathbbCP^n+1/mathbbCP^n$ is homeomorphic to $S^2n+2$. That said, as a general rule, for $Nsubseteq M$ closed manifolds, there is no reason for $M/N$ to be homotopy equivalent to a manifold.
– Jason DeVito
Aug 1 at 0:43
thanks - if this is correct - this counts as an answer.
– wonderich
Aug 1 at 14:24
The notation $mathbbCP^n+1/mathbbCP^n$ is typically understood to be the quotient space where we identify $mathbbCP^n$ to a point. (But not other identifications are made). Is this what you meant? If so, for the usual embedding of $mathbbCP^n$ into $mathbbCP^n+1$, $mathbbCP^n+1/mathbbCP^n$ is homeomorphic to $S^2n+2$. That said, as a general rule, for $Nsubseteq M$ closed manifolds, there is no reason for $M/N$ to be homotopy equivalent to a manifold.
– Jason DeVito
Aug 1 at 0:43
The notation $mathbbCP^n+1/mathbbCP^n$ is typically understood to be the quotient space where we identify $mathbbCP^n$ to a point. (But not other identifications are made). Is this what you meant? If so, for the usual embedding of $mathbbCP^n$ into $mathbbCP^n+1$, $mathbbCP^n+1/mathbbCP^n$ is homeomorphic to $S^2n+2$. That said, as a general rule, for $Nsubseteq M$ closed manifolds, there is no reason for $M/N$ to be homotopy equivalent to a manifold.
– Jason DeVito
Aug 1 at 0:43
thanks - if this is correct - this counts as an answer.
– wonderich
Aug 1 at 14:24
thanks - if this is correct - this counts as an answer.
– wonderich
Aug 1 at 14:24
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
The notation $mathbbCP^n+1/mathbbCP^n$ is typically understood to be the quotient space where we identify $mathbbCP^n$ to a point. (But not other identifications are made).
For the usual embedding of $mathbbCP^n$ into $mathbbCP^n+1$, given by mapping homogeneous coordinates $[z_0:z_1:...:z_n]mapsto [z_0:z_1:...:z_n: z_n+1]$, $mathbbCP^n+1/mathbbCP^n$ is homeomorphic to $S^2n+2$. One way to see this is as follows.
First, let $Z_i = fracz_iz_n+1$ (assuming $z_n+1neq 0$). Also, let $Z = sum_i=0^n |Z_i|^2$.
Then, we map $mathbbCP^n+1rightarrow S^2n+2$ via $$(z_0:...:z_n+1)mapsto begincases left(frac2Z_01+Z, frac2Z_11+Z, ..., frac2Z_n1+Z, frac-1+Z1+Z right) & z_n+1neq 0 \ (0,0,..,0,1) & z_n+1 = 0 endcases.$$
The case that $z_n+1 = 0$ corresponds exactly to a point being in $mathbbCP^n$. As a result, $f$ is constant on $mathbbCP^n$ so gives a map $overlinef:mathbbCP^n+1/mathbbCP^nrightarrow S^2n+2$. This map is a homeomorphism.
Finally, typically, there is no reason for $M/N$ to be a manifold even when $M$ and $N$ are really nice. For example, if $N$ is the equator on $M = S^2$, then $M/N$ is homeomorphic to $S^2wedge S^2$, which is not a manifold due to the wedge point.
answered Aug 1 at 16:08
Jason DeVito
29.4k473129
I post a new question maybe you may be also able to answer.
– wonderich
Aug 1 at 22:51
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 notation $mathbbCP^n+1/mathbbCP^n$ is typically understood to be the quotient space where we identify $mathbbCP^n$ to a point. (But not other identifications are made).
For the usual embedding of $mathbbCP^n$ into $mathbbCP^n+1$, given by mapping homogeneous coordinates $[z_0:z_1:...:z_n]mapsto [z_0:z_1:...:z_n: z_n+1]$, $mathbbCP^n+1/mathbbCP^n$ is homeomorphic to $S^2n+2$. One way to see this is as follows.
First, let $Z_i = fracz_iz_n+1$ (assuming $z_n+1neq 0$). Also, let $Z = sum_i=0^n |Z_i|^2$.
Then, we map $mathbbCP^n+1rightarrow S^2n+2$ via $$(z_0:...:z_n+1)mapsto begincases left(frac2Z_01+Z, frac2Z_11+Z, ..., frac2Z_n1+Z, frac-1+Z1+Z right) & z_n+1neq 0 \ (0,0,..,0,1) & z_n+1 = 0 endcases.$$
The case that $z_n+1 = 0$ corresponds exactly to a point being in $mathbbCP^n$. As a result, $f$ is constant on $mathbbCP^n$ so gives a map $overlinef:mathbbCP^n+1/mathbbCP^nrightarrow S^2n+2$. This map is a homeomorphism.
Finally, typically, there is no reason for $M/N$ to be a manifold even when $M$ and $N$ are really nice. For example, if $N$ is the equator on $M = S^2$, then $M/N$ is homeomorphic to $S^2wedge S^2$, which is not a manifold due to the wedge point.
answered Aug 1 at 16:08
Jason DeVito
29.4k473129
I post a new question maybe you may be also able to answer.
– wonderich
Aug 1 at 22:51
add a comment |Â
up vote
1
down vote
accepted
The notation $mathbbCP^n+1/mathbbCP^n$ is typically understood to be the quotient space where we identify $mathbbCP^n$ to a point. (But not other identifications are made).
For the usual embedding of $mathbbCP^n$ into $mathbbCP^n+1$, given by mapping homogeneous coordinates $[z_0:z_1:...:z_n]mapsto [z_0:z_1:...:z_n: z_n+1]$, $mathbbCP^n+1/mathbbCP^n$ is homeomorphic to $S^2n+2$. One way to see this is as follows.
First, let $Z_i = fracz_iz_n+1$ (assuming $z_n+1neq 0$). Also, let $Z = sum_i=0^n |Z_i|^2$.
Then, we map $mathbbCP^n+1rightarrow S^2n+2$ via $$(z_0:...:z_n+1)mapsto begincases left(frac2Z_01+Z, frac2Z_11+Z, ..., frac2Z_n1+Z, frac-1+Z1+Z right) & z_n+1neq 0 \ (0,0,..,0,1) & z_n+1 = 0 endcases.$$
The case that $z_n+1 = 0$ corresponds exactly to a point being in $mathbbCP^n$. As a result, $f$ is constant on $mathbbCP^n$ so gives a map $overlinef:mathbbCP^n+1/mathbbCP^nrightarrow S^2n+2$. This map is a homeomorphism.
Finally, typically, there is no reason for $M/N$ to be a manifold even when $M$ and $N$ are really nice. For example, if $N$ is the equator on $M = S^2$, then $M/N$ is homeomorphic to $S^2wedge S^2$, which is not a manifold due to the wedge point.
answered Aug 1 at 16:08
Jason DeVito
29.4k473129
I post a new question maybe you may be also able to answer.
– wonderich
Aug 1 at 22:51
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The notation $mathbbCP^n+1/mathbbCP^n$ is typically understood to be the quotient space where we identify $mathbbCP^n$ to a point. (But not other identifications are made).
For the usual embedding of $mathbbCP^n$ into $mathbbCP^n+1$, given by mapping homogeneous coordinates $[z_0:z_1:...:z_n]mapsto [z_0:z_1:...:z_n: z_n+1]$, $mathbbCP^n+1/mathbbCP^n$ is homeomorphic to $S^2n+2$. One way to see this is as follows.
First, let $Z_i = fracz_iz_n+1$ (assuming $z_n+1neq 0$). Also, let $Z = sum_i=0^n |Z_i|^2$.
Then, we map $mathbbCP^n+1rightarrow S^2n+2$ via $$(z_0:...:z_n+1)mapsto begincases left(frac2Z_01+Z, frac2Z_11+Z, ..., frac2Z_n1+Z, frac-1+Z1+Z right) & z_n+1neq 0 \ (0,0,..,0,1) & z_n+1 = 0 endcases.$$
The case that $z_n+1 = 0$ corresponds exactly to a point being in $mathbbCP^n$. As a result, $f$ is constant on $mathbbCP^n$ so gives a map $overlinef:mathbbCP^n+1/mathbbCP^nrightarrow S^2n+2$. This map is a homeomorphism.
Finally, typically, there is no reason for $M/N$ to be a manifold even when $M$ and $N$ are really nice. For example, if $N$ is the equator on $M = S^2$, then $M/N$ is homeomorphic to $S^2wedge S^2$, which is not a manifold due to the wedge point.
answered Aug 1 at 16:08
Jason DeVito
29.4k473129
The notation $mathbbCP^n+1/mathbbCP^n$ is typically understood to be the quotient space where we identify $mathbbCP^n$ to a point. (But not other identifications are made).
For the usual embedding of $mathbbCP^n$ into $mathbbCP^n+1$, given by mapping homogeneous coordinates $[z_0:z_1:...:z_n]mapsto [z_0:z_1:...:z_n: z_n+1]$, $mathbbCP^n+1/mathbbCP^n$ is homeomorphic to $S^2n+2$. One way to see this is as follows.
First, let $Z_i = fracz_iz_n+1$ (assuming $z_n+1neq 0$). Also, let $Z = sum_i=0^n |Z_i|^2$.
Then, we map $mathbbCP^n+1rightarrow S^2n+2$ via $$(z_0:...:z_n+1)mapsto begincases left(frac2Z_01+Z, frac2Z_11+Z, ..., frac2Z_n1+Z, frac-1+Z1+Z right) & z_n+1neq 0 \ (0,0,..,0,1) & z_n+1 = 0 endcases.$$
The case that $z_n+1 = 0$ corresponds exactly to a point being in $mathbbCP^n$. As a result, $f$ is constant on $mathbbCP^n$ so gives a map $overlinef:mathbbCP^n+1/mathbbCP^nrightarrow S^2n+2$. This map is a homeomorphism.
Finally, typically, there is no reason for $M/N$ to be a manifold even when $M$ and $N$ are really nice. For example, if $N$ is the equator on $M = S^2$, then $M/N$ is homeomorphic to $S^2wedge S^2$, which is not a manifold due to the wedge point.
answered Aug 1 at 16:08
Jason DeVito
29.4k473129
answered Aug 1 at 16:08
Jason DeVito
29.4k473129
answered Aug 1 at 16:08
Jason DeVito
29.4k473129
answered Aug 1 at 16:08
Jason DeVito
29.4k473129
29.4k473129
I post a new question maybe you may be also able to answer.
– wonderich
Aug 1 at 22:51
add a comment |Â
I post a new question maybe you may be also able to answer.
– wonderich
Aug 1 at 22:51
I post a new question maybe you may be also able to answer.
– wonderich
Aug 1 at 22:51
I post a new question maybe you may be also able to answer.
– wonderich
Aug 1 at 22:51
add a comment |Â
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The notation $mathbbCP^n+1/mathbbCP^n$ is typically understood to be the quotient space where we identify $mathbbCP^n$ to a point. (But not other identifications are made). Is this what you meant? If so, for the usual embedding of $mathbbCP^n$ into $mathbbCP^n+1$, $mathbbCP^n+1/mathbbCP^n$ is homeomorphic to $S^2n+2$. That said, as a general rule, for $Nsubseteq M$ closed manifolds, there is no reason for $M/N$ to be homotopy equivalent to a manifold.
– Jason DeVito
Aug 1 at 0:43
thanks - if this is correct - this counts as an answer.
– wonderich
Aug 1 at 14:24