Proving Hopf's Fibration $pi: mathbb{S}^{2n+1}tomathbb{CP}^n$ is a submersion
$begingroup$
Prove that the following map is a smooth, surjective submersion:
begin{align*}
pi:mathbb{S}^{2n+1}&tomathbb{CP}^n\
(x_0,y_0,...,x_n,y_n)&mapsto [x_0+iy_0:...:x_n+iy_n]
end{align*}
The surjectivity is immediate. I've also verified smoothness by considering the map $mathbb{R}^{2n+2}setminus{0}to mathbb{CP}^n$ with the same law and using the fact that $mathbb{S}^{2n+1}subset mathbb{R}^{2n+2}setminus{0}$ is an embbededing.
Now to prove $pi$ is a submersion, I could use the standard approach of calculating the local formula for $dpi$ and prove it has maximal rank. But that is obviously not the most pleasant thing to do.
I wonder if there is a more clever way or some kind of shorcut for this.
smooth-manifolds projective-space smooth-functions
asked Dec 1 '18 at 14:49
rmdmc89rmdmc89
2,1071922
$endgroup$
add a comment |
$begingroup$
Prove that the following map is a smooth, surjective submersion:
begin{align*}
pi:mathbb{S}^{2n+1}&tomathbb{CP}^n\
(x_0,y_0,...,x_n,y_n)&mapsto [x_0+iy_0:...:x_n+iy_n]
end{align*}
The surjectivity is immediate. I've also verified smoothness by considering the map $mathbb{R}^{2n+2}setminus{0}to mathbb{CP}^n$ with the same law and using the fact that $mathbb{S}^{2n+1}subset mathbb{R}^{2n+2}setminus{0}$ is an embbededing.
Now to prove $pi$ is a submersion, I could use the standard approach of calculating the local formula for $dpi$ and prove it has maximal rank. But that is obviously not the most pleasant thing to do.
I wonder if there is a more clever way or some kind of shorcut for this.
smooth-manifolds projective-space smooth-functions
asked Dec 1 '18 at 14:49
rmdmc89rmdmc89
2,1071922
$endgroup$
add a comment |
$begingroup$
Prove that the following map is a smooth, surjective submersion:
begin{align*}
pi:mathbb{S}^{2n+1}&tomathbb{CP}^n\
(x_0,y_0,...,x_n,y_n)&mapsto [x_0+iy_0:...:x_n+iy_n]
end{align*}
The surjectivity is immediate. I've also verified smoothness by considering the map $mathbb{R}^{2n+2}setminus{0}to mathbb{CP}^n$ with the same law and using the fact that $mathbb{S}^{2n+1}subset mathbb{R}^{2n+2}setminus{0}$ is an embbededing.
Now to prove $pi$ is a submersion, I could use the standard approach of calculating the local formula for $dpi$ and prove it has maximal rank. But that is obviously not the most pleasant thing to do.
I wonder if there is a more clever way or some kind of shorcut for this.
smooth-manifolds projective-space smooth-functions
asked Dec 1 '18 at 14:49
rmdmc89rmdmc89
2,1071922
$endgroup$
Prove that the following map is a smooth, surjective submersion:
begin{align*}
pi:mathbb{S}^{2n+1}&tomathbb{CP}^n\
(x_0,y_0,...,x_n,y_n)&mapsto [x_0+iy_0:...:x_n+iy_n]
end{align*}
The surjectivity is immediate. I've also verified smoothness by considering the map $mathbb{R}^{2n+2}setminus{0}to mathbb{CP}^n$ with the same law and using the fact that $mathbb{S}^{2n+1}subset mathbb{R}^{2n+2}setminus{0}$ is an embbededing.
Now to prove $pi$ is a submersion, I could use the standard approach of calculating the local formula for $dpi$ and prove it has maximal rank. But that is obviously not the most pleasant thing to do.
I wonder if there is a more clever way or some kind of shorcut for this.
smooth-manifolds projective-space smooth-functions
smooth-manifolds projective-space smooth-functions
asked Dec 1 '18 at 14:49
rmdmc89rmdmc89
2,1071922
asked Dec 1 '18 at 14:49
rmdmc89rmdmc89
2,1071922
edited Dec 4 '18 at 11:51
rmdmc89
asked Dec 1 '18 at 14:49
rmdmc89rmdmc89
2,1071922
asked Dec 1 '18 at 14:49
rmdmc89rmdmc89
2,1071922
asked Dec 1 '18 at 14:49
rmdmc89rmdmc89
2,1071922
2,1071922
add a comment |
add a comment |
1 Answer
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$S^{2n+1}subset mathbb{C}^{n+1}$. Let $p:mathbb{C}^{n+1}-{0}$ be the quotient map, it is a submersion. For every $zin mathbb{C}^{n+1}-{0}$, $uin T_z(mathbb{C}^{n+1}-{0})$, $dp_zu=0$ if and only if $u=cz$. Suppose now $zin S^{2n+1}$, $T_zS^{2n+1}={u:langle z,urangle=0}$. This implies that $ker dp_zcap T_zS^{2n+1}=Vect_{mathbb{R}}{iz}$ and the dimension of the image of $dp_z(TS^{2n+1})$ is $2n$.
answered Dec 1 '18 at 15:30
Tsemo AristideTsemo Aristide
57.5k11444
$endgroup$
$begingroup$
I can't see why $dp_zu=0Leftrightarrow u=cz$. Besides, if $langle z,urangle=0$ and $dp_zu=0$, doesn't it follow that $u=cz$ with $c=0$?
$endgroup$
– rmdmc89
Dec 1 '18 at 19:59
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
$S^{2n+1}subset mathbb{C}^{n+1}$. Let $p:mathbb{C}^{n+1}-{0}$ be the quotient map, it is a submersion. For every $zin mathbb{C}^{n+1}-{0}$, $uin T_z(mathbb{C}^{n+1}-{0})$, $dp_zu=0$ if and only if $u=cz$. Suppose now $zin S^{2n+1}$, $T_zS^{2n+1}={u:langle z,urangle=0}$. This implies that $ker dp_zcap T_zS^{2n+1}=Vect_{mathbb{R}}{iz}$ and the dimension of the image of $dp_z(TS^{2n+1})$ is $2n$.
answered Dec 1 '18 at 15:30
Tsemo AristideTsemo Aristide
57.5k11444
$endgroup$
$begingroup$
I can't see why $dp_zu=0Leftrightarrow u=cz$. Besides, if $langle z,urangle=0$ and $dp_zu=0$, doesn't it follow that $u=cz$ with $c=0$?
$endgroup$
– rmdmc89
Dec 1 '18 at 19:59
add a comment |
$begingroup$
$S^{2n+1}subset mathbb{C}^{n+1}$. Let $p:mathbb{C}^{n+1}-{0}$ be the quotient map, it is a submersion. For every $zin mathbb{C}^{n+1}-{0}$, $uin T_z(mathbb{C}^{n+1}-{0})$, $dp_zu=0$ if and only if $u=cz$. Suppose now $zin S^{2n+1}$, $T_zS^{2n+1}={u:langle z,urangle=0}$. This implies that $ker dp_zcap T_zS^{2n+1}=Vect_{mathbb{R}}{iz}$ and the dimension of the image of $dp_z(TS^{2n+1})$ is $2n$.
answered Dec 1 '18 at 15:30
Tsemo AristideTsemo Aristide
57.5k11444
$endgroup$
$begingroup$
I can't see why $dp_zu=0Leftrightarrow u=cz$. Besides, if $langle z,urangle=0$ and $dp_zu=0$, doesn't it follow that $u=cz$ with $c=0$?
$endgroup$
– rmdmc89
Dec 1 '18 at 19:59
add a comment |
$begingroup$
$S^{2n+1}subset mathbb{C}^{n+1}$. Let $p:mathbb{C}^{n+1}-{0}$ be the quotient map, it is a submersion. For every $zin mathbb{C}^{n+1}-{0}$, $uin T_z(mathbb{C}^{n+1}-{0})$, $dp_zu=0$ if and only if $u=cz$. Suppose now $zin S^{2n+1}$, $T_zS^{2n+1}={u:langle z,urangle=0}$. This implies that $ker dp_zcap T_zS^{2n+1}=Vect_{mathbb{R}}{iz}$ and the dimension of the image of $dp_z(TS^{2n+1})$ is $2n$.
answered Dec 1 '18 at 15:30
Tsemo AristideTsemo Aristide
57.5k11444
$endgroup$
$S^{2n+1}subset mathbb{C}^{n+1}$. Let $p:mathbb{C}^{n+1}-{0}$ be the quotient map, it is a submersion. For every $zin mathbb{C}^{n+1}-{0}$, $uin T_z(mathbb{C}^{n+1}-{0})$, $dp_zu=0$ if and only if $u=cz$. Suppose now $zin S^{2n+1}$, $T_zS^{2n+1}={u:langle z,urangle=0}$. This implies that $ker dp_zcap T_zS^{2n+1}=Vect_{mathbb{R}}{iz}$ and the dimension of the image of $dp_z(TS^{2n+1})$ is $2n$.
answered Dec 1 '18 at 15:30
Tsemo AristideTsemo Aristide
57.5k11444
answered Dec 1 '18 at 15:30
Tsemo AristideTsemo Aristide
57.5k11444
answered Dec 1 '18 at 15:30
Tsemo AristideTsemo Aristide
57.5k11444
answered Dec 1 '18 at 15:30
Tsemo AristideTsemo Aristide
57.5k11444
57.5k11444
$begingroup$
I can't see why $dp_zu=0Leftrightarrow u=cz$. Besides, if $langle z,urangle=0$ and $dp_zu=0$, doesn't it follow that $u=cz$ with $c=0$?
$endgroup$
– rmdmc89
Dec 1 '18 at 19:59
add a comment |
$begingroup$
I can't see why $dp_zu=0Leftrightarrow u=cz$. Besides, if $langle z,urangle=0$ and $dp_zu=0$, doesn't it follow that $u=cz$ with $c=0$?
$endgroup$
– rmdmc89
Dec 1 '18 at 19:59
$begingroup$
I can't see why $dp_zu=0Leftrightarrow u=cz$. Besides, if $langle z,urangle=0$ and $dp_zu=0$, doesn't it follow that $u=cz$ with $c=0$?
$endgroup$
– rmdmc89
Dec 1 '18 at 19:59
$begingroup$
I can't see why $dp_zu=0Leftrightarrow u=cz$. Besides, if $langle z,urangle=0$ and $dp_zu=0$, doesn't it follow that $u=cz$ with $c=0$?
$endgroup$
– rmdmc89
Dec 1 '18 at 19:59
add a comment |
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