Submanifold of real projective space
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Would you like to tell me how to prove that ${[x_{0}:x_{1}:x_{2}]: x_{0}x_{1} + x_{2}^{2} = 0 } subset mathbb{R}mathbb{P}^{2}$ is a submanifold (of $mathbb{R}mathbb{P}^{2}$)?
differential-geometry differential-topology projective-space
asked Dec 12 '18 at 14:09
kNiEsSoKkkNiEsSoKk
6710
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add a comment |
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Would you like to tell me how to prove that ${[x_{0}:x_{1}:x_{2}]: x_{0}x_{1} + x_{2}^{2} = 0 } subset mathbb{R}mathbb{P}^{2}$ is a submanifold (of $mathbb{R}mathbb{P}^{2}$)?
differential-geometry differential-topology projective-space
asked Dec 12 '18 at 14:09
kNiEsSoKkkNiEsSoKk
6710
$endgroup$
add a comment |
$begingroup$
Would you like to tell me how to prove that ${[x_{0}:x_{1}:x_{2}]: x_{0}x_{1} + x_{2}^{2} = 0 } subset mathbb{R}mathbb{P}^{2}$ is a submanifold (of $mathbb{R}mathbb{P}^{2}$)?
differential-geometry differential-topology projective-space
asked Dec 12 '18 at 14:09
kNiEsSoKkkNiEsSoKk
6710
$endgroup$
Would you like to tell me how to prove that ${[x_{0}:x_{1}:x_{2}]: x_{0}x_{1} + x_{2}^{2} = 0 } subset mathbb{R}mathbb{P}^{2}$ is a submanifold (of $mathbb{R}mathbb{P}^{2}$)?
differential-geometry differential-topology projective-space
differential-geometry differential-topology projective-space
asked Dec 12 '18 at 14:09
kNiEsSoKkkNiEsSoKk
6710
asked Dec 12 '18 at 14:09
kNiEsSoKkkNiEsSoKk
6710
edited Dec 13 '18 at 8:39
kNiEsSoKk
asked Dec 12 '18 at 14:09
kNiEsSoKkkNiEsSoKk
6710
asked Dec 12 '18 at 14:09
kNiEsSoKkkNiEsSoKk
6710
asked Dec 12 '18 at 14:09
kNiEsSoKkkNiEsSoKk
6710
6710
add a comment |
add a comment |
1 Answer
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I believe you want to say $x_0x_1+x_2^2=0$. Let $V$ be that subset. Consider $U_0={[x_0,x_1,x_2],x_0neq 0}$ define $f([x_0,x_1,x_2])={x_1over x_0}+{x_2^2over x_0^2}$ is a submersion on $U_0$ this shows that $Vcap U_0$ is a submanifold.
answered Dec 12 '18 at 15:28
Tsemo AristideTsemo Aristide
58.7k11445
$endgroup$
$begingroup$
Thanks! I found the proposition $f:mathbb{R}mathbb{P}^{2} rightarrow mathbb{R}$ smooth and $qinmathbb{R}$ a regular value $implies f^{-1}(q)$ is a submanifold of $mathbb{R}mathbb{P}^{2}$. Is this what you are using?
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– kNiEsSoKk
Dec 13 '18 at 9:56
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
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active
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$begingroup$
I believe you want to say $x_0x_1+x_2^2=0$. Let $V$ be that subset. Consider $U_0={[x_0,x_1,x_2],x_0neq 0}$ define $f([x_0,x_1,x_2])={x_1over x_0}+{x_2^2over x_0^2}$ is a submersion on $U_0$ this shows that $Vcap U_0$ is a submanifold.
answered Dec 12 '18 at 15:28
Tsemo AristideTsemo Aristide
58.7k11445
$endgroup$
$begingroup$
Thanks! I found the proposition $f:mathbb{R}mathbb{P}^{2} rightarrow mathbb{R}$ smooth and $qinmathbb{R}$ a regular value $implies f^{-1}(q)$ is a submanifold of $mathbb{R}mathbb{P}^{2}$. Is this what you are using?
$endgroup$
– kNiEsSoKk
Dec 13 '18 at 9:56
add a comment |
$begingroup$
I believe you want to say $x_0x_1+x_2^2=0$. Let $V$ be that subset. Consider $U_0={[x_0,x_1,x_2],x_0neq 0}$ define $f([x_0,x_1,x_2])={x_1over x_0}+{x_2^2over x_0^2}$ is a submersion on $U_0$ this shows that $Vcap U_0$ is a submanifold.
answered Dec 12 '18 at 15:28
Tsemo AristideTsemo Aristide
58.7k11445
$endgroup$
$begingroup$
Thanks! I found the proposition $f:mathbb{R}mathbb{P}^{2} rightarrow mathbb{R}$ smooth and $qinmathbb{R}$ a regular value $implies f^{-1}(q)$ is a submanifold of $mathbb{R}mathbb{P}^{2}$. Is this what you are using?
$endgroup$
– kNiEsSoKk
Dec 13 '18 at 9:56
add a comment |
$begingroup$
I believe you want to say $x_0x_1+x_2^2=0$. Let $V$ be that subset. Consider $U_0={[x_0,x_1,x_2],x_0neq 0}$ define $f([x_0,x_1,x_2])={x_1over x_0}+{x_2^2over x_0^2}$ is a submersion on $U_0$ this shows that $Vcap U_0$ is a submanifold.
answered Dec 12 '18 at 15:28
Tsemo AristideTsemo Aristide
58.7k11445
$endgroup$
I believe you want to say $x_0x_1+x_2^2=0$. Let $V$ be that subset. Consider $U_0={[x_0,x_1,x_2],x_0neq 0}$ define $f([x_0,x_1,x_2])={x_1over x_0}+{x_2^2over x_0^2}$ is a submersion on $U_0$ this shows that $Vcap U_0$ is a submanifold.
answered Dec 12 '18 at 15:28
Tsemo AristideTsemo Aristide
58.7k11445
answered Dec 12 '18 at 15:28
Tsemo AristideTsemo Aristide
58.7k11445
answered Dec 12 '18 at 15:28
Tsemo AristideTsemo Aristide
58.7k11445
answered Dec 12 '18 at 15:28
Tsemo AristideTsemo Aristide
58.7k11445
58.7k11445
$begingroup$
Thanks! I found the proposition $f:mathbb{R}mathbb{P}^{2} rightarrow mathbb{R}$ smooth and $qinmathbb{R}$ a regular value $implies f^{-1}(q)$ is a submanifold of $mathbb{R}mathbb{P}^{2}$. Is this what you are using?
$endgroup$
– kNiEsSoKk
Dec 13 '18 at 9:56
add a comment |
$begingroup$
Thanks! I found the proposition $f:mathbb{R}mathbb{P}^{2} rightarrow mathbb{R}$ smooth and $qinmathbb{R}$ a regular value $implies f^{-1}(q)$ is a submanifold of $mathbb{R}mathbb{P}^{2}$. Is this what you are using?
$endgroup$
– kNiEsSoKk
Dec 13 '18 at 9:56
$begingroup$
Thanks! I found the proposition $f:mathbb{R}mathbb{P}^{2} rightarrow mathbb{R}$ smooth and $qinmathbb{R}$ a regular value $implies f^{-1}(q)$ is a submanifold of $mathbb{R}mathbb{P}^{2}$. Is this what you are using?
$endgroup$
– kNiEsSoKk
Dec 13 '18 at 9:56
$begingroup$
Thanks! I found the proposition $f:mathbb{R}mathbb{P}^{2} rightarrow mathbb{R}$ smooth and $qinmathbb{R}$ a regular value $implies f^{-1}(q)$ is a submanifold of $mathbb{R}mathbb{P}^{2}$. Is this what you are using?
$endgroup$
– kNiEsSoKk
Dec 13 '18 at 9:56
add a comment |
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