Proving projective space is a topological manifold
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I'm trying to prove that $mathbb{P}^n(mathbb{R})$ is a topological manifold of dimension n.
So I can define $f_i: U_i rightarrow mathbb{R}$ such that $f_i([x_0:x_1: dots : x_n])= (x_0, dots x_{i-1}, x_{i+1}, dots x_n)$ where $U_i={[x_0:dots : x_n] : x_i=1}$.
then I only need to prove this is an homeomorphism.
Open sets on $mathbb{P}^n(mathbb{R})$ are $pi(A)$ where $A$ is an open set of $mathbb{R}^{n+1}$.
For $f_0$:
Is open because if $A=A_0 times ... times A_n$
$f_0(pi(A))=f_0( .cup_{x_0in A_0} [1:frac{A_1}{x_0}: dots : frac{A_n}{x_0}]) = cup_{x_0 in A_0} frac{A_1}{x_0}times dots times frac{A_n}{x_0}$ which is open.
But I'm not sure how to see this is continuous. I would appreciate any help for doing this
Thanks
general-topology
asked Dec 10 '18 at 14:36
JohannaJohanna
599
$endgroup$
add a comment |
$begingroup$
I'm trying to prove that $mathbb{P}^n(mathbb{R})$ is a topological manifold of dimension n.
So I can define $f_i: U_i rightarrow mathbb{R}$ such that $f_i([x_0:x_1: dots : x_n])= (x_0, dots x_{i-1}, x_{i+1}, dots x_n)$ where $U_i={[x_0:dots : x_n] : x_i=1}$.
then I only need to prove this is an homeomorphism.
Open sets on $mathbb{P}^n(mathbb{R})$ are $pi(A)$ where $A$ is an open set of $mathbb{R}^{n+1}$.
For $f_0$:
Is open because if $A=A_0 times ... times A_n$
$f_0(pi(A))=f_0( .cup_{x_0in A_0} [1:frac{A_1}{x_0}: dots : frac{A_n}{x_0}]) = cup_{x_0 in A_0} frac{A_1}{x_0}times dots times frac{A_n}{x_0}$ which is open.
But I'm not sure how to see this is continuous. I would appreciate any help for doing this
Thanks
general-topology
asked Dec 10 '18 at 14:36
JohannaJohanna
599
$endgroup$
$begingroup$
You wrote "I only need to prove that this is an homeomorphism". A formula does not define a function, and hence does not define a homeomorphism. You must also specify the domain and range. Notice how this is done in the answer of @FedericoFallucca.
$endgroup$
– Lee Mosher
Dec 10 '18 at 15:24
$begingroup$
Oh sorry I forgot that line, already edited it.
$endgroup$
– Johanna
Dec 10 '18 at 15:30
add a comment |
$begingroup$
I'm trying to prove that $mathbb{P}^n(mathbb{R})$ is a topological manifold of dimension n.
So I can define $f_i: U_i rightarrow mathbb{R}$ such that $f_i([x_0:x_1: dots : x_n])= (x_0, dots x_{i-1}, x_{i+1}, dots x_n)$ where $U_i={[x_0:dots : x_n] : x_i=1}$.
then I only need to prove this is an homeomorphism.
Open sets on $mathbb{P}^n(mathbb{R})$ are $pi(A)$ where $A$ is an open set of $mathbb{R}^{n+1}$.
For $f_0$:
Is open because if $A=A_0 times ... times A_n$
$f_0(pi(A))=f_0( .cup_{x_0in A_0} [1:frac{A_1}{x_0}: dots : frac{A_n}{x_0}]) = cup_{x_0 in A_0} frac{A_1}{x_0}times dots times frac{A_n}{x_0}$ which is open.
But I'm not sure how to see this is continuous. I would appreciate any help for doing this
Thanks
general-topology
asked Dec 10 '18 at 14:36
JohannaJohanna
599
$endgroup$
I'm trying to prove that $mathbb{P}^n(mathbb{R})$ is a topological manifold of dimension n.
So I can define $f_i: U_i rightarrow mathbb{R}$ such that $f_i([x_0:x_1: dots : x_n])= (x_0, dots x_{i-1}, x_{i+1}, dots x_n)$ where $U_i={[x_0:dots : x_n] : x_i=1}$.
then I only need to prove this is an homeomorphism.
Open sets on $mathbb{P}^n(mathbb{R})$ are $pi(A)$ where $A$ is an open set of $mathbb{R}^{n+1}$.
For $f_0$:
Is open because if $A=A_0 times ... times A_n$
$f_0(pi(A))=f_0( .cup_{x_0in A_0} [1:frac{A_1}{x_0}: dots : frac{A_n}{x_0}]) = cup_{x_0 in A_0} frac{A_1}{x_0}times dots times frac{A_n}{x_0}$ which is open.
But I'm not sure how to see this is continuous. I would appreciate any help for doing this
Thanks
general-topology
general-topology
asked Dec 10 '18 at 14:36
JohannaJohanna
599
asked Dec 10 '18 at 14:36
JohannaJohanna
599
edited Dec 10 '18 at 15:29
Johanna
asked Dec 10 '18 at 14:36
JohannaJohanna
599
asked Dec 10 '18 at 14:36
JohannaJohanna
599
asked Dec 10 '18 at 14:36
JohannaJohanna
599
599
$begingroup$
You wrote "I only need to prove that this is an homeomorphism". A formula does not define a function, and hence does not define a homeomorphism. You must also specify the domain and range. Notice how this is done in the answer of @FedericoFallucca.
$endgroup$
– Lee Mosher
Dec 10 '18 at 15:24
$begingroup$
Oh sorry I forgot that line, already edited it.
$endgroup$
– Johanna
Dec 10 '18 at 15:30
add a comment |
$begingroup$
You wrote "I only need to prove that this is an homeomorphism". A formula does not define a function, and hence does not define a homeomorphism. You must also specify the domain and range. Notice how this is done in the answer of @FedericoFallucca.
$endgroup$
– Lee Mosher
Dec 10 '18 at 15:24
$begingroup$
Oh sorry I forgot that line, already edited it.
$endgroup$
– Johanna
Dec 10 '18 at 15:30
$begingroup$
You wrote "I only need to prove that this is an homeomorphism". A formula does not define a function, and hence does not define a homeomorphism. You must also specify the domain and range. Notice how this is done in the answer of @FedericoFallucca.
$endgroup$
– Lee Mosher
Dec 10 '18 at 15:24
$begingroup$
You wrote "I only need to prove that this is an homeomorphism". A formula does not define a function, and hence does not define a homeomorphism. You must also specify the domain and range. Notice how this is done in the answer of @FedericoFallucca.
$endgroup$
– Lee Mosher
Dec 10 '18 at 15:24
$begingroup$
Oh sorry I forgot that line, already edited it.
$endgroup$
– Johanna
Dec 10 '18 at 15:30
$begingroup$
Oh sorry I forgot that line, already edited it.
$endgroup$
– Johanna
Dec 10 '18 at 15:30
add a comment |
1 Answer
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$begingroup$
$U_i:={[x_0,dots , x_n] : x_ineq 0}$ is an open set of $mathbb{P}^n$ and you can define
$phi_i: U_ito mathbb{R}^n$ such that
$phi_i([x_0,dots , x_n] ):=(frac{x_0}{x_i},dots , frac{x_n}{x_i})$
That it is an omeomorphism but $cup_{i=0}^n U_i=mathbb{P}^n$ and for every $ineq j$
$phi_jcirc phi_i^{-1}(x_1,dots , x_n)= (frac{x_1}{x_j},dots ,frac{1}{x_j},dots , frac{x_n}{x_j}) $
and in $phi_i(U_icap U_j)={(x_1,dots , x_{n}) : x_jneq 0}$
Is infinitely differentiable
Your map is not well defined.. for example $f_0([1,1,dots , 0])=(1,0, dots ,0)$ but $[1,1,0,dots , 0]=[2,2,0 dots, 0]$so $f_0([2,2,0 dots, 0]=(2,0,dots, 0)$
answered Dec 10 '18 at 15:02
Federico FalluccaFederico Fallucca
1,95219
$endgroup$
1
$begingroup$
But I'm using the representative of the class that has a 1 on the i-th coordinate, so $[2,2,0...0] $ is in fact represented by $[1,1,0...0]$. I don't see why I can't do this
$endgroup$
– Johanna
Dec 10 '18 at 15:38
$begingroup$
You’re right sorry
$endgroup$
– Federico Fallucca
Dec 10 '18 at 15:46
add a comment |
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1 Answer
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1 Answer
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$begingroup$
$U_i:={[x_0,dots , x_n] : x_ineq 0}$ is an open set of $mathbb{P}^n$ and you can define
$phi_i: U_ito mathbb{R}^n$ such that
$phi_i([x_0,dots , x_n] ):=(frac{x_0}{x_i},dots , frac{x_n}{x_i})$
That it is an omeomorphism but $cup_{i=0}^n U_i=mathbb{P}^n$ and for every $ineq j$
$phi_jcirc phi_i^{-1}(x_1,dots , x_n)= (frac{x_1}{x_j},dots ,frac{1}{x_j},dots , frac{x_n}{x_j}) $
and in $phi_i(U_icap U_j)={(x_1,dots , x_{n}) : x_jneq 0}$
Is infinitely differentiable
Your map is not well defined.. for example $f_0([1,1,dots , 0])=(1,0, dots ,0)$ but $[1,1,0,dots , 0]=[2,2,0 dots, 0]$so $f_0([2,2,0 dots, 0]=(2,0,dots, 0)$
answered Dec 10 '18 at 15:02
Federico FalluccaFederico Fallucca
1,95219
$endgroup$
1
$begingroup$
But I'm using the representative of the class that has a 1 on the i-th coordinate, so $[2,2,0...0] $ is in fact represented by $[1,1,0...0]$. I don't see why I can't do this
$endgroup$
– Johanna
Dec 10 '18 at 15:38
$begingroup$
You’re right sorry
$endgroup$
– Federico Fallucca
Dec 10 '18 at 15:46
add a comment |
$begingroup$
$U_i:={[x_0,dots , x_n] : x_ineq 0}$ is an open set of $mathbb{P}^n$ and you can define
$phi_i: U_ito mathbb{R}^n$ such that
$phi_i([x_0,dots , x_n] ):=(frac{x_0}{x_i},dots , frac{x_n}{x_i})$
That it is an omeomorphism but $cup_{i=0}^n U_i=mathbb{P}^n$ and for every $ineq j$
$phi_jcirc phi_i^{-1}(x_1,dots , x_n)= (frac{x_1}{x_j},dots ,frac{1}{x_j},dots , frac{x_n}{x_j}) $
and in $phi_i(U_icap U_j)={(x_1,dots , x_{n}) : x_jneq 0}$
Is infinitely differentiable
Your map is not well defined.. for example $f_0([1,1,dots , 0])=(1,0, dots ,0)$ but $[1,1,0,dots , 0]=[2,2,0 dots, 0]$so $f_0([2,2,0 dots, 0]=(2,0,dots, 0)$
answered Dec 10 '18 at 15:02
Federico FalluccaFederico Fallucca
1,95219
$endgroup$
1
$begingroup$
But I'm using the representative of the class that has a 1 on the i-th coordinate, so $[2,2,0...0] $ is in fact represented by $[1,1,0...0]$. I don't see why I can't do this
$endgroup$
– Johanna
Dec 10 '18 at 15:38
$begingroup$
You’re right sorry
$endgroup$
– Federico Fallucca
Dec 10 '18 at 15:46
add a comment |
$begingroup$
$U_i:={[x_0,dots , x_n] : x_ineq 0}$ is an open set of $mathbb{P}^n$ and you can define
$phi_i: U_ito mathbb{R}^n$ such that
$phi_i([x_0,dots , x_n] ):=(frac{x_0}{x_i},dots , frac{x_n}{x_i})$
That it is an omeomorphism but $cup_{i=0}^n U_i=mathbb{P}^n$ and for every $ineq j$
$phi_jcirc phi_i^{-1}(x_1,dots , x_n)= (frac{x_1}{x_j},dots ,frac{1}{x_j},dots , frac{x_n}{x_j}) $
and in $phi_i(U_icap U_j)={(x_1,dots , x_{n}) : x_jneq 0}$
Is infinitely differentiable
Your map is not well defined.. for example $f_0([1,1,dots , 0])=(1,0, dots ,0)$ but $[1,1,0,dots , 0]=[2,2,0 dots, 0]$so $f_0([2,2,0 dots, 0]=(2,0,dots, 0)$
answered Dec 10 '18 at 15:02
Federico FalluccaFederico Fallucca
1,95219
$endgroup$
$U_i:={[x_0,dots , x_n] : x_ineq 0}$ is an open set of $mathbb{P}^n$ and you can define
$phi_i: U_ito mathbb{R}^n$ such that
$phi_i([x_0,dots , x_n] ):=(frac{x_0}{x_i},dots , frac{x_n}{x_i})$
That it is an omeomorphism but $cup_{i=0}^n U_i=mathbb{P}^n$ and for every $ineq j$
$phi_jcirc phi_i^{-1}(x_1,dots , x_n)= (frac{x_1}{x_j},dots ,frac{1}{x_j},dots , frac{x_n}{x_j}) $
and in $phi_i(U_icap U_j)={(x_1,dots , x_{n}) : x_jneq 0}$
Is infinitely differentiable
Your map is not well defined.. for example $f_0([1,1,dots , 0])=(1,0, dots ,0)$ but $[1,1,0,dots , 0]=[2,2,0 dots, 0]$so $f_0([2,2,0 dots, 0]=(2,0,dots, 0)$
answered Dec 10 '18 at 15:02
Federico FalluccaFederico Fallucca
1,95219
edited Dec 10 '18 at 15:09
answered Dec 10 '18 at 15:02
Federico FalluccaFederico Fallucca
1,95219
answered Dec 10 '18 at 15:02
Federico FalluccaFederico Fallucca
1,95219
answered Dec 10 '18 at 15:02
Federico FalluccaFederico Fallucca
1,95219
1,95219
1
$begingroup$
But I'm using the representative of the class that has a 1 on the i-th coordinate, so $[2,2,0...0] $ is in fact represented by $[1,1,0...0]$. I don't see why I can't do this
$endgroup$
– Johanna
Dec 10 '18 at 15:38
$begingroup$
You’re right sorry
$endgroup$
– Federico Fallucca
Dec 10 '18 at 15:46
add a comment |
1
$begingroup$
But I'm using the representative of the class that has a 1 on the i-th coordinate, so $[2,2,0...0] $ is in fact represented by $[1,1,0...0]$. I don't see why I can't do this
$endgroup$
– Johanna
Dec 10 '18 at 15:38
$begingroup$
You’re right sorry
$endgroup$
– Federico Fallucca
Dec 10 '18 at 15:46
1
1
$begingroup$
But I'm using the representative of the class that has a 1 on the i-th coordinate, so $[2,2,0...0] $ is in fact represented by $[1,1,0...0]$. I don't see why I can't do this
$endgroup$
– Johanna
Dec 10 '18 at 15:38
$begingroup$
But I'm using the representative of the class that has a 1 on the i-th coordinate, so $[2,2,0...0] $ is in fact represented by $[1,1,0...0]$. I don't see why I can't do this
$endgroup$
– Johanna
Dec 10 '18 at 15:38
$begingroup$
You’re right sorry
$endgroup$
– Federico Fallucca
Dec 10 '18 at 15:46
$begingroup$
You’re right sorry
$endgroup$
– Federico Fallucca
Dec 10 '18 at 15:46
add a comment |
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$begingroup$
You wrote "I only need to prove that this is an homeomorphism". A formula does not define a function, and hence does not define a homeomorphism. You must also specify the domain and range. Notice how this is done in the answer of @FedericoFallucca.
$endgroup$
– Lee Mosher
Dec 10 '18 at 15:24
$begingroup$
Oh sorry I forgot that line, already edited it.
$endgroup$
– Johanna
Dec 10 '18 at 15:30