My proof that the real projective space $P^n$ is locally euclidean.












1














I would like to show that the real projective space $P^n$ is locally homeomorphic to an open subset of $mathbb{R}^{n-1}$.



I would just like to sketch the proof before diving into details. Let $[x]$ be a linear subspace in $P^n$, then there exists at least one element $x in mathbb{R}^n$ and $x in [x]$ such that the $i$th component is equals to one.



Take an open ball $B subset mathbb{R}^n$ around the point $x$.
And if we restrict the quotient map $q: mathbb{R}^n times{0) rightarrow P^n$ that defines $P^n$ and its topology to $B cap (mathbb{R}^{n} cap (x in mathbb{R}^n |x_i = 1))$.



Then we can show that this restriction is a homeomorphism between $B cap (mathbb{R}^{n} cap (x in mathbb{R}^n |x_i = 1))$ and its image.



Is this feasible?



I checked and I don't think anyone else used a similar proof for showing $P^n$ is a manifold, so I want to know if my strategy is valid.



thank you






general-topology manifolds







share|cite|improve this question




















  • 1




    This is ok if you can show that the image of your restriction is open. That follows if the quotient map is open. And it is open because $q^{-1}(q(U))=bigcup_{lambdaneq 0}lambda U$. Also note that the ball $B$ has to be chosen with radius less then $1$. Otherwise you risk that $0in B$ (btw. your quotient map should be $q:mathbb{R}^nbackslash 0to P^n$). An alternative (and imo a bit simplier) approach is to consider $P^n$ as the quotient of $S^n$ under antipodal points identification.
    – freakish
    Nov 23 at 14:54












  • Thanks for noting the typo. However, does the radius of the ball matter? Since I am only looking at the intersection of the ball with $(mathbb{R}^{n} cap (x in mathbb{R}^n |x_i = 1))$, that shouldn't include $0$ even if it's in $B$ right?
    – Ecotistician
    Nov 23 at 21:04












  • What is your definition of $P^n$?
    – Paul Frost
    Nov 27 at 10:04
















1














I would like to show that the real projective space $P^n$ is locally homeomorphic to an open subset of $mathbb{R}^{n-1}$.



I would just like to sketch the proof before diving into details. Let $[x]$ be a linear subspace in $P^n$, then there exists at least one element $x in mathbb{R}^n$ and $x in [x]$ such that the $i$th component is equals to one.



Take an open ball $B subset mathbb{R}^n$ around the point $x$.
And if we restrict the quotient map $q: mathbb{R}^n times{0) rightarrow P^n$ that defines $P^n$ and its topology to $B cap (mathbb{R}^{n} cap (x in mathbb{R}^n |x_i = 1))$.



Then we can show that this restriction is a homeomorphism between $B cap (mathbb{R}^{n} cap (x in mathbb{R}^n |x_i = 1))$ and its image.



Is this feasible?



I checked and I don't think anyone else used a similar proof for showing $P^n$ is a manifold, so I want to know if my strategy is valid.



thank you






general-topology manifolds







share|cite|improve this question




















  • 1




    This is ok if you can show that the image of your restriction is open. That follows if the quotient map is open. And it is open because $q^{-1}(q(U))=bigcup_{lambdaneq 0}lambda U$. Also note that the ball $B$ has to be chosen with radius less then $1$. Otherwise you risk that $0in B$ (btw. your quotient map should be $q:mathbb{R}^nbackslash 0to P^n$). An alternative (and imo a bit simplier) approach is to consider $P^n$ as the quotient of $S^n$ under antipodal points identification.
    – freakish
    Nov 23 at 14:54












  • Thanks for noting the typo. However, does the radius of the ball matter? Since I am only looking at the intersection of the ball with $(mathbb{R}^{n} cap (x in mathbb{R}^n |x_i = 1))$, that shouldn't include $0$ even if it's in $B$ right?
    – Ecotistician
    Nov 23 at 21:04












  • What is your definition of $P^n$?
    – Paul Frost
    Nov 27 at 10:04














1












1








1







I would like to show that the real projective space $P^n$ is locally homeomorphic to an open subset of $mathbb{R}^{n-1}$.



I would just like to sketch the proof before diving into details. Let $[x]$ be a linear subspace in $P^n$, then there exists at least one element $x in mathbb{R}^n$ and $x in [x]$ such that the $i$th component is equals to one.



Take an open ball $B subset mathbb{R}^n$ around the point $x$.
And if we restrict the quotient map $q: mathbb{R}^n times{0) rightarrow P^n$ that defines $P^n$ and its topology to $B cap (mathbb{R}^{n} cap (x in mathbb{R}^n |x_i = 1))$.



Then we can show that this restriction is a homeomorphism between $B cap (mathbb{R}^{n} cap (x in mathbb{R}^n |x_i = 1))$ and its image.



Is this feasible?



I checked and I don't think anyone else used a similar proof for showing $P^n$ is a manifold, so I want to know if my strategy is valid.



thank you






general-topology manifolds







share|cite|improve this question















I would like to show that the real projective space $P^n$ is locally homeomorphic to an open subset of $mathbb{R}^{n-1}$.



I would just like to sketch the proof before diving into details. Let $[x]$ be a linear subspace in $P^n$, then there exists at least one element $x in mathbb{R}^n$ and $x in [x]$ such that the $i$th component is equals to one.



Take an open ball $B subset mathbb{R}^n$ around the point $x$.
And if we restrict the quotient map $q: mathbb{R}^n times{0) rightarrow P^n$ that defines $P^n$ and its topology to $B cap (mathbb{R}^{n} cap (x in mathbb{R}^n |x_i = 1))$.



Then we can show that this restriction is a homeomorphism between $B cap (mathbb{R}^{n} cap (x in mathbb{R}^n |x_i = 1))$ and its image.



Is this feasible?



I checked and I don't think anyone else used a similar proof for showing $P^n$ is a manifold, so I want to know if my strategy is valid.



thank you






general-topology manifolds




general-topology manifolds






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 23 at 21:04

























asked Nov 23 at 6:59









Ecotistician

31318




31318








  • 1




    This is ok if you can show that the image of your restriction is open. That follows if the quotient map is open. And it is open because $q^{-1}(q(U))=bigcup_{lambdaneq 0}lambda U$. Also note that the ball $B$ has to be chosen with radius less then $1$. Otherwise you risk that $0in B$ (btw. your quotient map should be $q:mathbb{R}^nbackslash 0to P^n$). An alternative (and imo a bit simplier) approach is to consider $P^n$ as the quotient of $S^n$ under antipodal points identification.
    – freakish
    Nov 23 at 14:54












  • Thanks for noting the typo. However, does the radius of the ball matter? Since I am only looking at the intersection of the ball with $(mathbb{R}^{n} cap (x in mathbb{R}^n |x_i = 1))$, that shouldn't include $0$ even if it's in $B$ right?
    – Ecotistician
    Nov 23 at 21:04












  • What is your definition of $P^n$?
    – Paul Frost
    Nov 27 at 10:04














  • 1




    This is ok if you can show that the image of your restriction is open. That follows if the quotient map is open. And it is open because $q^{-1}(q(U))=bigcup_{lambdaneq 0}lambda U$. Also note that the ball $B$ has to be chosen with radius less then $1$. Otherwise you risk that $0in B$ (btw. your quotient map should be $q:mathbb{R}^nbackslash 0to P^n$). An alternative (and imo a bit simplier) approach is to consider $P^n$ as the quotient of $S^n$ under antipodal points identification.
    – freakish
    Nov 23 at 14:54












  • Thanks for noting the typo. However, does the radius of the ball matter? Since I am only looking at the intersection of the ball with $(mathbb{R}^{n} cap (x in mathbb{R}^n |x_i = 1))$, that shouldn't include $0$ even if it's in $B$ right?
    – Ecotistician
    Nov 23 at 21:04












  • What is your definition of $P^n$?
    – Paul Frost
    Nov 27 at 10:04








1




1




This is ok if you can show that the image of your restriction is open. That follows if the quotient map is open. And it is open because $q^{-1}(q(U))=bigcup_{lambdaneq 0}lambda U$. Also note that the ball $B$ has to be chosen with radius less then $1$. Otherwise you risk that $0in B$ (btw. your quotient map should be $q:mathbb{R}^nbackslash 0to P^n$). An alternative (and imo a bit simplier) approach is to consider $P^n$ as the quotient of $S^n$ under antipodal points identification.
– freakish
Nov 23 at 14:54






This is ok if you can show that the image of your restriction is open. That follows if the quotient map is open. And it is open because $q^{-1}(q(U))=bigcup_{lambdaneq 0}lambda U$. Also note that the ball $B$ has to be chosen with radius less then $1$. Otherwise you risk that $0in B$ (btw. your quotient map should be $q:mathbb{R}^nbackslash 0to P^n$). An alternative (and imo a bit simplier) approach is to consider $P^n$ as the quotient of $S^n$ under antipodal points identification.
– freakish
Nov 23 at 14:54














Thanks for noting the typo. However, does the radius of the ball matter? Since I am only looking at the intersection of the ball with $(mathbb{R}^{n} cap (x in mathbb{R}^n |x_i = 1))$, that shouldn't include $0$ even if it's in $B$ right?
– Ecotistician
Nov 23 at 21:04






Thanks for noting the typo. However, does the radius of the ball matter? Since I am only looking at the intersection of the ball with $(mathbb{R}^{n} cap (x in mathbb{R}^n |x_i = 1))$, that shouldn't include $0$ even if it's in $B$ right?
– Ecotistician
Nov 23 at 21:04














What is your definition of $P^n$?
– Paul Frost
Nov 27 at 10:04




What is your definition of $P^n$?
– Paul Frost
Nov 27 at 10:04















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