Normal Vector Bundle of RPn
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From Hatcher:
In the normal bundle $N S^n$ the identification $(x, v) ∼ (−x, −v)$ can be written as $(x,tx) ∼ (−x,t(−x))$. This identification yields the product bundle $mathbb{RP}^n×mathbb{R}$ since the section $x to (−x, −x)$ is well-defined in the quotient.
What does this even mean? Is the section $S^n to NS^n$ or $mathbb{RP}^n to N mathbb{RP}^n$? What does the second one mean? I understand that the normal bundle on $S^1$ is trivial, and I see how when $n = 1$ this is different from the Mobius bundle, but I don't understand what Hatcher is trying to say here. Any insight would be appreciated.
general-topology geometry algebraic-topology vector-bundles
asked Nov 20 at 21:55
Emilio Minichiello
3297
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From Hatcher:
In the normal bundle $N S^n$ the identification $(x, v) ∼ (−x, −v)$ can be written as $(x,tx) ∼ (−x,t(−x))$. This identification yields the product bundle $mathbb{RP}^n×mathbb{R}$ since the section $x to (−x, −x)$ is well-defined in the quotient.
What does this even mean? Is the section $S^n to NS^n$ or $mathbb{RP}^n to N mathbb{RP}^n$? What does the second one mean? I understand that the normal bundle on $S^1$ is trivial, and I see how when $n = 1$ this is different from the Mobius bundle, but I don't understand what Hatcher is trying to say here. Any insight would be appreciated.
general-topology geometry algebraic-topology vector-bundles
asked Nov 20 at 21:55
Emilio Minichiello
3297
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
From Hatcher:
In the normal bundle $N S^n$ the identification $(x, v) ∼ (−x, −v)$ can be written as $(x,tx) ∼ (−x,t(−x))$. This identification yields the product bundle $mathbb{RP}^n×mathbb{R}$ since the section $x to (−x, −x)$ is well-defined in the quotient.
What does this even mean? Is the section $S^n to NS^n$ or $mathbb{RP}^n to N mathbb{RP}^n$? What does the second one mean? I understand that the normal bundle on $S^1$ is trivial, and I see how when $n = 1$ this is different from the Mobius bundle, but I don't understand what Hatcher is trying to say here. Any insight would be appreciated.
general-topology geometry algebraic-topology vector-bundles
asked Nov 20 at 21:55
Emilio Minichiello
3297
From Hatcher:
In the normal bundle $N S^n$ the identification $(x, v) ∼ (−x, −v)$ can be written as $(x,tx) ∼ (−x,t(−x))$. This identification yields the product bundle $mathbb{RP}^n×mathbb{R}$ since the section $x to (−x, −x)$ is well-defined in the quotient.
What does this even mean? Is the section $S^n to NS^n$ or $mathbb{RP}^n to N mathbb{RP}^n$? What does the second one mean? I understand that the normal bundle on $S^1$ is trivial, and I see how when $n = 1$ this is different from the Mobius bundle, but I don't understand what Hatcher is trying to say here. Any insight would be appreciated.
general-topology geometry algebraic-topology vector-bundles
general-topology geometry algebraic-topology vector-bundles
asked Nov 20 at 21:55
Emilio Minichiello
3297
asked Nov 20 at 21:55
Emilio Minichiello
3297
asked Nov 20 at 21:55
Emilio Minichiello
3297
asked Nov 20 at 21:55
Emilio Minichiello
3297
asked Nov 20 at 21:55
Emilio Minichiello
3297
3297
add a comment |
add a comment |
1 Answer
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If $M$ is a submanifold of $mathbb{R}^n$, the normal bundle of $M$ is the vector bundle $NM$ such that for every $xin M$, $NM_x$ the fibre of $NM$ at $x$ is the subspace of $mathbb{R}^n$ othogonal to $T_xM$.
If $M=S^n$, $NS_x=(x,tx),tinmathbb{R}$. You cannot define the normal bundle of $mathbb{R}P^n$ since it is not embedded in $mathbb{R}^n$, but $NS^n$ behaves well in respect to $(x,tx)rightarrow (-x,-tx)$ as Hatcher mentioned and induces a bundle on $mathbb{R}P^n$ which is trivial. Basically, you can lift the action of $mathbb{Z}/2$ of on $S^n$ and the quotient $NP^n$ is a trivial bundle. Another way to see this, is t consider the map $q:mathbb{R}P^ntimes mathbb{R}rightarrow NP^n$ defined by $q([x],t)rightarrow p(x,tx)$ where $p:S^nrightarrow mathbb{R}P^n$ is the canonical projection and $p(x)=[x]$. In fact Hatcher says that $p(x,t)=p(-x,t)$ thus $q$ is well defined.
answered Nov 20 at 22:11
Tsemo Aristide
55.2k11444
Shouldn't $p$ only accept one argument though? How do you write $p(x,tx)$ if $p: S^n to mathbb{RP}^n$?
– Emilio Minichiello
Nov 20 at 22:24
add a comment |
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1 Answer
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If $M$ is a submanifold of $mathbb{R}^n$, the normal bundle of $M$ is the vector bundle $NM$ such that for every $xin M$, $NM_x$ the fibre of $NM$ at $x$ is the subspace of $mathbb{R}^n$ othogonal to $T_xM$.
If $M=S^n$, $NS_x=(x,tx),tinmathbb{R}$. You cannot define the normal bundle of $mathbb{R}P^n$ since it is not embedded in $mathbb{R}^n$, but $NS^n$ behaves well in respect to $(x,tx)rightarrow (-x,-tx)$ as Hatcher mentioned and induces a bundle on $mathbb{R}P^n$ which is trivial. Basically, you can lift the action of $mathbb{Z}/2$ of on $S^n$ and the quotient $NP^n$ is a trivial bundle. Another way to see this, is t consider the map $q:mathbb{R}P^ntimes mathbb{R}rightarrow NP^n$ defined by $q([x],t)rightarrow p(x,tx)$ where $p:S^nrightarrow mathbb{R}P^n$ is the canonical projection and $p(x)=[x]$. In fact Hatcher says that $p(x,t)=p(-x,t)$ thus $q$ is well defined.
answered Nov 20 at 22:11
Tsemo Aristide
55.2k11444
Shouldn't $p$ only accept one argument though? How do you write $p(x,tx)$ if $p: S^n to mathbb{RP}^n$?
– Emilio Minichiello
Nov 20 at 22:24
add a comment |
up vote
2
down vote
If $M$ is a submanifold of $mathbb{R}^n$, the normal bundle of $M$ is the vector bundle $NM$ such that for every $xin M$, $NM_x$ the fibre of $NM$ at $x$ is the subspace of $mathbb{R}^n$ othogonal to $T_xM$.
If $M=S^n$, $NS_x=(x,tx),tinmathbb{R}$. You cannot define the normal bundle of $mathbb{R}P^n$ since it is not embedded in $mathbb{R}^n$, but $NS^n$ behaves well in respect to $(x,tx)rightarrow (-x,-tx)$ as Hatcher mentioned and induces a bundle on $mathbb{R}P^n$ which is trivial. Basically, you can lift the action of $mathbb{Z}/2$ of on $S^n$ and the quotient $NP^n$ is a trivial bundle. Another way to see this, is t consider the map $q:mathbb{R}P^ntimes mathbb{R}rightarrow NP^n$ defined by $q([x],t)rightarrow p(x,tx)$ where $p:S^nrightarrow mathbb{R}P^n$ is the canonical projection and $p(x)=[x]$. In fact Hatcher says that $p(x,t)=p(-x,t)$ thus $q$ is well defined.
answered Nov 20 at 22:11
Tsemo Aristide
55.2k11444
Shouldn't $p$ only accept one argument though? How do you write $p(x,tx)$ if $p: S^n to mathbb{RP}^n$?
– Emilio Minichiello
Nov 20 at 22:24
add a comment |
up vote
2
down vote
up vote
2
down vote
If $M$ is a submanifold of $mathbb{R}^n$, the normal bundle of $M$ is the vector bundle $NM$ such that for every $xin M$, $NM_x$ the fibre of $NM$ at $x$ is the subspace of $mathbb{R}^n$ othogonal to $T_xM$.
If $M=S^n$, $NS_x=(x,tx),tinmathbb{R}$. You cannot define the normal bundle of $mathbb{R}P^n$ since it is not embedded in $mathbb{R}^n$, but $NS^n$ behaves well in respect to $(x,tx)rightarrow (-x,-tx)$ as Hatcher mentioned and induces a bundle on $mathbb{R}P^n$ which is trivial. Basically, you can lift the action of $mathbb{Z}/2$ of on $S^n$ and the quotient $NP^n$ is a trivial bundle. Another way to see this, is t consider the map $q:mathbb{R}P^ntimes mathbb{R}rightarrow NP^n$ defined by $q([x],t)rightarrow p(x,tx)$ where $p:S^nrightarrow mathbb{R}P^n$ is the canonical projection and $p(x)=[x]$. In fact Hatcher says that $p(x,t)=p(-x,t)$ thus $q$ is well defined.
answered Nov 20 at 22:11
Tsemo Aristide
55.2k11444
If $M$ is a submanifold of $mathbb{R}^n$, the normal bundle of $M$ is the vector bundle $NM$ such that for every $xin M$, $NM_x$ the fibre of $NM$ at $x$ is the subspace of $mathbb{R}^n$ othogonal to $T_xM$.
If $M=S^n$, $NS_x=(x,tx),tinmathbb{R}$. You cannot define the normal bundle of $mathbb{R}P^n$ since it is not embedded in $mathbb{R}^n$, but $NS^n$ behaves well in respect to $(x,tx)rightarrow (-x,-tx)$ as Hatcher mentioned and induces a bundle on $mathbb{R}P^n$ which is trivial. Basically, you can lift the action of $mathbb{Z}/2$ of on $S^n$ and the quotient $NP^n$ is a trivial bundle. Another way to see this, is t consider the map $q:mathbb{R}P^ntimes mathbb{R}rightarrow NP^n$ defined by $q([x],t)rightarrow p(x,tx)$ where $p:S^nrightarrow mathbb{R}P^n$ is the canonical projection and $p(x)=[x]$. In fact Hatcher says that $p(x,t)=p(-x,t)$ thus $q$ is well defined.
answered Nov 20 at 22:11
Tsemo Aristide
55.2k11444
edited Nov 20 at 22:18
answered Nov 20 at 22:11
Tsemo Aristide
55.2k11444
answered Nov 20 at 22:11
Tsemo Aristide
55.2k11444
answered Nov 20 at 22:11
Tsemo Aristide
55.2k11444
55.2k11444
Shouldn't $p$ only accept one argument though? How do you write $p(x,tx)$ if $p: S^n to mathbb{RP}^n$?
– Emilio Minichiello
Nov 20 at 22:24
add a comment |
Shouldn't $p$ only accept one argument though? How do you write $p(x,tx)$ if $p: S^n to mathbb{RP}^n$?
– Emilio Minichiello
Nov 20 at 22:24
Shouldn't $p$ only accept one argument though? How do you write $p(x,tx)$ if $p: S^n to mathbb{RP}^n$?
– Emilio Minichiello
Nov 20 at 22:24
Shouldn't $p$ only accept one argument though? How do you write $p(x,tx)$ if $p: S^n to mathbb{RP}^n$?
– Emilio Minichiello
Nov 20 at 22:24
add a comment |
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