Is Projective space minus a point really homeomorphic to vector space?
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In his book Fundamentals of Algebraic Topology Weintraub claims on page 96:
For $k=mathbb R$ or $mathbb C$, $mathbb kP^nsetminus [1,0,cdots,0]$ is homeomorphic to $k^n$.
This look fishy to me, in particular because for $k=mathbb C, ngeq 2$ the assertion is completely false in the holomorphic category.
However I can neither prove nor disprove Weintraub's statement in the topological category.
So, is the displayed assertion true or false?
algebraic-topology
asked Dec 21 '18 at 9:30
doloreshazedoloreshaze
141
$endgroup$
add a comment |
$begingroup$
In his book Fundamentals of Algebraic Topology Weintraub claims on page 96:
For $k=mathbb R$ or $mathbb C$, $mathbb kP^nsetminus [1,0,cdots,0]$ is homeomorphic to $k^n$.
This look fishy to me, in particular because for $k=mathbb C, ngeq 2$ the assertion is completely false in the holomorphic category.
However I can neither prove nor disprove Weintraub's statement in the topological category.
So, is the displayed assertion true or false?
algebraic-topology
asked Dec 21 '18 at 9:30
doloreshazedoloreshaze
141
$endgroup$
3
$begingroup$
Put the standard CW structure on $mathbb{K}P^n$. Then you can assume up to homotopy that the point $x$ being removed lies in the open top cell. Then $mathbb{K}P^{n}setminus{x}$ deformation retracts onto $mathbb{K}P^{n-1}$, and this is clearly not homotopy equivalent to a (contractible) vector space. Actually $mathbb{K}P^{n}setminus{x}$ is a tubular neighbourhood of the canonical embedding $mathbb{K}P^{n-1}hookrightarrow mathbb{K}P^{n}$. Perhaps this is what Weintraub has meant? (Although I checked the book, and he has written exactly what you say.)
$endgroup$
– Tyrone
Dec 21 '18 at 10:26
2
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Note that the projective space minus a point is diffeomorphic to the total space of the tautological bundle of projective space of dimension one lower. Or in other words the projective space is the Thom space of said bundle.
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– ThorbenK
Dec 21 '18 at 14:30
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@Thorbenk You are absolutely right, and your diffeomorphism is even an analytic isomorphism.
$endgroup$
– doloreshaze
Dec 21 '18 at 16:35
add a comment |
$begingroup$
In his book Fundamentals of Algebraic Topology Weintraub claims on page 96:
For $k=mathbb R$ or $mathbb C$, $mathbb kP^nsetminus [1,0,cdots,0]$ is homeomorphic to $k^n$.
This look fishy to me, in particular because for $k=mathbb C, ngeq 2$ the assertion is completely false in the holomorphic category.
However I can neither prove nor disprove Weintraub's statement in the topological category.
So, is the displayed assertion true or false?
algebraic-topology
asked Dec 21 '18 at 9:30
doloreshazedoloreshaze
141
$endgroup$
In his book Fundamentals of Algebraic Topology Weintraub claims on page 96:
For $k=mathbb R$ or $mathbb C$, $mathbb kP^nsetminus [1,0,cdots,0]$ is homeomorphic to $k^n$.
This look fishy to me, in particular because for $k=mathbb C, ngeq 2$ the assertion is completely false in the holomorphic category.
However I can neither prove nor disprove Weintraub's statement in the topological category.
So, is the displayed assertion true or false?
algebraic-topology
algebraic-topology
asked Dec 21 '18 at 9:30
doloreshazedoloreshaze
141
asked Dec 21 '18 at 9:30
doloreshazedoloreshaze
141
asked Dec 21 '18 at 9:30
doloreshazedoloreshaze
141
asked Dec 21 '18 at 9:30
doloreshazedoloreshaze
141
asked Dec 21 '18 at 9:30
doloreshazedoloreshaze
141
141
3
$begingroup$
Put the standard CW structure on $mathbb{K}P^n$. Then you can assume up to homotopy that the point $x$ being removed lies in the open top cell. Then $mathbb{K}P^{n}setminus{x}$ deformation retracts onto $mathbb{K}P^{n-1}$, and this is clearly not homotopy equivalent to a (contractible) vector space. Actually $mathbb{K}P^{n}setminus{x}$ is a tubular neighbourhood of the canonical embedding $mathbb{K}P^{n-1}hookrightarrow mathbb{K}P^{n}$. Perhaps this is what Weintraub has meant? (Although I checked the book, and he has written exactly what you say.)
$endgroup$
– Tyrone
Dec 21 '18 at 10:26
2
$begingroup$
Note that the projective space minus a point is diffeomorphic to the total space of the tautological bundle of projective space of dimension one lower. Or in other words the projective space is the Thom space of said bundle.
$endgroup$
– ThorbenK
Dec 21 '18 at 14:30
$begingroup$
@Thorbenk You are absolutely right, and your diffeomorphism is even an analytic isomorphism.
$endgroup$
– doloreshaze
Dec 21 '18 at 16:35
add a comment |
3
$begingroup$
Put the standard CW structure on $mathbb{K}P^n$. Then you can assume up to homotopy that the point $x$ being removed lies in the open top cell. Then $mathbb{K}P^{n}setminus{x}$ deformation retracts onto $mathbb{K}P^{n-1}$, and this is clearly not homotopy equivalent to a (contractible) vector space. Actually $mathbb{K}P^{n}setminus{x}$ is a tubular neighbourhood of the canonical embedding $mathbb{K}P^{n-1}hookrightarrow mathbb{K}P^{n}$. Perhaps this is what Weintraub has meant? (Although I checked the book, and he has written exactly what you say.)
$endgroup$
– Tyrone
Dec 21 '18 at 10:26
2
$begingroup$
Note that the projective space minus a point is diffeomorphic to the total space of the tautological bundle of projective space of dimension one lower. Or in other words the projective space is the Thom space of said bundle.
$endgroup$
– ThorbenK
Dec 21 '18 at 14:30
$begingroup$
@Thorbenk You are absolutely right, and your diffeomorphism is even an analytic isomorphism.
$endgroup$
– doloreshaze
Dec 21 '18 at 16:35
3
3
$begingroup$
Put the standard CW structure on $mathbb{K}P^n$. Then you can assume up to homotopy that the point $x$ being removed lies in the open top cell. Then $mathbb{K}P^{n}setminus{x}$ deformation retracts onto $mathbb{K}P^{n-1}$, and this is clearly not homotopy equivalent to a (contractible) vector space. Actually $mathbb{K}P^{n}setminus{x}$ is a tubular neighbourhood of the canonical embedding $mathbb{K}P^{n-1}hookrightarrow mathbb{K}P^{n}$. Perhaps this is what Weintraub has meant? (Although I checked the book, and he has written exactly what you say.)
$endgroup$
– Tyrone
Dec 21 '18 at 10:26
$begingroup$
Put the standard CW structure on $mathbb{K}P^n$. Then you can assume up to homotopy that the point $x$ being removed lies in the open top cell. Then $mathbb{K}P^{n}setminus{x}$ deformation retracts onto $mathbb{K}P^{n-1}$, and this is clearly not homotopy equivalent to a (contractible) vector space. Actually $mathbb{K}P^{n}setminus{x}$ is a tubular neighbourhood of the canonical embedding $mathbb{K}P^{n-1}hookrightarrow mathbb{K}P^{n}$. Perhaps this is what Weintraub has meant? (Although I checked the book, and he has written exactly what you say.)
$endgroup$
– Tyrone
Dec 21 '18 at 10:26
2
2
$begingroup$
Note that the projective space minus a point is diffeomorphic to the total space of the tautological bundle of projective space of dimension one lower. Or in other words the projective space is the Thom space of said bundle.
$endgroup$
– ThorbenK
Dec 21 '18 at 14:30
$begingroup$
Note that the projective space minus a point is diffeomorphic to the total space of the tautological bundle of projective space of dimension one lower. Or in other words the projective space is the Thom space of said bundle.
$endgroup$
– ThorbenK
Dec 21 '18 at 14:30
$begingroup$
@Thorbenk You are absolutely right, and your diffeomorphism is even an analytic isomorphism.
$endgroup$
– doloreshaze
Dec 21 '18 at 16:35
$begingroup$
@Thorbenk You are absolutely right, and your diffeomorphism is even an analytic isomorphism.
$endgroup$
– doloreshaze
Dec 21 '18 at 16:35
add a comment |
1 Answer
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As mentioned by Tyrone in his comment (which I upvoted) , Weintraub is indeed completely wrong because $X=kmathbb P^nsetminus [1,0,cdots,0]$ is homotopic to $kmathbb P^{n-1}$, which of course is not homotopic to the contractible space $k^n$ for $ngeq 2$.
A different from Tyrone's way to see the homotopy is by noticing that $kmathbb P^{n-1}$ (identified to the the hyperplane $x_0=0$ of $kmathbb P^n$) is a strong deformation retract of $X$ under the homotopy $(t,[x_0,x_1,cdots,x_n])mapsto [tx_0,x_1,cdots,x_n]$.
answered Dec 21 '18 at 12:15
doloreshazedoloreshaze
141
$endgroup$
add a comment |
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As mentioned by Tyrone in his comment (which I upvoted) , Weintraub is indeed completely wrong because $X=kmathbb P^nsetminus [1,0,cdots,0]$ is homotopic to $kmathbb P^{n-1}$, which of course is not homotopic to the contractible space $k^n$ for $ngeq 2$.
A different from Tyrone's way to see the homotopy is by noticing that $kmathbb P^{n-1}$ (identified to the the hyperplane $x_0=0$ of $kmathbb P^n$) is a strong deformation retract of $X$ under the homotopy $(t,[x_0,x_1,cdots,x_n])mapsto [tx_0,x_1,cdots,x_n]$.
answered Dec 21 '18 at 12:15
doloreshazedoloreshaze
141
$endgroup$
add a comment |
$begingroup$
As mentioned by Tyrone in his comment (which I upvoted) , Weintraub is indeed completely wrong because $X=kmathbb P^nsetminus [1,0,cdots,0]$ is homotopic to $kmathbb P^{n-1}$, which of course is not homotopic to the contractible space $k^n$ for $ngeq 2$.
A different from Tyrone's way to see the homotopy is by noticing that $kmathbb P^{n-1}$ (identified to the the hyperplane $x_0=0$ of $kmathbb P^n$) is a strong deformation retract of $X$ under the homotopy $(t,[x_0,x_1,cdots,x_n])mapsto [tx_0,x_1,cdots,x_n]$.
answered Dec 21 '18 at 12:15
doloreshazedoloreshaze
141
$endgroup$
add a comment |
$begingroup$
As mentioned by Tyrone in his comment (which I upvoted) , Weintraub is indeed completely wrong because $X=kmathbb P^nsetminus [1,0,cdots,0]$ is homotopic to $kmathbb P^{n-1}$, which of course is not homotopic to the contractible space $k^n$ for $ngeq 2$.
A different from Tyrone's way to see the homotopy is by noticing that $kmathbb P^{n-1}$ (identified to the the hyperplane $x_0=0$ of $kmathbb P^n$) is a strong deformation retract of $X$ under the homotopy $(t,[x_0,x_1,cdots,x_n])mapsto [tx_0,x_1,cdots,x_n]$.
answered Dec 21 '18 at 12:15
doloreshazedoloreshaze
141
$endgroup$
As mentioned by Tyrone in his comment (which I upvoted) , Weintraub is indeed completely wrong because $X=kmathbb P^nsetminus [1,0,cdots,0]$ is homotopic to $kmathbb P^{n-1}$, which of course is not homotopic to the contractible space $k^n$ for $ngeq 2$.
A different from Tyrone's way to see the homotopy is by noticing that $kmathbb P^{n-1}$ (identified to the the hyperplane $x_0=0$ of $kmathbb P^n$) is a strong deformation retract of $X$ under the homotopy $(t,[x_0,x_1,cdots,x_n])mapsto [tx_0,x_1,cdots,x_n]$.
answered Dec 21 '18 at 12:15
doloreshazedoloreshaze
141
answered Dec 21 '18 at 12:15
doloreshazedoloreshaze
141
answered Dec 21 '18 at 12:15
doloreshazedoloreshaze
141
answered Dec 21 '18 at 12:15
doloreshazedoloreshaze
141
141
add a comment |
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$begingroup$
Put the standard CW structure on $mathbb{K}P^n$. Then you can assume up to homotopy that the point $x$ being removed lies in the open top cell. Then $mathbb{K}P^{n}setminus{x}$ deformation retracts onto $mathbb{K}P^{n-1}$, and this is clearly not homotopy equivalent to a (contractible) vector space. Actually $mathbb{K}P^{n}setminus{x}$ is a tubular neighbourhood of the canonical embedding $mathbb{K}P^{n-1}hookrightarrow mathbb{K}P^{n}$. Perhaps this is what Weintraub has meant? (Although I checked the book, and he has written exactly what you say.)
$endgroup$
– Tyrone
Dec 21 '18 at 10:26
2
$begingroup$
Note that the projective space minus a point is diffeomorphic to the total space of the tautological bundle of projective space of dimension one lower. Or in other words the projective space is the Thom space of said bundle.
$endgroup$
– ThorbenK
Dec 21 '18 at 14:30
$begingroup$
@Thorbenk You are absolutely right, and your diffeomorphism is even an analytic isomorphism.
$endgroup$
– doloreshaze
Dec 21 '18 at 16:35