Cell complex structure of real projective plane
$begingroup$
A Cell Complex structure of $RP^2$, the real protective plane, is $e^0∪e^1∪e^2$. But I am unable to find this cell complex structure for $RP^2$ from iterative method: What should be the set $X^0$ and then how the set $X^1$ will look like ? Anybody please.
algebraic-topology
edited Jan 11 at 15:41
Paul Frost
11.5k3934
asked Jan 11 at 12:55
Prince ThomasPrince Thomas
620311
$endgroup$
add a comment |
$begingroup$
A Cell Complex structure of $RP^2$, the real protective plane, is $e^0∪e^1∪e^2$. But I am unable to find this cell complex structure for $RP^2$ from iterative method: What should be the set $X^0$ and then how the set $X^1$ will look like ? Anybody please.
algebraic-topology
edited Jan 11 at 15:41
Paul Frost
11.5k3934
asked Jan 11 at 12:55
Prince ThomasPrince Thomas
620311
$endgroup$
add a comment |
$begingroup$
A Cell Complex structure of $RP^2$, the real protective plane, is $e^0∪e^1∪e^2$. But I am unable to find this cell complex structure for $RP^2$ from iterative method: What should be the set $X^0$ and then how the set $X^1$ will look like ? Anybody please.
algebraic-topology
edited Jan 11 at 15:41
Paul Frost
11.5k3934
asked Jan 11 at 12:55
Prince ThomasPrince Thomas
620311
$endgroup$
A Cell Complex structure of $RP^2$, the real protective plane, is $e^0∪e^1∪e^2$. But I am unable to find this cell complex structure for $RP^2$ from iterative method: What should be the set $X^0$ and then how the set $X^1$ will look like ? Anybody please.
algebraic-topology
algebraic-topology
edited Jan 11 at 15:41
Paul Frost
11.5k3934
asked Jan 11 at 12:55
Prince ThomasPrince Thomas
620311
edited Jan 11 at 15:41
Paul Frost
11.5k3934
asked Jan 11 at 12:55
Prince ThomasPrince Thomas
620311
edited Jan 11 at 15:41
Paul Frost
11.5k3934
edited Jan 11 at 15:41
Paul Frost
11.5k3934
edited Jan 11 at 15:41
Paul Frost
11.5k3934
11.5k3934
asked Jan 11 at 12:55
Prince ThomasPrince Thomas
620311
asked Jan 11 at 12:55
Prince ThomasPrince Thomas
620311
asked Jan 11 at 12:55
Prince ThomasPrince Thomas
620311
620311
add a comment |
add a comment |
1 Answer
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$RP^2$ is the quotient space obtained from $S^2$ by identifying antipodal points $x,-x$. Let $p : S^2 to RP^2$ denote the quotient map. Now let us give $S^2$ the following CW-structure:
Two $0$-cells $d^0_pm = { (pm 1,0,0) }$.
Two open $1$-cells $d^1_pm = { (x,y,0) mid x^2 + y^2 = 1, (-1)^{pm 1} y > 0 }$.
Two open $2$-cells $d^2_pm = { (x,y,z) mid x^2 + y^2 + z^2 = 1, (-1)^{pm 1} z > 0 }$.
Attaching maps for $d^1_pm$ are $phi^1_pm : D^1 to S^2, phi^1_pm(x) = pm (sin(pi(x+1)/2), cos((pi(x+1)/2),0)$ and those for $d^2_pm$ are $phi^2_pm : D^2 to S^2, phi^2_pm (x,y) = pm (x,y, sqrt{1- x^2 - y^2})$.
For $i = 0,1,2$ we have $p(d^i_+) = p(d^i_-) =: e^i$. Obviously $p$ maps both $d^i_pm$ hoemomorphically onto $e^i$. Now by construction $e^0,e^1,e^2$ form an open cell decomposition of $RP^2$. Attaching maps $psi^i$ are induced by those of the cells of $S^2$, i.e. we have $psi^i = pphi^i_pm$.
answered Jan 11 at 15:39
Paul FrostPaul Frost
11.5k3934
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add a comment |
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1 Answer
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1 Answer
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active
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$begingroup$
$RP^2$ is the quotient space obtained from $S^2$ by identifying antipodal points $x,-x$. Let $p : S^2 to RP^2$ denote the quotient map. Now let us give $S^2$ the following CW-structure:
Two $0$-cells $d^0_pm = { (pm 1,0,0) }$.
Two open $1$-cells $d^1_pm = { (x,y,0) mid x^2 + y^2 = 1, (-1)^{pm 1} y > 0 }$.
Two open $2$-cells $d^2_pm = { (x,y,z) mid x^2 + y^2 + z^2 = 1, (-1)^{pm 1} z > 0 }$.
Attaching maps for $d^1_pm$ are $phi^1_pm : D^1 to S^2, phi^1_pm(x) = pm (sin(pi(x+1)/2), cos((pi(x+1)/2),0)$ and those for $d^2_pm$ are $phi^2_pm : D^2 to S^2, phi^2_pm (x,y) = pm (x,y, sqrt{1- x^2 - y^2})$.
For $i = 0,1,2$ we have $p(d^i_+) = p(d^i_-) =: e^i$. Obviously $p$ maps both $d^i_pm$ hoemomorphically onto $e^i$. Now by construction $e^0,e^1,e^2$ form an open cell decomposition of $RP^2$. Attaching maps $psi^i$ are induced by those of the cells of $S^2$, i.e. we have $psi^i = pphi^i_pm$.
answered Jan 11 at 15:39
Paul FrostPaul Frost
11.5k3934
$endgroup$
add a comment |
$begingroup$
$RP^2$ is the quotient space obtained from $S^2$ by identifying antipodal points $x,-x$. Let $p : S^2 to RP^2$ denote the quotient map. Now let us give $S^2$ the following CW-structure:
Two $0$-cells $d^0_pm = { (pm 1,0,0) }$.
Two open $1$-cells $d^1_pm = { (x,y,0) mid x^2 + y^2 = 1, (-1)^{pm 1} y > 0 }$.
Two open $2$-cells $d^2_pm = { (x,y,z) mid x^2 + y^2 + z^2 = 1, (-1)^{pm 1} z > 0 }$.
Attaching maps for $d^1_pm$ are $phi^1_pm : D^1 to S^2, phi^1_pm(x) = pm (sin(pi(x+1)/2), cos((pi(x+1)/2),0)$ and those for $d^2_pm$ are $phi^2_pm : D^2 to S^2, phi^2_pm (x,y) = pm (x,y, sqrt{1- x^2 - y^2})$.
For $i = 0,1,2$ we have $p(d^i_+) = p(d^i_-) =: e^i$. Obviously $p$ maps both $d^i_pm$ hoemomorphically onto $e^i$. Now by construction $e^0,e^1,e^2$ form an open cell decomposition of $RP^2$. Attaching maps $psi^i$ are induced by those of the cells of $S^2$, i.e. we have $psi^i = pphi^i_pm$.
answered Jan 11 at 15:39
Paul FrostPaul Frost
11.5k3934
$endgroup$
add a comment |
$begingroup$
$RP^2$ is the quotient space obtained from $S^2$ by identifying antipodal points $x,-x$. Let $p : S^2 to RP^2$ denote the quotient map. Now let us give $S^2$ the following CW-structure:
Two $0$-cells $d^0_pm = { (pm 1,0,0) }$.
Two open $1$-cells $d^1_pm = { (x,y,0) mid x^2 + y^2 = 1, (-1)^{pm 1} y > 0 }$.
Two open $2$-cells $d^2_pm = { (x,y,z) mid x^2 + y^2 + z^2 = 1, (-1)^{pm 1} z > 0 }$.
Attaching maps for $d^1_pm$ are $phi^1_pm : D^1 to S^2, phi^1_pm(x) = pm (sin(pi(x+1)/2), cos((pi(x+1)/2),0)$ and those for $d^2_pm$ are $phi^2_pm : D^2 to S^2, phi^2_pm (x,y) = pm (x,y, sqrt{1- x^2 - y^2})$.
For $i = 0,1,2$ we have $p(d^i_+) = p(d^i_-) =: e^i$. Obviously $p$ maps both $d^i_pm$ hoemomorphically onto $e^i$. Now by construction $e^0,e^1,e^2$ form an open cell decomposition of $RP^2$. Attaching maps $psi^i$ are induced by those of the cells of $S^2$, i.e. we have $psi^i = pphi^i_pm$.
answered Jan 11 at 15:39
Paul FrostPaul Frost
11.5k3934
$endgroup$
$RP^2$ is the quotient space obtained from $S^2$ by identifying antipodal points $x,-x$. Let $p : S^2 to RP^2$ denote the quotient map. Now let us give $S^2$ the following CW-structure:
Two $0$-cells $d^0_pm = { (pm 1,0,0) }$.
Two open $1$-cells $d^1_pm = { (x,y,0) mid x^2 + y^2 = 1, (-1)^{pm 1} y > 0 }$.
Two open $2$-cells $d^2_pm = { (x,y,z) mid x^2 + y^2 + z^2 = 1, (-1)^{pm 1} z > 0 }$.
Attaching maps for $d^1_pm$ are $phi^1_pm : D^1 to S^2, phi^1_pm(x) = pm (sin(pi(x+1)/2), cos((pi(x+1)/2),0)$ and those for $d^2_pm$ are $phi^2_pm : D^2 to S^2, phi^2_pm (x,y) = pm (x,y, sqrt{1- x^2 - y^2})$.
For $i = 0,1,2$ we have $p(d^i_+) = p(d^i_-) =: e^i$. Obviously $p$ maps both $d^i_pm$ hoemomorphically onto $e^i$. Now by construction $e^0,e^1,e^2$ form an open cell decomposition of $RP^2$. Attaching maps $psi^i$ are induced by those of the cells of $S^2$, i.e. we have $psi^i = pphi^i_pm$.
answered Jan 11 at 15:39
Paul FrostPaul Frost
11.5k3934
edited Jan 11 at 17:47
answered Jan 11 at 15:39
Paul FrostPaul Frost
11.5k3934
answered Jan 11 at 15:39
Paul FrostPaul Frost
11.5k3934
answered Jan 11 at 15:39
Paul FrostPaul Frost
11.5k3934
11.5k3934
add a comment |
add a comment |
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