0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

$RP^2$ 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 $\psi^i = p\phi^i_\pm$.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.