Cell structure of connected sum of complex projective planes
$begingroup$
I am trying to find a cell structure for the connected sum $mathbb{C}P^2# mathbb{C}P^2$.
My approach is the following:
- Find a cellular structure for $mathbb{C}P^2- text{int}(D^4)$ where $D^4$ is a $4$-disc.
I know that $mathbb{C}P^2= mathbb{C}P^1 cup_{pi} D^4$, where $pi$ is the projection map. $mathbb{C}P^1$ is homeomorphic to $S^2$, so it has $2$ $0$-cells, $2$ $1$-cells and $2$ $2$-cells.
So $mathbb{C}P^2$ has $2$ $0$-cells, $2$ $1$-cells, $2$ $2$-cells and $1$ $4$-cell.
I think that by assuming that $D^4$ is in the interior of the $4$-cell, $mathbb{C}P^2- text{int}(D^4)$ would have the cell structure of $mathbb{C}P^1$ by gluing $e_{4}-text{int}(D^4)$, where $e_{4}$ is the $4$-cell.
$e_{4}-text{int}(D^4)$ has $4$ $0$-cells, $6$ $1$-cells, $6$ $2$-cells, $6$ $3$-cells and $2$ $4$-cells (I think).
My first problem is: when gluing that to $mathbb{C}P^1$, which cells do I loose? Do I loose all the cells of $mathbb{C}P^1$?
If yes, $mathbb{C}P^2- text{int}(D^4)$ would have $2$ $0$-cells, $4$ $1$-cells, $4$ $2$-cells, $6$ $3$-cells and $2$ $4$-cells.
- In this case, when gluing the boundaries $S^3$, would I loose the cells of one of the $S^3$'s? (because they are identified).
So would I have {$2$ $0$-cells, $4$ $1$-cells, $4$ $2$-cells, $6$ $3$-cells and $2$ $4$-cells} $cup$ {$2$ $0$-cells, $4$ $1$-cells, $4$ $2$-cells, $6$ $3$-cells and $2$ $4$-cells}- {$2$ $0$-cells, $2$ $1$-cells, $2$ $2$-cells, $2$ $3$-cells}? (I know that the notation is horrible, but it is to explain myself better)
Thank you in advance!!
geometry homology-cohomology projective-space cw-complexes
asked Jan 25 at 20:09
KarenKaren
946
$endgroup$
add a comment |
$begingroup$
I am trying to find a cell structure for the connected sum $mathbb{C}P^2# mathbb{C}P^2$.
My approach is the following:
- Find a cellular structure for $mathbb{C}P^2- text{int}(D^4)$ where $D^4$ is a $4$-disc.
I know that $mathbb{C}P^2= mathbb{C}P^1 cup_{pi} D^4$, where $pi$ is the projection map. $mathbb{C}P^1$ is homeomorphic to $S^2$, so it has $2$ $0$-cells, $2$ $1$-cells and $2$ $2$-cells.
So $mathbb{C}P^2$ has $2$ $0$-cells, $2$ $1$-cells, $2$ $2$-cells and $1$ $4$-cell.
I think that by assuming that $D^4$ is in the interior of the $4$-cell, $mathbb{C}P^2- text{int}(D^4)$ would have the cell structure of $mathbb{C}P^1$ by gluing $e_{4}-text{int}(D^4)$, where $e_{4}$ is the $4$-cell.
$e_{4}-text{int}(D^4)$ has $4$ $0$-cells, $6$ $1$-cells, $6$ $2$-cells, $6$ $3$-cells and $2$ $4$-cells (I think).
My first problem is: when gluing that to $mathbb{C}P^1$, which cells do I loose? Do I loose all the cells of $mathbb{C}P^1$?
If yes, $mathbb{C}P^2- text{int}(D^4)$ would have $2$ $0$-cells, $4$ $1$-cells, $4$ $2$-cells, $6$ $3$-cells and $2$ $4$-cells.
- In this case, when gluing the boundaries $S^3$, would I loose the cells of one of the $S^3$'s? (because they are identified).
So would I have {$2$ $0$-cells, $4$ $1$-cells, $4$ $2$-cells, $6$ $3$-cells and $2$ $4$-cells} $cup$ {$2$ $0$-cells, $4$ $1$-cells, $4$ $2$-cells, $6$ $3$-cells and $2$ $4$-cells}- {$2$ $0$-cells, $2$ $1$-cells, $2$ $2$-cells, $2$ $3$-cells}? (I know that the notation is horrible, but it is to explain myself better)
Thank you in advance!!
geometry homology-cohomology projective-space cw-complexes
asked Jan 25 at 20:09
KarenKaren
946
$endgroup$
add a comment |
$begingroup$
I am trying to find a cell structure for the connected sum $mathbb{C}P^2# mathbb{C}P^2$.
My approach is the following:
- Find a cellular structure for $mathbb{C}P^2- text{int}(D^4)$ where $D^4$ is a $4$-disc.
I know that $mathbb{C}P^2= mathbb{C}P^1 cup_{pi} D^4$, where $pi$ is the projection map. $mathbb{C}P^1$ is homeomorphic to $S^2$, so it has $2$ $0$-cells, $2$ $1$-cells and $2$ $2$-cells.
So $mathbb{C}P^2$ has $2$ $0$-cells, $2$ $1$-cells, $2$ $2$-cells and $1$ $4$-cell.
I think that by assuming that $D^4$ is in the interior of the $4$-cell, $mathbb{C}P^2- text{int}(D^4)$ would have the cell structure of $mathbb{C}P^1$ by gluing $e_{4}-text{int}(D^4)$, where $e_{4}$ is the $4$-cell.
$e_{4}-text{int}(D^4)$ has $4$ $0$-cells, $6$ $1$-cells, $6$ $2$-cells, $6$ $3$-cells and $2$ $4$-cells (I think).
My first problem is: when gluing that to $mathbb{C}P^1$, which cells do I loose? Do I loose all the cells of $mathbb{C}P^1$?
If yes, $mathbb{C}P^2- text{int}(D^4)$ would have $2$ $0$-cells, $4$ $1$-cells, $4$ $2$-cells, $6$ $3$-cells and $2$ $4$-cells.
- In this case, when gluing the boundaries $S^3$, would I loose the cells of one of the $S^3$'s? (because they are identified).
So would I have {$2$ $0$-cells, $4$ $1$-cells, $4$ $2$-cells, $6$ $3$-cells and $2$ $4$-cells} $cup$ {$2$ $0$-cells, $4$ $1$-cells, $4$ $2$-cells, $6$ $3$-cells and $2$ $4$-cells}- {$2$ $0$-cells, $2$ $1$-cells, $2$ $2$-cells, $2$ $3$-cells}? (I know that the notation is horrible, but it is to explain myself better)
Thank you in advance!!
geometry homology-cohomology projective-space cw-complexes
asked Jan 25 at 20:09
KarenKaren
946
$endgroup$
I am trying to find a cell structure for the connected sum $mathbb{C}P^2# mathbb{C}P^2$.
My approach is the following:
- Find a cellular structure for $mathbb{C}P^2- text{int}(D^4)$ where $D^4$ is a $4$-disc.
I know that $mathbb{C}P^2= mathbb{C}P^1 cup_{pi} D^4$, where $pi$ is the projection map. $mathbb{C}P^1$ is homeomorphic to $S^2$, so it has $2$ $0$-cells, $2$ $1$-cells and $2$ $2$-cells.
So $mathbb{C}P^2$ has $2$ $0$-cells, $2$ $1$-cells, $2$ $2$-cells and $1$ $4$-cell.
I think that by assuming that $D^4$ is in the interior of the $4$-cell, $mathbb{C}P^2- text{int}(D^4)$ would have the cell structure of $mathbb{C}P^1$ by gluing $e_{4}-text{int}(D^4)$, where $e_{4}$ is the $4$-cell.
$e_{4}-text{int}(D^4)$ has $4$ $0$-cells, $6$ $1$-cells, $6$ $2$-cells, $6$ $3$-cells and $2$ $4$-cells (I think).
My first problem is: when gluing that to $mathbb{C}P^1$, which cells do I loose? Do I loose all the cells of $mathbb{C}P^1$?
If yes, $mathbb{C}P^2- text{int}(D^4)$ would have $2$ $0$-cells, $4$ $1$-cells, $4$ $2$-cells, $6$ $3$-cells and $2$ $4$-cells.
- In this case, when gluing the boundaries $S^3$, would I loose the cells of one of the $S^3$'s? (because they are identified).
So would I have {$2$ $0$-cells, $4$ $1$-cells, $4$ $2$-cells, $6$ $3$-cells and $2$ $4$-cells} $cup$ {$2$ $0$-cells, $4$ $1$-cells, $4$ $2$-cells, $6$ $3$-cells and $2$ $4$-cells}- {$2$ $0$-cells, $2$ $1$-cells, $2$ $2$-cells, $2$ $3$-cells}? (I know that the notation is horrible, but it is to explain myself better)
Thank you in advance!!
geometry homology-cohomology projective-space cw-complexes
geometry homology-cohomology projective-space cw-complexes
asked Jan 25 at 20:09
KarenKaren
946
asked Jan 25 at 20:09
KarenKaren
946
asked Jan 25 at 20:09
KarenKaren
946
asked Jan 25 at 20:09
KarenKaren
946
asked Jan 25 at 20:09
KarenKaren
946
946
add a comment |
add a comment |
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