Open sets in complex projective space
Clash Royale CLAN TAG#URR8PPP
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Let $BbbP^1$ be the complex projective line, with the topology given by the quotient topology of $Bbb C^2-0$ under scaling by $Bbb C^times$. Meaning as a set we have:
$$Bbb P^1 = BbbC^2-0/Bbb C^times.$$
We will denote the orbit of $(a,b)$ under this action by $[a:b]$.
Is it true that for any $Usubset Bbb C$ open, the set:
$$[1:z]mid zin U,$$
is open in $Bbb P^1$?
I can't seem to show this. I've tried to take the preimage under the quotient map:
$$q^-1([1:z]mid zin U)=lambda(1,z)mid lambdainBbb C^times,zin U,$$
$$=bigcup_lambdainBbb C^times(lambdatimeslambda U)$$
and then thinking of the product topology, but I can't seem to finish this off.
algebraic-geometry projective-space
asked Sep 8 at 12:00
user591163
82
add a comment |Â
up vote
1
down vote
favorite
Let $BbbP^1$ be the complex projective line, with the topology given by the quotient topology of $Bbb C^2-0$ under scaling by $Bbb C^times$. Meaning as a set we have:
$$Bbb P^1 = BbbC^2-0/Bbb C^times.$$
We will denote the orbit of $(a,b)$ under this action by $[a:b]$.
Is it true that for any $Usubset Bbb C$ open, the set:
$$[1:z]mid zin U,$$
is open in $Bbb P^1$?
I can't seem to show this. I've tried to take the preimage under the quotient map:
$$q^-1([1:z]mid zin U)=lambda(1,z)mid lambdainBbb C^times,zin U,$$
$$=bigcup_lambdainBbb C^times(lambdatimeslambda U)$$
and then thinking of the product topology, but I can't seem to finish this off.
algebraic-geometry projective-space
asked Sep 8 at 12:00
user591163
82
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $BbbP^1$ be the complex projective line, with the topology given by the quotient topology of $Bbb C^2-0$ under scaling by $Bbb C^times$. Meaning as a set we have:
$$Bbb P^1 = BbbC^2-0/Bbb C^times.$$
We will denote the orbit of $(a,b)$ under this action by $[a:b]$.
Is it true that for any $Usubset Bbb C$ open, the set:
$$[1:z]mid zin U,$$
is open in $Bbb P^1$?
I can't seem to show this. I've tried to take the preimage under the quotient map:
$$q^-1([1:z]mid zin U)=lambda(1,z)mid lambdainBbb C^times,zin U,$$
$$=bigcup_lambdainBbb C^times(lambdatimeslambda U)$$
and then thinking of the product topology, but I can't seem to finish this off.
algebraic-geometry projective-space
asked Sep 8 at 12:00
user591163
82
Let $BbbP^1$ be the complex projective line, with the topology given by the quotient topology of $Bbb C^2-0$ under scaling by $Bbb C^times$. Meaning as a set we have:
$$Bbb P^1 = BbbC^2-0/Bbb C^times.$$
We will denote the orbit of $(a,b)$ under this action by $[a:b]$.
Is it true that for any $Usubset Bbb C$ open, the set:
$$[1:z]mid zin U,$$
is open in $Bbb P^1$?
I can't seem to show this. I've tried to take the preimage under the quotient map:
$$q^-1([1:z]mid zin U)=lambda(1,z)mid lambdainBbb C^times,zin U,$$
$$=bigcup_lambdainBbb C^times(lambdatimeslambda U)$$
and then thinking of the product topology, but I can't seem to finish this off.
algebraic-geometry projective-space
algebraic-geometry projective-space
asked Sep 8 at 12:00
user591163
82
asked Sep 8 at 12:00
user591163
82
asked Sep 8 at 12:00
user591163
82
asked Sep 8 at 12:00
user591163
82
asked Sep 8 at 12:00
user591163
82
82
add a comment |Â
add a comment |Â
1 Answer
1
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oldest
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0
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accepted
Hint : the set $A = [1:z] : z in Bbb C$ is open in $Bbb P^1$, so open sets in $A$ are also open in $Bbb P^1$.
answered Sep 8 at 12:10
Nicolas Hemelsoet
5,225417
You are implicitly using the homeomorphism $U_0=Acong Bbb C$ given by $zmapsto [1:z]$?
– user591163
Sep 8 at 12:40
Excellent! Thank you
– user591163
Sep 8 at 13:20
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
Hint : the set $A = [1:z] : z in Bbb C$ is open in $Bbb P^1$, so open sets in $A$ are also open in $Bbb P^1$.
answered Sep 8 at 12:10
Nicolas Hemelsoet
5,225417
You are implicitly using the homeomorphism $U_0=Acong Bbb C$ given by $zmapsto [1:z]$?
– user591163
Sep 8 at 12:40
Excellent! Thank you
– user591163
Sep 8 at 13:20
add a comment |Â
up vote
0
down vote
accepted
Hint : the set $A = [1:z] : z in Bbb C$ is open in $Bbb P^1$, so open sets in $A$ are also open in $Bbb P^1$.
answered Sep 8 at 12:10
Nicolas Hemelsoet
5,225417
You are implicitly using the homeomorphism $U_0=Acong Bbb C$ given by $zmapsto [1:z]$?
– user591163
Sep 8 at 12:40
Excellent! Thank you
– user591163
Sep 8 at 13:20
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
Hint : the set $A = [1:z] : z in Bbb C$ is open in $Bbb P^1$, so open sets in $A$ are also open in $Bbb P^1$.
answered Sep 8 at 12:10
Nicolas Hemelsoet
5,225417
Hint : the set $A = [1:z] : z in Bbb C$ is open in $Bbb P^1$, so open sets in $A$ are also open in $Bbb P^1$.
answered Sep 8 at 12:10
Nicolas Hemelsoet
5,225417
answered Sep 8 at 12:10
Nicolas Hemelsoet
5,225417
answered Sep 8 at 12:10
Nicolas Hemelsoet
5,225417
answered Sep 8 at 12:10
Nicolas Hemelsoet
5,225417
5,225417
You are implicitly using the homeomorphism $U_0=Acong Bbb C$ given by $zmapsto [1:z]$?
– user591163
Sep 8 at 12:40
Excellent! Thank you
– user591163
Sep 8 at 13:20
add a comment |Â
You are implicitly using the homeomorphism $U_0=Acong Bbb C$ given by $zmapsto [1:z]$?
– user591163
Sep 8 at 12:40
Excellent! Thank you
– user591163
Sep 8 at 13:20
You are implicitly using the homeomorphism $U_0=Acong Bbb C$ given by $zmapsto [1:z]$?
– user591163
Sep 8 at 12:40
You are implicitly using the homeomorphism $U_0=Acong Bbb C$ given by $zmapsto [1:z]$?
– user591163
Sep 8 at 12:40
Excellent! Thank you
– user591163
Sep 8 at 13:20
Excellent! Thank you
– user591163
Sep 8 at 13:20
add a comment |Â
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