Non split real form of projective space
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On the complex projective space $mathbb{P}^1_mathbb{C}$ we have an involution $zmapsto -frac{1}{bar{z}}$. Using this as descent datum we should end up with a real form, which is not split (this can be seen for example by the fact that this real form will have no real points, as the involution mentioned above has no fixed points).
I have tried to calculate this real form naively by looking at the map $sigma: mathbb{C}[x,y]to mathbb{C}[x,y]$ given by $sigma(lambdacdot x) = bar lambda cdot (-y )$ and $sigma(lambdacdot y) = bar lambda cdot (-x )$ , and then looking at the scheme $Proj(mathbb{C}[x,y]^sigma)$ i.e. the graded ring of invariants. Is this approach valid?
My result is that $mathbb{C}[x,y]^sigma cong mathbb{R}[x-y, i(x+y)] $, which to me seems to be isomorphic to the standard real projective space, yielding a contradiction, so I am wondering where the mistake possibly is.
algebraic-geometry complex-geometry projective-space projective-schemes descent
asked Dec 9 '18 at 17:25
NotoneNotone
7611413
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add a comment |
$begingroup$
On the complex projective space $mathbb{P}^1_mathbb{C}$ we have an involution $zmapsto -frac{1}{bar{z}}$. Using this as descent datum we should end up with a real form, which is not split (this can be seen for example by the fact that this real form will have no real points, as the involution mentioned above has no fixed points).
I have tried to calculate this real form naively by looking at the map $sigma: mathbb{C}[x,y]to mathbb{C}[x,y]$ given by $sigma(lambdacdot x) = bar lambda cdot (-y )$ and $sigma(lambdacdot y) = bar lambda cdot (-x )$ , and then looking at the scheme $Proj(mathbb{C}[x,y]^sigma)$ i.e. the graded ring of invariants. Is this approach valid?
My result is that $mathbb{C}[x,y]^sigma cong mathbb{R}[x-y, i(x+y)] $, which to me seems to be isomorphic to the standard real projective space, yielding a contradiction, so I am wondering where the mistake possibly is.
algebraic-geometry complex-geometry projective-space projective-schemes descent
asked Dec 9 '18 at 17:25
NotoneNotone
7611413
$endgroup$
2
$begingroup$
What you should get is the projective quadric with equation $X^2+Y^2+Z^2=0$.
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 17:38
add a comment |
$begingroup$
On the complex projective space $mathbb{P}^1_mathbb{C}$ we have an involution $zmapsto -frac{1}{bar{z}}$. Using this as descent datum we should end up with a real form, which is not split (this can be seen for example by the fact that this real form will have no real points, as the involution mentioned above has no fixed points).
I have tried to calculate this real form naively by looking at the map $sigma: mathbb{C}[x,y]to mathbb{C}[x,y]$ given by $sigma(lambdacdot x) = bar lambda cdot (-y )$ and $sigma(lambdacdot y) = bar lambda cdot (-x )$ , and then looking at the scheme $Proj(mathbb{C}[x,y]^sigma)$ i.e. the graded ring of invariants. Is this approach valid?
My result is that $mathbb{C}[x,y]^sigma cong mathbb{R}[x-y, i(x+y)] $, which to me seems to be isomorphic to the standard real projective space, yielding a contradiction, so I am wondering where the mistake possibly is.
algebraic-geometry complex-geometry projective-space projective-schemes descent
asked Dec 9 '18 at 17:25
NotoneNotone
7611413
$endgroup$
On the complex projective space $mathbb{P}^1_mathbb{C}$ we have an involution $zmapsto -frac{1}{bar{z}}$. Using this as descent datum we should end up with a real form, which is not split (this can be seen for example by the fact that this real form will have no real points, as the involution mentioned above has no fixed points).
I have tried to calculate this real form naively by looking at the map $sigma: mathbb{C}[x,y]to mathbb{C}[x,y]$ given by $sigma(lambdacdot x) = bar lambda cdot (-y )$ and $sigma(lambdacdot y) = bar lambda cdot (-x )$ , and then looking at the scheme $Proj(mathbb{C}[x,y]^sigma)$ i.e. the graded ring of invariants. Is this approach valid?
My result is that $mathbb{C}[x,y]^sigma cong mathbb{R}[x-y, i(x+y)] $, which to me seems to be isomorphic to the standard real projective space, yielding a contradiction, so I am wondering where the mistake possibly is.
algebraic-geometry complex-geometry projective-space projective-schemes descent
algebraic-geometry complex-geometry projective-space projective-schemes descent
asked Dec 9 '18 at 17:25
NotoneNotone
7611413
asked Dec 9 '18 at 17:25
NotoneNotone
7611413
asked Dec 9 '18 at 17:25
NotoneNotone
7611413
asked Dec 9 '18 at 17:25
NotoneNotone
7611413
asked Dec 9 '18 at 17:25
NotoneNotone
7611413
7611413
2
$begingroup$
What you should get is the projective quadric with equation $X^2+Y^2+Z^2=0$.
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 17:38
add a comment |
2
$begingroup$
What you should get is the projective quadric with equation $X^2+Y^2+Z^2=0$.
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 17:38
2
2
$begingroup$
What you should get is the projective quadric with equation $X^2+Y^2+Z^2=0$.
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 17:38
$begingroup$
What you should get is the projective quadric with equation $X^2+Y^2+Z^2=0$.
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 17:38
add a comment |
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What you should get is the projective quadric with equation $X^2+Y^2+Z^2=0$.
$endgroup$
– Lord Shark the Unknown
Dec 9 '18 at 17:38