Mixing
Parametrise $z_0^2 + cdots + z_3^2 = 0$
$begingroup$
Find a function $f:mathbb{CP}^1 times mathbb{CP}^1 longrightarrow mathbb{CP}^3$ with image
$$
{[z_0:z_1:z_2:z_3] in mathbb{CP}^3 mid z_0^2 + cdots + z_3^2 = 0}.
$$
I have got no idea how to get the solution. I know it has to involve some imaginary $i$ because otherwise, when we plug in purely real input $([x_0:x_1],[y_0:y_1])$, it's not possible to obtain $z_0^2 + cdots + z_3^2 = 0$.
geometry algebraic-geometry algebraic-curves
asked Jan 22 at 21:30
BernoulliBernoulli
33618
$endgroup$
add a comment |
$begingroup$
Find a function $f:mathbb{CP}^1 times mathbb{CP}^1 longrightarrow mathbb{CP}^3$ with image
$$
{[z_0:z_1:z_2:z_3] in mathbb{CP}^3 mid z_0^2 + cdots + z_3^2 = 0}.
$$
I have got no idea how to get the solution. I know it has to involve some imaginary $i$ because otherwise, when we plug in purely real input $([x_0:x_1],[y_0:y_1])$, it's not possible to obtain $z_0^2 + cdots + z_3^2 = 0$.
geometry algebraic-geometry algebraic-curves
asked Jan 22 at 21:30
BernoulliBernoulli
33618
$endgroup$
2
$begingroup$
HINT: Over $Bbb C$, the quadratic form $z_0^2+dots+z_3^2$ is equivalent to the quadratic form $z_0z_3pm z_1z_2$.
$endgroup$
– Ted Shifrin
Jan 22 at 21:37
$begingroup$
Is it like un-diagonalisation?
$endgroup$
– Bernoulli
Jan 22 at 21:41
$begingroup$
LOL, sure. It's just making an obvious change of basis in $Bbb C^4$.
$endgroup$
– Ted Shifrin
Jan 22 at 21:44
add a comment |
$begingroup$
Find a function $f:mathbb{CP}^1 times mathbb{CP}^1 longrightarrow mathbb{CP}^3$ with image
$$
{[z_0:z_1:z_2:z_3] in mathbb{CP}^3 mid z_0^2 + cdots + z_3^2 = 0}.
$$
I have got no idea how to get the solution. I know it has to involve some imaginary $i$ because otherwise, when we plug in purely real input $([x_0:x_1],[y_0:y_1])$, it's not possible to obtain $z_0^2 + cdots + z_3^2 = 0$.
geometry algebraic-geometry algebraic-curves
asked Jan 22 at 21:30
BernoulliBernoulli
33618
$endgroup$
Find a function $f:mathbb{CP}^1 times mathbb{CP}^1 longrightarrow mathbb{CP}^3$ with image
$$
{[z_0:z_1:z_2:z_3] in mathbb{CP}^3 mid z_0^2 + cdots + z_3^2 = 0}.
$$
I have got no idea how to get the solution. I know it has to involve some imaginary $i$ because otherwise, when we plug in purely real input $([x_0:x_1],[y_0:y_1])$, it's not possible to obtain $z_0^2 + cdots + z_3^2 = 0$.
geometry algebraic-geometry algebraic-curves
geometry algebraic-geometry algebraic-curves
asked Jan 22 at 21:30
BernoulliBernoulli
33618
asked Jan 22 at 21:30
BernoulliBernoulli
33618
asked Jan 22 at 21:30
BernoulliBernoulli
33618
asked Jan 22 at 21:30
BernoulliBernoulli
33618
asked Jan 22 at 21:30
BernoulliBernoulli
33618
33618
2
$begingroup$
HINT: Over $Bbb C$, the quadratic form $z_0^2+dots+z_3^2$ is equivalent to the quadratic form $z_0z_3pm z_1z_2$.
$endgroup$
– Ted Shifrin
Jan 22 at 21:37
$begingroup$
Is it like un-diagonalisation?
$endgroup$
– Bernoulli
Jan 22 at 21:41
$begingroup$
LOL, sure. It's just making an obvious change of basis in $Bbb C^4$.
$endgroup$
– Ted Shifrin
Jan 22 at 21:44
add a comment |
2
$begingroup$
HINT: Over $Bbb C$, the quadratic form $z_0^2+dots+z_3^2$ is equivalent to the quadratic form $z_0z_3pm z_1z_2$.
$endgroup$
– Ted Shifrin
Jan 22 at 21:37
$begingroup$
Is it like un-diagonalisation?
$endgroup$
– Bernoulli
Jan 22 at 21:41
$begingroup$
LOL, sure. It's just making an obvious change of basis in $Bbb C^4$.
$endgroup$
– Ted Shifrin
Jan 22 at 21:44
2
2
$begingroup$
HINT: Over $Bbb C$, the quadratic form $z_0^2+dots+z_3^2$ is equivalent to the quadratic form $z_0z_3pm z_1z_2$.
$endgroup$
– Ted Shifrin
Jan 22 at 21:37
$begingroup$
HINT: Over $Bbb C$, the quadratic form $z_0^2+dots+z_3^2$ is equivalent to the quadratic form $z_0z_3pm z_1z_2$.
$endgroup$
– Ted Shifrin
Jan 22 at 21:37
$begingroup$
Is it like un-diagonalisation?
$endgroup$
– Bernoulli
Jan 22 at 21:41
$begingroup$
Is it like un-diagonalisation?
$endgroup$
– Bernoulli
Jan 22 at 21:41
$begingroup$
LOL, sure. It's just making an obvious change of basis in $Bbb C^4$.
$endgroup$
– Ted Shifrin
Jan 22 at 21:44
$begingroup$
LOL, sure. It's just making an obvious change of basis in $Bbb C^4$.
$endgroup$
– Ted Shifrin
Jan 22 at 21:44
add a comment |
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$begingroup$
HINT: Over $Bbb C$, the quadratic form $z_0^2+dots+z_3^2$ is equivalent to the quadratic form $z_0z_3pm z_1z_2$.
$endgroup$
– Ted Shifrin
Jan 22 at 21:37
$begingroup$
Is it like un-diagonalisation?
$endgroup$
– Bernoulli
Jan 22 at 21:41
$begingroup$
LOL, sure. It's just making an obvious change of basis in $Bbb C^4$.
$endgroup$
– Ted Shifrin
Jan 22 at 21:44