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Apr 7, 2020 at 18:24 comment added abx Just apply both sides to $\dfrac{\partial }{\partial z_1}\wedge \ldots \wedge \dfrac{\partial }{\partial z_n}$, this gives the equality of the Griffiths-Harris determinant with the Jacobian determinant of your map.
Apr 7, 2020 at 13:15 review Close votes
Apr 22, 2020 at 1:48
Apr 7, 2020 at 13:04 comment added abx Hint: $\theta _0^{n+1}d\left(\dfrac{\theta _1}{\theta _0}\right)\wedge\ldots \wedge d\left(\dfrac{\theta _n}{\theta _0}\right)=$ $\sum_i (-1)^i \theta _id\theta_0\wedge\ldots \wedge \widehat{d\theta _i}\wedge\ldots \wedge d\theta _n$, where the $\widehat{d\theta _i}$ means that this term is deleted. Note that this is a general fact about maps into projective space, nothing particular to do with Jacobians.
Apr 7, 2020 at 12:34 history asked Partha CC BY-SA 4.0