Consider the Veronese surface $S$ defined by the map: $$\phi: \mathbb{P}^2\rightarrow \mathbb{P}^5$$ where $\phi(x_0,x_1,x_2)=(x_0^2,x_0x_1,x_0x_2,x_1^2,x_1x_2,x_2^2)$.

The problem asks to show that a line in $\mathbb{P}^2$ is mapped to a conic in $\mathbb{P^5}$

I will use the following notation The definition of the Veronese map will be taken from the question above. I will denote the coordinates in $\mathbb{P}^{5}$ by $[z_{0}:\ldots:z_{5}]$ and the coordinates in $\mathbb{P}^{2}$ by $[x_{0}:x_{1}: x_{2}]$.

Claim The Veronese surface is cut out by three quadrics: $$C_{1}: z_{0}z_{3} - z_{1}^{2} = 0$$ $$C_{2}: z_{0}z_{5} -z_{2}^{2}= 0 $$ $$C_{3}: z_{3}z_{5} - z_{4}^{2} = 0. $$

Suppose a line in $\mathbb{P}^2$ is defined by $ax_0+bx_1+cx_2=0$. So can the image of the line in $\mathbb{P}^5$ be thought as the intersection of $S$ and something else? I have the following questions:

  • How do we show that it is a conic?
  • $\begingroup$ At least for the line defined by $x_0 =0$, it is not hard to believe the conclusion holds. The image is $(0,0,0,x_1^2,x_1x_2,x_2^2)$ which is $C_3$ intersected with $z_0=z_1=z_2=0$. I think the general case can be reduced to this, but I have not worked out the details. $\endgroup$ – Youngsu Mar 30 at 16:56
  • $\begingroup$ Can you please give me some more hint? Any plane can indeed be transformed into $x_0=0$ by rotation but what will be the effect on the conic. Again is there any other direct way to find out? $\endgroup$ – Gimgim Mar 30 at 23:58

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