This is part of Exercise 5.13 from Undergraduate Algebraic Geometry by Reid:

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 $\mathbb{P^5}$ and a conic in $\mathbb{P}^2$ is mapped to a quartic in $\mathbb{P}^5$.

My attempt:

Suppose a line in $\mathbb{P}^2$ is defined by $ax_0+bx_1+cx_2=0$. Then we have also $V: ay_0+by_1+cy_2=0$ in $\mathbb{P}^5$. So the image of the line in $\mathbb{P}^5$ is the intersection of $V$ and $S$. I have the following questions:

  • How do we show that it is a conic?
  • How do we decide we don't need more equations? For example, $ay_1+by_3+cy_5=0$ can also define it. So does $ay_2+by_4+cy_5=0$.

I read some pages by Harris. It has some nice description of the Veronese surface, but my questions are not solved. I have similar questions then about a conic mapped to quartic.

Thank you for your help!


Let $x_0^2+x_1^2+x_2^2=0$ be a conic in $\mathbb{P}^2$. Its image in $\mathbb{P}^5$ is the intersection of $y_0+y_3+y_5=0$ and the surface $S$. Making a change of variable so $y_5=0$ and plugging $-y_0-y_3$ into $y_5$ of the three defining equations of $S$, I got $$y_1^2=-(y_3^2+y_4^2)\\ y_1^2=-(y_0^2-y_2^2)\\ y_1^2=y_0y_3$$

The pullback of the first two are union of the conic and a line ($x_0=0$ and $x_1=0$, respectively). How to write it as a single quartic so that the pullback does not contain the extra line?

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    $\begingroup$ I think you would avoid some potential confusion if you use the coordinates $[y_0, \dots, y_5]$ on $\mathbb{P}^5$. Here's one example of the first phenomena would want to observe. Consider the line $x_2 = 0$ in $\mathbb{P}^2$, then $[x_0, x_1, 0] \mapsto [x_0^2, x_0x_1, 0, x_1^2, 0, 0]$. So the image of the line under the Veronese embedding is contained in the conic in $\mathbb{P}^5$ defined by the equation $y_1^2 - y_0y_3 = 0$. $\endgroup$ – Michael Albanese Feb 6 '16 at 12:42
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    $\begingroup$ As @MichaelAlbanese points out you should use different notation for the coordinates in $\mathbb{P}^{2}$ and $\mathbb{P}^{5}$. The first thing you should note is that the image of veronese surface is cut out by three quadric hypersurfaces in $\mathbb{P}^{5}$. A line in $\mathbb{P}^{2}$ imposes one more constraint on the coordinates and because the veronese map is quadratic ideal generated by the corordinates on this line is an extra quadratic equation. $\endgroup$ – DBS Feb 6 '16 at 13:11
  • $\begingroup$ @MichaelAlbanese: Thank you for your reply! I've changed my notation. And I'll think about how to apply your example to my case. $\endgroup$ – KittyL Feb 6 '16 at 13:28
  • $\begingroup$ @DBS: Thank you for your reply! I'm still confused. What do you mean by "cut out by three" and which three? $\endgroup$ – KittyL Feb 6 '16 at 13:31
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    $\begingroup$ Cut out by three: the image of veronese is a closed sub variety (scheme). So it is defined by an ideal in P^5 what would that ideal be? If we use coordinates [z0:....:z-5] in P^5 then we see that (in your notation of veronese map) z0z3 =z1^2 for example. Similarly there are two other quadrics. The line in P^2 is defined by a quadrics equation in P^5 what you should show is that this quadric equation imposes an extra condition I.E. it is not already contained in the ideal defining the veronese. $\endgroup$ – DBS Feb 6 '16 at 14:09

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: $$C_{1}: z_{0}z_{3} - z_{1}^{2} = 0$$ $$C_{2}: z_{0}z_{5} -z_{2}^{2}= 0 $$ $$C_{3}: z_{3}z_{5} - z_{4}^{2} = 0. $$

It should be checked that they are indeed enough to define the locus but we will not get into that.

We consider a quadric $Q$ in $\mathbb{P}^{2}$ given by $Q: a_{0}x^{2}_{0} -(a_{1}x^{2}_{1} + a_{2}x^{2}_{2}) = 0 $. We can and will assume that $a_{0} \neq 0$ and after rescaling $a_{0} = 1$.

Clearly the image of $Q$ is contained in the hyperplane $H: z_{0} - a_{1}z_{3} - a_{2}z_{5} = 0$.

Our goal: Understand the intersection of this hyperplane with quadrics $C_{i}$.

$$C_{1} \cap H : (a_{1}z_{3} + a_{2}z_{5})z_{3}-z_{1}^{2}= 0$$
$$C_{2} \cap H : (a_{1}z_{3} + a_{2}z_{5})z_{5}-z_{2}^{2}= 0$$

Multiplying the first equation above by $z_{5}$ and the second one by $z_{3}$ and subtracting them we get the equation $E:z_{1}^{2}z_{3} - z_{2}^{2}z_{5} = 0.$ This equation $E$ represents the locus of the intersection of $H$ with two of the three quadrics. The justification for this ad-hoc process is the the following: Inside the open (quasi-affine) variety $z_{3}z_{5} \neq 0$ we are allowed to multiply by non-zero functions $z_{3}$ and $z_{5}$. Outside this locus one has to check that this equation is still valid.

Next we study the intersection of $E$ and $C_{3}$. Again one multiplies $E$ by $z_{5}$ and $C_{3}$ by $z_{1}^{2}$ subtract the resulting equations and get $z_{1}^{2}z_{4}^{2}- z_{2}^{2}z_{5}^{2} = 0$ - a quartic.

The computation of the image of a cubic in $\mathbb{P}^{2}$ under the Veronese in Harris's book, page 2 here , is a great illustration of the complexity of figuring out the intersections.

  • $\begingroup$ Thank you so much! One more question: does it matter that the quartic is degenerate? $\endgroup$ – KittyL Feb 7 '16 at 14:59
  • $\begingroup$ I am not sure why it would matter. The singularity locus is the point of intersection of the two quadrics defining the quartic. It is possible to traceback what is happening at those singular points. I haven't checked it carefully but I am guessing that if you look at the projective lines in the conic (set a variable to zero) then they should map to conics in the Veronese. The nonsingular points are going to be the intersections of those conics. $\endgroup$ – DBS Feb 7 '16 at 15:17

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