[Math] Image of a line or conic on Veronese surface.

algebraic-curvesalgebraic-geometryprojective-geometry

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!

Edit:

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?

Best Answer

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 $E:z_{1}^{2}z_{5} - z_{2}^{2}z_{3} = 0.$ 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 $z_{3}$ and $C_{3}$ by $z_{1}^{2}$ subtract the resulting equations and get $z_{1}^{2}z_{4}^{2}- z_{2}^{2}z_{3}^{2} = 0$ - 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.

Related Solutions

Algebraic Geometry – Veronese Surface and Intersection of Quadratics

There are probably tricks one can use to solve this particular problem, but one can solve this sort of problem in general by using Gröbner bases. I highly recommend Cox, Little, and O'Shea's Ideals, Varieties, and Algorithms: they treat this exact problem in Chapter 3 Elimination, particularly in $\S3$ Implicitization. The main result is the following theorem.

Theorem 1 (Polynomial Implicitization) If $k$ is an infinite field, let $F: k^m \to k^n$ be the function determined by the polynomial parametrization \begin{align*} x_1 &= p_1(t_1, \ldots, t_m)\\ &\ \, \vdots\\ x_n &= p_n(t_1, \ldots, t_m) \, . \end{align*} Let $I = \langle x_1 - p_1, \ldots, x_n - p_n \rangle$ and let $I_m = I \cap k[x_1, \ldots, x_n]$ be the $m^\text{th}$ elimination ideal. Then $\mathbb{V}(I_m)$ is the smallest variety in $k^n$ containing $F(k^m)$.

This deals with affine varieties, and the results in $\S$5 of Chapter 8 can be used to extend to projective varieties.

Let $z_0, \ldots, z_5$ be the coordinates on $\mathbb{P}^5$. The image of $v$ is given parametrically by the equations $$ z_0 = x_0^2 \qquad z_1 = x_1^2 \qquad z_2 = x_2^2 \qquad z_3 = x_0 x_1 \qquad z_4 = x_0 x_2 \qquad z_5 = x_1 x_2 \, . $$ Let $I = \langle z_0-x_0^2, z_1-x_1^2, z_2-x_2^2, z_3-x_0 x_1, z_4-x_0 x_2, z_5-x_1 x_2 \rangle \trianglelefteq k[x_0,x_1,x_2,z_0, \ldots, z_5]$ be the ideal corresponding to this parametrization. I computed a Gröbner basis for $I$ in Sage with respect to the graded lexicographic monomial ordering $x_0 > x_1 > x_2 > z_0 > \cdots > z_5$ as follows:

R.<x0,x1,x2,z0,z1,z2,z3,z4,z5> = PolynomialRing(QQ,9,order = "deglex")
gens = [z0-x0^2, z1-x1^2, z2-x2^2, z3-x0*x1, z4-x0*x2, z5-x1*x2]
I = Ideal(gens)
G = I.groebner_basis()

This outputs a Gröbner basis containing $20$ generators. Those only involving the $z_i$ are $$ z_0 z_1 - z_3^2, \qquad z_0z_2 - z_4^2, \qquad z_0 z_5 - z_3 z_4, \qquad z_1 z_2 - z_5^2, \qquad z_1 z_4 - z_3 z_5 $$ and these are the quadrics defining the image of $v$.

Property of Veronese Embedding

I'll outline the general situation. Firstly, note the general fact that we can write $V(F) = V(x_0F, \dots, x_nF)$ because the $x_i$ cannot all simultaneously vanish (we are working in projective space).

Now consider the Veronese embedding $\nu_d : \mathbb{P}^n \rightarrow \mathbb{P}^m $. If $S \subset \mathbb{P}^n$ is a projective variety, then $S= V(F_1, \dots, F_N)$ for some homogeneous $F_i$; their degrees may be different. By the fact mentioned above, $S = V(G_1, \dots, G_M)$ where the $G_i$ are homogeneous and their degrees are the same, in fact $\deg(G_i) = c \cdot d $ for some $c$. Therefore we can write $G_i = H_i \circ \nu_d$ for homogeneous $H_i$ where $\deg(H_i) = c$. Then we have the map $$ \mathbb{P}^n \supset S = V(G_1, \dots, G_M) \xrightarrow{\nu_d} V(H_1, \dots, H_M) \subset \mathbb{P}^m. $$ Thus $\nu_d(S) = \nu_d(\mathbb{P}^n) \cap V(H_1, \dots, H_M) $.

In your case, $Y= \nu_d(\mathbb{P}^n)$, $Z = \nu_d(S)$ and $V = V(H_1, \dots, H_M)$ is the hypersurface you're looking for.

Best Answer

Related Solutions

Algebraic Geometry – Veronese Surface and Intersection of Quadratics

Property of Veronese Embedding

Related Question