4

In need help in my homework assignment:

let $X =\left \{ (x_{0}:x_{1}:x_{2}:x_{3}:x_{4})\in \mathbb{P}^{4} : rk\begin{pmatrix} x_{0} & x_{1} & x_{2}\\ x_{2} & x_{3} & x_{4} \end{pmatrix} < 2\right \} $

  1. show that there is a morphism $\varphi :X\rightarrow \mathbb{P}^2 $ which on the open subset of $X$ where $(x_{0},x_{1},x_{2}) \neq(0,0,0)$ is given by the projection $ (x_{0}:x_{1}:x_{2}:x_{3}:x_{4}) \mapsto (x_{0},x_{1},x_{2})$

  2. for each point $P \in \mathbb{P}^{2} $ determine the inverse image $\varphi ^{-1}(P)$

I thought at first to display X like V(I), but I got no intuition of what that could be, can you help?

GAJO
  • 155

2 Answers2

3

Here are some facts which you should prove and which solve your exercise:

  1. Yes $X=V(I)$ for $I=(x_0x_3-x_1x_2,x_ox_4-x_2^2,x_1x_4-x_2x_3)$ .

  2. The complete definition of the morphism $\phi:X\to \mathbb P^2$ is:
    $$\phi (x_0:x_1:x_2:x_3:x_4)=(x_0:x_1:x_2) \quad \text{if} \: (x_{0},x_{1},x_{2}) \neq(0,0,0) $$ $$\phi (x_0:x_1:x_2:x_3:x_4)=(x_2:x_3:x_4) \quad \text{if} \: (x_{2},x_{3},x_{4}) \neq(0,0,0)$$ The rank $\lt2$ condition ensures that these definitions are compatible on $X$ and thus that $\phi$ is a well defined morphism $X\to \mathbb P^2$.

  3. Points in $\mathbb P^2$ have exactly one preimage except for $(0:1:0)$ which has as preimage the whole projective line $x_0=x_2=x_4=0$ [parametrically the points $(0:x_1:0:x_3:0) $ ] included in $X$.

Conclusion
The variety $X$ is isomorphic to $\mathbb P^2$ blown-up at $(0:1:0)$.
Quite a pretty problem that!

Georges Elencwajg
  • 150,790
  • I just find this variety the projection of veronese surface $\mathbb{P}^2\rightarrow \mathbb{P}^5$ at a point on it, since this veronese surface contains no lines (otherwise the pull-back of this line must have degree $1/2$!), it must be isomorphic to $\mathbb{P}^2$ blown-up at a point. Here is a question: is it the simplest embedding of $\mathbb{P}^2$ blown-up at a point into some projective space? – Yuchen Liu May 10 '13 at 12:49
  • Dear @jerry: no, you can't embed the projective plane blown-up at a point in $\mathbb P^3$. Quadrics of course won't do and smooth cubic surfaces have 27 lines with self-intersection $-1$, which is 26 too many! Cubic surfaces are however isomorphic to the plane blown-up in six points, not one. Surfaces of degree $\gt 3$, won't do either because they aren't even rational, since they have non-zero arithmetic genus. – Georges Elencwajg May 10 '13 at 14:12
  • @Jerry: By the way, could you please briefly describe the equations in coordinates for your projection of the Veronese surface? – Georges Elencwajg May 10 '13 at 14:19
  • 1
    Well, I first construct a rational map from $\mathbb{P}^2$ to $\mathbb{P^4}$, as $[x,y,z]\mapsto [x,y,z,yz/x,z^2/x]=[x^2,xy,xz,yz,z^2]$, and it is simply omitting the $y^2$ term for degree 2 Veronese embedding! – Yuchen Liu May 10 '13 at 14:35
  • Ah, I see, and your rational map is not defined at $[0:1:0]$, which you blow up and then you embed the blown-up surface into $\mathbb P^4$ , thus obtaining $X$. Very nice: +1. Why didn't you give that construction as an answer to the question? Come to think of it, it's not too late! – Georges Elencwajg May 10 '13 at 15:00
  • OK, I just love to chat with you (and I totally agree with your comment "Quite a pretty problem that!") :) I will write a more thorough answer. – Yuchen Liu May 10 '13 at 15:05
2

This answer is a complement to the conclusion in Georges' answer.

First, construct a rational map $\sigma:\mathbb{P}^2\dashrightarrow\mathbb{P}^4$ defined by $$(x_0:x_1:x_2)\mapsto(x_0^2:x_0 x_1:x_0 x_2:x_1 x_2:x_2^2) $$

Obviously $\sigma$ is not defined only at $(0:1:0)$, and it is easy to check that the closure of the image of $\mathbb{P}^2$ under $\sigma$ is exactly $X$ (actually $\sigma=\varphi^{-1}$!).

Claim: $X$ is isomorphic to $\mathbb{P}^2$ blown-up at $(0:1:0)$, and $\sigma$ is exactly the blow-up map.

Consider the Veronese embedding $\bar\sigma:\mathbb{P}^2\rightarrow\mathbb{P}^5$ defined by $$ (x_0:x_1:x_2)\mapsto(x_0^2:x_0 x_1:x_0 x_2:x_1 x_2:x_2^2:x_1^2) $$ The projection map $\pi:\mathbb{P}^5\dashrightarrow \mathbb{P}^4$ is defined by omitting the last homogeneous coordinate of $\mathbb{P}^5$, then obviously $\sigma=\pi\circ\bar\sigma$. Denote $\bar X=\bar\sigma(\mathbb{P}^2)$, thus we get a rational map $\pi:\bar X\dashrightarrow X$.

The center $p$ of projection $\pi$ is $p=(0:0:0:0:0:1)=\bar\sigma(0:1:0)\in \bar X$, hence $X$ is the projection of $\bar X$ with center $p\in\bar X$. Since $\bar X$ contains no lines (otherwise the preimage of this line must have degree $1/2$ in $\mathbb{P}^2$!) and every line through $p$ passes at most one other point in $\bar X$ (counting multiplicity, because $\bar X$ is the intersection of quadratic hypersurfaces), $\pi:\bar X\dashrightarrow X$ is exactly the blow-up map of $\bar X$ at $p$. Therefore, $\sigma=\pi\circ\bar\sigma$ is the blow-up map of $\mathbb{P^2}$ at $(0:1:0)$, and $\varphi$ is the blow-down map.

Yuchen Liu
  • 2,714
  • Dear jerry, thanks for developing your comments into an answer:+1. – Georges Elencwajg May 11 '13 at 06:17
  • Dear jerry: questions in algebraic geometry related to commutative algebra, categories, etc. seem to attract more attention on this site than those about classical, Italian-style geometry: Veronese, Segre, ...You seem to be a welcome exception: is that your personal taste or is it the point of view generally adopted in your courses? What books are used in these courses? Griffiths taught a course in Beijing in 1982 and wrote a book in the classical style from his notes afterwards: Introduction to Algebraic Curves. Have you used it? – Georges Elencwajg May 11 '13 at 07:12
  • Dear @GeorgesElencwajg, I am an exception not only on this site but also around my schoolmates, so it is of course my personal taste. I love algebraic geometry because it has many many interesting examples, from long long ago until now. My first book on algebraic geometry is F. Kirwan's "Complex algebraic curves", and of course I have read Griffiths' book. There is a period in my undergraduate years that I was totally addicted to classical algebraic geometry, 27 lines, quadratic transformation, Newton's classification of cubic curves, singularity of plane curves, etc. – Yuchen Liu May 11 '13 at 14:36
  • Dear @jerry, thanks for answering . I sincerely wish you much deserved success in your studies. – Georges Elencwajg May 11 '13 at 14:40
  • @GeorgesElencwajg: Some books I read are Brieskorn's "Plane Algebraic Curves", Newton's Classification of Cubic curves (Modern English translation), Riemann's paper on abelian functions, Semple and Roth's "Introduction to Algebraic Geometry", Poincare's "Papers on Fuchsian Functoins", Weyl's "The Concept of Riemann Surfaces", etc. At that time, I thought Hartshorne is terrible to read and those BEAUTIFUL geometric objects are replaced by ABSTRACT algebraic concepts which I cannot see them! So I postponed my reading of Hartshorne until now... – Yuchen Liu May 11 '13 at 14:47
  • @GeorgesElencwajg: When I begin my master study two years ago, I am forced to solve a problem related with some advanced concepts in algebraic geometry (like log canonical threshold), so I must read some modern books. Since I want to study algebraic surfaces (because they are so beautiful and richful), at first Beauville's book made me confused: why is sheaf cohomology used everywhere? I started to learn sheaf cohomology, and after some time I found it quite useful to solve many classical problems. Also, I felt unconfortable of the lack of rigor, so I started to learn scheme theory recently. – Yuchen Liu May 11 '13 at 14:57
  • Dear jerry: you seem to be a hard worker and have a strong mathematical personality: congratulations! – Georges Elencwajg May 11 '13 at 18:13