4
$\begingroup$

Let $\psi: \mathbb{P}^r \times \mathbb{P}^s \to \mathbb{P}^N$ be the Segre embedding with $N = rs + r + s$, as in Hartshorne exercise I.2.14. To be explicit: the image of the pair $([a_0 : \ldots : a_r], [b_0 : \ldots : b_s])$ is $[\ldots : a_ib_j : \ldots]$ in lexicographic order. Let $[z_{00} : \ldots : z_{rs}]$ be homogeneous coordinates on $\mathbb{P}^N$. From the proof that $\operatorname{im} \psi$ is a subvariety of $\mathbb{P}^N$ I know that it is the zero locus of the polynomials $z_{ij}z_{kl} - z_{il}z_{kj}$. But what are the equations for $\psi(X) \subset \mathbb{P}^N$ where $X$ is a subset of $\mathbb{P}^r \times \mathbb{P}^s$ defined in an algebraic way?

More concretely: I have two conics $C = Z(f) \subset \mathbb{P}^2$ and $D^* = Z(g^*) \subset \mathbb{P}^2$ ($D^*$ is the dual conic of the conic $D = Z(g)$). I want to investigate the structure of the incidence correspondence $X = \{(p,\ell) \in C \times D^* : p \in \ell\}$. This can also be described as the set of pairs $([x:y:z],[a:b:c]) \in \mathbb{P}^2 \times \mathbb{P}^2$ such that $f(x,y,z) = 0$, $g^*(a,b,c) = 0$, and $ax + by + cz = 0$. This embeds into $\mathbb{P}^8$ via the Segre embedding. What are the equations for $\psi(X)$ in $\mathbb{P}^8$?

I know that $\psi(([x:y:z],[a:b:c])) = [ax : bx : cx : ay : by : cy : az : bz : cz]$. Hence (I believe) $\psi(X) = \psi(C \times D^*) \cap Z(z_{00} + z_{11} + z_{22})$. But how do I express $\psi(C \times D^*)$ in terms of these new coordinates? For $C$ do I need $f(z_{00}, z_{10},z_{20}) = 0$, or $f(z_{01},z_{11},z_{21}) = 0$, or $f(z_{02}, z_{12}, z_{22}) = 0$, or something completely different?

Thanks in advance.

Edit: Using the fact that all conics are isomorphic and the 2-uple embedding I can actually assume $C = D^* = \mathbb{P}^1$ (Hartshorne exercise I.3.1(c)). What does the incidence correspondence become then, and after the Segre embedding how can it be described in homogeneous coordinates on $\mathbb{P}^3$?

$\endgroup$
  • 2
    $\begingroup$ Dear Ricardo, Not sure if this helps, but I found evaluating the image of the diagonal under the Segre embedding to be helpful in seeing what it does. Regards, $\endgroup$ – user38268 Apr 21 '13 at 9:08
  • $\begingroup$ Thanks @BenjaLim. I see that on the image of the diagonal some coordinates are equal everywhere, so it's isomorphic to some lower dimensional variety, but unfortunately I still don't have a clear enough picture of the Segre embedding to answer my question. $\endgroup$ – Ricardo Buring Apr 21 '13 at 9:34
3
$\begingroup$

I'll just tell you the equations for cutting out $\psi(C\times \mathbb P^2) $ from the Segre variety $S\subset \mathbb P^8$.
There is a similar result for $\psi(\mathbb P^2 \times D^*)$ and since $\psi(C \times D^*)=\psi(C\times \mathbb P^2)\cap \psi(\mathbb P^2 \times D^*)$ you will be done by adding your perfectly correct equation $z_{00} + z_{11} + z_{22}=0$ .

So let's find out $\psi(C\times \mathbb P^2)\subset S=\psi(\mathbb P^2\times \mathbb P^2)\subset \mathbb P^8 \:.$
The diabolical trick is to replace the equation $f(x,y,z)=0$ by the three equations $$f(x,y,z)\cdot a^2=0,f(x,y,z)\cdot b^2=0,f(x,y,z)\cdot c^2=0 $$ The equivalence of this system with $f(x,y,z)=0$ is due to the fact that we can't have simultaneously $a^2=b^2=c^2=0$. The equations in the system can then be translated in equations in the variables $z_{ij}$.
Let's look at an example :
Suppose $f(x,y,z)=xy+yz+zx$. Then we say that $xy+yz+zx=0$ is equivalent to the system $$(xy+yz+zx)\cdot a^2=0,(xy+yz+zx)\cdot b^2=0,(xy+yz+zx)\cdot c^2=0 \quad (SYST) $$
Writing $(xy+yz+zx)\cdot a^2=axay+ayaz+azax$ etc. the above system $(SYST)$ becomes $$z_{00} z_{10} + z_{10}z_{20} +z_{20}z_{00}=0,z_{01} z_{11} + z_{11}z_{21} +z_{21}z_{01}=0, z_{02} z_{12} + z_{12}z_{22} +z_{22}z_{02}=0 $$ These three equations, joined of course to the equations $z_{ij}z_{kl} - z_{il}z_{kj}=0$ for the Segre embedding, define the subvariety $$\psi(C\times \mathbb P^2)\subset S=\psi(\mathbb P^2\times \mathbb P^2)\subset \mathbb P^8 \:.$$

$\endgroup$
  • $\begingroup$ Thank you, Georges. That's very diabolical; I like it. Where did you learn this trick? Would you happen to know what my incidence correspondence $X$ would become if I chose $C = D^* = \mathbb{P}^1$ (motivated by the edit to my question)? I could then embed $X$ in $\mathbb{P}^3$. Right now I'm not sure if $C = D^* = \mathbb{P}^1$ even makes any sense. $\endgroup$ – Ricardo Buring Apr 22 '13 at 9:20
  • $\begingroup$ Dear Ricardo, I learned the trick long ago, I don't remember where. You might ask the question in your comment as a genuine question, maybe reformulated so as to be self-contained, but unfortunately I have no time for it now. $\endgroup$ – Georges Elencwajg Apr 22 '13 at 10:41
  • $\begingroup$ I'll do that. Thanks again. $\endgroup$ – Ricardo Buring Apr 22 '13 at 11:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.