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 $\mathbb{P}^3$?

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    $\begingroup$ 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

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 $$\psi(C\times \mathbb P^2)\subset S=\psi(\mathbb P^2\times \mathbb P^2)\subset \mathbb P^8 \:.$$

  • $\begingroup$ 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

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