# A very simple question about a Harris' exercise

In the book of Harris Algebraic Geometry, a First Course, exercise 2.29, it is asked to show that the composition of $1\times \nu: \mathbb{P}^1\times\mathbb{P}^1 \to \mathbb{P}^1\times\mathbb{P}^2$ where $\nu$ is the Veronese embedding with the Segre embedding $\sigma:\mathbb{P}^1\times\mathbb{P}^2 \to \mathbb{P}^5$ can be represented (after identification of $\mathbb{P}^1\times\mathbb{P}^1$ with its image the quadric $Q: Z_0Z_3-Z_1Z_2=0$ in $\mathbb{P}^3$ by the Segre mapping), by the map $$\phi: [Z_0,Z_1,Z_2,Z_3]\to [F_0(Z),\cdots,F_1(Z)]$$ where $(Q,F_0,\cdots,F_5)$ is a basis of the 7 dimensional space of homogenous quadratic polynomials that contains the line $L: Z_0=Z_1=0$ of $Q$.

This seems strange to me because $\phi$ is not even defined on the line $L$. What did I miss ?

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The composition $\sigma\circ(1\times\nu)$ will map $((x:y),(z:w))\mapsto ((x:y),(z^2:zw:w^2))\mapsto (xz^2:xzw:xw^2:yz^2:yzw:yw^2).$

Identifying $\Bbb P^1\times\Bbb P^1\cong Q\subseteq\Bbb P^3$ takes $((x:y),(z:w))\mapsto (xz:xw:yz:yw)$ or conversely, $(a:b:c:d)\mapsto ((1:c/a),(1:b/a))=((a:c),(a:b))$ where we suppose that $a\neq 0.$ Thus, on the affine patch $a\neq 0,$ the composition is determined by

$$(a:b:c:d)\mapsto (a^3:a^2b:ab^2:ca^2:cab:cb^2)=(a^3:a^2b:ab^2:ca^2:a^2d:adb) =(a^2:ab:b^2:ac:ad:bd),$$ where we've used $ad=bc.$

Thus, we wish to show that $\langle a^2,ab,b^2,ac,ad,bd,ad-bc\rangle$ is a basis of the homogeneous quadratics containing $a=b=0.$ This part should be routine.

Edit: The identification on the patch $c\neq 0$ is given by $(a:b:c:d)\mapsto ((a/c:1),(1:d/c))=((a:c),(c:d)).$ On this patch the composition is determined by $$(a:b:c:d)\mapsto (ac^2:acd:ad^2:c^3:c^2d:cd^2)=(ac^2:bc^2:bcd:c^3:c^2d:cd^2)=(ac:bc:bd:c^2:cd:d^2).$$

On this patch we find a similar expression, but now our forms vanish on $c=d=0.$

I think the question is not saying that the first expression should be valid everywhere on $Q,$ since it does not make sense to evaluate it on $a=b=0,$ as you say. On $c\neq 0$ or $d\neq 0$ we have this alternate expression, which should coincide with the original one on the intersection $a,c\neq 0$ for example. This way we'll get a globally defined map on $Q.$ Unfortunately, the way the question seems to be stated doesn't make this precise at all.

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thank you very much ! I found exactly what you found and then got stuck precisely because the identification can only be done on the affine patches $a \neq 0$ or $b\neq 0$. Globally the composition is well defined on the line $(0:1)\times \mathbb{P}^1$ but not the globally defined $\phi$ on $\mathbb{P}^3$ using the six polynomials as given in the text of the exercise. So there is still something I do not understand well. – brunoh Aug 26 '12 at 18:43
Dear @brunoh, you're welcome! I think you are correct that the expression is not globally defined. In order to define the map $Q\to\mathbb P^5$ I believe we need to consider a different expression along the line $a=b=0.$ – Andrew Aug 26 '12 at 19:40
Actually, it is not difficult to show that whatever six homogenous same degree polynomials we choose to represent the $(F_i)$ they will intersect at least on one point in Q. So the composition ϕ can never be represented globally by six independant polynomials, which means this rational normal scroll can not be parametrized globally by homogenous polynomials defined on Q. Precisely the contrary of what is asked to be proven in the fist part of the exercise, but in line with the second part. So your answer should be correct: I misunderstood the question asked in the first part. – brunoh Aug 26 '12 at 22:16