# image of Segre-Veronese as a tuple of polynomials

This question shares the same context as pullback of rational normal curve under Segre map, but it is otherwise independent. It relates to Exercise 2.29 in Harris (AG-first course).

So we begin with the Segre-Veronese map \begin{align} \phi:P^1 \times P^1 \stackrel{id \times \nu}{\rightarrow} P^1 \times P^2 \stackrel{\sigma}{\rightarrow} P^5 \end{align} given by \begin{align} [X_0:X_1] \times[Y_0:Y_1]\stackrel{\nu}{\mapsto} [X_0:X_1]\times[Y_0^2 : Y_0Y_1 : Y_1^2]\stackrel{\sigma}{\mapsto}[X_0 Y_0^2: X_0 Y_0 Y_1 : X_0 Y_1^2:X_1 Y_0^2: X_1 Y_0 Y_1 : X_1 Y_1^2]. \end{align} The map $\phi$ can be viewed as a map on the surface $Q$ of $P^3$ given by $Z_0 Z_3 - Z_1 Z_2=0$, which is the identification of $P^1 \times P^1$ inside $P^3$ under the segre map $P^1 \times P^1 \rightarrow P^3$.

$\phi$ is morphism of the projective variety $Q$, because its restriction to each affine open set $U_i$ is given by a $6$-tuple of polynomials. In particular, on the open set $X_0 \neq 0 = U_0 \cup U_1$, $\phi$ is given by \begin{align} (A): \, \phi([Z_0:Z_1:Z_2:Z_3]) = [Z_0^2:Z_0 Z_1 : Z_1^2 : Z_0 Z_2:Z_0 Z_3: Z_1 Z_3]. \end{align} Now, notice that on the line $L:=\left\{Z_0=Z_1=0\right\} \subset Q$ all of these polynomials vanish at the same time so that we can not use this $6$-tuple as a description of $\phi$, except on the open set of $Q$ given by $Q-L$.

The question now is can we find a description of $\phi$ as a $6$-tuple of homogeneous polynomials well-defined on the entire $Q$? In his Exercise 2.29, Harris claims that this is not possible if these polynomials have the same degree.

But i seem to have proved that this is not possible for any $6$-tuple of homogeneous polynomials. Here is my argument: Suppose \begin{align} (B): \, \phi(Z) = [F_0(Z):\cdots:F_5(Z)], \, \, \, \forall Z:=[Z_0 : Z_1 : Z_2 : Z_3] \in Q. \end{align} Then the $6$-tuple of $(A)$ will agree with with the $6$-tuple of $(B)$ on $Q-L$ and so on every open set $(Q-L)\cap U_i$. Since the restriction of $\phi$ on $U_i$ is a morphism of an affine variety, the two $6$-tuples will agree on the entire $U_i \cap Q$ and so on the entire $Q$. But this is a contradiction since $(A)$ is not defined on the line $L$.

Question: Any comments on my reasoning above? If i am right, then $\phi$ can not be represented by a $6$-tuple of homogeneous polynomials irrespectively of their degree.

## 1 Answer

If $F_0, F_1, \ldots, F_5$ are homogeneous polynomials of varying degrees, then the notation $$[F_0:\cdots:F_5]$$ isn't even meaningful. The $F_i$ all have to be homogeneous of the same degree to make any sense of it.

Remember, homogeneous coordinates for points in projective space are only defined up to a scalar multiple. Consequently, if you want to define a function from a projective space to a projective space, you need to give something where every output coordinate scales the same way when you scale the input coordinates.

• Ah, that's right, i missed that. Many thanks! Any objections to my argument (assuming all $F_i$ have the same degree)? – Manos Apr 21 '15 at 4:08