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.


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.

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

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