# Degree of the projection of a projective variety

Let $$X\subseteq\mathbb{P}^n\times\mathbb{P}^m\subseteq \mathbb P^{(n+1)(m+1)-1}$$ be a projective variety of dimension $$p$$ and degree $$d$$ defined over an algebraically closed field $$k$$, where $$\mathbb{P}^n\times\mathbb{P}^m\subseteq \mathbb P^{(n+1)(m+1)-1}$$ via the Segre embedding. Let $$X'\subseteq \mathbb{P}^n$$ be the projection of $$X$$ on $$\mathbb{P}^n$$, which we will assume having dimension equal to that of $$X$$. What can we say about the degree $$d'$$ of $$X'$$? Is there some formula linking the degrees $$d$$ and $$d'$$. Since an analytical charcaterization is also good for my purposes, feel free to assume $$k=\mathbb C$$.

• What do you mean by degree $d$? With respect to what ample line bundle? – R. van Dobben de Bruyn Oct 9 '18 at 23:17
• @R.vanDobbendeBruyn I mean the classical one: Number of points in the intersection of $X$ with $p$-hyperplanes in general position. – Vincenzo Zaccaro Oct 9 '18 at 23:50
• @VincenzoZaccaro: Your question is ill-posed. Without specifying more details, there is no such thing as a hyperplane in $\mathbb{P}^n \times \mathbb{P}^m$. Subvarieties such as $X$ come with bidegrees, not degrees. – Ozob Oct 10 '18 at 3:01
• Let $h_1,h_2$ be the hyperplane class in $CH^1(\mathbb{P}^n)$ and $CH^1(\mathbb{P}^m)$, and $q_1,q_2$ the two projections. Then $d=(q_1^*h_1+q_2^*h_2)^p$, and $d'=q_1^*h_1^p$. Obviously you cannot say more than $d'\leq d$. – abx Oct 10 '18 at 4:10
• One can say more if $X$ is irreducible. Would you be willing to add that hypothesis? Or do you need more general results, for possibly reducible varieties? – Zach Teitler Nov 10 '18 at 13:35

As abx says in his comment, all you can say is $$d' \le d$$. Here are some explicit examples with $$m = 2$$ and $$n = 1$$.
(1) To see that $$d'$$ can be arbitrarily smaller than $$d$$, fix a line $$L \subseteq \mathbb{P}^2$$ and set $$X = L \times S$$, where $$S \subseteq \mathbb{P}^1$$ is a finite set of points. Then $$X$$ has degree $$\lvert S\rvert$$ while has $$X' = L$$ has degree 1.
(2) To see that we can also have $$d' = d$$, take a degree $$d$$ curve $$C \subseteq \mathbb{P}^2$$ and set $$X = C \times \{*\}$$ for some point $$* \in \mathbb{P}^1$$. Then $$X$$ has degree $$d$$, and $$X' = C$$ also has degree $$d$$.