Degree of a smooth projective variety

Question

Let $i_1:X \hookrightarrow \mathbb{P}^n$ and $i_2:Y \hookrightarrow \mathbb{P}^N$ be two projective schemes. Let $f:X \to Y$ be a surjective projective morphism between smooth projective varieties over $\mathbb{C}$. Denote by $g$ the composition of $f$ with $i_2$. Under what condition on $g$ can we conclude that the degree of $X$ is equal to the degree of $Y$ in $\mathbb{P}^N$ added to the degree of the generic fiber of $g$?

Answer 1

Jana, the current formulation of your question is still not right. Everything else fixed one can still choose a different $i_1$ with which one can change the degree of $X$ (if $g$ is finite, this will not change the degree of the general fiber).

To get a sensible formulation you should do this: Indeed, degree of a projective variety is determined by an embedding, but an embedding is determined by the choice of a very ample line bundle. So, one may actually associate the notion of degree with the given very ample line bundle. Then, if you allow rational numbers as degree, then you can even associate a notion of degree to any ample line bundle.

This set-up allows for a formulation of what you want without a problem: Let $f:X\to Y$ be a morphism between projective varieties, $\mathscr L$ a very ample line bundle on $Y$, $\mathscr M$ an $f$-very ample line bundle on $X$ and set $\mathscr N=f^*\mathscr L\otimes \mathscr M$. (Note that it is an easy exercise to prove that $\mathscr N$ is very ample).

Now you can ask if it is true that the degree associated to $\mathscr N$ on $X$ is the sum of the degree associated to $\mathscr L$ on $Y$ and the degree associated to $\mathscr M|_F$ on $F$, the general fiber of $f$.

It turns out that this still has a snowball's chance in hell to be true, but if you asked whether it is true that the degree associated to $\mathscr N$ on $X$ is the product of the degree associated to $\mathscr L$ on $Y$ and the degree associated to $\mathscr M|_F$ on $F$, the general fiber of $f$, then it is a slightly better question.

Let's see what we can say about this. It is easy to see that the degree associated to a line bundle $\mathscr L$ is just $c_1(\mathscr L)^d$ where $d$ is the dimension of the variety. So, all you need to do is to compare the Chern classes of these line bundles. For simplifying the typing let me use divisors: $L$ will be a divisor for $\mathscr L$, $M$ will be a divisor for $\mathscr M$, and $N$ will be a divisor for $\mathscr N$. In fact, let's choose them so $N=f^*L+M$. Further let $n=\dim X$, $d=\dim Y$ and $F$ still the general fiber of $f$, so $\dim F=n-d$.

Let's see the case of $f$ being finite first. So, $n=d$ and $F$ is a finite set of points. The obvious choice for $\mathscr M$ in this case is just $\mathscr O_X$ so we would get something sensible. In other words $N=f^*L$ and we want to compare $N^n$ and $L^n$. It is easy to see that if one represents $L^n$ by as many (general) points, then $N^n=f^*L^n$ is represented by the same number of fibers, each of which consists of $\deg f$ many points. In other words, the degree on $X$ is the degree on $Y$ multiplied by the degree of the map. This is the result Igor was referring to.

Next assume that $n>d$ and we'll see that we cannot expect anything like this. So we want to compare $N^n$ with $L^d$ and $(M|_F)^{n-d}=M^{n-d}\cdot F$. Since $N=f^*L+M$ we have that $$ N^n = (f^*L+M)^n = \dots + {n\choose d} f^*L^d \cdot M^{n-d} + \dots $$ As before, one may think about $L^d$ as a bunch of points on $Y$ and hence $f^*L^d$ as a bunch of fibers. On each of these fibers the effect of $(\_ )\cdot M^{n-d}$ is the same as restriction, so we get that $$ f^*L^d \cdot M^{n-d} = (L^d) \cdot ((M|_F)^{n-d}) $$ (the right hand side is a product of numbers!)

So, we get that $$ \deg X = {n\choose d} \deg Y \cdot \deg F + {\rm more\ terms} $$

Note that these extra terms are all non-negative by the choice of our line bundles.(Of course, $L^j=0$ for $d>j$, but $M^l$ is not necessarily $0$ for $l>n-d$).

So, on one hand we already get the multiplier $n\choose d$, but more importantly we may also get additional terms that are hard to account for based on just those degrees. These terms come from the fact that while we required that $\mathscr M$ is $f$ very ample we did not require (and we cannot!) that it has absolutely no positivity in the "horizontal" direction. If it does, that will produce more positive terms. So, the best one can hope for is a product formula in case $f$ is finite. Otherwise one could say that there is an estimate that $\deg X\geq \deg Y \cdot \deg F$ and most of the time the inequality is strict.

Answer 2

This seems to be addressed in this paper by Derksen and Kraft, Prop. 8.3