# Degree of a smooth projective variety

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 $g$?

• What definition of degree are you using for $X$ and $Y$? Have you fixed a choice of embedding into projective space? – Daniel Loughran Nov 20 '13 at 12:49
• Dear Jana, A projective variety does not come with a fixed projective embedding. Given a projective variety, you only know that such a projective embedding exists, but there is not a canonical choice. – jmc Nov 20 '13 at 14:59
• @Jana would you mind making it more precise then? Do the embeddings have anything to do with the map $f$? – Karl Schwede Nov 20 '13 at 18:17
• @Jana If there is no relation between $f$ and the respective projective embeddings then the question is absolutely ridiculous. – Dan Petersen Nov 20 '13 at 19:00
• @Jana: it seems that you don't get the problem with your question, pointed out by the people above (which are right). Let us make it more precise: you want a relation between the degrees of $X$, of $Y$ and of a general fibre. If you compose your map with any other closed embedding of $Y$ into a projective space, you change the degree of $Y$ but not the two others degrees. Hence, asked like this the question has a trivial answer: No relation. So you should probably add something to explain that the degree is related to the morphism. – Jérémy Blanc Nov 20 '13 at 21:53

## 2 Answers

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.

• @Kovacs: Thank you very very much for your answer. It is very very informative and helpful. – Jana Nov 21 '13 at 2:42

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

• That only seems to discuss finite morphisms, though. – user5117 Nov 20 '13 at 14:27
• @ArtiePrendergast-Smith true, but one has to start somewhere :) – Igor Rivin Nov 20 '13 at 14:48
• Igor: Indeed! My previous comment came out more negative than I intended it; I just meant to point out that the Derksen-Kraft result doesn't give a complete solution. – user5117 Nov 20 '13 at 14:55
• @Rivin: Thanks for the answer. Do you have a bit more general solution in mind? – Jana Nov 20 '13 at 15:53
• In fact, the result is about finite morphisms given by a linear projection. This is a very precise case, but probably one of the only ones where the result works. – Jérémy Blanc Nov 20 '13 at 21:47