Degree of an embedded algebraic variety - MathOverflow

Degree of an embedded algebraic variety

2009-12-24T21:00:06Z

Let $X$ be an algebraic variety and $A$ is a ample divisor on $X$. Let $m$ be a sufficiently large natural number such that $X \overset{\varphi_{mA}}{\to} \mathbf{P}H^0(X, \mathcal{O}_X(mA))$ defined by the linear system $|mA|$ is an embedding. Denote by $Y$ the image. What's a effective upper bound of $Y$ in terms of $\text{dim}X$, $A^{\text{dim}X}$ and $m$, or other invariants of $X$ and $A$.

This may not be a concrete question. But I am wondering how can we compute the degree of the image of a variety under the (bi)rational morphism given by a linear system.

Answer by Felipe Voloch for Degree of an embedded algebraic variety

2009-12-24T21:58:48Z

The degree of $Y$ is $m^{\dim X}A^{\dim X}$ almost by definition.

Answer by Charles Siegel for Degree of an embedded algebraic variety

2009-12-27T06:13:03Z

I'll expand on Felipe's answer. The degree is defined to be $(\dim X)!$ times the lead term of the Hilbert polynomial of the variety. Now, given an ample line bundle $A$, the Hilbert polynomial of the embedding given by $A$ is $n\mapsto\chi(X,A^n)$. First off, this function is in fact a polynomial (check any standard book, Hartshorne for instance). This term will then be the self intersection $A^{\dim X}$, which can be computed in any number of ways (see any book on intersection theory), and so, for the line bundle $mA$, we'll get $(mA)^{\dim X}=m^{\dim X}A^{\dim X}$.

But as Felipe said, it's pretty much by definition. In a real sense, the definition of $A^{\dim X}$ can be taken to be $(\dim X)!$ times the lead term of the Hilbert polynomial.

Answer by Emerton for Degree of an embedded algebraic variety

2010-01-06T05:00:22Z

There is a more geometric way to explain Felipe's answer. To compute the degree of a closed subvariety $Y$ of projective space, of dimension $n$ say, you intersect it with $n$ different hyperplanes in sufficiently general position (with respect to each other and also with respect to $Y$), so that the result is a collection of points, and you add up the number of points. This is the degree of $Y$.

In other words, degree has three properties that serve to define it:

(a) it is additive with respect to unions (EDIT: of distinct varieties of the same dimension, say, to avoid scheme-theoretic issues).

(b) the degree of a point is one.

(c) degree is preserved by taking sufficiently general hyperplane sections.
(I could omit the caveat sufficiently general here is I was willing to work with scheme structures, and not just varieties).

(It is then an exercise, using these conditions, to relate the degree as defined this way to the degree defined via the Hilbert polynomial.)

Now if $Y$ is the image of $X$ under the emdedding given by $|m A|$, then the intersection of a hyperplane with $Y$ pulls back, under the isomorphism $X \buildrel \sim \over \to Y$, to a member of the linear system $|m A|.$ (This is by the very definition of the embedding given by $|m A|$.) When you intersect $n$ such divisors, the number of points you get is then (m A)^n. (Because intersection is invariant under deformation of the cycles being intersected, you can replace all the divisors by the linearly equivalent divisor $m A$.)