Degree of an embedded algebraic variety - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T12:23:14Z http://mathoverflow.net/feeds/question/9703 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9703/degree-of-an-embedded-algebraic-variety Degree of an embedded algebraic variety Fei YE 2009-12-24T21:00:06Z 2010-01-07T03:05:54Z <p>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 $A$. </p> <p>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.</p> http://mathoverflow.net/questions/9703/degree-of-an-embedded-algebraic-variety/9707#9707 Answer by Felipe Voloch for Degree of an embedded algebraic variety Felipe Voloch 2009-12-24T21:58:48Z 2009-12-24T21:58:48Z <p>The degree of $Y$ is $m^{\dim X}A^{\dim X}$ almost by definition.</p> http://mathoverflow.net/questions/9703/degree-of-an-embedded-algebraic-variety/9862#9862 Answer by Charles Siegel for Degree of an embedded algebraic variety Charles Siegel 2009-12-27T06:13:03Z 2009-12-27T06:13:03Z <p>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 $(mA)^{\dim X}=m^{\dim X}A^{\dim X}$.</p> <p>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.</p> http://mathoverflow.net/questions/9703/degree-of-an-embedded-algebraic-variety/10884#10884 Answer by Emerton for Degree of an embedded algebraic variety Emerton 2010-01-06T05:00:22Z 2010-01-07T03:05:54Z <p>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 $Y$.</p> <p>In other words, degree has three properties that serve to define it:</p> <p>(a) it is additive with respect to unions (EDIT: of distinct varieties of the same dimension, say, to avoid scheme-theoretic issues).</p> <p>(b) the degree of a point is one.</p> <p>(c) degree is preserved by taking sufficiently general hyperplane sections.<br /> (I could omit the caveat <I>sufficiently general</I> here is I was willing to work with scheme structures, and not just varieties).</p> <p>(It is then an exercise, using these conditions, to relate the degree as defined this way to the degree defined via the Hilbert polynomial.)</p> <p>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 $m A$.)</p>