Is a complex manifold projective just because its blow-up at a point is ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T02:15:41Z http://mathoverflow.net/feeds/question/54867 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/54867/is-a-complex-manifold-projective-just-because-its-blow-up-at-a-point-is Is a complex manifold projective just because its blow-up at a point is ? Georges Elencwajg 2011-02-09T10:19:34Z 2011-03-02T19:49:05Z <p>Consider a compact connected complex manifold $X$ of dimension $n$. Siegel proved in 1955 that its field of meromorphic functions $\mathcal M (X)$ has transcendence degree over $\mathbb C$ at most $n$. Moishezon studied those complex manifolds for which the degree is $n$, and consequently these manifolds are now called Moishezon manifolds.</p> <p>In dimension 2, every Moishezon surface is projective algebraic according to a theorem of Chow-Kodaira proved in 1952, long before Moishezon formally introduced his concept. However in dimensions 3 and more, there exist nonprojective Moishezon manifolds (you can see an example in Shafarevich's book <em>Basic Algebraic geometry</em>). </p> <p>Nevertheless a Moishezon manifold $X$ is close to projective: Moishezon's main result is that after a finite number of blow-ups with smooth centers, $X$ becomes algebraic projective. So if a blow-up of $X$ is projective, you can't deduce that $X$ was projective. However this main result says nothing about the dimensions of the manifolds you blow up. I've heard it claimed that if only one point is blown-up, you can't get from a non-projective to a projective manifold, but I could obtain neither precise reference nor proof. Hence my question :</p> <p><strong>If a compact complex manifold becomes projective algebraic after blowing-up a point, was it already projective algebraic?</strong></p> http://mathoverflow.net/questions/54867/is-a-complex-manifold-projective-just-because-its-blow-up-at-a-point-is/54883#54883 Answer by Artie Prendergast-Smith for Is a complex manifold projective just because its blow-up at a point is ? Artie Prendergast-Smith 2011-02-09T13:56:05Z 2011-02-10T08:03:45Z <p>Dear Georges, maybe the following argument works. (It's quite possible a sign went wrong somewhere, though.)</p> <p>Let $\pi: Y \rightarrow X$ be the blowup. By assumption Y is projective, so it carries an ample line bundle A say. Let E denote the exceptional divisor of the blowup, and consider line bundles of the form $A+nE$ (for positive integers n). If C is any curve in Y which is not contained in E, then $(A+nE).C = A.C + nE.C$ is positive, for any n. On the other hand, let L be a line in E: then $E.L = -1$. (Note that all other curves in E are numerical multiples of this one.) So if we set n=A.L (positive, by ampleness of A) then we have $(A+nE).L = 0$. So the line bundle A+nE is nef, and has degree 0 exactly on those curves which lie in E.</p> <p>I claim that $A+nE$ is in fact basepoint-free. To see this, it suffices (by the Basepoint-free Theorem, see e.g. Koll\'ar--Mori Chapter 3) to show that the line bundle $m(A+nE)-K_Y$ is nef and big, for some positive integer m. Now A is ample and E is effective, so $A+nE$ is big for all positive n (ample+effective=big --- this is also in Koll\'ar--Mori). Moreover, bigness is an open condition, so for m sufficiently large, $m(A+nE)-K_Y$ is still big. So it remains to prove nefness. </p> <p>Recall that we were free to choose A to be any ample line bundle, so choose it to satisfy the condition that $A-K_Y$ is itself ample (again, using the fact that ampleness is open). Then $m(A+nE)-K_Y = (mA-K_Y) +mnE$, so in particular it has positive degree on any curve C which is not contained in E. On the other hand, if L is a line in E, then $(m(A+nE)-K_Y).L = -K_Y.L$ (by the calculations in the first paragraph.). Moreover, $-K_Y=\pi^\ast(-K_X)-(dim X-1)E$, so $-K_Y.L=-(dim X-1)E.L>0$. So $m(A+nE)-K_Y$ has nonnegative degree on all curves, i.e. it is nef.</p> <p>Putting all this together, we have that $m(A+nE)$ is basepoint-free for suitable positive integers m and n. So it defines a contraction morphism $p: Y \rightarrow Z$ to another projective variety Z. But the morphism p contracts exactly those curves on which A+nE has degree 0, which by construction are exactly the curves contained in E. Therefore Z is exactly the blow-down of E to a point, hence isomorphic to X. Since Z is projective, so is X.</p> http://mathoverflow.net/questions/54867/is-a-complex-manifold-projective-just-because-its-blow-up-at-a-point-is/57126#57126 Answer by Parsa for Is a complex manifold projective just because its blow-up at a point is ? Parsa 2011-03-02T16:31:13Z 2011-03-02T16:31:13Z <p>Can anyone please tell me if there's something wrong with the following reasoning?</p> <p>Sticking to the notation in the answer above, let $A$ be an ample divisor on $Y$. Then we can write $A \sim \pi^*D - kE$ for some divisor $D$ on $X$. Since $A \cdot l>0$, $\pi^*D \cdot l =0$, and $E \cdot l =-1$ we must have $k>0$. In fact, we can take $k=1$ since for any curve $C$ on $Y$ not contained in $E$, $(\pi^*D-kE)\cdot C \leq (\pi^*D-E)\cdot C$ because $E\cdot C \geq 0$, and we still have $(\pi^*D-E)\cdot l =1$. So take $A \sim \pi^*D -E$ to be our ample divisor on $Y$. By Seshadri's criterion, there exists an $0&lt; \epsilon &lt;1$ such that for any curve $C'$ on $Y$, $A \cdot C' \geq \epsilon ~ m(C')$, where $m(C')=sup_{q \in C'} m_q(C')$ is the multiplicity of the curve. </p> <p>Now let $C$ be any curve on $X$ and denote by $C'$ its strict transform under the blowup. If $C$ does not pass through the center $p \in X$, then $C' \cong C$ and $C'$ does not meet $E$, so $E \cdot C' =0$. Then $A \cdot C' = \pi^*D \cdot C' = D \cdot C \geq \epsilon ~ m(C') = \epsilon ~ m(C)$.</p> <p>If $C$ does pass through $p$ with multiplicity $m$, then $E \cdot C' =m$ and we have $A \cdot C' = \pi^*D \cdot C' -E \cdot C' = D \cdot C - m \geq \epsilon ~ m(C')$. But this means that $D \cdot C \geq \epsilon ~ m(C') + m \geq \epsilon ~ m(C)$. Hence $D$ is ample on $X$ by Seshadri's criterion (which still holds on complete non-projective schemes) so $X$ is projective.</p> <p>The problem is I'm not quite sure where I'm using the smoothness of $X$, since blowing up points on a (singular) non-projective surface will make the blowup projective. (Maybe in the first line where it's assumed that $Pic(Y) \cong Pic(X) \oplus \mathbb Z$.) In any case, the above answer (and this one if it's correct) gives a quick proof of the Chow-Kodaira theorem, which says that any smooth complete algebraic surface must be projective.</p> <p>I expect the statement to hold for singular complete algebraic varieties in dimension $\geq 3$. That is, if $X$ is any complete algebraic variety of dimension $\geq 3$, then $X$ is projective if and only if $Bl_p(X)$ is projective.</p> http://mathoverflow.net/questions/54867/is-a-complex-manifold-projective-just-because-its-blow-up-at-a-point-is/57162#57162 Answer by inkspot for Is a complex manifold projective just because its blow-up at a point is ? inkspot 2011-03-02T19:49:05Z 2011-03-02T19:49:05Z <p>There is a sledgehammer available for this particular nut: Mori's results on extremal rays, in his paper (Annals 1979?) on smooth projective varieties $Y$ where $K_Y$ is not nef: if $Y=Bl_PX$, with exceptional divisor $E$, then any line in $E$ will span an extremal ray and then can be contracted in the category of projective varieties. Since all curves in $E$ lie in the same ray, $E$ must be contracted to a point, so the result of the contraction must be $Y$.</p>