The Relationship between Complex and Algebraic Geomety - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T07:48:40Z http://mathoverflow.net/feeds/question/4895 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4895/the-relationship-between-complex-and-algebraic-geomety The Relationship between Complex and Algebraic Geomety Jean Delinez 2009-11-10T17:31:27Z 2009-11-17T07:37:14Z <p>I have recently begun to study algebraic geometry, coming from a differential geometry background. It seems that there is a deep link between complex manifolds and complex varieties. For example, one often hears that they are two different ways of looking at the same thing. Can anybody give a precise statement of this relationship?</p> http://mathoverflow.net/questions/4895/the-relationship-between-complex-and-algebraic-geomety/4899#4899 Answer by Csar Lozano Huerta for The Relationship between Complex and Algebraic Geomety Csar Lozano Huerta 2009-11-10T18:38:07Z 2009-11-17T07:37:14Z <p>There is no enough room here to write the answer to your question. However, you can take a look at the appendix of Hartshorne and if you have time, a classical reference is Griffiths and Harris: Principles in algebraic geometry. However roughly speaking, the concept of positivity will tell you when a compact Kahler complex manifold is algebraic.</p> http://mathoverflow.net/questions/4895/the-relationship-between-complex-and-algebraic-geomety/4902#4902 Answer by Greg Kuperberg for The Relationship between Complex and Algebraic Geomety Greg Kuperberg 2009-11-10T19:06:20Z 2009-11-11T17:32:44Z <p>The Wikipedia article is more technical than it should be, and for the reader in a hurry not all that well written. Here is a summary of the main points as best I understand them:</p> <p>Complex manifolds are analogous to smooth complex algebraic varieties, not to the singular ones. But that discrepancy is surmountable, because you can also have complex analytic varieties which can have similar singularities. Then the first and most important relation is that every complex algebraic variety is a complex analytic variety. Every Zariski open set is analytically open; analytic gluing maps are more general than algebraic ones; and the allowed analytic charts are more general than the allowed algebraic charts. Also every algebraic morphism is an analytic morphism, so you get a morphism between categories.</p> <p>But the connection is better than that because of the GAGA principle (globally analytic implies globally algebraic). My understanding of GAGA is very sketchy, but I think that the following is correct. Among other consequences of GAGA, a closed analytic subvariety of a proper (equivalent to compact) algebraic variety is algebraic. An analytic isomorphism between two proper algebraic varieties is algebraic. I would suppose that there is a similar principle for compact fibrations as well.</p> <p>So, <s>if you make compact analytic varieties algebraically, you can't escape from the algebraic class</s> some of the main constructions of complex manifolds do not escape from the algebraic class. (But not all: deformations and infinite group actions can escape.) All projective analytic varieties are algebraic, and in dimension 1 all compact curves are projective. Moreover, there are limited ways for a compact analytic manifold to avoid being projective, by Moishezon's theorem and Kodaira's theorem. In practice, then, most of the complex manifolds that people make are algebraic. Also, most of the analytic calculations on a proper algebraic variety are algebraic: Many global calculations are algebraic by GAGA, and many local calculations are algebraic just by truncating Taylor series.</p> <p>Contrast all this with real algebraic vs real analytic. It is still true that (the real points of) a smooth real algebraic variety is a real analytic manifold. More strongly than in the complex case, although it is highly non-trivial, every compact real analytic manifold is real algebraic. But the real algebraic structure is massively not unique, even for a circle, and that makes all the difference.</p> <p><hr /></p> <p>The other answerers in this thread, who are more expert in this topic than I am, had more information about why a compact complex manifold might not be a smooth projective variety. Just for clarity, I will restrict attention to the compact, smooth case. Also, you say "proper" rather than "compact" in the algebraic category because every algebraic variety is "compact" in the extremely coarse Zariski topology. An algebraic variety is proper if and only if it is analytically compact. The main use of the word proper is to emphasize that it is more general than projective, which means given by polynomial equations in complex projective space.</p> <p>There are two very different initial reasons that an analytic complex manifold might not be projective. It might not be Moishezon: A complex $n$-manifold is Moishezon if it has $n$ algebraically independent meromorphic functions. (The number of algebraically independent elements or the transcendence degree of a field is called the Krull dimension. The meromorphic Krull dimension of a compact complex $n$-manifold is at most $n$.) Or it might not be Kähler: A complex $n$-manifold is Kähler if it has a Riemannian metric such that the covariant derivative of the complex structure vanishes. So to summarize what people said about compact complex manifolds (much of which is in the back of Hartshorne's book):</p> <p>projective ⇒ algebraic ⇒ Moishezon ⟺ bimeromorphically projective<br /> projective ⇒ Kähler ⇒ symplectic ⇒ non-zero $H^2$<br /> algebraic ⇒ non-zero $H^2$ (<a href="http://sbseminar.wordpress.com/2008/02/14/complex-manifolds-which-are-not-algebraic/" rel="nofollow">exposited</a> by David Speyer)<br /> Moishezon and Kähler ⟺ projective (Moishezon)<br /> Kähler and integrally symplectic ⟺ projective (Kodaira)</p> <p>In addition, projective and algebraic structure and the Moishezon property are all unstable with respect to analytic deformation. And bimeromorphic equivalence preserves <code>$\pi_1$</code>. Taubes found compact complex manifolds that have the wrong <code>$\pi_1$</code> to be Kähler; indeed they can have any <code>$\pi_1$</code>. Voisin found compact Kähler manifolds with the wrong homotopy type to be projective, disproving Kodaira's conjecture that every compact Kähler manifold can be deformed to projective. And, way out in left field, a left-invariant complex structure (LICS) on a compact simple Lie group is a compact complex manifold that has no $H^2$ and can be simply connected too.</p> <p>Still, despite these beautiful non-projective compact complex manifolds, it's generally easier to study projective examples. It's generally easier to sidestep analysis and do algebra instead.</p> http://mathoverflow.net/questions/4895/the-relationship-between-complex-and-algebraic-geomety/4922#4922 Answer by Kevin Lin for The Relationship between Complex and Algebraic Geomety Kevin Lin 2009-11-10T20:52:41Z 2009-11-10T20:52:41Z <p>The Kodaira embedding theorem states that a compact complex manifold is projective -- and thus algebraic by Chow's theorem -- if and only if it has a Kaehler structure with integral Kaehler class. Kaehler manifolds are symplectic, and integral Kaehler class means that the cohomology class of the corresponding symplectic form $\omega$ lives in the second cohomology with $\mathbb{Z}$ coefficients. Note that the Kaehler class of a Kaehler manifold must be nonzero, because the symplectic volume $\int_X \omega^{n}$ must be nonzero.</p> <p>So maybe one way to find compact things that are non-Kaehler and non-projective is to look for things with zero second cohomology, because then they wouldn't even be symplectic. Perhaps there are examples of compact toric varieties that satisfy this?</p> http://mathoverflow.net/questions/4895/the-relationship-between-complex-and-algebraic-geomety/4937#4937 Answer by jvp for The Relationship between Complex and Algebraic Geomety jvp 2009-11-10T22:16:53Z 2009-11-11T03:40:52Z <p>While every algebraic manifold is complex analytic the converse is far from true.</p> <p>If $M$ is a compact complex manifold then $a(M)$, the algebraic dimension of $M$, is defined as the transcendence degree of $\mathbb C(M)$ over $\mathbb C$, where $\mathbb C(M)$ is the field of meromorphic functions on $M$. </p> <p>The algebraic dimension is at most the dimension of $M$ and when the equality holds $M$ is called a Moishezon space. In general what one has is a variety $\hat M$ with $\dim \hat M = \dim M$, a projective variety $N$ with $\dim N = a(M)$, and analytic morphisms $\pi : \hat M \to M$ and $\varphi : \hat M \to N$ such that </p> <ul> <li>$\pi$ is bimeromorphic; </li> <li>the generic fiber of $\varphi$ is irreducible; </li> <li>the field of meromorphic functions on $M$ is the same as the one of $N$. More precisely: $\varphi^* \mathbb C(N) = \pi^* \mathbb C ( M)$.</li> </ul> <p><hr /></p> <p>It also has to be noted that from a topological point of view there are many more compact complex manifolds than compact algebraic manifolds. To be more precise, there are strong restrictions on the fundamental group of compact Kähler manifolds, while there is a deep result of Taubes that implies that every finitely generated group is the fundamental group of a compact complex $3$-fold. </p> <p><hr /></p> <p>Motivated by comments on Kuperberg's answwer, let me also mention that there were a conjecture by Kodaira claiming that any compact Kähler manifold is deformation equivalent to projective manifolds. While the conjecture is true in dimension two, as Kodaira has himself shown, it has been recently disproved by Claire Voisin. </p> <p><hr /></p> <p>Compact complex manifolds which are not algebraic are not artificial beasts. Let me recall a very natural example of algebraic nature. Consider the quotient $X=SL(2,\mathbb C)/\Gamma$, where $\Gamma$ is a discrete cocompact subgroup. I think it was Mostow who proved that $X$ has no analytic hypersurfaces, which implies in particular $a(X)=0$. To see that $X$ is far from an algebraic variety, despite being algebraically defined, notice that there are $1$-forms on $SL(2,\mathbb C)$ which are invariant by right translations and which are not closed. So they induce global holomorphic $1$-forms on the quotient $X$ which are not closed. But on compact Kähler manifolds, and more generally on compact manifolds bimeromorphic to them, every holomorphic $1$-form is closed.</p> http://mathoverflow.net/questions/4895/the-relationship-between-complex-and-algebraic-geomety/5098#5098 Answer by JS Milne for The Relationship between Complex and Algebraic Geomety JS Milne 2009-11-11T20:18:45Z 2009-11-11T20:18:45Z <p>No one seems to have mentioned that if a complex manifold is an algebraic variety, then it has a nice compactification. More precisely, by Hironaka it can be embedded in a compact manifold such that the boundary is a divisor with normal crossings. </p> <p>In dimension 1 this is the only condition: a Riemann surface is an algebraic curve if and only if it can be compactified by adding a finite number of points. In particular, bounded holomorphic functions on it are constant (so the complex upper half plane is not an algebraic curve because z-i/z+i is a bounded holomorphic function).</p> <p>So the two main obstructions to a complex manifold being an algebraic variety (or, at least, an algebraic space) are that it may not have a nice compactification (in the above sense) and it may not have enough meromorphic functions.</p> <p>To say that a complex manifold is an algebraic variety is a very strong statement.</p>