## Algebraic Geometry versus Complex Geometry

This question is motivated by this one.

I would like to hear about results concerning complex projective varieties which

1. have a complex analytic proof but no known algebraic proof; or
2. have an algebraic proof but no known complex analytic proof.

For example, I don't think there exists an equivalent of Mori's bend-and-break argument that avoids reduction to positive characteristic. So the existence of rational curves on Fano varieties would be an example of 2.

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I wonder, are there any theorems with analytic proofs for which one can prove there is no algebraic proof, or vice-versa? – Kevin Lin Nov 12 2009 at 19:14

To my knowledge, there is no characteristic-free proof of Grauert–Riemenschneider vanishing theorem: Let $\pi \colon \widetilde{X} \to X$ a desingularization of $X$, $\mathcal{F}$ an ample locally free sheaf on $X$ such that $\pi^{\ast}(\mathcal{F})$ is also locally free, then $R^p\pi_*(\pi^{\ast}(\mathcal{F}) \otimes \omega_{\widetilde{X}}) = 0$ for all $p \geq 1$, where $\omega_{\widetilde{X}}$ denotes the canonical sheaf on $\widetilde{X}$.

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Concerning your example, there is definitely no analytic proof of the existence of rational curves on Fano manifolds. This is one of the dream of complex geometers... You can also consider this weaker statement: given a Fano manifold $X$, can you construct an entire curve (i.e. a non constant holomorphic image of the complex plane) in it by analytic methods? Even this is not known...

On the other hand, no algebraic proof is known for invariance of plurigenera for varieties not of general type (the analytic proof is due to Siu, later refined by Paun, and the algebraic proof for varieties of general type is due to Kawamata).

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Ah, Jorge, it's you! Hi!! – diverietti Nov 19 2010 at 10:51
Ciao Simonetti! – jvp Nov 20 2010 at 3:03

Kevin are you aware of the Ran/Kawamata argument for this theorem? (1992 pair of articles in Journal of Algebraic Geometry) Kawamata's version is not only algebraic but shockingly cute. The only problem is the profs who told me about it couldn't really explain why having unobstructed formal deformations (the algebraic side) means also that the analytic obstruction vanishes/the Kuranishi space is smooth.

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Formal smoothness (unobstructedness) implies smoothness in the analytic category as well. Of course you need to prove that the Kuranishi family exists, but once it exists smoothness can be checked infinitesimally. – Barbara Nov 19 2010 at 7:56

If $X$ is a proper curve of genues $g$ over an algebraically closed field $K$ of characteristic $0$, and $U$ an open subset, say obtained by removing $n$ closed points from $X$, then by comparison with the complex topology (more precisely by the Riemann Existence Theorem) one can derive that $\pi_1^{et}(U)$ is isomorphic to the profinite completion of the group $$\lt a_1,\ldots,a_g, b_1,\ldots,b_g,c_1,\ldots,c_n|[a_1,b_1]\cdot\ldots\cdot[a_g,b_g]c_1\ldots c_n\gt$$ As far as I know, there is no algebraic proof for this fact.

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The Bogomolov-Tian-Todorov theorem states that deformations of Calabi-Yau manifolds (compact Kaehler manifolds with trivial canonical bundle) are unobstructed. This recent paper of Iacono and Manetti gives an algebraic proof of the theorem for algebraically closed fields of characteristic 0. As far as I know this is the only algebraic result in this direction.

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I thought Ran and Kawamata proved this in some algebraic case using something called "$T^1$-lifting property"? The only source I've found for this impression is a remark by Huybrecths in his book "Complex Geometry" (p. 273), so maybe I completely misunderstood him. – Gunnar Magnusson Nov 19 2010 at 12:17

I dunno if this counts, but for a long time there was only an analytic proof of Hodge decomposition for smooth complete (i.e. compact) complex algebraic varieties. Then Deligne-Illusie proved it by algebraic methods. There is still no characteristic zero algebraic proof of the Hodge decomposition. And it would be great if someone gives such a proof!

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Deligne-Illusie did not give an algebraic proof of the Hodge decomposition; what they did was to give an algebraic proof of the degeneration of the Hodge to De Rham spectral sequence (which implies that $b_r = \sum{p+q = r} h^{p,q}$). THe first algebraic proof of this is in fact due to Faltings (as a consequence of the existence of the Hodge-Tate decomposition) and his result motivated D-I to find a simpler algebraic proof. – ulrich Apr 13 2010 at 13:10

The following statement about holomorphic vectors bundles is in the same spirit as Dmitri's answer. The tensor product of two slope-polystable bundles is again slope stable. Assuming the Hitchin-Kobayashi correspondence, this is easy. Slope-polystablility is equivalent to the existence of a Hermitian-Yang-Mills connection whilst the tensor product of two HYM connections is easily seen to be again HYM. On the other hand, a purely algebraic proof of this fact is not so straightforward. (That said, neither is the proof of the Hitchin-Kobayashi theorem.)

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1) I don't know if this qualifies, but it seems to me that there is no non-analitic proof of the statement that a surface of general type with $c_1^2=3c_2$ is a quotient of a unit two-dimensional complex ball. This is a theorem of Yau.

2) There is a notion of K-stability of complex projective (polarised) varitieties (this is an algebraic notion). This K-stability holds for variteties that admit a costant scalar curvature Kahler (cscK) metric in the chosen polarisation (Donaldson). Usually it is hard to prove that a polarised variety is K-stable unless you know that is has a cscK metric.

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1. is not too surprising perhaps, given that the (open) unit ball is not an algebraic variety. So the question is: can one rephrase the fact that a given variety is a ball quotient in algebraic terms? – algori Jan 21 2010 at 22:23
Well, at least you can say that surfaces of general type with $c_1^2=3c^2$ have infinite residually finite fundamental groups. So, its profinite completion is infinite. In other words it has finite etale covers of any degree. This is an algebraic statement. Surelly algebraic proof is unknown. Notice that there exsit connected surfaces with $c_1^2/c_2=a$ with a arbirtary close to 3. So, nobody knows what are fundamental groups of these guys if $a$ is not 3. – Dmitri Jan 22 2010 at 0:06

Here is one I am curious about : Suppose X is a proper variety over $\mathbb{C}$. Then there are only finitely etale covers of X in each degree.

This is proven in SGA 1 by comparison with the classical fundamental group, but is there a purely algebraic proof?

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I've wondered if the following is an example of 1, but I'm not expert enough in algebra to know.

The set of smooth points of an irreducible complex projective variety is connected in the classical topology.

The argument I know goes like this: Suppose it were disconnected, a disjoint union of $A$ and $B$, say. These are locally analytic sets. Then, by a theorem of Remmert and Stein, their closures $\overline{A}$ and $\overline{B}$ are analytic sets. Then, by Chow's part of the GAGA principle, $\overline{A}$ and $\overline{B}$ are varieties, and the original variety is not irreducible.

I've always wondered if you can avoid the Remmert-Stein theorem in the middle (without using Hironaka's theorem).

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Thanks for this example! Do you know any reference where I could read the proof in more detail (i.e. where I can read the proof of Remmert-Stein)? – Konrad Voelkel Nov 12 2009 at 9:56
For Remmert-Stein, there's Remmert-Stein's paper (in German) is Math. Ann., 126:263–306, 1953, also proven in chapter 3 of the book "From holomorphic functions to complex manifolds" by Fritzsche and Grauert. The proof I sketched (which is essentially the whole proof) is written down in a paper of mine with Dumas: "Slicing, skinning, and grafting "American Journal of Mathematics 131 (2009), 1419–1429". – Richard Kent Nov 12 2009 at 12:18
I'm confused, why can't you argue like this: For any two smooth points, we can join them by a sequence of curves. For these curves, the intersection with non-smooth set is Zariski closed, so finite, and removing them does not disconnect. – moonface Nov 12 2009 at 15:32
The set of smooth points in an irreducible algebraic variety is (obviously) again an irreducible algebraic variety, and there is a theorem that X(C) is connected in the classical topology if X is an irreducible algebraic variety over C. There are two proofs of the theorem in Shafarevich Basic Algebraic Geometry VII 2, neither of which uses much. – JS Milne Nov 12 2009 at 15:45
If A contained in U is dense in V, then it is dense in U (definition of subspace topology). Thus the closure of a proper closed subset of U is a proper closed subset of V. – Tom Church Jan 22 2010 at 4:32