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In differential topology, there are some funny phenomena that can only happen in dimension 4. For example, only in dimension 4 you can have a closed topological manifold admitting infinitely many distinct smooth structures.

Is there something like this happening in algebraic geometry? Does there exist an integer $k>1$ and some interesting geometric statement that only holds for varieties of dimension$\neq k$?

P.S. It should be noted that a professional topologist has said that he has doubts about one important paper on manifold topology. He did not give a lot of details, and not being a topologist, I can not judge the veracity of his claims. I do not know if the existence statement in the second sentence of this post can be established independently of that paper. If somebody does know, let us know.

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    $\begingroup$ could the down-voters kindly explain the reason why this question is a bad one? I am aware that the fact that smooth topology is interesting in dimension 4 is a miracle. I am asking whether such a miracle happens in algebraic geometry. Asking for a miracle by itself is not a valid reason to be down-voted, I would think. $\endgroup$ – user137767 Apr 17 '19 at 20:17
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    $\begingroup$ There are many special things which happen in reals (for a silly example, polynomials of even degree with no roots), but algebraic geometry is best understood over complex numbers and such special things are rare and I know of no easy examples. One can always look for special things, for example, over $\mathbb{C}$ every unirational variety is rational if $\dim\leq 2$ and in positive characteristic true only for $\dim =1$. False otherwise. $\endgroup$ – Mohan Apr 18 '19 at 1:59
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Let $X,L$ be a smooth polarized projective variety of dimension $n$ with $K_X =\mathcal{O}_X$. Let $v \in H^{\bullet}(X,\mathbb{Q})$ a primitive vector and consider $M_{L,v}$ the moduli space of $L$-stable sheaves with Chern character equal to $v$. Assume that the actual dimension of $M_{L,v}$ is equal to its virtual dimension. Then $M_{L,v}$ has an open subset which is smooth, say $M_{L,v}^{sm}$.

Miracle when $\dim X =2$ : $M_{L,v}^{sm}$ carries a holomorphic symplectic form! This is because only when $\dim X = 2$ is there a symplectic isomorphism $\mathrm{Ext}_X^1(F,F) \simeq \mathrm{Ext}_X^{1}(F,F)^*$ (induced by Serre duality).

For the time being, there is no Theorem which implies that all moduli spaces of sheaves (or rather objects) on higer dimensional projective varieties that carry a holomorphic symplectic form are necessarily moduli spaces of objects on a $\mathrm{K}$-trivial surface.

In fact, we know examples of such moduli spaces which original constructions go via higher dimensional manifolds (the Fano scheme of lines on a cubic fourfold for instance). But in every such example known, there is a (possibly non commutative) $\mathrm{K}$-trivial surface hidden in the story. For instance, the Fano scheme of lines on a cubic fourfold can be realized as the moduli space of objects in the derived category a non-commutative $\mathrm{K}$-trivial surface which sits inside the derived category of the cubic fourfold.

So it seems that having moduli spaces carying holomorphic symplectic form is a miracle related to (possibly non-commutative) two-dimensional $\mathrm{K}$-trivial varieties of dimension 2.

Note the miraculous numerical coincidence of my answer with your question : complex dimension $2$ corresponds to real dimension 4 ;)

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  • $\begingroup$ Isn't this miracle in part explained by (or at least closely related to) results in shifted symplectic geometry, where the shiftedness is given by $2-n$ where $n$ is the dimension, so for surfaces one ends up with ordinary symplectic objects? $\endgroup$ – pbelmans Apr 18 '19 at 13:19
  • $\begingroup$ @pbelmans I think it is somewhat unfair from the historical point of view (because from my understanding, the field of shifted symplectic geometry came into existence because of such examples). For this particular case, it is probably enough to understand Serre duality well. Maybe shifted symplectic geometry provides a uniform point of view on such examples but I do not think that (at this point of time) it provides such a deep perspective that we would call it an explanation of this example. Would be happy if somebody explains whether/why I am wrong. $\endgroup$ – user137767 Apr 18 '19 at 16:06

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