Why Donaldson's Four-Six Conjecture? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T00:57:27Z http://mathoverflow.net/feeds/question/112417 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/112417/why-donaldsons-four-six-conjecture Why Donaldson's Four-Six Conjecture? Chris Gerig 2012-11-14T21:22:07Z 2012-11-15T20:55:01Z <p>Simon Donaldson apparently made the following conjecture: Two closed symplectic 4-manifolds $(X_1,\omega_1)$ and $(X_2,\omega_2)$ are diffeomorphic if and only if $(X_1\times S^2,\omega_1\oplus\omega)$ is deformation-equivalent to $(X_2\times S^2,\omega_2\oplus\omega)$. Here $\omega$ is a symplectic structure on $S^2$, and a <em>deformation-equivalence</em> is a diffeomorphism $\phi:X_1\times S^2\to X_2\times S^2$ such that $\omega_1\oplus\omega$ and $\phi^*(\omega_2\oplus\omega)$ can be joined by a path of symplectic forms.</p> <p>However, where I read this did not contain any background or the original source. <strong>Where did Donaldson make this claim? And why did he make this claim? What is the motivation / are there good examples where this holds?</strong> Ivan Smith showed (through examples) that this conjecture fails when we replace $S^2$ by $\mathbb{T}^2$, so the statement itself seems pretty rigid.</p> http://mathoverflow.net/questions/112417/why-donaldsons-four-six-conjecture/112457#112457 Answer by YangMills for Why Donaldson's Four-Six Conjecture? YangMills 2012-11-15T06:19:17Z 2012-11-15T06:19:17Z <p>About your first question, of where did Donaldson make this claim, you can look at Smith's paper <a href="http://arxiv.org/abs/math/0012096" rel="nofollow">here.</a> On page 3 where he states this conjecture he says "According to ([5]; p.437) Simon Donaldson has formulated..."</p> <p>His reference [5] is the book of McDuff-Salamon, "Introduction to symplectic topology" (2nd edition). There the authors say "Indeed, inspired by this fact and his results on the existence of symplectic submanifolds, Donaldson made the following conjecture".</p> <p>Judging from these sources, it seems quite likely that Donaldson did not put this conjecture in writing, and that it was indeed publicized by the book of McDuff-Salamon.</p> <p>And about motivation and examples, these two sources provide quite a lot of information.</p> http://mathoverflow.net/questions/112417/why-donaldsons-four-six-conjecture/112460#112460 Answer by Jonny Evans for Why Donaldson's Four-Six Conjecture? Jonny Evans 2012-11-15T07:04:17Z 2012-11-15T07:04:17Z <p>I think that YangMills is probably right that Donaldson never wrote the conjecture down. But there are some interesting circles of ideas surrounding the conjecture which deserve mention which, again he probably never wrote down, but I think motivated some of his work on symplectic manifolds: namely, the idea that one could define invariants of symplectic manifolds inductively by dimension. For instance, take a 4-dimensional symplectic (Donaldson) hypersurface in a symplectic 6-manifold. There is a sense (only an asymptotic sense) in which you can do this uniquely. Is that enough to use smooth invariants of the 4-manifold to define symplectic invariants of the six-manifold? No-one has ever succeeded, due to the complicated nature of the asymptotic uniqueness.</p> <p>The question about 4/6-manifolds which Chris Gerig is asking about is probably motivated by a more concrete phenomenon: smooth (i.e. Seiberg-Witten) invariants of symplectic 4-manifolds see the same information as symplectic (i.e. Gromov-Witten) invariants; after crossing with a sphere, homeomorphic but non-diffeomorphic symplectic 4-manifolds become diffeomorphic 6-manifolds, however symplectically you can still detect their Gromov-Witten invariants by counting curves in the 6-manifold (see the early papers of Ruan). The classic example is to compare the Barlow surface (a surface of general type) and a (homeomorphic) blow-up of the projective plane. One is minimal, the other has many -1-curves and you can still see these after crossing with a sphere.</p> <p>This also explains why 4 and 6 are the relevant dimensions: smooth geometry in dimension 4 and symplectic geometry in dimension 6 are both "hard" in the Gromov sense. There are elliptic PDEs whose moduli spaces can be used to distinguish exotic pairs. By contrast there's no hard smooth invariants for 6-manifolds, so the question doesn't generalise.</p> <p>I guess the conjecture Chris mentions is the most optimistic extrapolation of this observation, designed to encourage people to think about the circle of ideas.</p>