Timeline for Analog of Newlander–Nirenberg theorem for real analytic manifolds
Current License: CC BY-SA 3.0
14 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Jun 28, 2014 at 6:43 | comment | added | Robert Bryant | @Misha: Both your remarks are good ones, though they don't directly cast doubt on the existence of some higher-order system as I was describing. For example, something like 'the vanishing of the (rough) Laplacian of the Ricci curvature' (a $4$th order transversely elliptic system on the metric) could, as far as I know, have global solutions on any compact $n$-manifold. (This would not yield a finite dimensional real-analytic pseudo-group, even if it did work.) Can't any $k$-jet of a diffeomorphism in dimension $3$ can be achieved by a birational transformation (possibly of very high degree)? | |
| Jun 28, 2014 at 6:37 | comment | added | Robert Bryant | @AndySanders: In dimensions higher than $2$, metrics of constant scalar curvature need not be real-analytic in any local coordinates. That's far too weak a condition to yield analyticity. You need something that is at least transversely elliptic. | |
| Jun 27, 2014 at 20:05 | comment | added | Misha | @RobertBryant: I have two remarks on this: 1. In dimensions $\ge 4$ the associated real-analytic pseudogroup would have to be infinite-dimensional. 2. In 3d every closed 3-manifold admits a "birational structure", an atlas one where transition maps are birational transformations of $R^3$. (Maybe this also works in higher dimensions, I do not know.) Sadly, I do not know how to capture it via any finite order tensor since the degree is unbounded. | |
| Jun 27, 2014 at 16:14 | comment | added | Andy Sanders | @RobertBryant, in this direction, aren't metrics of constant scalar curvature real analytic? Although this isn't a "single geometric structure," there's only 3 cases of positive, zero or negative. In this vein, perhaps the resolution of the Yamabe conjecture is a step in the right direction. | |
| Jun 27, 2014 at 15:40 | comment | added | Robert Bryant | @IgorKhavkine: OK. However, as far as I know, it's an open question as to whether there is any analogous geometric structure in dimension $3$ that exists on all $3$-manifolds. For example, not every $3$-manifold carries a flat projective structure (though several of Thurston's 8 geometries do), which would, of course, define an underlying analytic structure. Whether there is some higher order differential equation on metrics in dimension $3$ whose solutions are all locally analytic and, moreover some metric satisfying these equations exists on every $3$-manifold is an interesting question. | |
| Jun 27, 2014 at 14:46 | comment | added | Igor Khavkine | @RobertBryant, I'd say that your examples are perfectly acceptable. Though, it would be nice if the answer had some universality to it (works in all or almost all dimensions, is capable of reproducing all or almost all analytic atlases). | |
| Jun 27, 2014 at 13:51 | comment | added | Robert Bryant | So that I can understand the question better, let me ask whether the following examples would count: In dimension $n=1$, let the geometric structure simply be a metric $g$ on $M$ (subject to no conditions). Then the local coordinates $t$ on $M$ for which $g = \mathrm{d}t^2$ generate a real-analytic atlas on $M$. In dimension $2$, let the geometric structure simply be a metric $g$ on $M$ subject to the (third-order) differential equation that the Gauss curvature of $g$ be (locally) constant. Again, the (local) Gauss normal exponential coordinates for $g$ generate an analytic atlas on $M$. | |
| Jun 27, 2014 at 6:28 | comment | added | Igor Khavkine | Thanks, @PedroLauridsenRibeiro. That's what led me to suspect that CR structure could be involved. What I'd really like to see is the reverse implication, something like an intrinsic "CR structure" allowing for an embedding and hence inducing a real analytic structure. | |
| Jun 27, 2014 at 1:38 | comment | added | Andy Sanders | I understand Igor, but seeing as there is not a differential equation which an analytic function on $\mathbb{R}^n$ satisfies which guarantees it's analytic (I don't think), I would be somewhat surprised that you could capture the real analytic condition via the vanishing of some tensor field. Not that this is all that helpful. | |
| Jun 27, 2014 at 1:06 | comment | added | Pedro Lauridsen Ribeiro | Hi Igor, one possibility perhaps is to use the fact that any real analytic manifold can be (uniquely?) embedded into its Grauert tube, which is a complex (in fact, Stein) manifold that can be identified with a deformation retract of the zero section of the tangent bundle. | |
| Jun 26, 2014 at 22:20 | comment | added | Igor Khavkine | Sorry, that doesn't really address my question. My question about analytic structure is not whether one exists, but how to specify one. I would prefer to see it done by specifying some tensor field that satisfies a special differential equation, in close analogy with the complex case. | |
| Jun 26, 2014 at 21:32 | comment | added | Ricardo Andrade | Perhaps it is important to note that analytic structures on a smooth manifold are unique only up to analytic diffeomorphisms. In fact, there are analytic structures on the real line which refine the usual smooth structure, yet do not give the same maximal analytic atlas. By way of contrast, any two complex structures on a given almost complex manifold are necessarily the same, i.e. induce the same maximal complex-holomorphic atlas. | |
| Jun 26, 2014 at 21:01 | history | answered | Andy Sanders | CC BY-SA 3.0 |