18
$\begingroup$

Usually one regards manifolds up to dimension 4 as a part of low dimensional topology. There are plenty of various results which work only in low dimensional topology; especially in dimension 4. However still there are phenomena which occur only up from certain dimensions above 4: For example the famous result of Milnor, which states that each $PL$ manifold of dimension $n$ is in fact smooth provided that $n \leq 7$. My question is the following: Could you give an example of the (reasonable) theorem of the type "each manifold of dimension $n$ have some property $P$ provided that $n \leq K$ (and for $n>K$ there are counterexamples)," where $K$ is some large number?

$\endgroup$
1
  • 1
    $\begingroup$ The Kervaire Invariant problem does not quite fit the parameters of the question, but I feel that it is worth mentioning as a near-example. $\endgroup$ Sep 30, 2014 at 14:23

4 Answers 4

32
$\begingroup$

The smallest example of a manifold that is homotopy equivalent to a topological group, but not rationally equivalent to a Lie group has dimension 1254.

$\endgroup$
3
  • $\begingroup$ +1. That's a very good example of a large K. Somehow I don't think the OP counts 6,7,8 as "large". I wouldn't either. $\endgroup$ Sep 30, 2014 at 22:30
  • 5
    $\begingroup$ Actually it seems the authors only prove that their example has minimal rank, not minimal dimension. Still a very nice answer. $\endgroup$ Oct 1, 2014 at 10:02
  • 6
    $\begingroup$ @DanPetersen -- indeed, they also mention a dimension-1250 example with rank 74. $\endgroup$ Dec 23, 2017 at 19:15
18
$\begingroup$

An important property of the cohomology of manifolds is the formality, which has been at the heart of a lot of work in rational homotopy theory. A simply connected topological space is said to be formal if its Sullivan minimal model is quasi-isomorphic (as an algebra) to its rational cohomology ring. Neisendorfer and Miller proved in their paper Formal and coformal spaces that every simply connected compact manifold of dimension lesser or equal to 6 is formal. There are counterexamples in dimension 7 and above, see for instance the papers Examples of nonformal closed $(k −1)$-connected manifolds of dimensions $4k − 1$ and more (Dranishnikov-Rudiyak) or On nonformal simply connected manifolds (Fernandez-Munoz) for instance.

$\endgroup$
2
  • 4
    $\begingroup$ As a supplement: More general is every $(k-1)$-connected compact manifold of dimension $\leq (4k-1)$ formal for $k>1$. The result mentioned by Sinan Yalin follows for $k=2$. To get a counterexample in dimension 7, one can (as an alternative to the mentioned papers) realize (via Sullivan-Barge-Realization) the minimal Sullivan algebra $(\bigwedge<v,w,x,y,z>,d)$ where $deg(v)=deg(w)=2$, $deg(x)=deg(y)=deg(z)=3$ and $d(v)=d(w)=0$, $d(x)=v^2$, $d(y)=vw$, $deg(z)=w^2$ as a compact simply connected 7-manifold. This result will be non-formal, since $<v,v,w>$ is a nontrivial Massey product. $\endgroup$ Sep 30, 2014 at 9:34
  • $\begingroup$ Here is a way to obtain archipelago's example geometrically: Consider the map $S^2\times S^2 \to S^4$ that collapses the complement of a small disk to a point. Then pull back the Hopf fiber bundle $S^3\to S^7 \to S^4$ over this map, and obtain a fiber bundle $S^3 \to X \to S^2\times S^2$. The manifold $X$ is the simply connected non-formal 7-manifold whose minimal model is described in the above comment. $\endgroup$ Jun 9, 2019 at 20:26
7
$\begingroup$

Theorem (Simons). Let $E\subset {\mathbb R}^n$ be of minimal perimeter. If $n\le7$, then $\partial E$ is a hyperplane.

If instead $n=2m\ge8$, then Simons provides the example of a minimal surface $$x_1^2+\cdots+x_m^2=x_{m+1}^2+\cdots+x_{2m}^2$$ whose mean curvature vanishes identically. Bombieri, De Giorgi & Giusti proved that this cone is a minimal surface. This shows that the limit $n\le7$ in the Theorem is sharp.

$\endgroup$
4
  • 2
    $\begingroup$ You might want to mention that $E$ is an open subset of $\mathbb R^n$ (presumably with smooth boundary), and also explain that "of minimal perimeter" refers to the fact that any local variation of $E$ that preserves the local volume of $E$ results in an increase of the perimeter (= area of $\partial E$). (Or should I only require the the first order variation of the perimeter be zero?) $\endgroup$ Sep 30, 2014 at 22:15
  • 3
    $\begingroup$ @AndréHenriques Hey, I apologize for reacting to such an old comment - the thread popped up because of a new answer. In any case, what you wrote is incorrect. The set $E$ need not be open, it certainly wouldn't have a smooth boundary - note that the example given has a singular point at the origin -, and the definition of 'minimal perimeter' that you give is completely wrong. I'm sorry for phrasing this so bluntly, and I'm certainly not trying to be rude, but I wanted to be emphatic to prevent any misunderstanding from somebody else reading your comment. $\endgroup$
    – Leo Moos
    Mar 2, 2023 at 23:41
  • $\begingroup$ @LeoMoos Thanks for pointing out my misunderstanding. Do you want to maybe elaborate on the correct meaning of 'minimal perimeter'? $\endgroup$ Mar 3, 2023 at 10:15
  • 1
    $\begingroup$ @AndréHenriques Sure! Being of 'minimal perimeter', in other words being area-minimizing, means that $E$ has less area than any competitor $T \subset \mathbf{R}^n$ with the same boundary (meaning $T$ and $E$ lie in the same homology class), and which has $T \Delta E$ compact. It's not a local notion, and it has nothing to do with enclosed volume. What you describe sounds a bit more like locally isoperimetric surfaces. For example, spheres have this property, but they're obviously not minimal, let alone area-minimizing. $\endgroup$
    – Leo Moos
    Mar 3, 2023 at 13:28
5
$\begingroup$

It's not quite as large as 1254, but there is a conjecture about the Yamabe problem which holds small dimensions but fails once the dimension is 25 or more.

When $(M^n, g)$ is a smooth compact Riemannian manifold of dimension $n \geq 3$ without boundary, the Yamabe Problem consists of finding a metric $\widetilde{g}$ which has constant scalar curvature and is conformal to g (i.e., satisfies $\widetilde{g} = u^{4/(n-2)} g$ for some smooth function $u$). The Compactness Conjecture stated that the set of solutions to this problem are compact in the $C^2$ topology unless $(M, g)$ is conformally equivalent to the round sphere. In 2008-09, there were a series of papers on this topic where it was shown that conjecture holds for $n \leq 24$ [1], but fails for all dimensions 25 and higher [2,3].

[1] Khuri, M. A.; Marques, F. C.; Schoen, R. M., A compactness theorem for the Yamabe problem, J. Differ. Geom. 81, No. 1, 143-196 (2009). ZBL1162.53029.

[2] Brendle, Simon, Blow-up phenomena for the Yamabe equation, J. Am. Math. Soc. 21, No. 4, 951-979 (2008). ZBL1206.53041.

[3] Brendle, Simon; Marques, Fernando C., Blow-up phenomena for the Yamabe equation. II, J. Differ. Geom. 81, No. 2, 225-250 (2009). ZBL1166.53025.[3]

$\endgroup$
3
  • $\begingroup$ Is there any conceptual explanation for 'why' things change at dimension 25? It seems a bit mysterious - I skimmed the last paper by Brendle and Marques, and they just mention that it ties in with the eigenvalues of some bilinear form becoming negative. $\endgroup$
    – Leo Moos
    Mar 3, 2023 at 8:58
  • 1
    $\begingroup$ @LeoMoos The paper to read is the first one, which introduces the bilinear form and reduces the proof to showing that it is positive definite. However, I don’t have a good conceptual explanation for why things change once you get to 25. $\endgroup$
    – Gabe K
    Mar 3, 2023 at 11:11
  • $\begingroup$ Thanks, I appreciate your comment! (It all seems a bit mysterious, but then again I really know very little about the Yamabe problem.) $\endgroup$
    – Leo Moos
    Mar 3, 2023 at 18:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.