Some time ago, a professor I know said that there are results in four dimensions that are harder to prove (for example, the Smooth Poincaré Conjecture, though I'm not specifically interested in it). Qualitatively, why and how can it be that in a problem, a specific dimension stands out from the others? Is there a nice, simpler example than Poincaré Conjecture and Supersymmetric Gauge stuff to explain that some specific $n$-dimensional space would be particularly more "stiff" to a result, and different from analogous spaces with just other dimensions?
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5There is a big thread about it on MathOverflow, though I can't claim to understand any of the answers. – Jun 05 '18 at 03:50
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Thanks for the resource, I'll look into it; however, aren't there some more "down to earth", rather simple examples that could reveal earlier the difference between such spaces? – WhiteLion Jun 05 '18 at 15:06
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2The smooth Poincare conjecture is not "valid for all but 4 dimensions", but rather "reduced to homotopy theory in all but dimensions $\leq 4$", and then solved for dimensions $\leq 3$. It is unknown but plausible whether every sphere of dimension larger than, say, $128$, has at least two smooth structures. – Jun 08 '18 at 00:51
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1Most closed, simply connected, smoothable 4-manifolds seem to carry infinitely many distinct (non-diffeomorphic) smooth structures. This is not possible in any other dimension: in dimension 3 there is a unique smooth structure on a topological manifold, in dimensions larger than 4 there are at most finitely many (this uses hard stuff, called "smoothing theory", to understand how many smooth structures there can be on a given PL manifold; it is the PL structures that are "clearly finite"). I could give examples of 4-manifolds with many smooth structures... – Jun 08 '18 at 00:53
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1and 5-manifolds with few but this is probably not what you're "getting at". Maybe an example is that the topological manifold $\Bbb R^k$ admits a unique smooth structure when $k \neq 4$, and uncountably many distinct ones when $k = 4$. But there is no down-to-earth reason for this; the proof uses not just gauge theory (hard enough to understand) but gauge theory on noncompact manifolds (much harder). – Jun 08 '18 at 00:54
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Thanks for the explanations, but I am not specifically interested in the Smooth Poincaré Conjecture itself; instead would want some "simpler" example of the same phenomenon of a strange dimension that stands out from the others, and why this should happen. I'll edit the question to emphasize and make this point clearer. – WhiteLion Jun 08 '18 at 00:58
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1I understood that, but it's very unclear to me that such an example exists. (I also am not quite sure what kind of property you're looking for, but I find it hard to imagine any such property that is comprehensible in a down-to-earth way.) The ramblings about smoothable or non-smoothable manifolds were just an attempt to explain things about 4-manifolds you might or might not care about. – Jun 08 '18 at 01:01
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1The one-sentence answer is that the Whitney trick and thus the $h$-cobordism theorem (and variants) fail spectacularly in dimension 4. (They also fail below dimension $4$, but dimension $1$ and $2$ are classical, and dimension $3$ has a very different flavor entirely.) – anomaly Jun 08 '18 at 01:08
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2A very vague but truthful answer is that in high dimensions you have a lot of room to move things around, so you can put things in a standard position more easily. In low dimensions, you don't have enough room for things to get too out of control in the first place. Four dimensions is right at the boundary where you have enough room for interesting things to happen, but you don't have enough room to untangle those problems. – Cheerful Parsnip Jun 08 '18 at 01:31
1 Answers
While the statements of the things that go wrong in 4 dimensions can be easy to state, it is very rare that the proofs do not involve a lot of high-tech machinery. The field only expanded very rapidly in 1982, when two very important theorems were published:
1) Freedman proved a theorem classifying all simply-connected closed topological 4-manifolds up to homeomorphism. Every such manifold has a unimodular (the defining matrix has determinant $\pm 1$) symmetric bilinear form $H^2(X;\Bbb Z) \otimes H^2(X;\Bbb Z) \to \Bbb Z$. They are determined by this pairing (up to isomorphism), as well as an additional number $\kappa \in \{0, 1\}$. Every unimodular symmetric bilinear form may be realized, and the values of $\kappa$ that may arise for a fixed bilinear form $\beta$ are known.
2) Donaldson proved that if $X$ is a closed 4-manifold whose intersection form is definite (this equivalently means that $\beta(x,x)$ is either negative or positive for all $x$), then the intersection form is diagonalizable (this means there is a basis of vectors $x_i \in H^2(X;\Bbb Z)$ so that $\beta(x_i, x_i) = \pm 1$, where $\pm$ is the same sign for all $i$, and $\beta(x_i, x_j) = 0$ for $i \neq j$).
These are especially astounding in combination: there are a great many definite intersection forms which are not diagonalizable over the integers (the number of them is exponential in the rank of $H^2(X)$), and thus an exponentially growing class of non-smoothable 3-manifolds, by Freedman's result.
These new tools provided exciting new directions for 4-manifolds (and the 3-manifolds they bound), which flourished and led to the many fascinating differences between 4D and high-D topology - but the fact that the techniques in (1) and (2) are themselves so complicated should warn you that most results will be.
Here is an example of the kind of question whose answer became better understood after 1982. You will see the appearance of both gauge theory and surgery theory.
Consider the class of closed oriented 3-manifolds with the same homology as $S^3$; these are the integer homology spheres. We may a define a relation on this class: we say that $Y \sim Y'$ if there is a compact 4-manifold $W$ with boundary $Y \sqcup -Y'$, so that both maps $H_*(Y) \to H_*(W) \leftarrow H_*(Y')$ are isomorphisms. These are called homology cobordisms, and the resulting quotient set is written $\Theta^3_{\Bbb Z}$; it has a group structure given by the connected-sum operation, and is called the homology cobordism group.
Given a closed simply connected 4-manifold $X$, whose intersection form has $\beta(x,x) \in 2\Bbb Z$ for all $x$ (we say $\beta$ is "even"), it is a theorem of Rokhlin that the signature of $\beta$ - the number of positive eigenvalues minus the number of negative values - is a multiple of 16. We write this as $\sigma(X) \in 16\Bbb Z$.
Given a homology 3-sphere $Y$, we may find a closed simply connected 4-manifold $W$ with even intersection form and $\partial W = Y$. It is an algebra theorem that $\sigma(W) \in 8\Bbb Z$. What Rokhlin's theorem promises, however, is that $\sigma(W)$ is independent of $W$ modulo $16$: given another bounding manifold $W'$, glue them together along the boundary to get a closed manifold $X = W \cup_Y -W'$, so we see $\sigma(W) - \sigma(W') = \sigma(X) \in 16\Bbb Z$.
Thus, sending a homology 3-sphere $Y$ to $\sigma(W)/8 \pmod{2}$, where $W$ is a bounding even simply connected 4-manifold, is well-defined; in fact, it descends to a homomorphism $\mu: \Theta^3_{\Bbb Z} \to \Bbb Z/2$. This is called Rokhlin's homomorphism.
I don't have a reference, but in sufficiently old papers you might occasionally see someone ask if $\mu$ is an isomorphism.
So far nothing is special. A variant of Rokhlin's theorem remains true for "spin $(8k+4)$-manifolds" (now Ochanine's theorem), and one can define a similar "signature defect" operation on spin $(8k+3)$-manifolds. (A homology 3-sphere has exactly one spin structure, so this adjective is not threatening.) Furthermore, the notion of homology cobordism makes sense in all dimensions; you can even stick words like "spin" in there if you want.
The thing that's special in 3/4 dimensions, unfortunately for you, is gauge theory. In 1990 Furuta pointed out that an invariant of Fintushel and Stern (a real number associated to a coprime sequence of integers $(a_1, \cdots, a_n)$ called the $R$-invariant of certain homology 3-spheres $\Sigma(a_1, \cdots, a_n)$) implies that $\Theta^3_{\Bbb Z}$ has an infinite subset of elements which are linearly independent; because this group contains a $\Bbb Z^\infty$ subgroup, it must be infinitely generated! That's a far cry from being isomorphic to $\Bbb Z/2$.
In high dimensions, there is, however, a theorem of Kervaire stating that $\Theta^n_{\Bbb Z} = 0$ for $n \geq 4$. It is a classic piece of 'surgery theory', the same tool that killed the Poincare conjecture,
To link this back up to Donaldson's theorem, here is a mildly explicit example; it is the standard one.
The simplest (in terms of rank) definite intersection form that is not diagonalizable is the $E8$ form. Freedman's result constructs a closed simply connected topological 4-manifold with intersection form E8 by a very inexplicit procedure; I cannot in the slightest draw this space for you. Donaldson's theorem says that there cannot be such a smooth closed 4-manifold.
However, there is such a 4-manifold with boundary. It is kind of hard to describe to someone not immersed in topology already, but it is defined as what's called a "plumbing construction" (a certain way of gluing simple 4-manifolds together to parts of their boundaries) based on the E8 form. Its boundary is called the Poincare homology sphere and written $\Sigma(2,3,5)$, the 3-manifold defined as a manifold as $SO(3)/A_5$, where $A_5$ is the symmetry group of the icosahedron sitting inside 3-dimensional space, centered at the origin. (Its fundamental group is called "$2I$", the binary icosahedral group, and is order 120.) Because $\sigma(E_8) = 8$, this implies that $\mu(\Sigma(2,3,5)) = 1$. When you do this '$E8$ plumbing' in dimension $4k$, but where $k > 1$, then the boundary turns out to be an exotic sphere (so its fundamental group is also trivial, as opposed to just being a homology sphere): it is a different category of 'weirdness'.
Donaldson's theorem proves slightly more: the manifold $\#^n \Sigma(2,3,5)$ is never zero in $\Theta^3_{\Bbb Z}$. Proof: If it were, it would bound a space with the homology of a ball (being zero means there is a homology cobordism to $S^3$, which you fill in with a disc). But you know that on the other hand it bounds $\oplus^n E8$ (taking the connect sum of a bunch of the manifolds we discussed above). Gluing these two 4-manifolds together on their boundaries gives you a smooth closed simply-connected 4-manifold with intersection form $\oplus^n E8$, which is not diagonalizable. This contradicts Donaldson's theorem, and so $\Sigma(2,3,5)$ is non-torsion in $\Theta^3_{\Bbb Z}$.
Freedman's $E8$ manifold was the first example of a non-triangulable manifold. This was proved by Casson, using an invariant closely related to Rokhlin's.
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The punchline is that I do not believe that the statement "there should be at least one interesting/instructive case to treat that doesn't require [some of the background of] a PhD", even though it is quite unfortunate; I hope that the two big theorems hint at why this might be true. – Jun 08 '18 at 02:12
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As for the punchline, I think Rokhlin's theorem is accessible (the original paper was only about two pages and contains two other significant results, though it does require knowledge of cobordism theory). Also, although its proof is very non-trivial, the statement of the $h$-cobordism is not that difficult and leads almost immediately to Smale's proof of the Poincare conjecture in dimension $\geq 5$. – anomaly Jun 08 '18 at 02:39
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1@anomaly I worry that Rokhlin's result is not so special to 4D (as it has higher-dimensional generalizations), but rather that the special thing is the understanding of the homology cobordism group (which is to some degree a 4-dimensional object, given the relation). I agree that the h-cobordism principle is pretty good, and probably deserves its own answer. – Jun 08 '18 at 03:04
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Thanks for the insightful answer; I'm already studying it in more detail. :) – WhiteLion Jun 11 '18 at 23:19