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69 votes
Accepted

Not all manifolds can be triangulated: In which dimensions?

In dimensions up to three, every manifold is triangulable (this is classical). In dimension 4, there are simply connected non-triangulable manifolds (such as the E8 manifold); in fact, a closed 4-...
mme's user avatar
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40 votes
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Exotic $R^4$ as the universal covering space

This is problem 4.79(A) of Kirby's 1995 problem list, contributed by Gompf. It is my impression that it is still open. There is some small progress: Remark 7.2 in this article observes that their ...
mme's user avatar
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23 votes
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What can we say about the Cartesian product of a manifold with its exotic copy?

Your question seems to be about simply connected exotic 4-manifolds, for which the answer is yes. That's because $M$ and $M^E$ are h-cobordant (by Wall), say via an h-cobordism W. Then $M \times W$ is ...
Danny Ruberman's user avatar
22 votes
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About Simon Donaldson's book on four dimensional manifold

Please do not ignore the other author, Peter Kronheimer. Based on all of the material I've read, I do not agree with your belief about the book. I think it is more detailed than you will find ...
Chris Gerig's user avatar
  • 17.2k
22 votes

Very particular kind of 4-manifolds. Classification

I would say no. If M is simply connected, then it is contractible and hence determined topologically by its boundary. But there's no current smooth classification; the case when the boundary is $S^3$ ...
Danny Ruberman's user avatar
19 votes
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Is a 4-dimensional submanifold of a spin manifold always spin?

Let $i$ denote an immersion $N \to M$. There is an exact sequence of vector bundles on $N$ given by $$0 \to TN \to i^*TM \to \nu \to 0$$ where $\nu$ is the normal bundle. As total Stiefel-Whitney ...
Michael Albanese's user avatar
19 votes

Homotopy groups of Diff(X) and Homeo(X)

No, this is not true, not even for spheres. Consider the following commutative diagram: $\require{AMScd}$ \begin{CD} \text{Diff}_{\partial}(D^d) @>>> \text{Homeo}_{\partial}(D^d) \sim *\\ @V ...
Jens Reinhold's user avatar
18 votes
Accepted

Is there a closed non-smoothable 4-manifold with zero Euler characteristic?

The Kirby-Siebenmann invariant in $H^4(M;\Bbb Z/2)$, an obstruction to smoothability, is additive under connected sum in dimension 4. In even dimensions, $\chi(M \# N) = \chi(M) + \chi(N) -2$. To ...
mme's user avatar
  • 9,398
18 votes
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Very particular kind of 4-manifolds. Classification

There are plenty of such manifolds, but as Danny indicates in his answer, there is not a known classification. Take any acyclic group $G$ with a finite aspherical 2-complex $C$ with $\pi_1(C)=G$. Then ...
Ian Agol's user avatar
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17 votes
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4-dimensional cohomology $\mathbb{CP}^2$'s

No. If $\Sigma$ is any homology 4-sphere with non-trivial fundamental group, $\mathbb{CP}^2 \# \Sigma$ is a homology $\mathbb{CP}^2$ with non-trivial fundamental group. (Here $\#$ denotes connected ...
Marco Golla's user avatar
  • 10.5k
17 votes
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Does the Hodge star operator determine the metric?

As Malkoun observed, your problem is pointwise. So let $V$ be a 4-dimensional vector space, and $g$ and $\hat{g}$ be inner products. Since we are working with two inner products, by the spectral ...
Willie Wong's user avatar
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16 votes
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Homotopy groups of Diff(X) and Homeo(X)

No, the statement about the kernel and cokernel being finite is not true. For a closed $d$-manifold, $d \neq 4$, smoothing theory identifies the homotopy fibre of $$B\mathrm{Diff}(M) \longrightarrow B\...
Oscar Randal-Williams's user avatar
15 votes
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Example of closed 4 manifold with $\mathbb{S}^1$ action with 1 fixed point and free away from it

Such a closed $4$-manifold does not exist, and this follows from: Church, P., & Lamotke, K. (1974). Almost free actions on manifolds. Bulletin of the Australian Mathematical Society, 10(2), 177-...
KSackel's user avatar
  • 1,111
15 votes
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Atiyah's proof of the moduli space of SD irreducible YM connections

Hopefully I remember this well. My adviser explained this computation to me I don't even want to think how many years ago. The deformation complex of the SD equation is $\DeclareMathOperator{\Ad}{...
Liviu Nicolaescu's user avatar
15 votes
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Exotic $\mathbb{R}^4$ with a complex structure?

It is a result of Gromov that an open manifold of dimension six or less admits a complex structure if and only if it admits an almost complex structure; see the corollary on page 103 of his book ...
Michael Albanese's user avatar
15 votes
Accepted

Characteristic class that cannot be represented by disjoint tori

In $H_2(CP^2)$, every class $nH$ where $H$ is a generator and n is odd is characteristic. However, if $n >3$, then such a class is not represented by a torus. It is not represented by a disjoint ...
Danny Ruberman's user avatar
14 votes

How to specify a compact topological 4-manifold with a finite amount of data

I understand what you're getting at, but I think the statement "any compact topological 4-manifold can be specified by a finite amount of data" has a trivial answer (modulo the literature). Cheeger ...
Ian Agol's user avatar
  • 66.9k
14 votes
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Does $(S^1\times S^3)\#(S^1\times S^3)\#(S^2\times S^2)$ admit a symplectic form?

No, $M$ is not symplectic. Consider a double cover $\tilde{M}$ of $M$ along one of the $S^1$ components. Then it is not hard to prove that $\tilde{M}$ is diffeomorphic with $(S^1\times S^3)\#2(S^1\...
Anubhav Mukherjee's user avatar
14 votes

4-manifold $M$ with intersection form of Leech lattice

If you're assuming $M$ is simply-connected, then it would be spin (since the Leech lattice is even). So a smooth manifold would violate Rokhlin's theorem. In the topological case (still assuming ...
Danny Ruberman's user avatar
14 votes

Does the Hodge star operator determine the metric?

First remark is that the statement you are trying to prove reduces to the pointwise statement and thus to a linear algebra statement. Let us say we are given 2 inner products $g$ and $\tilde{g}$ on $\...
Malkoun's user avatar
  • 5,021
13 votes
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Stable torus that is not a torus

Suppose $M\times S^1$ is homeomorphic to $T^{n+1}$. Then $\pi_1(M\times S^1) \cong \pi_1(T^{n+1})$, so $\pi_1(M)\oplus\mathbb{Z} \cong \mathbb{Z}^{n+1}$, and hence $\pi_1(M) \cong \mathbb{Z}^n$. ...
Michael Albanese's user avatar
12 votes

Behavior of genus function on a 4-manifold for sums

Sometimes the function $G$ can be constantly 0: consider the class $x = [S^2\times\{p\}]$ in $H_2(S^2\times F)$, where $F$ is a surface. Then $G(nx)$ can be realised by an embedded sphere for all $n$: ...
Marco Golla's user avatar
  • 10.5k
12 votes

Is there a closed non-smoothable 4-manifold with zero Euler characteristic?

Alternatively, one can start with the $E_8$-manifold and connect sum with (five copies of) $S^1\times S^3$, and appeal to Donaldson's diagonalisation theorem instead. More precisely, the (negative) $...
Marco Golla's user avatar
  • 10.5k
12 votes
Accepted

Rational slice knot that is not slice

Yes. The figure-eight knot is an example: it bounds a smooth slice disk in a rational homology ball. This has been proven in a bunch of different ways, going back to the 1980s. Here are a couple of ...
Adam Levine's user avatar
12 votes
Accepted

Unknotted $S^{n-2}$ in $S^n$

My understanding is this remains an open problem in the smooth category. I believe there have been a few claims of proofs of this statement in the literature over the years, but as far as I know none ...
Ryan Budney's user avatar
  • 43.1k
11 votes
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Construction of invariants of 4-manifolds with the Kirby calculus

Disclaimer: Shameless self-advertising. Yes, it can be done, and it's really beautiful! You can define the Crane-Yetter invariant with Kirby calculus, and possibly other TQFTs ("dichromatic models"). ...
Manuel Bärenz's user avatar
11 votes

About Simon Donaldson's book on four dimensional manifold

It's been a while but I remember that I found John Morgan's "An introduction to gauge theory" quite helpful when I was first trying to read about 4-manifolds and gauge theory. While it doesn't go ...
Stefan Behrens's user avatar
11 votes
Accepted

Behavior of genus function on a 4-manifold for sums

In the case that $x\cdot x \neq 0$, topological methods based on the G-signature show that the genus goes to infinity more or less quadratically in $n$. (I'll be more specific below.) This goes back ...
Danny Ruberman's user avatar
11 votes

Is there a closed non-smoothable 4-manifold with zero Euler characteristic?

You can also get this from Donaldson's theorem by a similar device. Take a non-diagonalizable definite form with even rank $2n$, and realize it (Freedman again) by a simply connected manifold. Then $W\...
Danny Ruberman's user avatar
11 votes
Accepted

Which 3-manifolds are known to admit exotic pairs of bounding 4-manifolds?

Here is a list of 3-manifolds $Y$ that are boundaries of exotic 4-manifolds https://arxiv.org/pdf/1901.07964.pdf If either $Y$ or $-Y$ (i.e with reverse orientation) has a contact structure with non-...
Anubhav Mukherjee's user avatar

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