71
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-...
- 9,581
41
votes
Accepted
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 ...
- 9,581
23
votes
Accepted
About Donaldson-Kronheimer'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 ...
- 17.5k
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 ...
- 19.4k
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 $S^3$ ...
- 19.4k
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 ...
- 19.3k
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 ...
- 11.9k
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 ...
- 9,581
18
votes
Accepted
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 ...
- 68.8k
17
votes
Accepted
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 ...
- 10.9k
17
votes
Accepted
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 ...
- 39k
16
votes
Accepted
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\...
- 18.2k
15
votes
Accepted
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-...
- 1,111
15
votes
Accepted
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}{...
- 34.7k
15
votes
Accepted
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 ...
- 19.3k
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 ...
- 19.4k
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 ...
- 68.8k
14
votes
Accepted
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\...
- 3,751
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 ...
- 19.4k
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 $\...
- 5,215
14
votes
Accepted
Low dimensional homotopy groups of $\operatorname{Top}(4)$
It is $\mathbb{Z}/2 \oplus \mathbb{Z}/2$, see this note.
- 8,167
13
votes
Accepted
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 $\pi_1(M) \cong \mathbb{Z}^n$. ...
- 19.3k
12
votes
About Donaldson-Kronheimer'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 ...
- 724
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 $n$: ...
- 10.9k
12
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 ...
- 19.4k
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) $...
- 10.9k
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 ...
- 779
12
votes
Accepted
Homotopy equivalent smooth 4-manifolds which are not stably diffeomorphic?
$\newcommand{\Z}{\mathbb Z}\newcommand{\RP}{\mathbb{RP}}$ $\RP^4$ and Capell-Shaneson's fake $\RP^4$, which I'll
denote $Q$, are an example with fundamental group $\Z/2$. I don't know if this ...
- 6,881
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 ...
- 44.3k
11
votes
Accepted
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"). ...
- 5,565
Only top scored, non community-wiki answers of a minimum length are eligible