Questions tagged [4-manifolds]

A smooth 4-manifold is a 4-manifold with a smooth structure. In dimension four, in marked contrast with lower dimensions, topological and smooth manifolds are quite different.

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Is the Lipschitz structure on $\mathbb{S}^4$ unique?

Sullivan [S] proved that on any $n$-dimensional, $n\neq 4$, topological manifold there is a unique Lipschitz structure. Although it is generally accepted result, it seems that the proof lacks some ...
Piotr Hajlasz's user avatar
12 votes
0 answers
248 views

Excluding exotic PL structures on S^4

Suppose you have a finite group $G<SO(5)$ such that $S^4/G$ is homeomorphic to $S^4$ and such that $S^4/G$ is a PL manifold with respect to a PL structure induced by a standard structure on $S^4$. ...
Christian Lange's user avatar
11 votes
3 answers
3k views

About Simon Donaldson's book on four dimensional manifold

Recently I'm reading Donaldson's Geometry of four manifolds. It seems to me that the book requires a lot for background. Additionally, the proof in the book is too sketchy without too much detail. I ...
Jack's user avatar
  • 397
11 votes
1 answer
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Is there a smooth $4$-manifold homeomorphic but not diffemorphic to $CP^2$? [closed]

Is there a smooth $4$-manifold homeomorphic but not diffemorphic to $CP^2$? Are there known non-smooth examples homeomorphic $CP^2$?
Yang's user avatar
  • 129
11 votes
2 answers
817 views

Existing proofs of Rokhlin's theorem for PL manifolds

I'm looking for a comprehensive reference to existing proofs of Rokhlin's theorem that a 4-dimensional closed spin PL manifold has signature divisible by 16. I'm specifically interested in direct ...
Vitali Kapovitch's user avatar
11 votes
1 answer
802 views

Is a 4-dimensional submanifold of a spin manifold always spin?

Let $M^d$ be a $d$-dimensional orientable spin manifold, and $N^4$ is a closed $4$-dimensional orientable submanifold of $M^d$. Is $N^4$ always spin? If $d=5$, is $N^4$ always spin? If $N^4$ is a ...
Xiao-Gang Wen's user avatar
11 votes
3 answers
1k views

Alexander polynomial or Reidemeister torsion for knotted surfaces?

An important invariant of a knot in $S^3$ is its Alexander polynomial, related also to Reidemeister torsion. Is there something like that for knotted surfaces in $S^4$? If not, what are the ...
John Vrem's user avatar
  • 374
11 votes
1 answer
877 views

Smooth four-manifolds with contractible universal cover

Let $X$ be a smooth compact four-manifold with definite non-trivial intersection form. Can the universal cover of $X$ be contractible? It semms to me that the answer is negative when $X$ is simply ...
Matei's user avatar
  • 123
11 votes
1 answer
1k views

Gromov-Witten invariants counting curves passing through two points

Let us say that a closed symplectic manifold $X$ is $GW_g$-connected if there is a nonvanishing Gromov-Witten invariant of the form $GW_{g,n}^{X,A}(\beta,point, point,\alpha_3,\ldots,\alpha_n)$ --in ...
Mike Usher's user avatar
  • 2,907
11 votes
0 answers
349 views

Fox re-imbedding theorem in dimension four

Fox re-imbedding theorem states the following: A compact 3-manifold $M$ with boundary that embeds in the three-sphere $S^3$, can be re-imbedded in $S^3$ so that its complement is a union of ...
Bruno Martelli's user avatar
10 votes
3 answers
1k views

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

I will just repeat the title: Is there a closed non-smoothable 4-manifold with zero Euler characteristic? I am guessing yes simply based on other existence theorems I have seen for 4-manifolds.
Cihan's user avatar
  • 1,596
10 votes
2 answers
1k views

Does the Hodge star operator determine the metric?

Let $M$ be a closed oriented smooth $4$-manifold with two metrics $g$ and $\tilde g$. Consider the Hodge star operators on $2$-forms $$ \star:\Omega^2(M) \to \Omega^2(M) \quad \text{and} \quad \tilde \...
Zhiqiang's user avatar
  • 881
10 votes
2 answers
677 views

4-dimensional cohomology $\mathbb{CP}^2$'s

Let $M$ be a closed, smooth $4$-manifold with integral cohomology ring isomorphic to that of $\mathbb{CP}^2$, is it diffeomorphic to it?
Nick L's user avatar
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10 votes
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When is the connected sum of manifolds orientation-independent?

Given $M$ and $N$, two connected orientable manifolds of the same dimension, when is $M$ # $N$ diffeomorphic to $M$ # $\overline{N}$, where $\overline{N}$ is $N$ with the orientation reversed? If $N$ ...
zygund's user avatar
  • 921
10 votes
3 answers
614 views

Doubles of 2-handlebodies

Let $X$ denote a $4$-manifold with boundary obtained by adding $k_1$ $1$-handles to $B^4$ and $k_2$ many $2$-handles to the resulting manifold i.e. $X$ is an arbitrary $4$-dimensional $2$-handlebody. ...
ThorbenK's user avatar
  • 1,175
10 votes
2 answers
630 views

Homotopy equivalent smooth 4-manifolds which are not stably diffeomorphic?

Recall that two 4-manifolds $M$ and $N$ are stably diffeomorphic if there exist $m,n$ such that $$M \#_n (S^2 \times S^2) \cong N \#_n (S^2 \times S^2).$$ That is, they become diffeomorphic after ...
Chris Schommer-Pries's user avatar
10 votes
1 answer
744 views

What are Kirby diagrams of candidate exotic 4-manifolds?

It is an open problem whether there exist smooth manifolds homeomorphic, but not diffeomorphic to the standard $S^4$. The same is true for the 4-torus and several other manifolds. Handle ...
Manuel Bärenz's user avatar
10 votes
2 answers
576 views

Embedding the product of three circles in the 4-sphere.

Let $M$ be a smooth submanifold of the 4-sphere $S^4$. I'm going to demand that $M$ be diffeomorphic to $S^1 \times S^1 \times S^1$. By Jordan-Brouwer separation, $M$ separates the 4-sphere into ...
Ryan Budney's user avatar
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10 votes
1 answer
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Can an exotic diffeomorphism of the 4-ball change the isotopy class of an embedded surface?

Let $W$ be $B^4$ or $S^3 \times I$. Let $Y$ be a properly embedded surface in $W$. Let $f : W \to W$ be a diffeomorphism which is the identity near $\partial W$. Very little is known about $\pi_0(\...
Kevin Walker's user avatar
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10 votes
1 answer
761 views

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

Let $M$ be a compact connected three manifold. By an exotic pair of bounding 4-manifolds, I mean two smooth 4-manifolds $X_1,X_2$ such that $X_1$ and $X_2$ are homeomorphic but not diffeomorphic, and ...
user101010's user avatar
  • 5,319
10 votes
1 answer
421 views

Is a gluing of homeomorphic Mazur manifolds diffeomorphic to $S^4$?

A recent paper proves the existence of homeomorphic but not diffeomorphic Mazur manifolds (see also examples of exotic pairs of contractible Stein manifolds). Let's call them $M_1$ and $M_2$. If we ...
Ian Agol's user avatar
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10 votes
1 answer
583 views

Index of Dirac operator and Chern character of symmetric product twisting bundle

I am having trouble understanding a couple of lines of computation from Theorem 13.30 in Besse's Einstein Manifolds text We are twisting the spinor bundle (on Einstein 4-manifold) $\Sigma$ with an ...
Guest123412341234's user avatar
10 votes
1 answer
518 views

Relation between the Casson-Gordon invariants $\sigma(M, \chi)$ and $\sigma_r(M, \chi)$

Setting: There are two objects in knot theory that are commonly referred to as the Casson-Gordon invariants: the invariant $\sigma$, and the invariant $\tau$ (see for example A. Conway’s notes ...
user331406's user avatar
10 votes
2 answers
880 views

slice=ribbon generalization to higher genus + potential counterexamples to slice=ribbon.

I have two questions about the slice=ribbon conjecture. (1) If a knot $K \hookrightarrow S^3$ has smooth slice genus $g$, you can ask if it bounds a smooth genus $g$ surface in $S^3 \times [0, -\...
Andrew Lobb's user avatar
10 votes
1 answer
473 views

Visualising locally flat embeddings of surfaces in R^4

As far as I understand it follows from the work of M. Freedman that there exist locally flat embeddings of two dimensional surfaces in $\mathbb R^4$ that can not be smoothed in the class of locally ...
aglearner's user avatar
  • 14k
10 votes
0 answers
319 views

Isotopy on embedded 3-manifolds in 4-manifolds

Suppose that $X$ be an oriented, closed four-manifold. Suppose that $Y$ is an oriented, closed three-manifold smoothly embedded in $X$. Suppose also that $f:X \to X$ is a diffeomorphism that fixes $Y$...
Anubhav Mukherjee's user avatar
10 votes
0 answers
203 views

Find a 4-manifold bounding a 3-torus with any abelian representation in SL_2

Fix $x,y,z\in \mathbb{C}^*$ and let $M=S^1\times S^1\times S^1$ with $\rho:\pi_1(M)\to \operatorname{SL}_2(\mathbb{C})$ mapping the three generators to diagonal matrices with entries $(x,x^{-1})$, $(y,...
Julien Marché's user avatar
9 votes
2 answers
520 views

Obtain 4-manifolds by repeating surgeries of submanifolds in $S^4$

In his paper QFT and Jones Polynomials, Witten states: "It is a not too deep result that every 3-manifold can be obtained from or reduced to $S^3$ (or any other desired 3-manifold) by repeated ...
miss-tery's user avatar
  • 745
9 votes
2 answers
520 views

Rational slice knot that is not slice

Does there exists a knot $K\subset \mathbb{S}^3$ such that $K$ is not slice $\exists W^4$, $\partial W = \mathbb{S}^3$ rational homology ball $\exists $ properly embedded smooth disk $(D,\partial D)\...
Overflowian's user avatar
  • 2,523
9 votes
2 answers
451 views

A knot in the solid torus and a Mazur manifold

Part 1: The following picture is from Saveliev's book Lectures on Topology of 3-manifolds, page 130: He indicates that the knot drawn in the solid torus $S^1 \times D^2$ is homologous to $S^1 \times \...
Terry Black's user avatar
9 votes
2 answers
309 views

Gordon's approach: slice knots and contractible $4$-manifolds

Let $K \subset S^3$ be a slice knot. Then it bounds a smooth embedded disk $D \subset B^4$. Let $S^3_{p/q}(K)$ denote a $3$-manifold obtained by $p/q$-surgery on $K \subset S^3$. The following theorem ...
Max Schumann's user avatar
9 votes
1 answer
330 views

Essential Klein bottle in simply connected symplectic 4 manifolds

Consider the following question: Let $X$ be a simply connected, symplectic 4-manifold. Does there exists a smoothly embedded Klein bottle $K\subset X$ such that the following conditions are both ...
user44651's user avatar
  • 1,049
9 votes
1 answer
615 views

Complexity of surfaces bounding knots in 4-ball and 3-sphere respectively

I'm interested in a complexity question related to problems like the slice-ribbon problem. To be specific, if $K \subset S^3$ is a knot, it might be non-trivial yet still bound a smoothly-embedded ...
Ryan Budney's user avatar
  • 43.1k
9 votes
1 answer
476 views

Are 4-dimensional mapping tori always spin?

We know that all compact orientable manifolds of dimension 3 are spin. In 4 dimensions, $CP^2$ is not spin. I would like to ask if all 4-dimensional compact orientable mapping tori are spin? See ...
Xiao-Gang Wen's user avatar
9 votes
1 answer
393 views

Non-isotopic homology spheres in $S^4$ with equal complements?

Are there two diffeomorphic smoothly embedded homology 3-spheres $M_1^3, M_2^3 \subset S^4$ that have diffeomorphic complements but such that $M_1$ and $M_2$ are not isotopic? I would be interested in ...
user101010's user avatar
  • 5,319
9 votes
1 answer
289 views

The works of González-Acuña and Duchon from 70s and 80s

I would like to access the following two works of González-Acuña from 1970: González-Acuña, F. Dehn’s construction on knots. Bol. Soc. Mat. Mexicana (2) 15 (1970), 58–79. and González-Acuña, F. On ...
Oğuz Şavk's user avatar
  • 1,292
9 votes
1 answer
306 views

Genus 2 3-manifolds bounding only $X^4$ with $b_2(X^4)$ big?

The genus of a closed orientable 3-manifold $M^3$ is the minimum genus among all Heegaard splitting surfaces for $M$. Every such 3-manifold bounds a compact 4-manifold. Let $I(M)$ denote the ...
user101010's user avatar
  • 5,319
9 votes
0 answers
167 views

Configuration space of 4 points as an orbifold

Setup: Consider the braid group $B_n$. One way to define this is as the fundamental group of the unordered configuration space $UC_n(\mathbb{C}) = \{\{z_1,\dotsc,z_n\}\subset \mathbb{C} \mid z_i \not= ...
Ethan Dlugie's user avatar
  • 1,267
8 votes
2 answers
547 views

Presentations of exotic 4-manifolds

TLDR I want to see more examples of exotic $4$-manifold (hopefully connected, simply connected, oriented, and closed). Are there known presentations of $4$-manifolds $M$ with exotic structures, ...
Student's user avatar
  • 5,038
8 votes
1 answer
2k views

Relation of SW and Donaldson Invariant

My question is: I am request for the reference that Is there any relationship between the Seiberg-Witten Invariant and Donaldson's Invariant? Or the relationship between Seiberg-Witten Moduli Space ...
Siqi He's user avatar
  • 693
8 votes
2 answers
1k views

A problem/conjecture related to 4-manifolds that deserves a name. What name does it deserve?

There's an old problem in 4-manifold theory that, as far as I know, doesn't have a name associated with it and really deserves a name. Let $M$ be a smooth 4-manifold with boundary. Let $S$ be a ...
Ryan Budney's user avatar
  • 43.1k
8 votes
1 answer
210 views

If $M$ is contractible manifold and $X\subset \partial M$, does the cone over $X$ embed in $M$?

Let $M$ be a compact contractible manifold, $X\subset\partial M$ and $C_X$ the cone over $X$. Question: Is it true that $C_X$ embeds in $M$ with its boundary $\partial C_X$ mapped to $X\subset \...
M. Winter's user avatar
  • 12.6k
8 votes
1 answer
957 views

A remark by Gromov on 4-manifolds

Gromov remarks in a a survey on manifolds (p.12) that "it is hard to imagine that there are infinitely many non-diffeomorphic, but mutually homeomorphic, quotients of the hyperbolic 4-space by ...
Thomas Riepe's user avatar
  • 10.7k
8 votes
2 answers
336 views

Open book decompositions in dimension 4

The question about the existence of open book decompositions for a closed oriented $n$-dimensional manifold seems to be answered in all dimensions except dimension $4$, where as far as I can tell the ...
ThorbenK's user avatar
  • 1,175
8 votes
1 answer
448 views

Topological mapping class groups of 4-manifolds

It is a classical result of Quinn that for a simply-connected closed $4$-manifold $X$ the isometries of its intersection form are in one-to-one correspondence with $\pi_0 \text{Homeo}(X)$. (Isotopy ...
Enumerator's user avatar
8 votes
1 answer
558 views

Fundamental groups of non-orientable closed four-manifolds

The fundamental group of a closed orientable manifold is finitely presented, and every finitely presented group arises as the fundamental group of a closed orientable four-manifold; see this question. ...
Michael Albanese's user avatar
8 votes
1 answer
650 views

Surgery along an embedded surface in a 4-manifold

Let $X$ be a 4-manifold and $\Sigma_g\subset X$ be an embedded closed orientable genus = $g$ surface. Suppose $\Sigma_g\subset X$ has a trivial closed normal bundle $N(\Sigma_g) = \Sigma_g\times D^2$. ...
Shinichiro Nakamura's user avatar
8 votes
1 answer
721 views

Almost parallelizable 4-manifolds

On the first page of Milnor-Kervaire's paper "Bernoulli numbers, homotopy groups, and a theorem of Rohlin", they assert without proof or reference that if $M$ is a compact connected oriented ...
Julia's user avatar
  • 155
8 votes
1 answer
598 views

Are all 4-manifolds $Pin^{\tilde{c}}$?

It's known that all oriented 4-manifolds admit a $Spin^c$ structure, ie. a spin structure on $TX\oplus\mathcal{L}$ for some complex line bundle $\mathcal{L}$. A usual generalization of this ...
user404153's user avatar
8 votes
1 answer
619 views

Are there symplectic 4-folds with $b_+>1$, $b_-=0$?

This is the question. Is it known that a symplectic $4$-fold with $b_2>1$ should have a homology class $C$ with $C^2<0$?
Dmitri Panov's user avatar
  • 28.8k