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Topology of cell complexes and manifolds, classification of manifolds (e.g. smoothing, surgery), low dimensional topology (e.g. knot theory, invariants of 4-manifolds), embedding theory, combinatorial and PL topology, geometric group theory, infinite dimensional topology (e.g. Hilbert cube manifolds, theory of retracts).
3
votes
Accepted
Fixed points of diffeomorphisms of tori isotopic to identity and their traces under isotopies
I believe this is not always possible: let $d\colon \mathbb R \to [0,1/2]$ send a real number to the distance to the nearest integer. Consider the map
$$F\colon \mathbb R^2, (x,y) \mapsto (x+2d(y),y), …
8
votes
Accepted
Why is the mapping class group of a surface with nonempty boundary torsion-free?
I think the reason that Dehn twists enter is that we can take the differential of a (orientation-preserving) diffeomorphism $f$ of $S_g$ that fixes a chosen basepoint $\ast \in S_g$ at this point, an …
18
votes
1
answer
1k
views
Simply connected finite CW-complex with only finitely many nontrivial homotopy and homology ...
Let $X$ be a simply connected finite CW-complex such that all but finitely many of its homotopy groups and its homology groups (with $\mathbb Z$ coefficients) are 0.
Is $X$ then necessarily contracti …
7
votes
2
answers
781
views
Automorphism of genus 2 surface with 5 fixed points
Is there a self-homeomorphism of a genus 2 (closed, orientable) surface, which has finite order and exactly 5 fixed points?
Of course, the same question can be asked replacing 2 by $g$ and $5$ by any …
7
votes
3
answers
326
views
Cyclic groups acting on balls, and interior fixed points
Let a finite cyclic group $G = \mathbb Z/n$ act continuously on an open $d$-ball $B^d$. Suppose further that this action extends to the closed ball $\overline{B^d}$. Is there necessarily a fixed point …
6
votes
3
answers
657
views
Smooth circle action, $\chi(M^{S^1}) = \chi(M)$
I need a reference for the following result (which I can prove myself, but my proof is rather ugly and I would prefer to just cite the statement instead or re-proving it):
Let $M$ be a closed smooth …
9
votes
1
answer
226
views
Is Homeo($M,D^n$) torsion-free?
Let $M$ is a connected smooth $n$-manifold, and suppose that $f$ is a self-homeomorphism of $M$ that has finite order, i.e. $f^k = \text{id}$ for some $k\geq 1$. Suppose moreover that $f$ fixes a non- …
71
votes
Independent evidence for the classification of topological 4-manifolds?
The answer to this question might have changed since it was first asked nine years ago: a book is now available whose goal it is to give a detailed elaboration on Freedman's work:
The Disc Embedding T …
13
votes
4
answers
616
views
Is $\mathrm{Diff}_0(S_g)$ torsion-free?
Let $S_g$ be a closed oriented smooth surface of genus $g>1$, and let us consider $\text{Diff}_0(S_g)$, the identity component of the diffeomorphism group of orientation preserving diffeomorphisms of …
25
votes
1
answer
2k
views
Is every degree 1 self-map a homotopy equivalence?
In a rather obscure article, I found (without proof) the following statement:
If $M$ is a closed orientable manifold, every degree $1$ map $f: M \rightarrow M$ is a homotopy equivalence.
Is this rea …
9
votes
1
answer
4k
views
Universal covering of compact surfaces
Is there any elementary (i.e. without using analytical methods like the theory of Riemann surfaces or more elaborate results from differential geometry) way to show that the universal covering of the …
19
votes
1
answer
786
views
Which cohomology classes are detected by tori?
Given a space $X$, I am looking for a characterization of classes $\alpha \in H^n(X;\bf Q)$ such that there is a map $f\colon T^n \to X$ so that $f^{\ast} \alpha$ pairs non-trivially against the funda …
6
votes
2
answers
318
views
Almost free actions on simply-connected spaces
Let $G$ be a simply-connected compact topological group (you can think of $SU(n)$ if you like it more concrete), and let $X$ be a finite-dimensional simply-connected $G$-CW-complex. If we know that al …
2
votes
Accepted
Almost free actions on simply-connected spaces
Since it was requested, here is the answer (from the comment) again.
The statement is wrong. For each $k$, one can start with a unique $0$-cell $G/(\mathbb Z/k)$ and attach a $2$-cell $G \times D^2$ …
2
votes
Accepted
Connected manifold without connected regular level set admits exactly two connected components
Since $M$ is connected and a manifold, it is path-connected. Thus, any two points $x,y \in M$ such that $f(x), f(y) > 0$ can be joined by a path $\gamma$. Suppose $\gamma$ does not lie entirely in $f^ …