25
votes
Accepted
Does anyone know a basepoint-free construction of universal covers?
I think that homotopy-theorists often fall into the habit of working mainly with based spaces, even when they don't need to. It can be instructive to notice when the use of a basepoint is unnecessary, ...
- 54.4k
24
votes
What happens to a closed manifold to ensure it is homeomorphic to a torus $T^{n}$?
For dimensions $n \geq 5$, the answer is yes. First, note that $M$ is homotopy equivalent to a torus since it must be a $K(\mathbb{Z}^n,1)$. Second, Hsiang-Wall show in "On Homotopy Tori II" ...
- 5,201
22
votes
Accepted
If the universal cover of a manifold is spin, must it admit a finite cover which is spin?
No, this is not true: for each dimension $d \geq 4$, there is a closed, oriented $d$-manifold which is not spin, whose universal cover is spin, but which does not have a finite cover that is spin.
The ...
- 20.6k
19
votes
What happens to a closed manifold to ensure it is homeomorphic to a torus $T^{n}$?
This answer is intended to give references of the cases for the case $n \leq 4$. In dimensions $n \leq 2$ this is covered in a first topology course so there are two interesting cases.
In dimension 3,...
- 9,388
18
votes
Accepted
Is there a closed manifold whose universal cover is $\mathbb{R}^n\setminus\{x_1, \dots, x_k\}$ for some $k > 1$?
If we demand that the universal cover is homeomorphic / diffeomorphic to $\mathbb{R}^n \setminus \{x_1,\ldots,x_k\}$ with $k>1$ the answer is no, there are no such closed manifolds. Each missing ...
- 1,356
17
votes
Self-covering spaces
Here are some pretty examples of self covering manifolds: Suppose that $F$ is a manifold and $f \colon F \to F$ is a periodic homeomorphism of period $k$. We define $M_f = F \times I \,/\, f$ to be ...
- 26.4k
15
votes
Accepted
what is this simple topological space?
These spaces $M_{p,q}=M_p\cup M_q$ are discussed in Examples 1.24 and 1.35 of my algebraic topology book. I don't know that they have a standard name, apart from $M_{2,2}$ which is the Klein bottle. ...
- 19.4k
14
votes
For which spaces $S^n$ ($n\geq 2$) is a universal covering space?
The answer is quite complicated. To begin with, the universal cover of your space $X$ is a sphere $S^n$ with a free action of a finite group $G=\pi_1(X)$. The group $G$ has to have periodic cohomology....
- 11.8k
13
votes
Self-covering spaces
A discussion (and partial classification with stronger assumptions on the cover) is given in a paper of van Limbeek. As Neil Hoffman points
out, a necessary condition is that the fundamental group is ...
- 66.8k
12
votes
For which spaces $S^n$ ($n\geq 2$) is a universal covering space?
A nice and quick survey on the groups acting freely on the sphere is given in chapter 3 of https://webusers.imj-prg.fr/~bernhard.keller/ictp2006/lecturenotes/skowronski.pdf .
Theorem 3.26 gives a nice ...
- 26.1k
10
votes
Spaces that are finitely covered by manifolds
Indeed, C. Thomas, 3-Manifolds and PD(3)-groups, in "Novikov conjectures", Vol. 2, gives a reference to Swan's example. Namely, there is a finite complex $X$ which is a homotopy 3-sphere, a free $S_3$...
- 31k
10
votes
Self-covering spaces
Adding further counter-examples to the discussion, compact hyperbolic manifolds in all dimensions are ruled out by the fact that Gromov's "simplicial volume" is nonzero (being proportional to ...
- 15.3k
10
votes
Self-covering spaces
EDIT: The following edits (in bold) are in response to Francesco
Polizzi astute comment.
To address question 1, there are some obvious necessary conditions For example, the Euler characteristic of ...
- 5,231
10
votes
Accepted
The Classification of all spaces for which $X$ is a covering space
In general, I would expect this to be a quite intractable problem. For instance, let's assume we are only interested in the category of manifolds, and we ask the question which $3$-manifolds are ...
- 2,890
10
votes
Does anyone know a basepoint-free construction of universal covers?
Part of 10.5.8 of Topology and Groupoids is, in a more usual notation, essentially the following, in which $\sigma, \tau$ are the source and target maps, $St_G x$ is $\sigma ^{-1} x$, by $N$ is ...
- 12.2k
9
votes
Coverings of a space and coverings of a groupoid
As you will see this is a very rich area, so my answer can do no more than give a sketch. The basic idea is that a good category of coverings, Cov(X), say, has certain good categorical properties and ...
- 9,197
9
votes
Does anyone know a basepoint-free construction of universal covers?
[UPDATE: As Tom Goodwillie points out, this is much more complicated than necessary and misunderstands the line of argument that he had in mind. Still, it has some interesting features so I will ...
- 55k
9
votes
Explicit description of the group of deck transformations acting on the universal cover of a Riemann Surface
Looking at the link you give, it seems that you want an explicit representation of $\pi_1 = \pi_1(S)$, the fundamental group of the genus two surface, into $\mathrm{PSL}(2, \mathbb{C})$ so that the ...
- 26.4k
9
votes
Minimum number of generators for quotients of congruence subgroups of SL(2, Z)
$\DeclareMathOperator\SL{SL}$Not always. For simplicity, take $N$ a large prime $p$. Following user44191's suggestion in the comments, take $r$ to be $T(N)-1$.
Since any congruence subgroup contains a ...
- 137k
9
votes
Covering of a knot complement
For this answer I will consider knots to be links (with one component).
In general the answer is "no". For example, consider $K = 4_1$, the figure-eight knot. Let $X = S^3 - K$. Neither ...
- 26.4k
9
votes
Accepted
What relationship is there between repeated roots of discriminants and orders of roots of the original polynomials?
$\def\P{{\mathbb{P}}}
\def\A{{\mathbb{A}}}
\newcommand{\O}{\mathcal{O}}
\DeclareMathOperator{\Disc}{Disc}$I think the story goes like this. The multiplicity of a zero of the discriminant counts ...
- 5,528
9
votes
What relationship is there between repeated roots of discriminants and orders of roots of the original polynomials?
(1) There is the following indirect explanation:
For a generic curve neither of these phenomena happen - the discriminant has no repeated roots and the branch points all have ramification index two. ...
- 137k
8
votes
Accepted
Galois categories for topological spaces?
The answer is yes (with mild hypothesis on the space). Moreover the topological situation is simpler, and this was very likely Grothendieck's inspiration.
To see this you need two facts.
First ...
- 3,948
8
votes
Nonpathological nonnormal covering space
They arise naturally all the time. The Galois correspondence tells us that a covering space is normal if and only if the corresponding subgroup of the fundamental group is normal. So whenever you ...
- 24.1k
8
votes
Accepted
induced group actions and covering maps on Eilenberg-Maclane space
If $M$ is connected, then @MarkGrant's fibration sequence gives a long exact sequence on homotopy groups showing that $\pi_1(M/\Sigma_k)\to \Sigma_k$ is surjective. Now apply the $K(-,1)$-functor and ...
- 666
8
votes
Accepted
Covering with Deck group $\mathfrak{S}_3$
Here is a picture from Topology and Groupoids
It is meant to show in (i) the Cayley graph of the presentation $\mathcal P$ of $G=S3$, $\{x,y:x^3,y^2,xyxy\}$. The Cayley graph is the $1$-skeleton ...
- 12.2k
8
votes
Accepted
Monodromy groups from Galois's viewpoint
All of the following can be found in the third chapter of Szamuley's "Galois Groups and Fundamental Groups" but I will try to sum it up a bit:
I think the starting point of making this precise is the ...
- 96
8
votes
Accepted
Construction of the universal covering space via compact-open topology
Here is the key step you need to finish the proof: We are supposing $X$ is locally path-connected and semilocally simply connected, $\pi:P(X,x_0)\to \widetilde{X}$ is the quotient map identifying path-...
- 6,661
7
votes
Homology of the universal cover
If we replace the field $k$ with the ring of integers $\Bbb Z$, then no.
There are non-trivial high dimensional knots $K: S^n \to S^{n+2}$, whose complements $X = S^{n+2}-K(S^n)$ have $\pi_1(X) \...
- 18.6k
7
votes
The homology of the universal covering space, why so difficult to compute
Perhaps an example could illuminate.
Let $k \in H^4(K(\mathbb{Q},2);\mathbb{Q}) \cong \mathbb{Q}$. Let also $a,b \in \mathbb{Q}^\times$ satisfy $k (a^2 - b) = 0$, and let $G := \mathbb{Z}$ act on ...
- 71
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