## $\omega$-nilpotent cover of a recurrent surface

Theorem. Any $$\omega$$-nilpotent cover of a recurrent Riemannian manifold is Liouville.

$$\omega$$-nilpotent ($$\Gamma=\bigcup_{i=1}^{\infty}Z_{i}$$, $$Z_{i}$$ normal in $$\Gamma$$, where $$Z_{n+1}$$ maps to the center of $$\Gamma/Z_{n}$$.)

The above theorem comes from a paper of T. Lyons and D. Sullivan. function theory random paths and covering spacesre (Theorem 2).

How to understand the $$\omega$$-nilpotent cover?

The second derived subgroup of the fundamental group of an open Riemann surface $$X$$ determines a covering space, is it $$\omega$$-nilpotent? (i.e. let $$G=\pi_{1}(X)$$, $$G’=[G,G]$$, $$G”=[G’,G’]$$, is the group $$G/G”$$ $$\omega$$-nilpotent?)

## Concerning the Spanier group relative to an open cover

Let $$\mathcal{U} = \{ U_i \; |\; i\in I \}$$ be an open covering of $$X$$‎. Spanier defined $$\pi (\mathcal{U}‎, ‎x)$$ to be the subgroup of $$\pi_1 (X‎, ‎x)$$ which contains all homotopy classes having representatives of the following type‎: $$‎\prod_{j=1}^{n}u_j *v_j * u^{-1}_{j}‎, ‎$$ ‎where $$u_j$$‘s are paths (starting at the base point $$x$$) and each $$v_j$$ is a loop inside one of the neighbourhoods $$U_i \in \mathcal{U}$$‎.

‎If an open cover $$‎\mathcal{U}$$ is a refinement of an open cover $$‎\mathcal{V}$$‎, then $$\pi (‎\mathcal{U}‎, ‎x) \subset \pi (‎\mathcal{V}‎, ‎x)$$‎.

My question is that:

If $$[f][g]\in \pi (‎\mathcal{U}‎, ‎x)$$ for $$[f],[g]\in \pi_1 (X,x)$$, then is there any refinement $$\mathcal{V}$$ of $$\mathcal{U}$$ so that $$[f]\in \pi (\mathcal{V},x)$$?

## Reduction to a vertex cover problem with weighted vertices and edges

### Description

Let us define a new problem with an instance $$I = (G = (V, E), K, L)$$, whereas:

• $$G$$ is an undirected graph
• $$K \le |V|$$
• $$L > 0$$ is the maximum limit
• Each vertex $$v \in V$$ has a weight $$W(v)$$
• Each edge $$e \in E$$ has a weight $$W(e)$$

Let $$P(v)$$ be a function that returns the minimum total weight for vertex $$v$$ to reach a vertex in $$V’$$.

The decision question is whether there exists a vertex set $$V’ \subseteq V, |V’| \le K$$, such that:

$$\sum_{v \in V} W(v) \cdot P(v) \le L$$

### Example

Consider the following graph $$G$$, with $$K = 1$$ and $$L = 9$$:

Taking the set $$\{v_3\}$$ as $$V’$$ would be the solution to the question, because the total cost is:

$$3 \cdot 0 + 2 \cdot 2 + 1 \cdot 5 = 9$$

Therefore, this is a yes-instance.

### Question

How do I prove that this problem is in $$\mathsf{NPC}$$? I tried reducing a $$\text{VC}$$-instance to this, but that does not seem to work.

What I have tried as well is by converting the above undirected graph to a directed graph with the weighted paths already computed (this is polynomial computable):

$$\begin{array}{l|l|l} & v_1 & v_2 & v_3 \ \hline v_1 & 0 & 3 & 5 \ \hline v_2 & 6 & 0 & 4 \ \hline v_3 & 15 & 6 & 0 \end{array}$$

However, I’m not entirely sure what $$\mathsf{NPC}$$ problem to reduce from. I got the tip to use the $$\text{CLIQUE}$$ problem from this answer, but I do not see how to perform the reduction, so that seems unlikely. To me, this looks more like something that can be reduced from the $$\text{KNAPSACK}$$-problem or the $$\text{SET COVER}$$-problem, using the weights from the transformed directed graph, but I fail to see how exactly to this. I am starting to think that the transformed directed graph is of no use here.

What $$\mathsf{NPC}$$-problem can I use for the reduction?

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## What is a simple counterexample by a factor of 2 minimal cover?

There is a greedy algorithm for minimal cover that finds the minimal cover in a biepertize graph by going from highest degree vertex . How can I show a simple counterexample of factor of 2 that the algorithm is slower by a factor of 2.

## Homology of universal abelian cover of a manifold

If one define the universal abelian cover $$M_0$$ of a manifold $$M$$ as the abelian cover (i.e. normal cover with abelian group of deck transformations) that covers any other abelian cover, then what can one say about $$H_1(M_0)$$ ? Note that Hurewicz Theorem gives us a group isomoprhism between $$H_1(M_0)$$ and the abelianization of $$[\pi_1(M),\pi_1(M)]$$.

In particular, I would like to understand why is the following integral independent of the choice of the $$C^1$$ curve $$\tau$$ in $$M_0$$: $$\int_\tau \overline{\omega},$$ where $$\overline{\omega}$$ is the lift of a closed 1-form $$\omega$$ on $$M$$.

## Pullback of etale cover along itself

I would like know if there’s anything wrong with the following proof.

Claim. Let $$f : X \rightarrow S$$ be a finite etale of degree $$n$$. Then there is a noncanonical isomorphism $$f^*X \cong \amalg_n X$$.

Proof by induction on $$n$$. For $$n = 1$$, a finite etale morphism of degree 1 must be an isomorphism $$S \rightarrow S$$, so we are done. For the induction step, assume the claim for all etale morphisms of degree $$n-1 \ge 1$$ over $$S$$ and let $$f : X \rightarrow S$$ be finite etale of degree $$n$$. The diagonal $$X \rightarrow f^* X$$ is a clopen immersion that splits off a component isomorphic to $$X$$, hence $$f^* X \cong X \amalg X’$$ where $$X’ \rightarrow X$$ is a finite etale morphism of degree $$n-1$$. By induction hypothesis we are done.

The above proof is a slight alteration of the proof of Stacks Project Lemma 40.18.3, which claims that there IS a trivializing etale cover for $$X$$. I was surprised that I couldn’t find a single source that remarks that we can just take the trivializing cover to be $$X$$ itself. Is there something wrong with my claim?

## Does a metal box with eye holes count as total cover?

I wish to play as a necromancer however despite the undead hordes protecting me if someone were to snipe me i would be knocked out of the fight due to my low health as a squishy wizard. As a solution I was thinking of getting myself an iron box with eye holes and command my undead horde from the box.

Is this actually possible with RAW or does the existence of the eye holes prevent it from becoming total cover?

## Reducing the vertex cover problem to a variation of the vertex cover problem

The following variation on the vertex cover problem was given:

Given is an instance of graph $$G = (V, E)$$. Does $$G$$ have a vertex cover of size at most $$\frac{|V|}{4}$$?

I was asked to prove that this problem is also $$\mathsf{NP}$$-hard. I am pretty sure that this can be done by using Karp-reduction from the original vertex cover problem to this variation.

That means that I would need some transformation function $$f$$ that would transform an instance of the vertex cover problem, $$I = (V, E, k)$$, to an instance of the above described problem, $$f(I) = (V’, E’)$$.

I am not sure how to come up with the proper transformation function, but this is what I think have figured out:

• Let $$V = \{v_1, v_2, …, v_n\}$$.
• Let $$V’ = V \cup V^+$$, where $$V^+ = \{v_{n+1}, v_{n+2}, …, v_{4n}\}$$, because this would mean that the total size $$|V’| = 4 \cdot |V|$$.
• Let $$E’ = E \cup E^+$$, where $$E^+ = \{\{v_i, v_j\} \mid v_i \in V, v_j \in V^+, j \in \{n +i, 2n+i, 3n+i\}\}$$. Basically, this means that each vertex in $$V$$ is connected to three different vertices in $$V^+$$ such that all vertices in $$V^+$$ are connected to exactly one vertex in $$V$$.

However, I am having a hard time proving the correctness of this reduction. I think I messed up the generation of $$E’$$, which would make the transformation function wrong. Anyone have an idea what the actual transformation function could be?