A real number is a Liouville number if it satisfies the following condition:

0<|x-p/q|<1/q^n, with p and q being both positive integers and q being greater than 0. Is it possible to modify this condition by replacing 1/q^n with summation (1/q^k) with k going from 1 to n? Next, is it possible to replace q^k with the product(q_k)= q_1*q_2*q_k, with q= n+1?

Given a topological space $X$, we have the notion of the fundamental groupoid $\Pi_1(X)$.

Here, the fundamental groupoid $\Pi_1(X)$ is made into a topological groupoid giving a topology on the morphism set.

One can then talk about $\Pi_1(\Pi_1(X))$. Is this related to (same as) the fundamental $2$-groupoid?

Let $F : \mathbb{R}[X] \rightarrow \mathbb R[X]$ be a linear map and let $H \in \mathbb{R}[u,x,y,z]$ be a polynomial. Suppose that

$$ F(P \times Q) = H(F(P),F(Q),P,Q)$$

for all $P, Q \in \mathbb{R}[X]$. Two obvious solutions for $F$ and $H$ are

$F = I$, the identity function, and $H(u,x,y,z) = \alpha u x + \beta u z + \gamma yx + \delta yz$ where $\alpha,\beta,\gamma,\delta \in \mathbb{R}$ and $\alpha +\beta+\gamma+\delta = 1$;

$F = D$, the derivative, and $H(u,x,y,z) = u z + x y$, from the Leibniz rule.

Do there exist solutions $F$ and $H$ in which $F$ is not a linear combination of $I$ and $D$?

Let $xy$ be an edge in a planar triangulation $T$ that is at least 4-connected and let $uxvy$ be the 4-cycle delineating the 4-face of the near-triangulation $G_{xy}$ obtained by deleting $xy$ from $T$; $T$ is said to be **Kempe-locked** with respect to $xy$ if there are exactly three Kempe chains including both $x$ and $y$ in every 4-coloring of $G_{xy}$ in which $x$ and $y$ have the same color. If $xy$ is a Kempe-locked edge in $T$, the proper subgraph $K_{xy}$ of $G_{xy}$ obtained by deleting $u$ and $v$ (and their incident edges) is said to be the **Kempe-locking configuration** for that edge; a Kempe-locking configuration is said to be **fundamental** if it contains no smaller Kempe-locking configuration as a proper subgraph. The Birkhoff diamond of order 10 is the lowest-order fundamental Kempe-locking configuration and it might be the only fundamental Kempe-locking configuration. Any ideas on how that could be proved? I have checked all isomorphism classes of 4-connected triangulations through order 17 and all isomorphism classes of 5-connected triangulations through order 24 and have found no other fundamental Kempe-locking configurations.

Peter Scholze formulated several conjectures about $q-$de Rham complex here. Especially intriguing to me is Conjecture $3.2$: Betti-de Rham comparison isomorphism.

In the paper the author emphasized the lack of nontrivial examples and suggests to do a computation for the universal family of elliptic curves.

**Question 1:** Has anyone done this computation (or any other interesting computations)? I will be especially interested in those involving basic hypergeometric functions.

**Question 2:** What does conjecture $3.2$ mean in practice? Naively one might hope that it leads to some $q-$deformation of periods.

Consider the set $[n]=\{1,2,\ldots,n\}$. Suppose for each set $A\subseteq [n]$ I have a $p_A \in [0,1]$. I now create a random collection $\mathcal{W}\subseteq\mathcal{P}([n])$ of subsets of $[n]$ by including each $A$ with probability $p_A$, independently. What is the probability that their union covers $[n]$, that is, that $$\bigcup_{W\in\mathcal{W}} W = [n]$$

This seems like a problem that absolutely has been considered before -- it's easy to state and seems natural -- and the answer should "just" be some suitably symmetric multivariate polynomial. Unfortunately, I can't figure it out on my own, and my google-fu hasn't availed me either.

Do you know a reference about mean curvature flow for curves in the plane with fixed Dirichlet boundary conditions?

I start with a curve connecting two points and I let it evolve by mean curvature flow keeping the endpoints fixed. Assuming the original curve is smooth, I guess the flow should exist for every time and in the limit it should converge to the straight line segment between the endpoints. Is this correct? Can I weaken the assumptions on the starting curve (say only Lipschitz) and get the same result?

Do you know if anything has been done for general Riemannian manifolds?

Consider the equation $$\displaystyle Ax^p + By^q = Cz^r, A,B,C \in \mathbb{Z}, \gcd(x,y,z) = 1, p,q,r \geq 2.$$

When $p^{-1} + q^{-1} + r^{-1} > 1$, the above equation is called *spherical* and satisfies an identity of the form

$$\displaystyle x = f(u,v), y = g(u,v), z = h(u,v)$$

where $f,g,h$ are binary forms with complex coefficients. Beukers showed that all solutions to the equation are parametrized by a finite family of such binary forms with integer coefficients.

We now consider the equation

$$\displaystyle x^r + y^2 = Cz^2, \gcd(x,y,z) = 1, C \in \mathbb{Z}, r \geq 3.$$

This equation is spherical, and if it has one integer solution then it will have infinitely many due to the existence of a polynomial parametrization (by binary forms). Suppose that it does indeed have a solution and we have a parametrization $x = f, y = g, z = h$ as above. What can we say about the discriminants of $f,g,h$?

I've a problem where I have to prove the following statements:

(i) if $SPACE(n) \subseteq P \implies SPACE(n^2) \subseteq P$

(ii) if $P = PSPACE(n) \implies SPACE(n) = SPACE(n^2)$

For the Space Hierarchy Theorems we have that $\forall f(n) = o(g(n)) \implies SPACE(g(n)) \nsubseteq SPACE(f(n))$

Using this result and the statements (i) and (ii) I have to prove that $P \neq SPACE(n)$.

I'm using the following post as reference: How do we know that P != LINSPACE without knowing if one is a subset of the other?

I suppose that I can use Thomas Klimpel's proof in the abovementioned link to prove the statement (i). But how can I prove the statements (ii) and (iii)?

Accept my apologies for this question, but I'm a rookie in Theory of Complexity :)

The Gabriel-Popescu theorem tells us that every Grothendieck category with a generator is a left exact localization of a module category. I'm interested in a slightly different way of "representing" such categories:

**Question:** Let $\mathcal C$ be a Grothendieck category with a generator.

Does there exist a Grothendieck topos $\mathcal X$ such that $\mathcal C \simeq Ab(\mathcal X)$ is equivalent to the category of abelian group objects in $\mathcal X$?

Slightly less naively, does there exist a ringed Grothendieck topos $(\mathcal X, \mathcal O_{\mathcal X})$ such that $\mathcal C \simeq \mathcal O_{\mathcal X}\textrm{-}Mod$ is equivalent to the category of $\mathcal O_\mathcal X$-modules?

Same as (2), but with $(\mathcal X, \mathcal O_{\mathcal X})$ being locally ringed?

I am simulating an agent navigating through space, where the agent's navigation strategy changes over time as a Markov chain with transition probabilities dependent on its position in space.

Currently, I model space as a finite 2D grid such that the whole system can be described by a homogeneous DTMC with states corresponding to (strategy, x, y). This model has the useful properties of being well-suited to MCMC simulations and even formal verification with temporal logic, but it is limiting in several ways:

- Adding time-dependence to the spatial map causes complexity to blow up, leading to infeasible computation loads
- Discrete space limits the type and magnitude of noise I can inject into the system, since making the step size arbitrarily small relative to the extent of the space also results in unmanageable computation loads

I am considering modeling the agent's movement through space as a dynamical system, with the strategy still represented by a DTMC--now inhomogeneous, with transition probabilities dependent on the state of the dynamical system at a given time step. This would dramatically increase the flexibility of the simulation, but I would lose access to the analytical properties of a memoryless MC.

Is there any rigorous way to get statistical properties of an arbitrary inhomogeneous DTMC? Or will I have to resort to experimental statistics on my simulations if I go this route? Any recommendations of alternative approaches are also appreciated.

Edit: It appears that Gyori, *et al.* (2015) showed that a hybrid system H can be approximated by a Markov chain M such that M|=a (M satisfies specification a) iff H|=a. This method is limited to deterministic dynamical systems, but it seems like a good start for where I am trying to go.

I am confused about the notion of good reduction. Let $R$ be a DVR, let $K$ be its fraction field. If we have a smooth proper $K$-scheme $V$, then I believe $V$ is said to have good reduction at the unique non-zero prime ideal if there exists a smooth proper $R$-scheme whose generic fiber is $V$.

I tend to dislike the word "exists". I think for abelian varieties, a condition equivalent to good reduction can be formulated in terms of the 1st cohomology. Can a condition not involving the existential quantifier and equivalent to good reduction be given for general smooth proper $K$-schemes?

A second question, if $F$ is a number field, then what is the right notion of good reduction modulo a non-zero prime ideal of the ring of integers of $F$ for smooth proper $F$-schemes? The issue for me is the compatibility between different prime ideals: if a smooth proper $F$-scheme has "good reduction everywhere", does it mean that there is a single integral model that has smooth fibers over every prime ideal, or just that for any prime ideal you can find a (proper flat, or I don't know what should be required really) model that has smooth fiber over that prime ideal?

Third question: given a smooth proper scheme, is there some functorially constructed "best" integral model so that all questions of reduction can be just answered using that particular model? I have heard something about Neron models but I think they only work for abelian varieties.

I apologize for these naive questions but all references I found so far refer to good reduction without giving a definition. If there is a reference addressing the above questions I will gladly study it.

Let $h: \mathbb{R}^3 \to \mathbb{R}$ be the usual height function (i.e. $h(x,y,z) = z$). One way that Morse functions on $S^2$ are often described is by picking an embedding $i: S^2 \to \mathbb{R}^3$ and then considering $h \circ i$ which, for a generic embedding will be a Morse function.

Does there exist a Morse function $f : S^2 \to \mathbb{R}$ so that there is no embedding $i: S^2 \to \mathbb{R}^3$ with $f = h \circ i$?

I think that this shows that every Morse function on $S^2$ can be factored through an immersion, but I am interested in embeddings.

I am in the context of simplicial sets. I have a square diagram that I want to show to be homotopy pullback. What I can grasp is that it is a pullback up to some equivalences in the middle of my reasoning, and I would like to promote this to a real statement.

So let $p:X \to Z, q:Y \to Z$ map of ssets, and $f:P \to X, g:P\to Y$ other two maps from the supposed-to-be homotopy pullback. Suppose that for every $x \in X$, the induced map on fibers $P \to Y$ is an equivalence. Furthermore, you can additionally suppose that $p,q$ are Kan fibrations. Is it true that $P$ is the homotopy pullback of the diagram?

Related question: Can homotopy pullbacks of spaces be checked on fibers?

The difference is that we are not dealing with spaces (=Kan complexes) but with general simplicial sets. Moreover, I'd like to use actual and not homotopical fibers.

Reference: http://palmer.wellesley.edu/~ivolic/pdf/Papers/CubicalHomotopyTheory.pdf

Proposition 3.3.18. Thanks!

**Definition** (informal) A *normal edge-5-coloring* of a bridgeless cubic graph $G$ is a proper 5 coloring of the edges of the graph, so that for each edge $e\in E(G)$, either $e$ and the four edges adjacent to it are colored with five colors, or they are colored with three colors.

Several decades ago, Jaeger proposed the following conjecture, which implies several (two) of the most notorious outstanding problems in graph theory (for example, the Berge-Fulkerson conjecture, and the cycle double cover conjecture).

**Conjecture 1** [Jaeger, 1980, 1985] Every bridgeless cubic graph has a normal edge-5-coloring.

Our question concerns a special case of this conjecture.

**Question** Let $G$ be a bridgeless cubic graph that has a Hamiltonian path. Does $G$ have a normal edge-5-coloring? As always, the intent of the question is to provide proof or counterexample.

Clearly, Hamiltonian graphs have a normal edge-5-coloring (in fact, an edge-3-coloring). Here, we relax the hamiltonicity condition to be one that requires Hamiltonian paths, and not necessarily Hamiltonian cycles. We believe that the condition is strong enough that a proof of it might be within reach, perhaps an algorithmic one. It is also a light enough condition that many graphs, including snarks, satisfy, in particular if their cyclic connectivity is high.

For an interesting paper on normal colorings, see [1].

[1] ˘Sámal, R., New approach to Petersen coloring, *Electronic Notes in Discrete Mathematics* 38 (2011) 755-760

**Definitions.**

- A
*Polish group*is a topological group $G$ that is homeomorphic to a separable complete metric space. - A group $G$ has
*no small subgroups*if there exists a neighborhood $U$ of the identity that contains no nontrivial subgroup of $G$.

**Example.** All finite dimensional Lie groups with a countable number of components are Polish groups, and it is well-known (see here) that a Lie group $G$ has no small subgroups. (Note that the proof of this fact crucially relies on the exponential map.)

I am interested in understanding when certain Polish groups will have no small subgroups. In particular, I have two questions.

**Question 1**. Are there examples of non-locally compact Polish groups which have no small subgroups?

**Question 2**. If the answer to Question 1 is yes, are there conditions on a non-locally compact Polish group which ensure that it has no small subgroups?

**Note.** Due to results of Gleason and Montgomery--Zippin, a locally compact group with no small subgroups is a Lie group.

Thanks in advance for the help!

If I understand correctly, the motives attached to eigenforms of weight $2$ are the motives of elliptic curves. For eigenforms of higher weight, can we achieve geometric understanding of what motives they correspond to (in particular cases)? Hodge--Tate weights will not satisfy Griffiths transversality so it is not entirely obvious what they should look like.

I am not 100% set on what "geometric understanding" should mean, but if you can give an explicit representative (i.e. a smooth projective variety) of the motives in question, that would be great. I understand that the choice would probably be non-canonical but in my situation that is an appreciated non-canonicity.

Firstly, a bit of notation. Let $C$ be a simplicial set. We define, for $x,y \in C$ vertices in $C$

$$Map(x,y) = \{x\}\times_{\Delta^{\{0\}}}Map_{sSet}(\Delta^1,C) \times_{\Delta^{\{1\}}} \{y\} $$

the sSet of arrows from x to y. You can imagine a $n$ simplex here as a cylinder with basis $\Delta^n$, with all $1_x$ on the left base and all $1_y$ on the right base. From this description it is evident that $Map(x,y) \simeq (C_{/y})_x$, the cones over y with all $1_x$ at the basis. Symmetrically, it is equivalent to $(C_{x/})_y$.

As the two sSet are fibers of a right and a left fibration, they are respectively left inner fibrant and right inner fibrant. Thus, the map sets are Kan complexes.

I am searching to generalize this to an arbitrary inner fibration $p:C \to D$, and show that $Map(x,y) \to Map(px,py)$ is a Kan fibration.

Thanks!

I can't find a calculation of earth's maximum seasonal change in irradiation by the sun. An earth-diameter disk gets the solar constant of 1360 watts per square meter on the lit side, the polar tilt or obliquity is 23.5 degrees, and this should result in a hemispheric minimal circular segment of that disk at the winter solstice, but how much less is that? It's just a tricky geometry problem.

This is a similar question to one that I posted in MSE a few days ago.

I recently came across this paper from Alahmedi, Alsulami, Jain and Zelmanov, which quoted the following result for $M_\infty(K)$, the (non-unital) $K$-algebra of countably indexed infinite matrices with finitely many nonzero entries.

Theorem 12: For an arbitrary $K$-algebra automorphism $\alpha$ of $M_\infty(K)$, there exists some invertible column finite matrix $T$ such that $$\alpha(a) = T^{-1} a T \text{ for all } a \in M_\infty(K)$$

Nathan Jacobson's *The Structure of Rings* was cited as the location of the result, but no specific location within the book was given. Unfortunately, I was unable to find the specific result from Jacobson that the theorem in the above paper was referencing.

Over the past few days, I've been able to put together a proof for this result on my own, but I was wondering if anybody knew the specific location of this result within the book or in another resource. I want to see how far my proof differs from the proof from "The Book" so to speak.

Edit: When I say "$K$-algebra," in the above, I'm invoking a slightly weaker definition of an "algebra." Being a vector space generated by an infinite collection of matrix units, $M_\infty(K)$ has zero center, hence it is impossible to fulfill the traditional requirement that $K \subseteq Z(M_\infty(K))$