# Recent MathOverflow Questions

### Is there an easy way to prove the function is non-negative in a compact domain?

Math Overflow Recent Questions - Sat, 02/10/2018 - 18:48

I encounter a function $$f(x,y)=-4xy \pi( cos(y) sin^2(x\pi))+sin(y)(-2y^2 x \pi +2x^3 \pi^3+(y^2+x^2 \pi^2)sin(2x\pi))$$ defined for $x\times y\in[0,2]\times[0,\pi]$. This function seems to be nonnegative in this compact domain. However, I just can prove this result after tedious analysis on the local property of $f$ around the zero points. Since this function seems to be simple, I wonder whether there is a straightforward technique to do so .Any hints or references in this respect would be greatly appreciated!

### Is the Giry Monad also a Comonad and if not, is there a probability measures (Co)monad?

Math Overflow Recent Questions - Sat, 02/10/2018 - 18:41

The Giry monad captures probability measures. Is it also a comonad? If so, what are the natural transformations?

If not, is there a monad that is also a comonad that captures probability measures?

### Cutting Lemma in Discrete Geometry

Math Overflow Recent Questions - Sat, 02/10/2018 - 18:26

I'm looking for a survey or a source for Cutting Lemma. I looked at Matusek's Discrete Geometry textbook, but it only proved Cutting Lemma for lines in $\mathbb{R}^2.$ I need to know the proof in $\mathbb{R}^d$ and other variations of Cutting Lemma.

Thanks in advance for any help!

### Descent of isomorphisms between irreducible closed subschemes

Math Overflow Recent Questions - Sat, 02/10/2018 - 17:27

Let $S$ be an affine scheme, $X$ be a projective $S$-scheme, $W,Z\to X$ two reduced, irreducible closed $S$-subschemes, flat over $S$. Let $S'\to S$ be a faithfully flat map, with $S'$ affine.

Assume there exists an $S'$-isomorphism $g: W_{S'}\xrightarrow{\simeq} Z_{S'}$, commuting with the closed immersions into $X_{S'}$.

When does there exist an $S$-morphism $f : W\to Z$ such that $f_{S'} = g$, and (hence) $f$ is an $S$-isomorphism?

### A first order ODE problem

Math Overflow Recent Questions - Sat, 02/10/2018 - 17:05

Fix $a>0$ and $b>0$. Does the following ODE $$G(x)^2+2axG(x)G'(x)+2aG'(x)(x-b)=0 \tag{*}$$ have a solution, say, $F(x)$, that satisfies $F(x)>0$ and $F'(x)<0$ on $(b,\infty)$?

I tried to solve it by Mathematica, and it gives $G$ as solution to $$x=e^{-2a\left(\text{log}\,G(x)-\frac{1}{G(x)}\right)}\left(C_0+b(2a)^{2a}\text{Gamma}\left(1-2a,\frac{2a}{G(x)}\right)\right) \tag{**}$$

for some free parameter $C_0$, where Gamma$(\cdot,\cdot)$ is the upper incomplete gamma function. However I'm not sure if this is a proper inverse function or if $G$ given by this equation has domain $(b,\infty)$.

Thanks!!

### Region of attraction of simple ODE with perturbation

Math Overflow Recent Questions - Sat, 02/10/2018 - 16:18

Consider the following simplest example:

$$\dot{x} = x(x-1)(x+1)$$ $[-1,1]$ is the ROA.

Now consider the two dimensional case:

\begin{aligned} &\dot{x} = x(x-1)(x+1)\\ &\dot{y} = y(y-1)(y+1) \end{aligned}Obviously, ROA is a square. However, if I consider the following coupled ODE:

\begin{aligned} &\dot{x} = x(x-1)(x+1) + \epsilon (y-x)\\ &\dot{y} = y(y-1)(y+1) + \epsilon (x-y) \end{aligned} where $\epsilon$ is a very small number. Or

\begin{aligned} &\dot{x} = x(x-1)(x+1) + \epsilon (-y+x)\\ &\dot{y} = y(y-1)(y+1) + \epsilon (-x+y) \end{aligned} Then I have the following ROAs: (blue line-case three, black line-case two, red line-case one)

My questions are:

1. There are two different tilt directions for case two and three. I know this is because of the slope of the coupling term (for case two, the slope of $x$ and $y$ in the coupling terms are $-1$). But how could I analyze this formally?

2. Is it a good way to analyze 1. by perturbation method? (observe the sign of the leading order term of the solution obtained from perturbation method?) and how could I proceed it for the coupling term?

3. Are there any reference about my questions?

Note: It is simple to check that if you just use the linearization method to find the Jacobian matrix (w.r.t the point $(0,0)$), the ROA will be the whole $\mathbb{R}^2$ , which is not correct.

### Stratification of space of labelled circles in the plane

Math Overflow Recent Questions - Sat, 02/10/2018 - 15:50

Consider the space of $n$ round circles in the plane to be the open subset of $\mathbb R^{3n}$:

$$C_n = \{ (v_1, v_2, \cdots, v_n, r_1, r_2, \cdots, r_n ) : v_i \in \mathbb R^2, r_i \in (0, \infty) \ \ \forall i \}$$

The $i$-th circle corresponding to a point in the above space would be the solutions to the equation $|x-v_i| = r_i$. With this convention, you can think of $C_n$ as the space of $n$ labelled circles in the plane, where the circles are allowed to intersect or even coincide.

There is a natural stratification of this space induced by the co-dimension one subvarieties consisting of subspaces where two circles are tangent, or where three (or more) circles intersect in a non-empty set. This is a fairly natural stratification since the complement of this stratification in $C_n$ is what you might call the "isotopy classes" of "regular" circles, as the combinatorics of their intersections do not change in the path-components and it is a dense open subspace of $C_n$.

Is there much known about the basic combinatorics of this stratification? For example, labelling the path-connected co-dimension $k$ strata for $k=0,1,2, \cdots$ and enumerating them? It looks like a souped-up version of the partition problem from a fairly naive perspective.

It seems like a natural problem and there is much closely related to it (spaces of arrangements in homotopy theory). But I've never come across quite this problem before.

Any insights would be appreciated -- especially if there's papers out there on this topic that I'm unaware of.

### Shift of zeta functions

Math Overflow Recent Questions - Sat, 02/10/2018 - 14:46

Let $k$ be a global field, and $X$ a smooth projective variety over $K$.

Let $Z(X,s)$ be the completed Zeta function of $X$, defined via geometric $\ell$-adic cohomology if the characteristic of $K$ is positive, and by product of local factors, including Serre's Archimedean factors, if the characteristic of $K$ is zero.

Suppose we want to relate, for an integer $n$, the shifted Zeta function $Z(X,s-n)$ to $Z(X,s)$ this way: there is a smooth projective variety $P_n$ over $K$, such that

$$Z(X,s-n) = Z(X\times_KP_n, s)$$ for every $n$.

Does $P_n$ exist and what is it? More generally, how to "geometrically" modify $X$ to get $Z(X,s-n)$ from $Z(X,s)$?

### On minimality of semitopological and quasitopological groups

Math Overflow Recent Questions - Sat, 02/10/2018 - 10:47

The phenomemnon of minimality is well-studied in the realm of topological groups.

Let us recall that a topological group $X$ is minimal if each bijecive continuous homomorphism $h:X\to Y$ to a topological group $Y$ is a topological isomorphism.

I am interested if the notion of minimality was studied in the realm of semi-topological or quasi-topological groups.

A semi-topological group is a group $G$ endowed with a Hausdorff topology making the group operation $G\times G\to G$, $(x,y)\mapsto xy$, separately continuous.

A semitopological group is called a quasi-topological group if the operation of inversion $G\to G$, $x\mapsto x^{-1}$, is continuous.

Let us define a semi-topological group $X$ to be semi-minimal if each continuous bijective homomorphism $h:X\to Y$ to a semitopological group $Y$ is a topological isomorphism.

By analogy we can define a quasi-topological group $X$ to be quasi-minimal if each continuous bijective homomorphism $h:X\to Y$ to a quasi-topological group $Y$ is a topological isomorphism.

It is clear that each semi-minimal quasi-topological group is quasi-minimal and each quasi-minimal topological group is minimal.

It can be shown that each compact Hausdorff semitopological group is semi-minimal (topological group).

Problem 1. Is each semi-minimal semi-topological group compact?

Problem 2. Is each quasi-minimal quasi-topological group compact?

I even cannot prove or disprove the following

Conjecture. No countable Boolean semi-topological group is semi-minimal.

Remark. By the answer to this MO problem, for some submonoid $M$ of the monoid $\omega^\omega$ of self-maps of a countable set $X$ there is no minimal Hausdorff topology on $\omega$ in which all self-maps $f\in M$ are continuous. At the moment this is the unique (known to me) example of an algebraic systems with unary operations, admitting no minimal Hausdorff topology.

### Continuity of a real function defined on an orbit space

Math Overflow Recent Questions - Sat, 02/10/2018 - 10:12

I am reading some paper, where the authors claim the following:

Let $G$ be a compact (Hausdorff) group and let $X$ be a locally, compact, Hausdorff space. Assume that $G$ acts on $X$ continuously. Denote by $C_0(X)$ the continuous functions on $X$ with complex values. Denote by $X/G$ the orbit space (it is Hausdorff as well), and let $\pi: X\to X/G$ be the canonical quotient map.

claim: Let $f\in C_0(X)$. The map $\pi(x)\mapsto sup_{g\in G}|f(g\cdot x)|$ is a continuous map $X/G\to \mathbb{R}$.

The proof starts as follows: Let $f\in C_0(X)$, let $\epsilon>0$ and let $x\in X$. Use continuity of $f$ to find, for every $g\in G$, an open neighborhood $W_g$ of $g\cdot x$ such that $|f(g\cdot x)-f(y)|<\epsilon$ for all $y\in W_g$. By compactness of $G$, there exists an open neighborhood $W$ of $x$ such that $g\cdot W\subseteq W_g$ for all $g\in G$.

I can not see why the last claim is true or why it follows from compactness... it seems like they actually claim that $\bigcap\limits_{g\in G}g^{-1}\cdot W_g$ contains an open neighborhood of $x$ (of course, it contains $x$, but I don't see why more than that).

An alternative approach to the proof or counter example for the claim would be appreciated as well.

Thank for any help!

### Smooth functions which come from densely defined holomorphic functions

Math Overflow Recent Questions - Sat, 02/10/2018 - 08:23

The actual question might seem convoluted without context so let me first give a motivating example.

Consider the holomorphic function $f:\mathbb{C} \setminus \{0\} \to \mathbb{C}$ given by $f(z)=e^{-z^{-2}}$. Consider now $f$ restricted to the line $\mathbb{R} \setminus \{0\}$. It has a (unique) extension to a continuous function on $\mathbb{R}$ which is moreover smooth.

The aim of the question is to understand which functions arise in the above manner.

I'll try to give two versions of the question, one general and one specific.

Specific version:

Let $f: X \to \mathbb{C}$ where $X$ is either the punctured plane or the punctured disk. Let $\gamma:\mathbb{R} \setminus \{0\} \to X$ be a line s.t. $f \circ \gamma : \mathbb{R} \setminus \{0\} \to \mathbb{C}$ has a continuous (smooth) extension to $\mathbb{R}$.

Question: What kind of smooth functions $g: \mathbb{R} \to \mathbb{C}$ can be written as $g= f \circ \gamma$ for some $f$ and $\gamma$ as above? Is there some direct description of this class?

General version:

Let $M$ be an real analytic (resp. algebraic) manifold. Let $N$ be a complex analytic (resp. algebraic) manifold and $F:N \dashrightarrow \mathbb{C}$ be a holomorphic function on $N$ with singularities (meaning it is defined and holomorphic on some dense open subset of $N$). Let $j: M \to N$ be an analytic (resp. algebraic) embedding (considering $N$ as a real manifold) satisfying that $f: F \circ j: M \dashrightarrow \mathbb{C}$ is defined on a dense open subset of $M$ and has a continuous extension to all of $M$ (Must $f$ be smooth? if not we will assume from here on that it is).

Definition: For a given $M$ as above the functions which are obtained from the above procedure (lets stick with the analytic version of the above) for some choices of $(N,F,j)$ will be called "Really Smooth" functions.

Questions:

1. Is there a different (more direct perhaps) description of this class of "Really Smooth" functions? (maybe via hyperfunctions?)

2. Are "Really Smooth" functions closed under differentiation?

3. Are "Really Smooth" functions closed under pre-composition? Meaning: Let $\pi: M \to N$ be an analytic map between real analytic manifolds and let $f$ be a "Really Smooth" function on $N$, is $f \circ \pi$ "Really Smooth"?

4. Are "Really Smooth" real-valued functions on $\mathbb{R}$ closed under composition?

Aside: I have no idea how to tag this question so any help with this will be welcome, thanks.

### Computational Geometric Aspects of Greedy Tour Expansion

Math Overflow Recent Questions - Sat, 02/10/2018 - 05:38

Has the following problem already been investigated from the Computational Geometry point of view and what are the results regarding worst case complexity?

Given

• a finite set $\mathcal{P}$ of $n$ distinct points in the Euclidean plane,
• its convex hull $\mathcal{T_{\text{card(CH(}\mathcal{P}\text{))}}}\ :=\ \text{CH(}\mathcal{P}\text{)}$ as the initial tour.

while $m<n$

• chose $\ \left( p\in\mathcal{P}\setminus\mathcal{T_m},\ t_i\in\mathcal{T_m}\right):$
$\quad\quad\quad\|p-t_i\|+\|t_{\text{succ(}i\text{)}}-p\|-\|t_{\text{succ(}i\text{)}}-t_i\|$ $\quad\le\quad\|q-t_k\|+\|t_{\text{succ(}k\text{)}}-q\|-\|t_{\text{succ(}k\text{)}}-t_k\|$
$\forall q\in\mathcal{P}\setminus\mathcal{T_m},\quad\left(t_k,t_{\text{succ(}k\text{)}}\right)\in\mathcal{T_m}$
• $\mathcal{T_{m+1}} := \lbrace\mathcal{T_m}\setminus\left(t_{\text{i}},t_{\text{succ(}i\text{)}}\right)\rbrace\cup\lbrace\left(t_i,p\right),\left(p,t_{\text{succ(}i\text{)}}\right)\rbrace$

in this context, $t_{\text{succ(}i\text{)}}$ shall denote the tour-vertex that is encountered immediately after vertex $t_\text{i}$ when traversing the tour w.l.o.g in counter clockwise order.

Please note, that the objective of the task is not to create a simple polygon through all points or even an optimal tour; it is rather to determine the point, whose integration into the tour incurs the least length-increase, as fast as possible.

I am looking for techniques from Computational Geometry, i.e. algorithms and datastructures, that bring about provable improvements beyond updating least detour information w.r.t. the newly generated edges after the insertion of a further point.

Here is an example of a selfintersecting tour with 50 points, that was generated with the greedy insertion algorithm, answering a question of Joseph O'Rourke:

to indicate, that geometric concepts may indeed be promising, consider the problem of deciding, whether a point $p\notin \mathcal{T_i}$ is closer (in the sense of incurred elongation) to tour edge $(t_i,t_j)$ or to $(t_j,t_k)$.

in that special case, where the tour edges are adjacent, the separator equals the radical axis of the circles with center and radius $(t_i,\|t_j-t_i\|)$ and $(t_k,\|t_k-t_j\|)$; whether that is an actual improvement, depends of course on the computational model and the limitations on available resources - for very large instances it may not be viable to store all distances between pairs of points, whence deciding on which side of the radical line a point lies, is cheaper than calculating and comparing two detours.

for non-adjacent tour edges the situation is not so simple and yields non-linear algebraic curves as separators in general.

### Can the "Bisector" be represented by a holomorphic function?

Math Overflow Recent Questions - Sat, 02/10/2018 - 01:54
Note:

In this question, a complex number is counted as a vector initiated from the origin.

______________________________________________________________-

Is there a holomorphic function $B:\mathbb{C}^2 \to \mathbb{C}$ such that for every two non zero complex numbers $z,w$ with $z/w \notin \mathbb{R},$ the vector $B(z,w)$ is a non zero vector indicating to the direction of the bisector of the angle $\angle (z,w)$?

Motivation:

The initial formula for the "Bisector" of $\angle (z,w)$ is $B'(z,w)=|z|w+|w|z$. But it is not a holomorphic function.(It is not even smooth at $z=0$ or $w=0$). So we search for a holomorphic remedy, a holomorphic function $B$ defined on whole $\mathbb{C}^2$ such that $B(z,w)$ is real proportional to $(|z|w+|w|z)$ via a non constant real function $\lambda$.

What about if we require that such $\lambda$ be positive(Non negative)?

### Schur's Lemma for Quantized Universal Enveloping Algebra

Math Overflow Recent Questions - Fri, 02/09/2018 - 20:01

Let $U_q(\mathfrak{g})$ (defined over $\mathbb{C}(q)$) be the quantized universal enveloping algebra of a simple Lie algebra $\mathfrak{g}$. Let $M$ a finite-dimensional simple left $U_q(\mathfrak{g})$-module. Is it true that $\dim_{\mathbb{C}(q)}\mathrm{End}_{U_q(\mathfrak{g})}(M)=1$? How to sketch a proof? The problem is that the field $\mathbb{C}(q)$ is not algebraically closed.

Math Overflow Recent Questions - Fri, 02/09/2018 - 12:27

Let $E$ be an infinite-dimensional complex Hilbert space.

For $A = (A_1,\cdots,A_d)\in\mathcal{L}(E)^d$, the algebraic spectral radius of $A$ was given by $$r_a(A)=\lim_{n\to+\infty}\left\|\sum_{f\in F(n,d)} A_f^* A_f\right\|^{\frac{1}{2n}} ,$$ where $F(n,d):=\{f:\,\{1,\cdots,n\}\longrightarrow \{1,\cdots,d\}\}$ and $A_f:=A_{f(1)}\cdots A_{f(n)}$, for $f\in F(n,d)$.

Further the geometric spectral radius of $A$ was given by $$r_g(A)=\max\{\|\lambda\|_2,\;\lambda=(\lambda_1,\cdots,\lambda_d) \in \sigma(A)\},$$ with $\sigma(A)$ denotes the Taylor spectrum of $A$. Note that it was shown that $r_g(A)$ doesn't depend of the choice of the spectra.

Moreover, it was shown in (1) that if $A_iA_j=A_jA_i$ for all $i,j$, we have

$$r_g(A)=r_a(A)=\displaystyle\lim_{n\to \infty}\left\|\displaystyle\sum_{|\alpha|=n}\frac{n!}{\alpha!}{A^*}^{\alpha}A^{\alpha}\right\|^{\frac{1}{2n}}.$$

If $A_iA_j=A_jA_i$ for all $i,j$, it is true that $$r_a(A)\leq w_e(A):=\sup\left\{\bigg(\displaystyle\sum_{i=1}^d|\langle A_ix,x\rangle|^2\bigg)^{\frac{1}{2}},\;x\in E,\;\|x\|=1\;\right\}\;?$$ If $d=1$, the claim is true. For its proof one can see Theorem 1.1 in Moshe Goldberg, Eitan Tadmor: On the numerical radius and its applications, doi: 10.1016/0024-3795(82)90155-0.

Notice that Gelu Popescu has a big paper about multivariable operators (Memoirs of the AMS, arXiv) and he proves many things about $w_e(A)$ in section 2 of his paper.

My attempt: We can see that $$w_e(A):= \displaystyle\sup_{(\lambda_1,\cdots,\lambda_d)\in B_d}w(\lambda_1A_1+\cdots+\lambda_dA_d),$$ with $B_d$ is the open unit ball of $\mathbb{C}^d$. I try to show that, if $A_iA_j=A_jA_i$ for all $i,j$, we have $r_a(A)=r_e(A)$ where $$r_e(A):= \displaystyle\sup_{(\lambda_1,\cdots,\lambda_d)\in B_d}r(\lambda_1A_1+\cdots+\lambda_dA_d).$$ Finally we apply the well known result: for $T\in\mathcal{L}(E)$ we have $$r(T)\le w(T).$$

### What is the probability that a plane curve crosses a segment?

Math Overflow Recent Questions - Fri, 02/09/2018 - 12:01

Given

• a straight segment of length $l$ in the plane, and
• a random curve that starts at one end of this segment, that is smooth/differentiable everywhere, that has everywhere a minimum curvature radius $r<l$, and that reaches spatial infinity,

what is the probability $p$ that the random curve crosses the segment? Given that the random curves all start at one end of the segment, some curves will cross the segment later on, whereas others will not. The crossing probability $p$ will depend on the segment length $l$ and on the radius $r$. What is the dependence?

Possibly, the problem is not specific enough. What would be necessary to make it well defined?

Problem 2: What would the probability be if the segment is not a segment, but a full straight line across the plane (say the x-axis), and if all the curves start at the origin (0,0)? Is this problem well-defined?

Problem 3: What would the probability be if the "segment" is taken to be the positive x-axis, and if again, all curves start at the origin?

### Exercise 1.1.(c) in Hartshorne's Deformation Theory

Math Overflow Recent Questions - Fri, 02/09/2018 - 00:36

Exercise 1.1.(c) in Hartshorne's Deformation Theory:

Over an algebraically closed field $k$, we define a curve in $\mathbb P^2_k$ to be the closed subscheme, defined by a homogeneous polynomial $f(x,y,z)$ of degree $d$ in the coordinate ring $S=k[x,y,z]$.

(c) For any finitely generated $k$-algebra $A$, we define a family of curves of degree $d$ in $\mathbb P^2$ over $A$ to be a closed subscheme $X\subseteq\mathbb P^2_A$, flat over $A$, whose fibers above closed points of $\mathrm{Spec}\,A$ are curves in $\mathbb P^2$. Show that the ideal $I_X\subseteq A[x,y,z]$ is generated by a single homogeneous polynomial $f$ of degree $d$ in $A[x,y,z]$.

My attempts: the condition on fibers above closed points is equivalent to the condition that for any $\mathfrak m\in\mathrm{Specm}\,A$ the ideal $(I_X,\mathfrak m)/\mathfrak m$ is principal. Equivalently, for any $\mathfrak m$ there is $f_{\mathfrak m}\in I_X$ such that $I_X\subset (f_{\mathfrak m},\mathfrak m)$. Also, it is clearly sufficient to prove that $I_X$ contains a homogeneous element of degree $d$.

### What are the transitive extensions of finite representations of cyclic groups?

Math Overflow Recent Questions - Thu, 02/08/2018 - 20:41

This question is a generalisation of this one. Let $H$ be a finite, transitive permutation group of degree $n$. If the point stabiliser subgroup $H_n$ of degree $n-1$ is some faithful permutation representation of a cyclic group $C_k$ (with $k$ not necessarily equal to $n-1$), what can $H_n$ and $H$ be? For example, must $k$ divide $n-1$?

### Sheaf-theoretic Grothendieck groups

Math Overflow Recent Questions - Thu, 02/08/2018 - 19:55

Let $S$ be a scheme, $M\to S$ a commutative monoid object in algebraic $S$-spaces, ie. an algebraic $S$-space such that, functorially on $S$-schemes $T$, $M(T)$ is a commutative monoid with neutral element.

Each commutative monoid $M(T)$ has its own group completion $M(T)^{\rm gp}$, given by the classical Grothendieck group construction.

Formation of $M(T)^{\rm gp}$ is functorial in $T$, and we may define the fppf sheaf $M^{\rm gp}$ to be the fppf sheafification of $T\mapsto M(T)^{\rm gp}$.

Is $M^{\rm gp}$ an algebraic $S$-space, hence a commutative group object in algebraic $S$-spaces? Is it at least when $M$ is cancellative?

-----------------------------------------------------

Example

Suppose $M$ is, in addition, cancellative (ie. $M(T)$ is a cancellative commutative monoid with zero for all $S$-schemes $T$).

In this case, $M^{\rm gp}$ is the fppf quotient sheaf of $M\times_SM\times_SM\times_SM$ by the following equivalence relation:

$$R := (M\times_SM\times_SM\times_SM)\times_{\mu, M\times_SM,\Delta_{M/S}}M$$

with:

• $\mu : M\times_SM\times_SM\times_SM\to M\times_SM$ the map defined functorially on $T$-sections by sending $(a_T, b_T, c_T, d_T)$ to $(a_T+d_T, b_T+c_T)$.
• $\Delta_{M/S} : M\to M\times_SM$ is the diagonal.
• $s,t : R\to M\times_SM$ are defined to be the pullback along $\mu$ of $\text{pr}_1\circ\Delta_{M/S}$ and $\text{pr}_2\circ\Delta_{M/S}$.

On $T$-sections, $R(T)$ is the equivalence relation that identifies the pairs $((a,b),(c,d))$ with $((a,d),(b,c))$ treating $(a,b)$ and $(c,d)$ as $"a-b"$ and $"c-d"$, and ensuring that $"a-b = c-d"$ if and only if $"a+d = b+c"$. This way, the monoid operation $+$ on $M(T)\times M(T)$ descends to a group operation, where the inverse of $(a,0)$ is $(0,a)$.

$R$ is not an étale equivalence relation, unless $M$ is étale over $S$, as the fiber of $M\times_SM\to M^{\rm gp}$ over $0$ is the diagonal copy of $M$ in $M\times_SM$.

This is always going to be the case, so either the question has negative answer, or one should come up with a better presentation for $M^{\rm gp}$, or verify Artin's axioms.

The case of interest is when $M\to S$ is locally of finite presentation and separated, $S$ is affine.

### Weak-convergence of probability measures implies the convergence of the measure of a continuity set

Math Overflow Recent Questions - Thu, 02/08/2018 - 00:54

Let $\Omega$ be a Polish space and $\mathcal{B}(\Omega)$ be its Borel $\sigma$-algebra. Let $\{\mu_n\}$ be a sequence of probability measures on $\mathcal{B}(\Omega)$ such that $\mu_n$ weak-converges to $\mu$. Then in general, $\mu_n(E) \nrightarrow \mu(E)$ for $E \in \mathcal{B}(\Omega)$.

But if for some $F \in \mathcal{B}(\Omega)$ we have $\partial F = F$ and $\mu_n(F) = 0$ for all $n$, can we say that $\mu_n(F) \rightarrow \mu(F)$? If not, are there any conditions on $\Omega$ that would make this true?