Take a convex polyhedron $P$ in $\mathbb R^3$ and remove all the faces, i.e. leave only the edges. Call this graph $E$. Let us now try to continuously deform $E$ in $\mathbb R^3$ so that all the edges of $E$ keep their length and remain straight (like metal sticks), but allow the change of angles between the edges.

**Question.** *Is it true that $E$ admits non-trivial deformations (not globally isomteric) if at least one face of $P$ is not a triangle?*

Note that if all the faces of $P$ are triangles it is not deformable by Cauchy's rigidity theorem: https://en.wikipedia.org/wiki/Cauchy%27s_theorem_(geometry)

The fundamental group of a closed orientable manifold is finitely presented, and every finitely presented group arises as the fundamental group of a closed orientable four-manifold; see this question.

Not every finitely presented group arises as the fundamental group of a closed non-orientable four-manifold. One necessary condition is that it contain an index two subgroup which corresponds to the orientation double cover. Is this the only restriction?

Let $G$ be a finitely presented group with an index two subgroup. Is there a closed non-orientable four-manifold $M$ with $\pi_1(M) \cong G$?

I am certain that the following result holds, but was not able to find a reference. Do you know one?

**Statement.** Let $(M^4,\omega)$ be a compact symlectic manifold and let $J_t$ be a smooth family of compatible almost complex structures, $t\in(-1,1)$. Suppose that $S_0$ is a smooth $J_0$-holomorphic sphere in $(M^4,J_0)$ with zero self-intersection.

Consider now $M^4\times (-1,1)$, introduce $J_t$ on each fibre $M^4\times t$, and take in it the surface $S^2_0\times 0$. Then there is a neighbourhood $U$ of $S^2_0\times 0$ such that for any $(x,t)\in U$ there is a unique $J_t$-holomorphic sphere passing through $(x,t)$ contained in $U$. In other words $J_t$-holomorphic spheres produce on $U$ a structure of a smooth $S^2$-fibration (over a $3$-ball).

Yes, this is a combinatorics question, dealing with something like matchings where sometimes one vertex is allowed to match two other vertices instead of just one like usual! But it arises from Magic and so will need some setup to state. (I would certainly be interested in generalizations, too.)

Background: In Magic: The Gathering, there are 6 types of "mana", denoted {W}, {U}, {B}, {R}, {G} (the 5 colors of mana), and {C} (colorless mana). (We can also imagine a 7th type, {P}, representing 2 life; this is not an actual type of mana.) Mana comes in whole number amounts; by an "amount of mana" I mean a specified whole number amount of each type (and some amount of life, I suppose). A *mana cost* describes a set of amounts of mana sufficient to pay the cost. A mana cost consists of some number of the following symbols:

- {W}, {U}, {B}, {R}, {G} -- payable only by 1 mana of that color
- {C} -- payable only by 1 colorless mana
- {W/U}, {U/B}, {B/R}, {R/G}, {G/W}, {W/B}, {U/R}, {B/G}, {R/W}, {G/U} -- payable by 1 mana of either appropriate color (all 10 color pairs are listed here)
- {1} -- payable by 1 mana of any type, colored or colorless
- {W/P}, {U/P}, {B/P}, {R/P}, {G/P} -- payable by the appropriate color or by 2 life (which we can think of as a fictitious type of mana, {P})
- {2/W}, {2/U}, {2/B}, {2/R}, {2/G} -- payable by the appropriate color of mana or by any 2 mana (colored or colorless)

(Yes, for those familiar with Magic, I am ignoring {X} and I am ignoring the question of snow, since those don't seriously affect the problem. Indeed we didn't actually need all the above, but I figured I'd be thorough...)

We can put a partial ordering on the set of mana costs as follows: A≤B if any amount of mana sufficient to pay for B is sufficient to pay for A.

Then, the question is, given two mana costs, how can we determine whether one is less than equal to another or not?

If we ignore the existence of the symbols of the form {2/M}, then this problem is not hard; using Hall's marriage theorem, we can see that one just has to check a number of inequalities; specifically, that for each set S of types of mana (including the fictitious {P}), the number of mana symbols in A corresponding to a subset of S must be no more than the number of mana symbols in B corresponding to a subset of S. This gives us at most 2^7 inequalities to check (in fact given the symbols that actually exist one only needs to check 65 of them).

But the {2/M} symbols are more of a problem. Because they can correspond to 1 or 2 mana, the marriage theorem doesn't apply. One way to handle this is disambiguation -- each {2/M} symbol can be disambiguated to either {2} or {M}. Then A≤B iff for every disambiguation B' of B there's some disambiguation A' of A such that A'≤B'.

However, this is slow in the worst case, because it requires checking lots of disambiguations. My question is: Can we do better?

I hypothesize the following two statements that will help reduce the number of disambiguations needed when symbols of the form {2/M} are involved:

Cancellation applies. That is: If we define addition of mana costs in the obvious way, it's clear that $a\le b$ implies $a+c\le b+c$. If we don't allow symbols of the form {2/M}, the converse also holds, by the reasoning above. Question: Does it also hold when such symbols are allowed?

Suppose we have mana costs A and B and suppose M is a color such that the symbol {2/M} does not occur in A. Is it true then that A≤B if and only if A≤B', where B' consists of B but with all instances of {2/M} replaced by {1}? (

**EDIT**: Disproved in the comments by Pace Nielsen. It might still hold if we require A have no symbols of the form {2/M} for*all*colors M, but that weaker form doesn't form such a useful part of computing the order relation.)

(I don't think there's any way to get rid of the necessity of doing at least *some* disambiguations; if A has more of some {2/M} symbol than B, I think you are just going to have to handle those manually. But both the above statements at least would bring it down to only doing that.)

So far I haven't been able to prove these work, nor find any counterexample. Can anyone prove or disprove these?

In addition like I said I'd be interested to see generalizations. As long as each symbol corresponds to 1 mana the whole thing just comes down to the marriage theorem like I said above. But when symbols can correspond to varying amounts I wouldn't expect it to work so nicely in general. It does seem to still work nicely for the set of symbols that actually exist, as listed above (although I might be wrong!). Is there some abstract property of this set of symbols that causes this, so that we can say when this sort of thing happens? It would be interesting to see.

Thank you all!

Let $X$ be a smooth, projective variety and $Y \subset X$ a smooth, effective divisor. Consider now the natural map $i^\ast:H^2(X) \to H^2(Y)$. Then,

When is the image of $c_1(\mathcal{O}_X(Y)) \in H^2(X)$ under the map $i^*$ non-zero?

If $Z$ is another smooth projective variety containing $Y$ as an effective divisor (meaning $\dim Z=\dim Y+1$), is $j^*(c_1(\mathcal{O}_Z(Y)))=i^*(c_1(\mathcal{O}_X(Y)))$ where $j^*:H^2(Z) \to H^2(Y)$ is the natural pull-back map?

If ($2$) is not true in general, is it true if $Z$ is a deformation of $X$?

Any reference/idea regarding the questions will be most welcome.

Let $P$ be a convex polytope in $\mathbb{R}^3$ whose every vertex lies in the $\mathbb{Z}^3$ lattice.

**Question:** If $P$ contains exactly one lattice point in its interior, what is the maximum possible volume of $P$?

Notice that a convex lattice polytope with no interior lattice point can have arbitrarily large volume, but if the number of interior lattice points in it is a given finite number, then its volume cannot be arbitrarily large. In two dimensions the answer is known: A convex lattice polygon with exactly one interior point has volume at most $9/2$, which follows from a more general inequality by P.R. Scott (see *On Convex Lattice Polygons*, Bull. Austral. Math. Soc., vol 15; 1976, 395-399).

The best example I know in that respect is the tetrahedron with vertices $(0,0,0),\ (4,0,0),\ (0,4,0)$ and $(0,0,4)$, with volume ${32}\over3$. Is this perhaps the maximum volume? Of course, the same question can be asked in higher dimensions as well, with an analogous example to consider as a possible candidate for an answer.

In view of several examples presented in answers and comments below, it seems that the optimal polytope should be a simplex, in every dimension. Has this been conjectured already?

The purpose of this question is to express doubts about abstract methods in mathematics.

When I studied theoretical mathematics in Warsaw I participated in prof. Guzicki lectures "introduction to mathematics". I was having hard time listening to it. Probably on second lecture I didn't understand it. The reason is that the approach was to formalize everything which is defined in mathematics. Is it good approach ? In my opinion, no. What is the advantage of set theoretic definition of natural numbers ? Does it help ? In my opinion, no. I understand that somewhere in begining of twentieth century mathematicians have discovered that the ground is a bit shaky. Namely some theorems appeared to be independent of the axioms. Still in my opinion, we can do mathematics having hope that someone later will fill the gap of axiomatization (if it is needed at all).

I admit that some abstract ideas are useful. Nevertheless I would like to collect examples when focus on abstract idea make given subject less clear.

Second example is manifolds. It turned out that some facts about manifolds can be proved in PL (piecewise linear) category, some in smooth category and some in pure topological category. Still we can investigate the structure of manifold as it is without going to the PL or smooth category. Say, I want to organize 3-manifolds by its complexity using fundamental group or torsion. I can analyze some examples using surgery on links or by gluing faces of polyhedron. At this moment I do not need to refer specifically whether I work in PL or smooth category.

Next example is about classification of finite simple groups. According to wikipedia "The proof of the classification theorem consists of tens of thousands of pages in several hundred journal articles written by about 100 authors, published mostly between 1955 and 2004". I can work with finite simple groups assuming that they are classified without knowing the details of the proof. There are so many problems still to be solved. I am sure that someone in the future will find shorter proof for classification theorem. It seems that using pure group-theoretic methods is not enough. Should we rather look at finite groups as part of geometry or other algebraic structure ?

Last example is about algebraic topology. When I was about to graduate, I was offered subject about classifying spaces $BG$ by prof. Jackowski for my master thesis. I rejected for personal reasons. Why mathematics analyse such abstract objects as classifying spaces ? I can explain to my colleague - not mathematician - what is Lie group. I can talk about rotations of 3-dimensional space. I can explain what is manifold. I can say - look at the surface of ball or torus - this is manifold. How can I explain, what is $BG$ ? They say that the purpose of algebraic topology is to analyse topological properties by algebraic tools. Don't believe them ! This discipline flies away in unknown direction.

Therefore I am planning to open small workshop where you can do mathematics with hand tools like in old times :)

My question now is how abstract mathematics should be ?

As in title. Why is (-a/b) + c is not equal to (-a + bc) / b but to -(a - b c)/b ? I can't really see it.

Can anybody tell me how to choose 6 numbers that have a range of 4, a median of 9, a mean of 9 and a mode of 7?

so far, i have 7,7,9,11

thanks

Let $\mathcal{C} = \mathcal{C}(\mathbb{R}_+, \mathbb{R})$ be the set of real-valued continuous functions on $\mathbb{R}_+$ metrised by the convergence on any compact sets, that is: $$d(w, \tilde{w}) = \sum_{k = 0}^{\infty}{\frac{1}{2^k}\frac{\delta^k(w, \tilde{w})}{1+\delta^k(w, \tilde{w})}}, ~~\delta^k(w, \tilde{w}) = \sup_{t \in [0, k]}{|w_t - \tilde{w}_t|}$$

Let $g: \mathbb{R}_+ \rightarrow \mathbb{R}$ be a bounded borel function (we can assume much more regularity on $g$ if needed), $a \in \mathbb{R}$ and let $G$ be the subset of $\mathcal{C}$ defined by: $$ G := \{w \in \mathcal{C}: \lim_{T \rightarrow \infty}\frac{1}{T}\int_0^T{g(w_s)ds} = a \} $$

My question is the following: Is $G$ a closed set? If not, what is $\partial G$, the boundary of $G$?

Thanks a lot for any help!

Suppose $(P,\leq)$ is a poset without maximal elements. For $X\subseteq P$ we set $X^u = \{p\in P: p \geq x \text{ for all } x\in X\}$ and call this the *set of upper bounds of* $X$. We say that $B\subseteq P$ is *unbounded* if $B^u = \emptyset$. Moreover we say $D\subseteq P$ is *dominating* if for all $p\in P$ there is $d\in D$ such that $p\leq d$. We set

- ${\frak b}(P) = \min\{|B|: B\subseteq P\text{ is unbounded}\}$, and
- ${\frak d}(P) = \min\{|D|: B\subseteq P\text{ is dominating}\}$.

It is easy to see that for all posets $P$ without maximal elements we have ${\frak b}(P) \leq {\frak d}(P)$. What is an example of a poset $P$ in which we can prove in $\textsf{ZFC}$ that ${\frak b}(P) < {\frak d}(P)$?

It seems that as $n$ increases, the ratio $$\frac{\varphi(2^n-1)}{2^n-1},$$ where $\varphi$ denotes the Euler totient function, takes on values reasonably often in the interval $(.3,.4)$.

Is there anything known about $$\lim \inf_{n \rightarrow \infty}\frac{\varphi(2^n-1)}{2^n-1}?$$

Given two sets as follows:

for A, B, and C

H: [(A x B) -> C]

G: [A -> [B -> C]]

Is there bijective function in G -> H?

What are some known examples of finite-dimensional integrable systems with symplectic Fano phase space?

Here by integrable system we mean a symplectic manifold $(X, \omega)$ of dimension $2n$ with $n$ independent Poisson-commuting functions ('integrals').

The question is whether $X$ admits an almost complex structure $J$ compatible with the symplectic structure (i.e. $\omega(Jv, Ju)=\omega(u, v)$ and $\omega(u, Ju)>0$ for non-zero $u$) such that $\int_{A}c_1(TX)>0$ for homology classes $A \in H_2(X, \mathbb{Z})$ representable by $J$-holomorphic curves (I am quite ignorant of integrable systems with non-Kaehler phase space; that's why I used a symplectic definition without referring to Kaehler structure).

My motivation is largely geometric; namely such an integrable system would endow the phase space with a Lagrangian fibration (Liouville-Arnold).

https://arxiv.org/abs/1605.09736 One example comes from compactified Ruijsenaars-Schneider systems whose phase space is $\mathbb{C}P^n$.

https://arxiv.org/abs/1005.2006 Another example is Gelfand - Zeytlin system whose phase space is flag variety $F_3$.

Some examples of non-K\"ahler symplectic Fano manifolds were constructed by Fine-Panov: https://arxiv.org/abs/0905.3237

Suppose I have two orthogonal matrices $Q, R$ of size $n \times n$ each, so that each of the three matrices $Q, R, Q^\top R$ has the property that at least $(1 - \epsilon)n$ of its eigenvalues are real, i.e. are $\pm 1$. (Notice that for $Q^\top R$, an eigenvector $v$ of eigenvalue $\pm 1$ equivalently satisfies that $Qv = \pm Rv$.)

Is there a set of orthonormal $v_1, \dots, v_k$ (say, for $k$ a constant fraction of $n$) so that $Q v_i \approx \pm v_i$ and $R v_i \approx \pm v_i$, i.e. so that $Q$ and $R$, which are both reflections and both agree up to a sign flip on large subspaces, both act simultaneously as approximate reflections on a common partial orthonormal basis?

I have the following expression that I would like to find a tight (or as tight as possible) simplified lower bound in terms of $n$, $l$ and $p$:

$f(n, l, p) = \min_{D} \sum_{(x_1, \dots, x_m) \in D}\frac{x_i}{10}\left(p\left(\frac{p(n/p-1)}{p + 1 -\frac{x_i}{10}}\right)^{n/p - 1}\right)$

where $D$ is defined as follows:

$D = \left\{(x_1, \dots, x_m)|m>\frac{n}{100p}, 1 \leq x_i \leq 10p \text{ for } 1 \leq i \leq m \text{ and } \sum_{1 \leq i \leq m} x_i > n/10\right\}$.

The following are the bounds on the values of $n$, $l$, and $p$: $n > p \geq l \geq 1$.

It seems like the correct (trivial) answer is that this expression is minimized when $m > n/10$ and all $x_i = 1$ (since this satisfies the sum requirement for $x_i$) because this minimizes the denominator of the exponential expression. If this is the case, then the lower bound would be (approximately)

$f(n, l, p) \geq n/100\left(p\left(n/p-1\right)^{n/p-1}\right)$.

But I somehow cannot prove that this. Help greatly appreciated.

Let $R$ be a ring. An endomorphism $\alpha:R\to R$ is said to be right central reflexive if for all $a,b\in R,$ $aRb=(0)\implies bR\alpha (a)\subset Z(R)$, where $Z(R)$ denotes the centre of the ring.

The ring $R$ is called right $\alpha$-central reflexive if $\alpha$ is a right central endomorphism for $R$.

Let us consider a statement as follows:

For all $a,b\in R$, $aR\alpha (b)=(0)\implies bRa\subset Z(R)$.....(*)

It can be proved that if $\alpha:R\to R$ is onto, $R$ is right central $\alpha$-reflexive implies (*) is true. I want to show show that surjectivity of $\alpha$ is not superfluous. But I could not find an into endomorphism $\alpha$ such that $(*)$ holds but $R$ is not a right central $\alpha$-reflexive ring.

Please help me in this regard.

I have known that $b_2(CP^n\sharp CP^n)=2$, however I have no idear how to prove this fact ! I appreciate any help for this simple question! Thank you!

Consider two standard normal random variable, X and Y.

They both have mean 0, and variance 1. But we don't know their dependency. Is it possible for X+2Y to be nonsymmetric?

In another word, is it possible for P(X+2Y>0) = 1/2 to not hold.

I understand if they follow multivariate normal, the sum has to be normal and thus symmetric.

Is it possible to construct non-jointly Gaussians X and Y such that the equality does not hold.

$\DeclareMathOperator{\Spec}{Spec}$
The following is from Bruhat and Tits article *Groupes Reductifs sur un Corps Locale II*. $A$ is an integral domain. Here $A$-scheme means "affine $A$-scheme," and $A[\mathfrak X]$ denotes the coordinate ring of $\mathfrak X$. This is a definition of what it means for an $A$-scheme to be "smooth" (I believe this is how *lisse* is meant to translate).

I understand being flat and finitely presented. But I wanted to clarify the last condition about the fibers. If $\mathfrak p$ is a prime of $A$, then the fiber $\mathfrak X_{\mathfrak p}$ is the $\kappa(\mathfrak p)$-scheme $\mathfrak X \times_{\Spec A} \Spec \kappa(\mathfrak p)$, where $\kappa(\mathfrak p) = A_{\mathfrak p}/\mathfrak p A_{\mathfrak p}$.

A geometric point of $\mathfrak X_{\mathfrak p}$ is a morphism of schemes $\Spec (k) \rightarrow \mathfrak X_{\mathfrak p}$, where $k$ is an algebraically closed field. In other words, a geometric point is a point $x$ of $\mathfrak X_{\mathfrak p}$, together with an embedding of the residue field of $\mathfrak X_{\mathfrak p}$ at $x$ into an algebraically closed field.

What is the definition of a "simple geometric point" of $\mathfrak X_{\mathfrak p}$? If all the geometric points are simple, is this the same as saying that for an algebraic closure $\overline{\kappa(\mathfrak p)}$ of $\kappa(\mathfrak p)$, all the stalks at all the points of the scheme

$$\mathfrak X_{\mathfrak p} \times_{\Spec \kappa(\mathfrak p)} \Spec \overline{\kappa(\mathfrak p)}$$

are regular local rings? Is there any requirement that the $\kappa(\mathfrak p)$-scheme $\mathfrak X_{\mathfrak p}$ be reduced? Or geometrically reduced?