For any $k \ge 3$ construct a non hamiltonian, connected k-regular bipartite graph. I have tried to find such graphs for small $k$-s but i got nothing. Can anybody help?

The dimensions of some representations of the Janko group J1 coinside with dimensions of smallest representations of the Lie algebra of type E7 (56, 133). It seems to be natural that there is a subgroup of the Lie group of type E7 isomorphic to J1. Is it true?

I hope this is a question that fits here and is not duplicated. Also that is clear since it can be a little ambiguous.

I was wondering if you know deep expressions, theorems, isomorphisms or symmetries in mathematics involving *weird* constants or quantities.

For example, the only negative integers $d$ such that the ring of integers of $\mathbb{Q}(\sqrt{d})$ is a unique factorization domain cannot be different from the ones in the set $\{-1,-2,-3,-7,-11,-19,-43,-67,-163\}$.

Another example is that the group of automorphisms, via orientation-preserving conformal mappings, of a compact Riemann surface of genus $g > 1$, has cardinality less or equal than $84(g − 1)$.

One more, are the maximum number of singularities in quintic, sextic, heptic and octic surfaces being $31,65,104$ and $174$ respectively.

A friend just showed me this *Mumford isomorphism* involving a $13$-th, tensor product power of line bundles:

http://www.concinnitasproject.org/portfolio/gallery.php?id=Mumford_David

I would appreciate also some literature for this or maybe an idea of this isomorphism.

Let

- $(M,d)$ be a metric space
- $\Lambda\subseteq M$
- $E$ be a $\mathbb R$-Banach space

Moreover, let $$\left\|f\right\|_{B(K,\:E)}:=\sup_{x\in K}\left\|f(x)\right\|_E\;\;\;\text{for }f:\Lambda\to E$$ for $K\subseteq\Lambda$ and $$B_{\text{loc}}(\Lambda,E):=\left\{f:\Lambda\to E\mid\left\|f\right\|_{B(K,\:E)}<\infty\text{ for all compact }K\subseteq\Lambda\right\}$$ be equipped with the topology induced by $\left\{\left\|\;\cdot\;\right\|_{B(K,\:E)}:K\subseteq\Lambda\text{ is compact}\right\}$. Since $E$ is complete, any Cauchy sequence in $B_{\text{loc}}(\Lambda,E)$ is convergent. Moreover, $C^0(\Lambda,E)$ is a closed subspace of $B_{\text{loc}}(\Lambda,E)$.

Now, assume that $\Lambda$ is separable and let $\mathcal C^0(\Omega;\Lambda,E)$ denote the set of $F:\Omega\times\Lambda\to E$ with

- $F(\;\cdot\;,x)$ is $\mathcal A$-measurable for all $x\in\Lambda$
- $F(\omega,\;\cdot\;)\in C^0(\Lambda,E)$ for all $\omega\in\Omega$

Mimicking the proof of the result described above, we obtain that if $(F^n)_{n\in\mathbb N}\subseteq C^0(\Omega;\Lambda,E)$ with $$\left\|F^m-F^n\right\|_{B(K,\:E)}\xrightarrow{m,\:n\:\to\:\infty}0\;\;\;\text{in probability for all compact }K\subseteq\Lambda\tag1,$$ there is a $F\in C^0(\Omega;\Lambda,E)$ with $$\left\|F-F^n\right\|_{B(K,\:E)}\xrightarrow{n\:\to\:\infty}0\;\;\;\text{in probability for all compact }K\subseteq\Lambda\tag2.$$

**Question**: If $M=\mathbb R$, $\Lambda=[0,\infty)$ and each $F^n$ is a local martingale on a common filtered probability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\ge0},\operatorname P)$, are we able to conclude that $F$ is a local martingale as well?

If this question is a duplicate / the answer already exists I am very sorry.

I recently started working with vectors in 3 dimensions. My visualisation of the dot product is the following:

Lets say we have the vector v(0,0,1) with the origin O. The dot product between v and the vector with the origin in O and end in P is: positive if P is situated anywhere "above" O and negative otherwise. It's like O would be the centre of a sphere and P would be situated in the upper/lower hemisphere. The radius is infinite. It surely sounds pretty stupid to smarter people but this is how I see it.

Could you point me so something similar regarding the cross product?

Thank you very much!

Let $f: R \to R$ be a distribution function. Consider Fourier Transform $\hat f=\frac{1}{\sqrt{2 \pi}}\int_Rf(y)e^{-ixy}dy, \quad x \in R$. Let $F$ be an operator defined as $$ (Ff)(t)=\int_0^{t-1}f(s)ds, t\in R_+. $$ Show that boundless of $F$ depends on the boundless of $\hat f$.

My question is in the title, but here is a more detailed formulation:

Let Top be the category of all topological spaces and continuous maps, and let CGTop be the subcategory of compactly generated spaces, as defined in Strickland's notes. We have a functor $k: Top \rightarrow CGTop$ that replaces the topology on a space by the associated compactly generated topology.

If X is a simplicial topological space (a simplicial object in the category of all topological spaces) is the identity map $|kX|\rightarrow k|X|$ a homeomorphism?

Here $|-|$ denotes geometric realization. I'm actually most interested in the "thick" realization (where the defining equivalence relation uses only the face maps), but the question makes sense for both the thick and thin versions.

It follows from the results in the above notes that $|kX|$ is compactly generated (being a quotient of a disjoint union of compactly generated spaces), and that the identity map $|kX|\rightarrow k|X|$ is continuous.

Here is a more general question: if $X$ is a topological space and $\sim$ is an equivalence relation on $X$, then $(kX)/{\sim}$ is compactly generated and the identity map $(kX)/{\sim} \to k(X/{\sim})$ is continuous. Is this map always a homeomorphism?

Suppose $\mathbf{v},\mathbf{w} \in \mathbb{R}^n$ (and if it helps, you can assume they each have non-negative entries), and let $\mathbf{v}^2,\mathbf{w}^2$ denote the vectors whose entries are the squares of the entries of $\mathbf{v}$ and $\mathbf{w}$.

My question is how to prove that \begin{align*} \|\mathbf{v}^2\|\|\mathbf{w}^2\| - \langle \mathbf{v}^2,\mathbf{w}^2\rangle \leq \|\mathbf{v}\|^2\|\mathbf{w}\|^2 - \langle \mathbf{v},\mathbf{w}\rangle^2. \end{align*}

Some notes are in order:

- The Cauchy-Schwarz inequality tells us that both sides of this inequality are non-negative. Thus the proposed inequality is a strengthening of Cauchy-Schwarz that gives a non-zero bound on the RHS.
- I know that this inequality is true, but my method of proving it is extremely long and roundabout. It seems like it should have a straightforward-ish proof, or should follow from another well-known inequality, and that's what I'm looking for.

Let $D$ be a finitely connected, bounded domain in $\mathbb C$. Suppose that the boundary of $D$ consists of finitely many, pairwise disjoint, piecewise $C^1$-Jordan curves. Does there exist an exhaustion sequence $(K_n)$ of compacta in $D$

with $K_n\subseteq K_{n+1}^\circ$ and $C^1$-maps $\phi_n:\mathbb C\to [0,1]$ such that $\phi_n=1$ on $K_n$, $\phi_n=0$ on $\mathbb C\setminus K_{n+1}$ and
$$\iint_D|\overline\partial \phi_n| dx dy\leq C$$ for every $n$.

Where did I go wrong here?

As part of my worksheet for uni, I had to calculate the sum of the series $\sum_{k=2}^\infty (k-1)*(1-p)^{k-2}$. ($p \in (0,1)$ denotes a probablity.) According to the German Wiki page on Geometric Series (relevant formula anchored in link, language not relevant), the series can be solved by differentiating. The result turns out to be $\frac{1}{p^2}$, which I know to be correct.

My initial calculations as detailed below, however, led to the incorrect result of $\frac{1}{p*(1-p)}$. I would like to know where exactly I went wrong.

Let $S$ denote the sum of the series.

\begin{eqnarray} S &:=& \sum_{k=2}^\infty (k-1)*(1-p)^{k-2} \\ &=& (2-1)*(1-p)^{2-2} + (3-1)*(1-p)^{3-2} + (4-1)*(1-p)^{4-2} + \dots \\ &=& 1*(1-p)^0 + 2*(1-p)^1 + 3*(1-p)^2 + \dots \\ \Leftrightarrow (1-p)*S &=& 1*(1-p)^1 + 2*(1-p)^2 + 3*(1-p)^3 + \dots \\\\ \Leftrightarrow (1-p)*S &=& (1*(1-p)^0 - (1-p)^0) + (2*(1-p)^1 - (1-p)^1) \\ &&+ (3*(1-p)^2 - (1-p)^2) + (4*(1-p)^3 - (1-p)^3) + \dots \\\\ \Leftrightarrow (1-p)*S &=& (1*(1-p)^0 + 2*(1-p)^1 + 3*(1-p)^2 + 4*(1-p)^3 + \dots) \\\\ && -((1-p)^0 + (1-p)^1 + (1-p)^2 + (1-p)^3 + \dots) \\\\ \Leftrightarrow S*(1-p) &=& S - \sum_{i=0}^\infty (1-p)^i \\ &=& S - \frac{1}{1-p} \\ \Leftrightarrow S*((1-p)-1) &=& - \frac{1}{1-p} \\ \Leftrightarrow S &=& - \frac{1}{(1-p) * (-p)} = \frac{1}{p*(1-p)} \end{eqnarray}

Any help is greatly appreciated! Thanks in advance!

Let $K$ be an extension field of $\mathbb{Q}_p$, let $O$ be the ring of integers of $K$, and let $P$ be the maximal ideal of $O$.

If $K$ is a finite extension of $\mathbb{Q}_p$, there is the well-known algebraic isomorphism

$$O[[X]]\cong\varprojlim O[X]/((1+X)^{p^n}-1).$$

A proof of this fact can be found, for example, in Lang's "Cyclotomic fields", in Washington's "Introduction to Cyclotomic Fields", and in Neukirch et al's "Cohomology of number fields". In all of this proofs, at some point one needs a generator of the maximal ideal $M$, that is, a uniformizer element of $K$. Hence, *a priori*, this proofs do not work if we let $K$ to be a non discretely valued extension.

Now, in page 96 of Lang's "Cyclotomic fields I and II", he claims (and uses) that this isomorphism is true when one let $K=\mathbb{C}_p$, the completion of the algebraic closure of $\mathbb{Q}_p$. My question is: **where can I find a proof of this?**, or, **why is this obviously true?**, since he only proves this isomorphism for $K=\mathbb{Q}_p$.

Note: Since I do not know too much about (the foundations of) Iwasawa theory, I do not know if this is an elementary question. If it is, please let me know and I will ask it in MSE.

Let us assume that we work over the complex field and let $X$ be a smooth projective variety and $\pi: P \to X$ a projective bundle (i.e. a fibration in projective spaces of constant dimension). Let $s : X \dashrightarrow P$ be a rational section. When does $s$ extend to a regular section?

As far as I can see, if the image $s(X)$ is contained (by taking the closure?) in a "horizontal" closed subset $W\subset P$ which is bijectively sent by $\pi$ to $X$, then I am done since $\pi$ is a bijective birational morphism $W \to X$ and by Zariski's main theorem it is an isomorphism. I am not sure though whether $s(X)$ can be completed in such a way, maybe over a curve or a surface it is ok. Is there another way to show this?

As someone has remarked here below, the case of curves is clear. I am interested in a case where $S$ is a smooth surface.

Let $X \sim \text{Binom}(n, p)$ a binomial random variable. I want to show that : $$\forall n >1, \quad \forall 0 < t < 0.9, \quad \exists C, \quad \mathbb P\bigg(|X-np| \leq C\sqrt{n}\bigg) \geq t$$

I want to prove the result for all $n$, so pure asymptotic is not enough.

**1-** Using the normal approximation together with a Berry-Essen bound we can prove that : $$\mathbb P\bigg(|X-np| \leq x\sqrt{p(1-p)}\sqrt{n}\bigg) > \Phi(x) - \Phi(-x) - \frac{2C(p^2 + (1-p)^2)}{\sqrt{np(1-p)}}
$$
Taking $x = 3$ and $C = .4215$, this yield (part of) the result when $np(1-p)$ is not too small, for instance $np(1-p) \geq 1$ and $n \geq 1000$. Remains to prove for $n < 1000$ and small $p$.

**2-** When $np$ is small, we have a great bound on the absolute error for the Poisson approximation : $$\bigg| \text{Binom}(n, p) - \text{Pois}(np) \bigg| \leq p(1 - e^{-np})$$

But I can't show the result for a Poisson random variable.

Any help, either with the Poisson r.v. or using another approach would be greatly appreciated.

Footnotes

- The value for $C$ comes from Nagaev, Chebotarev (2010).
- I said $t < 0.9$ but anything that generalizes to $t < 0.99$ and beyond is great, obviously we can't go as far as $t = 1$.

Let $n>1$ be an integer. We say that two points $(x_1,\ldots,x_n),(y_1,\ldots,y_n)\in\mathbb{Z}^n$ are a member of the edge set $E_n$ if and only if $$\sum_{i=1}^n|x_i-y_i| = 1.$$

**Question.** Given an integer $n>1$, is there a maximal integer $m(n)$ such that the complete graph $K_{m(n)}$ is a minor of $(\mathbb{Z}^n, E_n)$? If yes, it would also be nice to know an explicit formula for $m(n)$.

**Remark.** I believe that for $n=2$ there is such a maximal integer, and $m(2) \in \{3,4\}$. I am fairly certain that $m(2) >4$ would allow to construct a counterexample to the 4-color theorem.)

**Edit.** In the comment section, Wojowu shows that $m(2)=4$.

If every nonzero $R$-module has a nonzero flat submodule, prove that $R$ is a von Neumann regular ring.

Thinking about the recent threads on structural consequences of the Axiom of Foundation (AF) over ZF-AF, I've been trying to find some conservativity result which explains why AF doesn't seem to have consequences in "ordinary mathematics." Here's a candidate that came to mind.

Let's call a sentence *fundamentally* $\Pi_2$ if it is of the form $T \models \varphi,$ where $T$ is a recursively enumerable theory in second order logic and $\varphi$ is in the language of $T.$ Note that over ZF, the fundamentally $\Pi_2$ sentences are just the $\Pi_2$ sentences.

Is ZF conservative over ZF-AF with respect to fundamentally $\Pi_2$ sentences?

Some simple observations:

- ZFC is conservative over ZFC-AF with respect to fundamentally $\Pi_2$ sentences, since any model of a second order theory is isomorphic to one in the well-founded part of the universe by the well-ordering theorem.
- ZF is conservative over ZF-AF with respect to fundamentally $\Sigma_2$ sentences (defined in the obvious manner), by considering the well-founded part of the universe. Same goes for ZFC over ZFC-AF.
- ZF+$\neg$ AC is
*not*conservative over ZF-AF+$\neg$ AC with respect to fundamentally $\Sigma_2$ sentences, since only the former proves that there is a linearly ordered set which is not well-orderable. - One cannot construct a counterexample to my conjecture using permutation models of ZFA, since fundamentally $\Sigma_2$ sentences can be transferred to a symmetric model of ZF by the Jech-Sochor Theorem.

Suppose we have a matrix $A \in \{0,1\}^{n \times n}$ where

$$A_{ij} = \begin{cases} 1 & \text{with probability} \quad p\\ 0 &\text{with probability} \quad 1-p\end{cases}$$

I would like to know the expected value

$$\mathbb E( \det (A) =0) \ \text{ for large }n,$$

i.e., the asymptotic probability of the determinant being zero as $n$ becomes large.

I know that the expected value

$$\lim_{n \rightarrow \infty} \mathbb E \left( \det (A) =0 \right) = 0$$ if $p \neq 0,1$ but could not find results for finite $n$/the asymptotic behavior.

Let $(M, g)$ be a compact Riemannian manifold.

Assume that $u_0$ is a positive smooth function on $M$ and let $u_t = e^{t \Delta} u_0$ be the solution to the heat equation on $(M, g)$ with initial data $u_0$.

Given $p > 1$, is it true that the function $$ t \mapsto \int_M (u_t)^p $$ is a convex function?

This question gives an answer to Functional decaying under the heat flow (?) in the case $p=2$.

Can an uncountable model of Peano Arithmetic be recursive?

What does it mean for an *uncountable* model to be recursive? Well, we represent the elements of the model using *real numbers* instead of natural numbers, and assume $+, \times, ^{-1}$ are computable functions and $\ge$ is a computable relation on the real numbers. (In particular, we could assume we are using lambda calculus, and add symbols for $+, \times, \ge, ^{-1}$, as well as a symbol for each computable real number. Or you could use some other model of Real computation.)

So an uncountable model $M$ of Peano Arithmetic is recursive if $\mathbb N_M \subseteq \mathbb R$ is a computable set, and $+_M, \times_M$ are computable.

In response to a request in my now closed post “Are there paradoxes in ZF + (the Axiom of Choice for
finite sets)?”, Alex Kruckman gave this reference to a proof that the Axiom of Choice (AC) for finite
index sets {i} is a theorem of Zermelo-Fraenkel set theory (ZF):
Theorem 8.1.2 on p. 139 of Hrabacek & Jech *Introduction to Set Theory*.
I looked at that (only
supposed?) proof and found that it was partially based on a simple, totally obvious observation which I had
mentioned in my post as possibly providing a proof in ZF of not only AC for finite index sets, but of
general AC, for arbitrary index sets. In my post, I rejected this possible proof as defective on the
grounds that it didn’t specify the values f(x) of the choice function f, so didn’t establish its existence.
Although in the post I surrendered to the intuitionist, constructivist requirement that an existence proof for such an f
must specify f constructively in order to be a valid proof, I am not really an intuitionist. I actually rejected
my possible proof mostly because it would contradict the Godel & Cohen theorems which together state
the independence of AC from ZF (whose proofs I haven’t read).

The proof(?) in H & J used ordinary induction on the (finite) size |S| of the index set {i} of the sets X $\in$ S, together with the fact that each non-empty set X, by definition of “non-empty”, contains some element x (this was the mentioned simple observation in my post). The proof assumed AC for S of size |S| $\leq$ n, and than let [S be such that] |S| = n + 1. (H & J neglected to initially show AC for |S| = 1, or any other n.) The proof said fix X $\in$ S; then the set S - {X} has n elements, so by the induction hypothesis, it has a choice function gx. Then Gx [my notation] = gx U {(X,x)} is a choice function for S (for any x $\in$ X). (This rendition of the proof ignores some typos and other oddities in the proof in my eBook copy of H & J from Amazon.)

In this “proof”, the induction on |S| is totally unnecessary; one could instead just point out that for each
X(i) $\in$ S there is an x(i) $\in$ X(i), so define f(x) = {(X(i),x(i))| X(i) $\in$ S, x(i) $\in$ X(i), i $\in$ {i}}. Then f is a
choice function for S. (This was the rejected possible proof in my now closed post.) However, this
“proof” doesn’t depend on |S| being finite. If it works at all, it works for arbitrary sets S. Assuming
Godel-Cohen is correct, it must be defective. How? Where? Why? (**Edit**: There seems to be some misunderstanding here. The "it" in the previous sentence refers to my supposed proof that AC is a theorem of ZF , not Godel-Cohen's that it isn't, which fact should have been obvious from my use of quotation marks around "proof" when referring to mine, not G-C's. Even more obviously, it is unlikely that my "Assuming Godel-Cohen is correct," would preface my saying that G-C's proof(s) must be defective. I was asking how, where, why **my** supposed proof is defective. Maybe I should have used a semi-colon instead of a period before the "assuming" of the sentence this parenthetical edit is immediately after, to link that sentence more closely with the sentence before it and make even clearer what the referent of the "it" is.) Also, if it (my supposed proof) is defective, surely the
supposed proof for finite |S| in H & J also is defective. The “proof” in H & J uses the fact that there is, for
the given |S| = n + 1, an x $\in$ X (without specifying the x) for just one X and one x, but this doesn’t make
that “proof” valid even though mine is invalid. One is allowed in ZF to universally quantify over sets X(i) $\in$ S
and existentially quantify over x(i) $\in$ X(i), as is done in the definition of f(x). It is necessary to
universally quantify over sets (systems) S and sets X $\in$ S, and existentially quantify over functions
f for each S, just to state AC.

(Asaf Karagila- Do you now understand that I **was** asking "Where was my mistake?" You seem not to understand English all that well. Also, I edited my post to use LaTex, just for you.)