Recall that a cardinal $\kappa$ is $(\lambda,\infty)$-almost-strongly-compact if every $\kappa$-complete filter can be refined to a $\lambda$-complete ultrafilter. A cardinal $\mu$ has the *tree property* if every $\mu$-sized tree with $\mu$-small levels has a branch of length $\mu$. (If in addition $\mu$ is inaccessible then $\mu$ is weakly compact.)

**Question:** Can the following constellation occur?

$\mu$ -- weakly inaccessible with the tree property

$\kappa$ -- a $(\mu^+,\infty)$-strongly-compact cardinal

every regular $\nu \in [\mu, \kappa)$ -- has the tree property.

I suspect this would be too good to be true. But I don't know much -- for all I know, maybe almost strong compactness implies inaccessibility, in which case of course the answer is *no*. But I'm having trouble tracking down even that information.

If $\kappa$ can be taken to be $(\mu,\infty)$-strongly-compact, that might be good enough for what I need. Also it should suffice for only the successor cardinals in $(\mu,\kappa)$ to have the tree property.

I apologize for the repeated changes to the question.

Let $D(\mathbb R) $ be the set of all differentiable functions $f: \mathbb R \to \mathbb R$. Then obviously $D(\mathbb R)$ forms a semigroup under usual function composition. Can we characterize (up to semigroup isomorphism) all finite subsemigroups of $D(\mathbb R)$ which do not contain any constant function ?

Let $Pr^L$ be the category of presentable categories and left adjoint functors (probably this should be at least a (2,1)-category; anyway ultimately I'm interested in the $(\infty,1)$-setting). For each regular cardinal $\kappa$, let $Pr^L_\kappa$ be the non-full subcategory of locally $\kappa$-presentable categories and left adjoint functors which preserve $\kappa$-presentable objects (equivalently, which have $\kappa$-accessible right adjoints).

I believe that limits (PIE limits in the ordinary setting) in $Pr^L$ are computed at the level of the underlying category, and that $Pr^L_\kappa$ is closed in $Pr^L$ under $\kappa$-small limits (at least for $\kappa > \omega$). Moreover, an arbitrary product of objects of $Pr^L_\kappa$ is again in $Pr^L_\kappa$ and the projection functors are even in $Pr^L_\kappa$. However, if $(F_\alpha : C \to D_\alpha)_\alpha$ is a family of functors in $Pr^L_\kappa$, the induced functor $F: C \to \Pi_\alpha D_\alpha$ is *not* typically in $Pr^L_\kappa$ when the product is $\kappa$-sized or larger.

Thus $Pr^L_\kappa$ is not closed in $Pr^L$ under small limits. This raises the following

**Question:** Is $Pr^L$ generated under small limits by $Pr^L_\kappa$ for some $\kappa$? How about $\kappa = \omega$?

Notes:

I believe that $Pr^L_0$ (the category of preseheaf categories)

*is*closed under limits in $Pr^L$, so I think the smallest candidate $\kappa$ in the above question is $\kappa = \omega$. This is what I meant in the question title, where "finitary left adjoint" is meant to indicate "morphism in $Pr^L_\omega$".Exactly what it means to be generated under limits by a non-full subcategory is a bit unclear. But let's at least stipulate that if every object of $Pr^L$ is a limit of a diagram in $Pr^L_\kappa$, then the answer to the question is

*yes*. I suppose I don't even know, for any $\kappa$, if the full subcategory of $Pr^L$ on the objects of $Pr^L_\kappa$ generates $Pr^L$ under limits in the usual sense, so figuring that out would be a good start.**EDIT:**Another consequence of an affirmative answer to the question (which might be easier to think about) would be the following: every presentable category would be coreflective in a $\kappa$-presentable category for the $\kappa$ in the question. This can't happen with $\kappa = 0$, but maybe it can happen for $\kappa = \omega$.**EDIT:**Here's a data point. Let $CMet \in Pr^L_{\omega_1}$ be the category of complete metric spaces and contractive maps. Let $PMet \in Pr^L_\omega$ be the category of pseudometric spaces and contractive maps. For each $\delta \geq 0$ there is a functor $\delta_\ast: PMet \to PMet$ preserving underlying sets with $d_{\delta_\ast X}(x,y) = min(d_X(x,y) - \delta, 0)$. These form an inverse system for $\delta > 0$, and I believe the limit of this inverse system is none other than $CMet$.For context, this question arose from reflecting on KotelKanim's question.

It is known that we may choose smooth $f:\mathbb R \to [0,1]$ such that $f(x)=1$ if $x\geq \frac{3}{4} $ and $f(x)=0$ if $x\leq -\frac{3}{4}+1.$

Define $h(x)= \sin (\frac{\pi}{2} f(x+1))$ if $x\leq 0$ and $h(x)= \cos (\frac{\pi}{2} f(x))$ if $x\geq 0.$

We note that support of $h$ is $[-3/4, 3/4]$ and $h^2(x)+ h^2(x-1)=1$ for all $x\in [0,1],$ $\|h\|_{L^2}=1,$ and this $h:\mathbb R \to \mathbb C$ is smooth.

My Question: Can we expect to choose $h:\mathbb R \to \mathbb C$ such that support of $h$ is $[-3/4, 3/4]$ and $h^2(x)+ h^2(x-1)=1$ for all $x\in [0,1],$ and $$\|h\|_{L^2}^2=\int_{-3/4}^{3/4} |h(x)|^2 dx =3/2?$$

Let $X= x_1 + x_2 + \ldots + x_m$, $Y=y_1 + y_2 + y_3 + \ldots + y_n$, and $Y' = y'_1 + y'_2 + \ldots + y'_n$, where

- Each $x_i$ is a Bernoulli variable which takes value $1$ with probability $p_i>0$.
- Each $y_i$ is a Bernoulli variable which takes value $1$ with probability $q_i>0$.
- For $i>2$, each $y'_i$ is a Bernoulli variable which takes value $1$ with probability $q_i>0$.
- $y'_1$ and $y'_2$ are two Bernoulli variables which take value $1$ with probability $(q_1 + q_2)/2$ (Suppose that $q_1 \neq q_2$).
- All the variables are independent.

Furthermore, suppose that $E(X) \geq E(Y)$ and $m>n$. Now, let $A = Pr(Y\geq X)$ and $A' = Pr(Y' \geq X)$. The question is whether $A>A'$ or $A'>A$?

Let $P$ be a compact connected set in the plane and $x,y\in P$.

Is it always possible to connect $x$ to $y$ by a path $\gamma$ such that the length of $\gamma\backslash P$ is arbitrary small?

**Comments:**

Be aware of pseudoarc --- it is a compact connected set which contains no nontrivial paths.

If $P$ contains an everywhere dense curve, then the answer is "yes"; the same holds if $P$ contains a dense countable collection of curves.

Theorem 9.3.1 in Hall's group theory says: Let $G$ be a solvable group and $|G|=m\cdot n$, where $% m=p_{1}^{\alpha _{1}}\cdot \cdot \cdot p_{r}^{\alpha _{r}}$, $(m,n)=1$. Let $% \pi =\{p_{1},...,p_{r}\}$ and $h_{m}$ be the number of $\pi -$Hall subgroups of $G$. Then $h_{m}=q_{1}^{\beta _{1}}\cdot \cdot \cdot q_{s}^{\beta _{s}}$ satisfies the following condition for all $i\in \{1,2,...,s\}$.

$% q_{i}^{\beta _{i}}\equiv 1$ (mod $p_{j}$), for some $p_{j}$.

This question arises now that if we replace assumption solvable group with $p$-solvable group whether again Theorem is true. In the other words: Is it true the following claim? Or is there any counterexample for the claim?

Let $G$ be a $p$-solvable group and $|G|=p^{\alpha }\cdot n$ such that $(p^{\alpha },n)=1$($p\neq 2$). Let $% h_{m}$ be the number of Sylow $p-$subgroups of $G$. Then $h_{m}=q_{1}^{\beta _{1}}\cdot \cdot \cdot q_{s}^{\beta _{s}}$ satisfies the following condition for all $i\in \{1,2,...,s\}$.

$q_{i}^{\beta _{i}}\equiv 1$ (mod $p$).

Is there a place to watch ICM 2018 plenary lectures (and other lectures if possible)?

Here is the official Youtube channel of the ICM but they don't seem to be posting the lectures.

https://www.youtube.com/channel/UCnMLdlOoLICBNcEzjMLOc7w

**Update:** The public lectures of Etienne Ghys, Cedric Villani, and Ingrid Daubechies are up!

Is it true that for every even integer $N > 2$, there exist positive integers $a,b$ such that $a + b = N$ and $\lambda(a) = \lambda(b) = -1$ ?

Here $\lambda$ is the Liouville function.

How do we explicitly compute the curvature form $\Omega$ of the Levi-Civita connection $\nabla^{L.C.}$ for the $n$-sphere $S^n$?

Thus, how do we calculate the trace over the matrix logarithm $\log((\sigma_2 \otimes I_{n/2})^T\cdot\Omega)$ for $\sigma_2$ the second Pauli matrix, and $I_{n/2}$ the identity matrix of dimension $n/2$?

Any help would be much appreciated. Thanks in advance! MSE link.

I am trying to find references for the following boundary value problem: Assume that $\Omega$ is a compact 3-dim spin manifold with Dirac operator $D$ such that the boundary consists of two smooth surfaces $\Sigma_1,\Sigma_2$ which meet orthogonally. For a spinor $\psi$, we consider the following boundary value problem

$$D(\psi)=\Psi \text{ in } \Omega \qquad P_{\pm}(\psi):=(id\pm i\gamma(N))\psi=P_{\pm}\eta \text{ on } \partial\Omega$$ for smooth spinors $\eta,\Psi$ on $\Omega$. Here, $N$ denotes the outward normal of the boundary and $\gamma$ the Clifford multiplication. Does anyone know a reference where existence, uniqueness and regularity of such boundary value problems has been considered? Thank you in advance.

Let $C$ be the following curve in $\mathbb{C}^2$. \begin{align} & 11664\, {c_1}^3\, {c_2}^2 + 536544\, {c_1}^3\, c_2 + 6170256\, {c_1}^3 + 67068\, {c_1}^2\, {c_2}^2 + 1542564\, {c_1}^2\, c_2 \\ & + 3085128\, c_1\, {c_2}^2 - 32393844\, c_1\, c_2 + 3085128\, c_1 + 17739486\, {c_2}^2 + 6941538\, c_2 = 0. \end{align} I checked that this curve has genus $1$ using Sage. Therefore it is an elliptic curve. How to change coordinates such that the equation of this curve is of the form $y^2 = f(x)$, where $f$ is some polynomial. Thank you very much.

Let F(x, y, z) =−xi−yj+(z^3)k, and let the surface S be the part of the cone z=(x2+y2)^(1/2) between the planes z= 2 and z= 3. Do the following: (a) Show that z=g(x, y) is a smooth graph, then it can be parametrized by r(x,y) so that rx * ry = -dg/dx i - dg/dy j + k (b) Find ∫∫S F·dS

I was wondering if there was a systematic approach to this question: What numbers will two formulas make in common?

Example: "The formula for the numbers of 5x that 4x will make, is 20x."

This example is trivial just to get the point across clearly. It's easy to answer when the degrees are both 1. I'm more interested in when the polynomials are of different degrees. Also, the restrictions I had in mind were to keep everything to whole numbers, including, coefficients and exponents and inputs and outputs, and even only addition between terms of the polynomials. I was hoping someone had already done this work and I could just read that. If it's been done I'd be grateful if anyone out there could let me know.

I'd be happy to clarify the question in any way if it's not already. Thank you for any response.

If $n$ is in the range of the Euler totient function, certain multiples of $n$ are likewise guaranteed to be totient values. The simplest nontrivial example of this is that, if $n$ is in the range of totient, so is $2n$:

Write $n = \phi(k)$.

If $k$ is odd, then $2n = \phi(4k)$. If $k$ is even, then $2n = \phi(2k)$.

More generally, for all positive integers $m \leq 27$, I can determine whether or not the range of totient is carried to itself by multiplication by $m$:

$m=1$: This trivially preserves being a totient value.

If $m \geq 3$ and $m$ is odd, multiplication by $m$ does not preserve being a totient value, since $1 = \phi(1)$ but $m$ is not in the range of $\phi$.

$m=2$: See above. Note that since multiplication by $2$ preserves being a totient value, so does multiplication by $4$, $8$, or $16$.

$m=6$: Multiplication by $6$ preserves being a totient value:

As before, write $n = \phi(k)$.

If $3 \nmid k$, $6n = \phi(9k)$.

If $3 \mid k$, $3n = \phi(3k)$. Then, since $3n$ is a totient value, so is $6n$.

$m=10$: $110 = \phi(11^{2})$ but $110 \cdot 10 = 1100$ is not in the range of $\phi$.

$m=12$: Multiplication by $6$ and $2$ will carry totient values to totient values, so multiplication by $12$ will too.

$m=14$: $14$ is not in the range of $\phi$, so multiplication by $14$ does not carry $1$ to a totient value.

$m=18$: Multiplication by $18$ preserves being a totient value:

As before, write $n = \phi(k)$.

If $3 \nmid k$, $18n = \phi(27k)$.

If $3 \mid k$, $9n = \phi(9k)$. Then, since $9n$ is a totient value, so is $18n$.

$m=20$: Multiplication by $20$ preserves being a totient value.

As before, write $n = \phi(k)$.

If $5 \nmid k$, $20n = \phi(25k)$.

If $5 \mid k$, $5n = \phi(5k)$. Then, since $5n$ is a totient value, so is $20n$.

$m=22$: $22 = \phi(23)$ but $22 \cdot 22 =484$ is not in the range of $\phi$.

$m=24$: Multiplication by $12$ and $2$ will carry totient values to totient values, so multiplication by $24$ will too.

$m=26$: $1$ is in the range of $\phi$ but $26$ is not.

So my question is: Does multiplication by $28$ always carry totient values to totient values?

I have tried submitting a sequence to the OEIS consisting of positive integers multiplication by which preserves being a totient value, but I have not had enough terms for the sequence (or an algorithm for determining membership in the sequence). (Should I mention what the sequence number would be if it were approved? I also don't think extending the sequence by this single term would make it admissible, but it's good to have this problem sized up by people who can devote more time to it than I can.)

Let for each $1 \leq i \leq l$ $V_i$ be a finite dimensional $k$-space, and \begin{equation} f : \times_{i=1}^l V_i \longrightarrow k \end{equation} a multihomogeneous polynomial map of degree $d_i$ in each space $V_i$. Is there a polynomial map \begin{equation} F : \times_{i=1}^l (V_i)^{\times d_i} \longrightarrow k, \end{equation} such that for each $1 \leq i \leq l$ $F$ is linear in each $V_i$, and such that we have \begin{equation} f(v_1, \dots v_l) = F(\underbrace{v_1, \dots v_1}_{d_1-many}, \underbrace{v_2, \dots v_2}_{d_2-many}, \dots, \underbrace{v_l, \dots v_l}_{d_l-many}) \ ? \end{equation}

Let $R =\mathbb{Z}/N \mathbb{Z}$. Let $f:R\to \mathbb{R}$, $\rho:R\to \lbrack 0,1\rbrack$. We assume that it takes trivial time to compute any given value $f(m)$ or $\rho(m)$.

Define $$S(\delta,m) = \sum_{n\in R: \rho(n)\geq \delta} f(n+m).$$

Is it possible to compute $S(\delta,m)$ efficiently? To be precise: say we are given $\delta_i\in \lbrack 0,1\rbrack$, $m_i\in R$ for $1\leq i\leq N$. Is it possible to compute $S(\delta_i,m_i)$ for all $1\leq i\leq N$ in roughly linear time on $N$ (or some other time substantially smaller than $N^2$)?

To see why this might not be an unreasonable request, consider the two following extreme cases.

If all $m_i=0$ (or all $m_i$ are equal), we can compute the $N$ sums $S(\delta_i,m_i)$ easily in linear time.

If all $\delta_i=0$ (or all $\delta_i$ are equal), we can compute the $N$ sums $S(\delta_i,m_i)$ in roughly linear time (more precisely: $O(N \log N)$) using FFT. The reason is that $S(\delta,m_i)$ equals $F(m_i)$, where $F$ is the convolution of $f$ with the function $g(n) = \begin{cases} 1 &\text{if $\rho(-m)>\delta$,}\\ 0 &\text{otherwise.}\end{cases}$.

Is the general problem feasible? Can it be shown not to be feasible?

Let $(M^n,g)$ be a Riemannian manifold with non-empty smooth boundary $\partial M$. For any two points $x,y\in M$, the distance between $x$ and $y$ may be defined as $$ d(x,y)=\inf_\gamma Length(\gamma), $$ where the infimum is taken over all $C^1$ curves lying in $M$. Can we prove there exists a path in the closure $\bar{M}$ which achieves $d(x,y)$? And the length-minimizing path is piecewise $C^1$? Note that since $\partial M$ is non-empty, the length-minimizing path (if exists) may intersect the boundary. Any reference for this question?

Let $sSet^2$ be the category of bisimplicial sets.

In the diagonal model structure on $sSet^2$ weak equivalences are diagonal weak equivalence (i.e.$ X \rightarrow Y$ is a weak equivalence if $dX \rightarrow dY$ is a weak equivalence of simplicial sets) and cofibrations are monomorphisms. Let us call the fibrations in this model structure as diagonal fibrations.

In Moerdijk model structure on $sSet^2$, $X \rightarrow Y$ is a fibration (weak equivalence) if $dX \rightarrow dY$ is a kan fibration (weak equivalence of simplicial sets).

Is there any condition known under which a Moerdijk fibrant $X$ is diagonal fibrant?

I am looking for the precise claim of this Bayes' theorem with proof (including probability space). I found this https://en.wikipedia.org/wiki/Bayes%27_theorem here. If X is continuous and Y is discrete, then $f_{X\mid Y}(x)=\frac{P(Y=y\mid X=x)f_{X}(x)}{P(Y=y)}$.