Some mathematics journals publish "research announcements", a class of publication that before today I had not heard of. An example is Electronic Research Announcements in Mathematical Sciences.

I presume that after publishing a research announcement in such a journal presenting a particular result, one can subsequently then publish a full research paper on the same result at a later date, generally in a different journal. Obviously, it is not ordinarily the case that journals will knowingly allow the same result to be published twice. Therefore, research announcements must have some defining properties which make this practice acceptable.

My question is the following: **what are the properties of the research announcement which allow the subsequent publication of the full research paper describing the same result?**

Such a question may be of importance; for example, to a journal editor who is handling a submission that describes a result which has been previously published as a research announcement. Perhaps the answer is simple: that the research announcement must contain no proofs. But perhaps the convention is more subtle than this, I'm not sure.

I am aware that the practice of publishing research announcements is not widespread, and I am not interested for the purposes of this question in discussing whether anyone *ought* to publish a research announcement in any particular situation.

I believe that this question is best suited to mathoverflow.net, rather than (for example) to academia.stackexchange.com, as I am asking specifically about publication practice in mathematics.

The number of positive real roots of a polynomial with real coefficients is strictly smaller than the number of nonzero coefficients of the polynomial. This is an immediate corollary of Descartes' rule of signs.

For a polynomial with complex coefficients of degree at most $p-1$, with $p$ prime, the number of distinct roots of the polynomial which are degree-$p$ roots of unity is strictly smaller than the number of nonzero coefficients of the polynomial. As observed by Tao, this is equivalent to an uncertainty inequality for prime-order groups.

For a polynomial over a field of zero characteristic, the multiplicity of any of its non-zero roots is strictly smaller than the number of nonzero coefficients of the polynomial. To my knowledge, this first appeared in a paper by Brindza.

Are these results reducible to each other? (Well, the ground fields are not quite the same, and yet...) Are there any other similar results known? And, ultimately, is there any "common parent" from which all these results can be derived?

Can you provide a proof or a counterexample for the following claim :

Let $P_m(x)=2^{-m}\cdot \left(\left(x-\sqrt{x^2-4}\right)^{m}+\left(x+\sqrt{x^2-4}\right)^{m}\right)$

Let $N=k\cdot 2^n+1$ such that $n>2$ , $0< k <2^n$ and

$\begin{cases} k \equiv 1,7 \pmod{30} ~ with ~ n \equiv 0 \pmod{4} ~,or \\ k \equiv 11,23 \pmod{30} ~ with ~ n \equiv 1 \pmod{4} ~,or \\ k \equiv 13,19 \pmod{30} ~ with ~ n \equiv 2 \pmod{4} ~,or \\ k \equiv 17,29 \pmod{30} ~ with ~ n \equiv 3 \pmod{4} \end{cases}$

Let $S_i=S_{i-1}^2-2$ with $S_0=P_k(8)$ , then $N$ is prime iff $S_{n-2} \equiv 0 \pmod N$ .

You can run this test here . A list of Proth primes sorted by coefficient $k$ can be found here . I have tested this claim for many random values of $k$ and $n$ and there were no counterexamples .

Note that for $k=1$ we have Inkeri's primality test for Fermat numbers . Reference : Tests for primality, Ann. Acad. Sci. Fenn. Ser. A I 279 (1960), 1-19.

Dear Colleagues and Friends,

Please let me know if you are aware of any references to the following question.

The classical result of Atiyah, Patodi and Singer tells us that if $W$ is a compact oriented Riemannian 4-manifold with boundary $M$ and, moreover, if we assume that near M the metric is isometric to a product, then $$ sign(W)= \frac{1}{3} \int_W p_1 - \eta(M),$$ where $p_1$ is the differential form representing the first Pontryagin class of $W$, and $\eta$ is the eta-invariant of $M$.

What about the case when both $W$ and $M$ are hyperbolic manifolds and are allowed to have cusps? Or, say, $W$ and $M$ are Riemannian as above, with infinite ends of finite volume, on which the metric is isometric to a product? (which will be the case if both are hyperbolic with cusps - I'm sure that this is not a very general setting :-))

Any information will be appreciated. Please excuse my ignorance as differential geometer.

Let $E$, $F$ be two complex Hilbert spaces and $\mathcal{L}(E)$ (resp. $\mathcal{L}(F)$) be the algebra of all bounded linear operators on $E$ (resp. $F$).

The algebraic tensor product of $E$ and $F$ is given by $$E \otimes F:=\left\{\xi=\sum_{i=1}^dv_i\otimes w_i:\;d\in \mathbb{N}^*,\;\;v_i\in E,\;\;w_i\in F \right\}.$$

In $E \otimes F$, we define $$ \langle \xi,\eta\rangle=\sum_{i=1}^n\sum_{j=1}^m \langle x_i,z_j\rangle_1\langle y_i ,t_j\rangle_2, $$ for $\xi=\displaystyle\sum_{i=1}^nx_i\otimes y_i\in E \otimes F$ and $\eta=\displaystyle\sum_{j=1}^mz_j\otimes w_j\in E \otimes F$.

The above sesquilinear form is an inner product in $E \otimes F$.

It is well known that $(E \otimes F,\langle\cdot,\cdot\rangle)$ is not a complete space. Let $E \widehat{\otimes} F$ be the completion of $E \otimes F$ under the inner product $\langle\cdot,\cdot\rangle$.

If $T\in \mathcal{L}(E)$ and $S\in \mathcal{L}(F)$, then the tensor product of $T$ and $S$ is denoted $T\otimes S$ and defined as $$\big(T\otimes S\big)\bigg(\sum_{k=1}^d x_k\otimes y_k\bigg)=\sum_{k=1}^dTx_k \otimes Sy_k,\;\;\forall\,\sum_{k=1}^d x_k\otimes y_k\in E \otimes F,$$ which lies in $\mathcal{L}(E \otimes F)$. The extension of $T\otimes S$ over the Hilbert space $E \widehat{\otimes} F$, denoted by $T \widehat{\otimes} S$, is the tensor product of $T$ and $S$ on the tensor product space, which lies in $\mathcal{L}(E\widehat{\otimes}F)$.

Let $\operatorname{Im} (X)$ and $\overline{\operatorname{Im} (X)}$ denote respectively the range of an operator $X$ and the closure of its range.

Let $T,M\in \mathcal{L}(E)$ and $S,N\in \mathcal{L}(F)$ be such that

$\operatorname{Im} (T)\subseteq \overline{\operatorname{Im} (M)}$.

$\operatorname{Im} (S)\subseteq\overline{\operatorname{Im} (N)}$.

I want to prove that $$\operatorname{Im}(T \widehat{\otimes} S)\subseteq \overline{\operatorname{Im}(M \widehat{\otimes} N)}.$$

Note that I show that $$\overline{\operatorname{Im} (M)}\otimes\overline{\operatorname{Im} (N)}\subseteq\overline{\operatorname{Im}(M \otimes N)}.$$

Suppose $P(\lambda, i)$ is the probability that a Poisson random variable with average $\lambda$ is equal to $i$, i.e. $\frac{\lambda^i}{e^{\lambda}i!}$

I think the following system of equations always has solution in $x$ and $y$, non-negative real numbers, for any $\alpha>0$ and $k\in \mathbb{N}_+$

\begin{cases} \alpha=\sum_{i=0}^{\infty}P(x, i)\cdot P(y, k+i) \\ \alpha=\sum_{i=0}^{\infty}P(x, i)\cdot P(y, k+i+1) \end{cases}

where the necessary condition $\alpha\leq P(k+1, k+1)$ holds. It is easy to prove that this is indeed a necessary condition, equivalent to the condition that $\alpha=P(\lambda,k+1)$ has a solution. It is also easy to see that solution $y$ of the system is smaller or equal to $\lambda$, the largest solution to the equation $\alpha=P(\lambda,k+1)$. Experiments show that for each fixed $\alpha$ and $k$, there is a solution, but I did not manage to prove it analytically.

Is there any analogue for mean value theorem for multidimensional functions? Any suggestion for the proof directions will be appreciated.

In Chriss and Ginzburg's fantastic book 'representation theory and complex geometry', they use the following Thom Isomorphism:

$\pi:E\rightarrow X$, is a G-equivariant affine linear bundle, then $\pi^{*}: K^{G}(X)\rightarrow K^{G}(F)$

It seems that the Thom isomorphism and cellular fibration can provide a lot of information of the equivariant K theory of flag varieties and Steinberg varieties.

In the book, the injectivity of the map is proved by using specialization which seems (for me) to be a topological method

Questions:

(1)Whether there is any algebraic proof of this isomorphism ?

(2)And does this still holds for other algebraically closed field?

I am trying to understand a basic computation of convolution. Throughout, $R$ is the real line as a topological group and $k$ is some base field. I would like to understand the computation of the convolution

$k_{(a,b)} \star k_{[0,\infty)}$.

I believe this convolution should be $k_{[a,\infty)}[-1]$, but I am unable to prove this.

In general, the stalk at $t \in R$ is (by Beck-Chevalley) the compactly supported cohomology of $k_{(a,b)} \boxtimes k_{[0,\infty)}$ restricted to the line $\{t_1 + t_2 = t\} \subset R^2$. In what follows, $A<B$ are real numbers:

- When $t>b$, the stalk is $R\Gamma_c(k_{(A,B)}) \simeq k[-1]$, the reduced cohomology of the circle, as one is computing the compactly supported cohomology of an open interval.
- When $a<t\leq b$, the stalk is $R\Gamma_c(k_{[A,B)})$. By the short exact sequence $k_{(-\infty,A)} \to k_{(-\infty,B)} \to k_{[A,B)}$, the stalk is thus zero. (The map $k_{(-\infty,A)} \to k_{(-\infty,B)}$ induces an isomorphism on compactly supported cohomology.)
- When $t=a$, the stalk is the compactly supported cohomology of $R$ with respect to a skyscraper sheaf. The skyscraper sheaf is flabby, so we conclude that the stalk at $t=0$ is a copy of $k$ in degree 0.
- When $t<a$, the stalk is zero.

Is this computation of stalks correct? I would like very much to know what I am doing wrong.

I recently got interested into this paper where the authors analyze the question of uniqueness of measure-valued solutions to the continuity equation. Although the paper is well written in my opinion, there is something that I can "understand" but quite not feel, i.e. I lack some intuition behind and I can't get it completely. Here is the main Theorem of the cited paper:

Although it is not important for this post, here is Cauchy problem (1.1) (to be understood in a distributional sense): $$ \partial_t \mu_t + \text{div} (b\mu_t) = 0, \qquad \mu_0 = \overline{\mu} $$ for some given measure $\overline{\mu}$ in $\mathbb R^d$. My question focuses on the assumptions and more precisely it is the following:

**Q.** What is condition (ii) *really* saying? What is this (rather mysterious) sequence $(V_k)_k$? Can someone get some intuition behind it?

As I said above, I can understand the paper, i.e. I see how this condition is used in the proof of the theorem and I see how this condition is verified in some examples: they usually choose $b_k$ to be the mollified vector fields (with some fixed convolution kernel) and then they choose either $V_k \equiv 1$ or in case $d=1$ they choose $V_k = 1/b_k^2$. I am completely lacking intuition behind this choices and I cannot figure out what this $V_k$ should stand for.

It is easily seen that the quantity $\langle A \xi, \xi \rangle$, being $A$ a matrix and $\xi \in \mathbb R^d$, depends only on the symmetric part $A^{sym}:=\frac{1}{2}(A + A^T)$ of the matrix $A$. All in all, recalling the definition of Rayleigh quotient condition (ii) is saying something like $$ \Lambda_k(x) V_k \le C V_k - \langle b_k, \nabla V_k \rangle, $$ where $\Lambda_k(x)$ is the maximum eigenvalue of the symmetric part of $\mathcal B_k(x)$. However, it does not seem to me a great progress... I am still missing the meaning of the formula.

Thanks.

Consider the following optimization problem in positive integers $n_1, n_2, n_3$.

$$\begin{array}{ll} \text{maximize} & n_1(n_2+n_3)\\ \text{subject to} & n_1+n_2+n_3 = N\end{array}$$

If $n_1, n_2, n_3$ were reals, the solution would be $n_1 = \frac N2$ and $n_2 = n_3$. However, in my problem, $n_1, n_2, n_3$ are positive integers. Please help me solve this quadratic integer optimization problem.

The function g is such that g(x) = kx2 where k is a constant. (b) Given that fg(2) = 12, work out the value of k I have no idea how to solve this any help? thanks :)

Let $v_1,v_2 \in \{0,1\}^n$. Denote $v_1v_2=((v_1)_1 (v_2)_1, \ldots, (v_1)_n (v_2)_n)$ and $|v|=\sum v_{i}$. \begin{align} {\scriptsize f(v_1, v_2) = \sum_{x_1=0}^{|v_1|} \sum_{x_2=0}^{|v_2|} \sum_{d=0}^{|v_1 v_2|} \frac{1}{2^{|v_1|+|v_2|-|v_1 v_2|}} \biggl| {|v_1| - |v_1 v_2| \choose x_1 - d} {|v_2| - |v_1 v_2| \choose x_2 - d} - {|v_1| - |v_1 v_2| \choose x_1 + 1 - d} {|v_2| - |v_1 v_2| \choose x_2 + 1 - d} \biggr|.} \end{align} I want to estimate $f(v_1, v_2)$ when $|v_1|, |v_2| \to \infty$.

As a first step, I obtain \begin{align} { \scriptsize f(v_1, v_2) = \sum_{x_1=0}^{|v_1|} \sum_{x_2=0}^{|v_2|} \sum_{d=0}^{|v_1v_2|} \frac{1}{2^{|v_1|+|v_2|-|v_1v_2|}} \biggl| \left( 1- \frac{(v_1-|v_1v_2|-x_1+d)(v_2-|v_1||v_2|-x_2+d)}{(x_1+1-d)(x_2+1-d)} \right) {|v_1| - |v_1v_2| \choose x_1 - d} \biggr|, } \end{align}

How to estimate $f(v_1,v_2)$? Thank you very much.

My question is the following. If I have a function $y(x)$, what is the expression that describes how it is transformed under a general homothetic transformation?

I need this because I'm trying to proof that given a particular solution of an homogeneous equation, if I apply a homothetic transformation to this particular solution I obtain a new one.

Thanks!

$\Omega$=Locally compact space, $L^1(H)$=Trace class operators on the Hilbert space $H$ and $\mu$= positive bounded Radon measure on $\Omega$.

Let $\phi:\Omega\to L^1(H)$ be a Borel measurable function. Does the following hold?

$$\sup_{u\in B(H)_{\|\cdot\|\leq1}}|\int tr(\phi(t)u)d\mu|=\int tr(|\phi(t)|)d\mu$$

Given the following differential equations:

\begin{equation} \begin{aligned} \dot{x}_1 &= f_1(x_1,\ldots,x_n) \\ \vdots \\ \dot{x}_n &= f_n(x_1,\ldots,x_n) \end{aligned} \end{equation}

In a compact way: $$\dot{\hat{x}} = \hat{F}(\hat{x})$$

Let the group $\Psi\subset S_n$, where $S_n$ is a symmetric group.

Suppose the following key property holds: $$\hat{F}(P_\sigma \hat{x})=P_\sigma \hat{F}(\hat{x}), \ \ \ \ \ \forall \sigma\in\Psi$$ where $P_\sigma$ is the permutation matrix corresponding to $\sigma\in \Psi$.

For example: \begin{equation} \begin{aligned} \dot{x}_1 &= x_1(x_1-1)(x_1+1) +x_2 \\ \dot{x}_2 &= x_2(x_2-1)(x_2+1) +x_1 + x_3 \\ \dot{x}_3 &= x_3(x_3-1)(x_3+1) +x_2\end{aligned} \end{equation}

In this case, we can choose $\sigma = (13)\in \Psi = \{(),(13)\}$, and the corresponding permutation matrix: $$P_\sigma = \begin{bmatrix}0 & 0 & 1\\0 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix}$$ It is easy to see that $\hat{F}(P_\sigma \hat{x})=P_\sigma \hat{F}(\hat{x})$

My question:

If $\hat{x}^*$ is a **stable** equilibrium point for $\dot{\hat{x}} = \hat{F}(\hat{x})$, can I say $P_\sigma \hat{x}^*$ is also a **stable** equilibrium point, for all $\sigma\in \Psi$?

Note: Easily to see that $P_\sigma \hat{x}^*$ is also an equilibrium point.

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 $A'>A$?