Suppose you have a projective manifold $M$, a very ample bundle $\scr L$ and a transverse holomorphic section $s \in H^0(\scr L)$. Then the zero set of $s$ is a complex submanifold $S_M$.

Can we have a embedding of the the projective manifold $M$ in some projective space such that image of $S_M$ will not be contained in a hyperplane? You can assume $M$ has complex dimension $2$.

If R(E/Q_\infty) is the fine Selmer group and Y(E/Q_\infty) is its dual, then we know that Y(E/Q_\infty) is a finitely generated \Lambda-module and by a theorem of Kato, it is also torsion. My understanding is that it should thus make sense to want to attach a p-adic L function.

Can we say what this p-adic L function is? Or does it just follow (trivially) from the Iwasawa main conjecture?

Let $M$ be a matroid with ground set $E$. Deletion and contraction in matroids commute with each other and with themselves, i.e. for all $e,f \in E$ one has

$(M/e)\setminus f = (M\setminus f)/e$, $\hspace{0.1cm}$ $(M\setminus e)\setminus f = (M \setminus f) \setminus e$ $\hspace{0.1cm}$ and $\hspace{0.1cm}$ $(M/e)/f = (M/f)/e$.

Are there any matroids, aside from uniform matroids, which have the following property:

For all $e,f\in E$

$(M/e)\setminus f = (M/f) \setminus e$ ?

Problem statement

Let $0 \in K \subseteq \mathbb R^n$ be a non-empty closed and convex set with non-empty interior and piecewise smooth boundary $\partial K$. Fix $\lambda > 1$ and define $\lambda K := \{ \lambda x : x \in K\}$. For $x \in \mathbb R^n$ denote by $P_C(x) := {\arg\min}_{w \in C} \|w - x\|_2$ the nearest-point projection onto the convex set $C$ (as measured in the standard $\ell^2$ norm).

For any $x \in \mathbb R^n \setminus (\lambda K)$, prove that $\|P_K(x)\|_2 \leq \|P_{\lambda K} (x)\|_2$.

ProgressFor notational ease, let $x_1 := P_K(x)$ and $x_\lambda = P_{\lambda K}(x)$. Define the normal cone of $K$ at $x_1$ by $$N_K(x_1) = \{ y \in \mathbb R^n : \langle y, w-x_1\rangle \leq 0 \text{ for all } w \in K\}$$ If $x_\lambda - x_1 \in N_K(x_1)$ then the result follows by examining $\frac{\mathrm d}{\mathrm d t}\|t x_\lambda + (1-t) x_1\|_2^2$ at $t = 0$. Indeed, this shows that the norm of $tx_\lambda + (1-t) x_1$ is increasing on $t \in [0, 1]$, hence $\|x_\lambda\|_2 \geq \|x_1\|_2$.

ExamplesIt is easy to show that the result holds if $K$ is the $\ell^1$ or $\ell^2$ ball, or if $K$ is an origin-centered ellipse. Finally, using that derivative trick again, I can show the result holds as long as $\langle x_1, x_\lambda - x_1\rangle \geq 0$, but have been unable to verify that this identity always holds. However, I have been unable to craft a toy set $K$ where it seems like $\langle x_1, x_\lambda - x_1\rangle < 0$ without it also seeming necessary that $x_1$ was not chosen optimally to be the $\arg\min$.

I am wondering if any convex geometers/probabilists have looked at the following question:

Given $n$ randomly distributed (not sure what assumption to put there) points in $\mathbb{R}^d$, for each point $x_i\in\{x_1,\ldots, x_n\}$, draw $N$ points uniformly on the sphere $S^{d-1}$ with radius $r>1$ centered at $x_i$, denote as $x_{i, 1}, \ldots, x_{i, N}$. Let $C$ be the convex hull of $x_{1,1}, \ldots, x_{1,N},\ldots, x_{n,1},\ldots,x_{n,N}$. What is the probability that $\forall i\in\{1,\ldots, n\}$, $B(x_i, 1)\in C$?

So in other words, for every original point $x_i$, we draw a unit ball around it, how likely that $C$ contains all these unit balls?

I found Probability that a convex shape contains the unit ball was asking a similar question. According to the comments, if $N$ is exponential in $d$, then my question holds with probability $1$, because for each $i$, the convex hull of $x_{i, 1},\ldots, x_{i, N}$ already contains a unit ball. My question is different that we have $n$ points, and I imagine the neighboring vertices help each other to enlarge the convex hull. So perhaps a tighter bound exists?

Has this problem been studied before? What are the assumptions that people put on the distribution of $x_i,\ldots, x_n$? Thanks for any comments/answers!

I need to derive the following expression: $$ y = C^{T}ACx $$ w.r.t the matrix $C$

where $C$ is a rotation matrix, $x$ is a vector in $R^{3}$ and $A$ is a selection matrix in $R^{3\times3}$

$$ A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \\ \end{bmatrix} $$

I've tried to directly apply the derivative of multiplication but i'm not sure if it applies and I end up with dimensionality errors. I'm quite stuck, and am checking the result against a numerical approximation of the jacobian.

Thanks for all the help

I am trying to understand the proof of weak Bezout theorem presented in Donu's algebraic geometry book. I wanted to ask where in the proof is he using the fact that we are working in projective space ? Is it when he assumes that translating the line at infinity if necessary ? Essentially the proof would work as well in complex plane if we have the point infinity in it as well right ? It would be nice if someone could also present couple of more intuition for this proof.

given $X_1,\cdots,X_n\overset{iid}\sim F$, where $F$ is a truncated normal, I wonder if there's something known about the distribution and specifically about the sub-Gaussianity of $X_i-\overline{X}$ and $\sum_{j=1}^k X_{i_j}-\overline{X}$? Thanks.

This question is about the 6d (2, 0) superconformal field theory (also called 'theory X' by some people). This SCFT, which can be considered as a relative quantum field theory (see here for a definition), exhibits a great deal of mathematical structure which is cogently expressed in the diagram below (taken from slides for D. Ben-Zvi's talk). In particular, its compactification on 2-torus is $N=4$ Yang--Mills theory. Note that the S-duality of the compactification of a topological twist of $N=4$ SYM (the so-called Kapustin--Witten TQFT) on a closed Riemann surface encodes geometric Langlands duality.

My question is: has there been any recent progress on understanding the compactification of theory X on general 4-manifolds (marked as '???' in the diagram)?

Let $W$ be a Weyl group generated by the simple reflections $s_i$, $i \in I$, where $I$ is the vertex set of the Dynkin diagram of $W$. For $J \subset I$, let $W_J$ be the subgroup of $W$ generated by $s_j$, $j \in J$. Let $w_0^J$ be the longest word in $W_J$.

Suppose that $K \subset J \subset I$. I think that the following identity is true $w_0^J w_0^K = w_0^{K^*} w_0^J$, where ${}^*: J \to J$ is defined by $s_{j^*} = w_0^J s_j w_0^J$, $j \in J$.

Are there some reference about the proof of $w_0^J w_0^K = w_0^{K^*} w_0^J$? Thank you very much.

Let $G=(V,E)$ be a simple, undirected and connected graph. We say that $S\subseteq V$ is a *cutting set* if $S\neq V$ and the induced subgraph on $V\setminus S$ is not connected any more.

If $S \subseteq V$ is a cutting set of $G$, is there a cutting set $S_0\subseteq S$ of $G$ such that for all $x\in S_0$ the set $S_0\setminus \{x\}$ is no longer a cutting set?

(This question has an easy positive answer for finite graphs, so it is only interesting for infinite graphs.)

The site percolatiom thresholds in a square lattice when allowing nearest neighbours (NN) and next nearest neighbour(NNN) are pretty well known. It is quite obvious that the percolation threshold will be lesser in the NNN case. However, I was looking for results related to how changing a certain parameter $q$ affects the percolation threshold in square lattices. I'll define $q$ below.

In site percolation of square lattices we generally consider lattice of square pixels. With probability $p$ a certain square pixel is black and with probability $1-p$ a certain square pixel is white. If any 4 of the nearest neighbours of a certain pixel are of the same colour as of the pixel then those two pixels are considered to belong to the same cluster (in the NN) case. Next nearest neighbours are not considered. Now, let us define another probability $q$. If the next-nearest neighbour of a certain pixel is of the same colour as of *that* certain pixel then with $q$ probability they are considered to belong to the same cluster. It can be easily seen that for the NN case $q=0$ and for the NNN case $q=1$.

My questions are:

Has there been any study regarding how variation of the probability $q$ affects site percolation threshold of infinite square lattices?

Mathematically, is it possible to show the nature of dependency and variation of the site percolation threshold with $q$? Are techniques like renormalization group useful here?

For a certain $q$ can site percolation threshold be accurately calculated (other than programming and extrapolating to find an approximate value)?

Consider Hex on an $n \times n$ board without a swap rule, so that the first player wins. Assume the first player tries to minimize the length of the game, and the second player tries to maximize the length. Call the resulting optimal length $f(n)$.

**Question:** What are the asymptotics of $f(n)$?

Trivially $f$ is at least linear and at most quadratic. Campbell, On optimal play in the game of Hex give a linear lower bound with constant 2.5 (as opposed to 2 if the second player helps the first to win).

Are there better bounds on $f(n)$, or conjectured bounds? I would expect it to be superlinear, but don’t know if it’s subquadratic. An intermediate exponent would be quite interesting.

It is known that there are finitely many square-free integers $d$ for which the ring of integers of $\mathbf{Q}(\sqrt{d})$ is Euclidean under the norm function.

Is there an analogous result for quadratic extensions of $K(t)$, $t$ an indeterminate and $K$ a (finite) field?

Also, can we classify the (imaginary) quadratic function fields that are PID in a fashion similar to the number field case?

I'm assuming the answers to these questions are well known, but my internet search has come up empty.

I want to ask about the observability inequality for the heat equation ( internal control), we consider the backward heat equation with Dirichlet boundary conditions:

\begin{array}{c} \varphi _{t}+\Delta \varphi =0\text{ in }\Omega \times (0,T) \\ \varphi =0\text{ on } \partial \text{}\Omega \times (0,T)\text{} \\ \varphi (T)=\varphi _{T}\text{ }% \end{array} The observability inequality in some books is given by: $$ \left\Vert \varphi (0)\right\Vert ^{2}\geq c\int_{\omega }\int_{0}^{T}\varphi (x,t)dtdx $$ and some times $$\left\Vert \varphi (T)\right\Vert ^{2}\geq c\int_{\omega }\int_{0}^{T}\varphi (x,t)dtdx$$ and $$\int_{\omega }\int_{T/4}^{3T/4}\varphi (x,t)dtdx\geq c\int_{\omega }\int_{0}^{T}\varphi (x,t)dtdx$$ To passe from the first to the second I guess that they made the substitution $s=T-t$, however, for the third it comes from the golobal Carleman estimates but I can not see the equivalence between them. Thank you.

Assuming we have a set of points $X=\{x_1,..,x_n\}$, all in $\mathbb{R}^d$, and construct the Vietoris-Rips-Complex $V_\epsilon (X)$ for some distance parameter $\epsilon > 0$. Is it possible to have non-trivial (simplicial) homology groups $H_k(V_\epsilon (X))$ in degree $k\geq d$?

If no, is there a proof? If yes, are there examples, e.g. for $d=2$?

The $j$-ivariant has the following Fourier expansion $$j(\tau)=\frac 1q +\sum_{n=0}^{\infty}a_nq^n=\frac{1}{q}+744+196884q+21493760q^2+\cdots.$$ Here is $q=e^{2\pi i \tau}$.

Is there some simple **effective** bound on the coefficients $a_n$?

**Backround.**

This question comes from On the “gap” in a theorem of Heegner. Let $D$ be a negative discriminant such that $h(D)=1$. We want to show that $J=j(\sqrt{D})$ generates a cubic extension of $\mathbf Q$. Since we have at our disposal a monic cubic polynomial with rational coefficients, the modular equation $\Phi_2(X,j)$, whose root is $J$, and the other two roots are non-real, it is sufficient to show that $J$ is not an integer.

In this case $j=j\left(\frac{-1+\sqrt D}{2} \right)$ is also an integer. Set

$$t=e^{2\pi i(-1+\sqrt D)/2}.$$

Then

$$J=\frac{1}{t^2}+744+196884t^2+O(t^4),$$

and

$$j^2-1488j+160512-J=42987520t+O(t^2).$$.

On the left there is an integer. However the right side tends to zero as $|D|$ gets large. Stark asserts that $|D|>60$ is enough for the RHS to be less than 1. **Why is it enough?**

Let $X$ be a (infinite) separable topological space and consider $C_p(X)$, the space of continuous functions on $X$ endowed with the point-wise convergence topology.

Q. I am looking for topological properties on $X$ which make $C_p(X)$ hereditary Lindelöf.

$$X=?\implies C_p(X)=\textrm{Hereditary Lindelöf}$$

Consider the mixed integer program $$Ax\leq b$$ $$By\leq c$$ $$\begin{bmatrix}x&y\end{bmatrix}C\begin{bmatrix}x\\y\end{bmatrix}+D\begin{bmatrix}x\\y\end{bmatrix}\leq d$$ where $x$ are integer variables of dimension $n$ and $y$ are real variables.

Because of the quadratic conditions this problems is in general $NP$ hard.

Is it possible convert this into exponentially larger mixed linear integer program **by blowing up the number of integer variables only to a polynomial in $n$**?

The remaining problem will be exponential time solvable as original one and would still be np hard.

**The Question**

I'll present my first two simple steps toward estimating certain Egyptians sums. I hope that someone (perhaps an expert in Analysis or Analytic Number Theory) will continue and will achieve essential results, possibly even about the baroque (multi perfect) numbers--and **this is the challenge** of this *Question*. (*See the end of this "step 2" section*, before the final comments).

*Remark: this post is a continuation of* Corner integrals of $\exp$

**A pre-approximation integral**

The said sums are related to an integral of $\ \exp(-\sum_{k=1}^n x_k)\ $ over the corner

$$ \Delta(n;\ S)\ :=\ \{(x_1\ldots x_n)\in[0;\infty)^n\, : \, \sum_{k=1}^n x_k\le A\} $$

The integral is

$$ I_{n;\ S}\ :=\ \int_{\Delta(n;\ S)} \exp\left(-\sum_{k=1}^n x_k\right)\cdot dx_1\ldots dx_n\,\ = \,\ 1 -\ \sum_{k=0}^{n-1} \frac{S^k}{k!} \cdot e^{-S} $$

**A connection with the Egyptian sums over selected primes** (*step 1*)

Let natural numbers $\ p_1\ \ldots\ p_n\ >\ 1$ be pairwise relatively, the main case would be $n$ different primes. They define a lattice

$$ L\ :=\ L(p_1\ldots p_n)\ :=\ (\log(p_1)\!\cdot\!\mathbb Z)\times\ldots \times(\log(p_n)\!\cdot\!\mathbb Z)\ \subseteq\mathbb R^n $$

We want to compute or realistically estimate

$$ \mathbf S(p_1\ldots p_n;\ S)\ :=\ \sum_{(x_1\ldots x_n)\,\in\, L\cap\Delta(n;\ S)} \,\ \frac 1{\prod_{k=1}^n p_k^{f_k}} $$

where $\ x_k\ :=\ f_k\cdot\log(p_k)\ $ for $\ k=1\ldots n$. We may visualize ech of the above fractions attached to the respective lattice point $\ (x_1\!\cdot\!\log(p_1)\ \ldots\ x_n\!\cdot\!\log(p_n)),\ $ as suggested by

$$ \mathbf S(p_1\ldots p_n;\ S)\ := \ \sum_{(x_1\ldots x_n)\,\in\, L\cap\Delta(n;\ S)} \,\ \exp\left(-\sum_{k=1}^n x_k\right) $$

**The first approximation and an overflow** *(step 2)*

Let $\ x\ :=\ (x_1\ \ldots\ x_n)\in \mathbb R^n.\ $ Define the respective cell $$ C(x)\ := \prod_{k=1}^n [x_k;\,x_k+\log(p_k)) $$

If $\ x\in L\ $ then $\ C(x)\ $ is called a lattice cell. In general,

$$ \int_{C(x)}\exp\left(-\sum_{k=1}^n t_k\right)\cdot dt_1\ldots dt_n\ = \ \exp\left(-\sum_{k=1}^n x_k\right)\cdot\prod_{k=1}^n \left(1-\frac 1{p_n}\right) $$

This applied to the lattice cells gives:

**Theorem**

$$ \mathbf S(p_1\ldots p_n;\ S)\ =\ \frac {\int_{\bigcup\{C(x)\,:\,x\in \Delta(n;\ S)\}} \ \exp\left(-\sum_{k=1}^n t_k\right)\cdot dt_1\ldots t_n} {\prod_{k=1}^n\left(1-\frac 1{p_k}\right)} $$

This desired value is approximated by $\ \frac{I_{n;\ S}}{\prod_{k=1}^n\left(1-\frac 1p\right)}.\ $ Of course, $\ \Delta(n;\ S)\ \subseteq \ \bigcup\{C(x)\,:\,x\in \Delta(n;\ S)\}.\ $ Unfortunately, the difference is challengingly annoying

$$ \Omega\,\ :=\,\ \bigcup_{x\in \Delta(n;\ S)} C(x)\ \ \setminus \,\ \Delta(n;\ S) $$

To estimate the overflow (call it an error) $$ E\ :=\ E(p_1\ldots p_N;\ S)\ :=\ \mathbf S(p_1\ldots p_n;\ S) - \frac{I_{n;\ S}}{\prod_{k=1}^n\left(1-\frac 1{p_k}\right)} $$

i.e.

$$ E(p_1\ldots p_N;\ S)\ =\ \frac {\int_E \exp\left(-\sum_{k=1}^n t_k\right)\cdot dt_1\ldots dt_n} {\prod_{k=1}^n\left(1-\frac 1{p_k}\right) } $$

should have implications for Egyptian sums (possibly, also for baroque numbers).

**Comments**

In the popular case of $\ p_1<\ldots<p_n\ $ being the first (smallest) $n$ primes, the Mertens' formula applies. Instead of its most common form about $\ \prod_{k=1}^n \left(1-\frac 1{p_k}\right)\ $ (as above), the less common upside down version fits here better: $$ \prod_{k=1}^n\frac {p_k}{p_k-1}\ =\ e^\gamma\cdot\log(p_n) + O(1) $$

On the top of the sums and integrals over the corners $\ \Delta(n;\ S),\ $ one could equally often consider their complements $\ [;\infty)^n\ \setminus\ \Delta(n;\ S)\ $ and general trapezoids $\ \Delta(n;\ T) \setminus\ \Delta(n;\ S)$.