TITLE: A result about Fermat's numbers. Is my proof correct? Is that result useful ?Can we generalize that result?
QUESTION [0 upvotes]: Let $n$ be an integer and $F_n:=2^{2^n}+1$.
$$n=2,3,4:F_n\equiv17\pmod{30}$$
$\mathbf{Result:}\;n>1:F_n\equiv17\pmod{30}$
$\mathbf{Proof:}$ Suppose $F_n - 1\equiv16\pmod{30}$. Then $2^{2^{n+1}}=({2^{2^n}})^2\equiv16^2\pmod{30}\equiv16\pmod{30}$ and $F_{n+1}\equiv17\pmod{30}$.
I found that result using primoradic (see stub OEIS: https://oeis.org/wiki/Primorial_numeral_system).
That's the way I found that result, which I did't know before and I wonder if there are other results with $\;210,\;2310,\;30030,\;\ldots$ (primorials)
P.S. : In primoradic, using Charles-Ange Laisant's notations for factoradic (with $A=10, B=11, C=12, D=13,\ldots$)
$$17=(000000.221)$$
$$257=(000011.221)$$
$$65.537=(2.240.221)$$
$$F_5=(J.5F1.721.221)$$
Perhaps, someone could give $F_6$, $F_7$ ...in primoradic, just for fun. Fill in the holes :$$F_6=(........0.221)$$
$$F_7=(......1:1:...)$$
$$F_8=(......4:0:...)$$
$$F_9=(......2:1:...)$$
$$F_{10}=(....0:0:...)$$
$$F_{11}=(......1:...)$$
$$F_{12}=(......0:...)$$
$$F_{13}=(......1:...)$$
$$2^{16384}+1=(......:0:...)$$
I have verified each of these results with my spreadsheet.
Maybe $F_n\equiv17\pmod{210}$ if n is even and $47$ if n is odd? A result appears clearly with $2310$ too. We need a proof. Perhaps here :I'm trying to generalize some simple results about $ 2 ^ n $. It's useful to write them in primoradic (see stub OEIS)., will be useful.

REPLY [2 votes]: Of course
$$F_n=F_0F_1\cdots F_{n-1}+2\equiv2\pmod{F_0F_1}$$
for $n\gt1$, and $F_n$ is odd, so there you have it.
Likewise, $F_n\equiv2\pmod{17}$ for $n\gt2$, $F_n\equiv2\pmod{257}$ for $n\gt3$, etc. This is why the Fermat numbers are pairwise relatively prime.


TITLE: existence of a finite free resolution
QUESTION [1 upvotes]: Problem 2.1.26 in Bruns and Herzog, CMR, reads as follows: "Let $R$ be a Cohen-Macaulay local ring of dimension $d$ and $M$ a finite $R$-module. Deduce that the $d$-th syzygy of $M$ in an arbitrary finite free resolution is either $0$ or a maximal Cohen-Macaulay module."

Question: Are the hypotheses of the problem enough to guarantee that $M$ has a finite free resolution?

Remark: As a reminder, see also this related question what classes of modules admit finite free resolutions?

REPLY [1 votes]: No, it is not enough. Take $M = R/\mathfrak m$, where $\mathfrak m$ is the maximal ideal. If $M$ has a finite free resolution, then $R$ is regular, and there are CM local rings which are not regular. For instance, take $R =\mathbb Z/(4)$ and $M =\mathbb Z/(2)$.
Problem 2.1.26 asks a property of a syzygy module which in general is not free.


TITLE: First Year ODE Exam Question
QUESTION [6 upvotes]: What are the eigenfunctions that satisfy
  $$xy''(x) + (1-x)y'(x) + y(x) = -\lambda y(x)$$
  on the interval $I = [0, \infty)$ given that the eigenvalues are natural numbers (positive integers) such that $y(0) = 1$?

This question was on my first-year university ODE final exam and not a single student was able to solve it. We are trying to have it curved from the final exam grade, do we have a strong case or is this fairly "ordinary" material?

REPLY [1 votes]: I believe, if you have encountered this as a Freshman, you are supposed find the solution as a power series. Quite generally when you have a differential equation of the form
$$
N(x) y'' + M(x)y'+P(x)y=0
$$
with $N,M,P$ polynomials, and $N(x)$ with no roots in the interior of the domain you want to solve your ODE in, then your solution can be written as a power series $y(x)=\sum_{n=0}^\infty a_n x^{n}$ (you don't know at this stage what the coefficient $a_n$ are going to be). I believe this is taught to students rather early, no?

Let's write the ODE slightly differently
$$
\underbrace{xy''}_A+\underbrace{y'}_{B}-\underbrace{xy'}_{C}+\underbrace{m y}_D=0
$$
with $m$ a positive integer (here $m=1+\lambda$). Insert the power series for $y(x)$. You get
$$
\underbrace{\sum_{n=2}^\infty a_n n(n-1)x^{n-1}}_A
+\underbrace{\sum_{n=1}^\infty a_n nx^{n-1}}_{B}-\underbrace{\sum_{n=1}^\infty a_n n x^n}_{C}+\underbrace{m\sum_{n=0}^\infty a_n x^n}_D=0
$$
Or upon manipulation and rearranging becomes
$$
m + a_1+\sum_{n=1}^\infty \Big[(n+1)^2a_{n+1}+(m-n)a_n\Big]x^n=0
$$
The condition $y(0)=1$ gives you $a_0=1$. To find the rest of the coefficients, note that a power series is zero if the coefficient of each power is zero. This means
$
a_1=-m$
and
$$
(n+1)^2a_{n+1}=-(m-n)a_n
$$
Now note that eventually (in fact for $n=m$) the right hand side of the above equation vanishes. This results in $a_{m+1}=a_{m+2}=\cdots = 0$. In other words, your solution is a polynomial. More explictly, you have a a recursive relation (for $n\leq m$)
$$
a_n = -\frac{m -n +1}{n^2}a_{n-1}\Longrightarrow
a_n = (-1)^{n-1}\frac{(m-1)!}{n!^2(m-n)!}a_1=(-1)^n {m\choose n}\frac{1}{n!}
$$
In other words, putting all of these together the solution is
$$
\boxed{
y(x):=L_m(x) = \sum_{n=0}^m (-1)^n {m\choose n}\frac{x^n}{n!}}
$$
This polynomial, as people mentioned in comments is called $m$th Laguerre polynomial, but you are probably not expected to know the name. Finding the above formula should have been enough to get you full credit probably.


TITLE: Infinite group whose every element is of order $4$?
QUESTION [1 upvotes]: I have constructed infinite group whose every element is of prime order by taking the set as set  of sequences whose elements are from integers modulo $p$ and operation is integers modulo $p$.
Now how can I get an infinite group whose every element is of order $4$ (non prime) except identity?.
Is there any general way of finding an infinite group whose every element is of order $n$?

REPLY [4 votes]: Suppose $\lvert a\rvert=4$ for some $a\in G$ for some group $G$. Then $(a^2)^2=e$ and $a^2$ is non-trivial, so  $$\lvert a^2\rvert=2.$$


TITLE: What does $v=c$ in the Lorentz transformation for time tell us?
QUESTION [4 upvotes]: For the simpler cases as boost in the x-direction, the time dilation formula following the Lorentz transformation for time is $$\Delta t'=\gamma(\Delta t-v\frac{\Delta x}{c^2})$$Now, we observe that as $v\to c$, $\gamma\to \infty$. And we also observe that if we increase $v$, the term in brackets beside $\gamma$ decreases more rapidly than the denominator of $\gamma$ i.e., $\sqrt{1-\frac{v^2}{c^2}}$. But if we put $v=c$ suddenly, $\gamma$ becomes undefined and the denominator of $\gamma$ and the term in brackets beside $\gamma$ becomes zero mathematically at the same time while the term in brackets was decreasing more rapidly. But while increasing $v$ to $c$ we see that $\Delta t'$ decreases and approaches zero and as we know $c$ is the maximum speed why don't we get $\Delta t'=0$ on putting $v=c$? Why do we get an indeterminate form? What is the explanation for this?

REPLY [1 votes]: Also, the fact that $\Delta t' \rightarrow 0$ if we formally let $v \rightarrow c$ can be interpreted as saying that no time at all passes for a particle moving at the speed of light.  Photons cannot "age" or in any other way change over time.


TITLE: Thermometric Properties
QUESTION [0 upvotes]: A thermocouple thermometer with 10 ohms of resistance is calibrated with its hot junction and cold junction immersed in steam and melting ice respectively. An EMF of 5.6mv is measured.
This thermocouple is then used in series with a voltmeter of resistance 100 ohm. If the voltmeter recorded 2.8 mV when the cold junction is in melting ice and its hot junction is in a liquid bath, what is the temperature of the liquid bath on the Centigrade scale?

I understand the calculations involved in finding the answer.  However, I am unsure of how either of the resistances affect the calculations.  Why are the resistances important?

REPLY [1 votes]: An ideal voltmeter has infinite resistance and will not allow current to flow through its leads. In this case the voltmeter has a rather low resistance (I would throw out such a voltmeter)
Below are two circuit diagrams, the first shows an ideal thermocouple and an ideal voltmeter. The second shows a more realistic version of each with the included resistance values. For the thermocouple, the ideal allows current to flow freely, so the resistor is added in series. For the voltmeter, the ideal allows no current to flow, so the resistor is added in parallel to allow flow.


text:
\begin{document}

\thispagestyle{empty}

\title[projections of zero length]
{{ The sharp Hausdorff measure condition for length of projections}}

\author{Yuval Peres}
\address{Yuval Peres,
Department of Statistics, University of California, Berkeley. \newline
 {\tt peres@stat.berkeley.edu}}

\author{Boris Solomyak}\thanks{Research of Peres was
partially supported by NSF grants \#DMS-0104073 and \#DMS-0244479.
Part of this work was done while he was visiting Microsoft Research.
Research of Solomyak was supported in part by NSF grant
 \#DMS-0099814.
}

\address{Boris Solomyak, Box 354350, Department of Mathematics,
University of Washington, Seattle, WA 98195.
{\tt solomyak@math.washington.edu}}

\subjclass{Primary: 28A80.
Secondary: 28A75,
           60D05,
           28A78
}

\begin{abstract}
In a recent paper, Pertti Mattila asked  which gauge functions $\varphi$ have the
property that
for any Borel set $A \subset \R^2$ with Hausdorff measure $\Hau^\varphi(A) > 0$, the
projection of $A$ to almost every line has positive length.
We show that finiteness of $\int_0^1\frac{\varphi(r)}{r^2} dr$,
 which is known to be sufficient for this property, is also necessary
for regularly varying $\varphi$.
Our proof is based on a random construction adapted to the gauge function.
\end{abstract}

\maketitle

\section{{\bf Introduction}} \label{sec:intro}

A classical theorem of Marstrand \cite{mars}
states that if a planar Borel set $A$ has Hausdorff dimension
strictly greater than $1$, then the orthogonal projection of $A$
to almost every line has positive length.

If $A$ has dimension 1, the situation is more delicate.
Recall that given a positive function $\varphi$ on $(0,\infty)$,
 the  Hausdorff measure
$\Hau^\varphi$ is defined
$
\Hau^\varphi(A)= \lim_{\eps  \downarrow 0} \inf \Bigl\{ \sum_i \varphi(\diam \: A_i) \, : \,
A \subset \bigcup_i A_i \, ,  \: \diam \:  A_i < \epsilon \Bigr\}.
$

In his definitive survey on Hausdorff dimension and projections,
Mattila \cite{mattila} asked  which gauge functions $\varphi$ have the property that
for any Borel set $A \subset \R^2$ with $\Hau^\varphi(A) > 0$, the projection
of $A$ to almost every line has positive length.
In this paper we settle Mattila's question, showing that an integral
 condition known to be sufficient for this property is also necessary; a partial result
 in this direction was  obtained in \cite{JM}. The solution also clarifies the relation
 between Hausdorff measures and  integral-geometric measure.

\medskip

\noindent{\bf Notation.} Let $\eta$ be the
isometry-invariant measure on the space of all lines in $\R^2$, and
 define the integral-geometric measure $\I^1$ on Borel sets in $\R^2$
by $\I^1(A)= \int \#(A \cap \ell) \, d\eta(\ell)$.
(See (\ref{integg}) and \cite{matbook}, Section 5.14.)
Let $p_\theta$ denote  the
 orthogonal projection from $\R^2$ onto the line through the origin making angle
 $\theta$ with the horizontal axis.
We denote by $\Leb^m$ the $m$-dimensional Lebesgue measure.
We write $f \asymp g$ if $f \le \const\cdot g$ and $g \le \const \cdot f$
for some
uniform constant.

\begin{theorem} \label{th-main} Let $\varphi:(0,\infty) \to  (0,\infty)$ be a
(weakly) increasing
function such that
\begin{equation} \label{regul}
\varphi(r)/r^2 \mbox{ \it is (weakly) decreasing.}
\end{equation}
Then the following are equivalent:

\item{\bf (i)} $\int_0^1 \frac{\varphi(r)}{r^2}\,dr < \infty$.
\item{\bf (ii)} If a Borel set $A\subset \R^2$ satisfies $\Hau^\varphi(A) > 0$,
then
\begin{equation} \label{faveq}
\Leb^1(p_\theta(A))>0 \mbox{ \rm for almost all  } \theta \in [0,\pi) \, .
\end{equation}
\item{\bf (iii)} $\Hau^\varphi$ is absolutely continuous to  $\I^1$ on
Borel sets in $\R^2$.

\end{theorem}
The implication (i)$\Rightarrow$(ii) is known:
Indeed, (i) gives $\int_0^1 \frac{1}{r}\, d\varphi(r) < \infty$ using integration by parts.
 By \cite{Ca}, Theorem IV.1,  this implies that for sets $A$ with
$\Hau^\varphi(A) > 0$, the one-dimensional capacity of $A$ is
positive; a theorem of Kaufman \cite{kauf}  (see also \cite{matbook}, Cor.\ 9.8)
then yields (\ref{faveq}). Clearly (ii)$\Rightarrow$(iii) by definition.
We establish the new implication (iii)$\Rightarrow$(i) by a
random construction; it is intriguing that so far, deterministic constructions
yield less sharp results. See Corollary \ref{cor-int} below
for an extension of Theorem \ref{th-main} to higher dimensions.
Note that for a Borel set $A \subset \R^2$, having positive
one-dimensional capacity is not necessary for (\ref{faveq});
rectifiable curves have zero capacity but certainly satisfy
(\ref{faveq}).

What if the gauge function does not satisfy the regularity condition (\ref{regul})?
In the following lemma we show that if $\varphi$ is increasing, then
we can find another gauge function $\varphi_1$, so that
$\varphi_1(r)/r^2 \downarrow$ and $\Hau^{\varphi_1} \asymp \Hau^\varphi$. (Note, however,
that it could happen that $\int_0^1 \frac{\varphi(r)}{r^2}\,dr=+\infty$, but
$\int_0^1 \frac{\varphi_1(r)}{r^2}\,dr < \infty$; in that case,
for $A$ with $\Hau^\varphi(A)>0$, almost every projection
 has positive length.)

\begin{lemma} \label{lem-gauge} Suppose that $\varphi:(0,\infty) \to (0,\infty)$
is weakly increasing. Then there exists a weakly increasing
$\varphi_1:(0,\infty) \to (0,\infty)$
such that  $\varphi_1(r)/r^2 \downarrow$, and
\begin{equation}\label{eq-gauge}
\Hau^{\varphi_1}(A) \le \Hau^\varphi(A) \le C\cdot \Hau^{\varphi_1}(A)
\end{equation}
for all Borel sets $A \subset \R^2$.
\end{lemma}

\noindent{\bf Remark.} Joyce and M\"orters \cite{JM} constructed a
set $A \subset \R^2$ such that
$0 < \Hau^\phi(A) < \infty$, for some $\phi$ satisfying
$\int_0^1 \frac{\phi(r)^\alpha}{r^{1+\alpha}}\,dr < \infty$ for every $\alpha>1$,
such that $\Leb^1(p_\theta(A))=0$ for {\em all} $\theta$.

Next, we discuss higher dimensions.
Let $\Hau^m$ denote $m$-dimensional Hausdorff measure in $\R^n$.
Recall from \cite{matbook}, Section 5.14,
that the integral-geometric measure $\I^m$ is defined on Borel sets
$A \subset \R^n$ by
\begin{equation} \label{integg}
\I^m(A)= \int \! \int \#\Bigl(A \cap p_V^{-1}(a)\Bigr) \, d\Hau^m(a) \, d\gamma_{n,m}(V) \,
\end{equation}
where $p_V$ is the orthogonal projection to $V$, and $\gamma_{n,m}$ is the
isometry-invariant measure on the Grassman manifold of $m$-dimensional subspaces
of $\R^n$. The next corollary generalizes Theorem \ref{th-main}.
\begin{cor} \label{cor-int}
 Let $m<n$  and suppose  $\psi$ is a  positive
function on $(0,\infty)$ such that $\psi(r)/r^{m-1}$ is weakly  increasing,
$\psi(r)/r^{m+1}$ is weakly decreasing (these regularity conditions could be relaxed).
Then the following are equivalent:
\item{\bf (i)} $\int_0^1 \frac{\psi(r)}{r^{m+1}}\,dr < \infty$.
\item{\bf (ii)} If a Borel set $\Lambda \subset \R^n$ satisfies $\Hau^\psi(A) > 0$,
then $\Leb^m(p_V(\Lambda))>0$ for $\gamma_{n,m}$ almost all $m$-dimensional subspaces
$V \subset \R^n$.
\item{\bf (iii)} $\Hau^\psi$ is absolutely continuous to  $\I^m$ on
Borel sets in $\R^n$.
\end{cor}
The implication (i)$\Rightarrow$(ii) is known:
it follows from combining \cite{Ca}, Theorem IV.1,  with \cite{matbook}, Cor.\ 9.8.
 Since (ii)$\Rightarrow$(iii) is obvious, we will only need to prove
 (iii)$\Rightarrow$(i), and this will follow from the same construction we
 use to establish the corresponding implication in Theorem \ref{th-main}.

\section{{\bf Outline of the construction}}
To prove the implication (iii)$\Rightarrow$(i) in Theorem \ref{th-main},
we establish the following.

\begin{prop} \label{prop-main}
 Let $\varphi$ be a weakly increasing function that satisfies (\ref{regul})
and $\int_0^1 \frac{\varphi(r)}{r^2} dr = \infty$. Then there
exists a  Borel set $A\subset \R^2$ such that $\Hau^\varphi(A) > 0$
and $\Leb^1(p_\theta(A))=0$ for almost all $\theta \in [0,\pi)$.
\end{prop}
The set $A$ is constructed as a random Cantor set in the plane; we will show
that it has all the desired properties almost surely.
This set is a (substantial)
modification of a stochastically self-similar set constructed
in \cite{buffon}.

Denote by $\Gk_k$ the collection of $4^k$ (closed) dyadic
subsquares of the unit square $[0,1]^2$ having  side length
$2^{-k}$. We consider all dyadic subsquares as a rooted tree,
with $[0,1]^2$ being the root and $\Gk_k$ being the set of nodes
at the $k$th level. For each node there are four edges leading to
nodes at the next level (its ``offspring'').

First we inductively define random subsets $\Fk_k \subset \Gk_k$ for $k\ge 0$,
and let $\Rk_k := \bigcup \{Q:\ Q\in \Fk_k\}$.
We start with $\Fk_0 = \Gk_0 = \{[0,1]^2\}$.
Passing from $\Fk_k$ to $\Fk_{k+1}$ is either deterministic or random,
depending on $k$.
Given $\Fk_{k}$, a set  of
dyadic squares of side length $2^{-k}$, we partition each of them into
four dyadic subsquares of side length $2^{-k-1}$.
Now, if the step is deterministic, then we keep all the squares, so that
$\#\Fk_{k+1} =4 \# \Fk_k$ and $\Rk_{k+1} = \Rk_k$.
If the step is random, then we choose one of the four
subsquares uniformly at random, all these choices being independent.
This way we obtain a subset $\Fk_{k+1}$ with the same cardinality as
$\Fk_k$.  The gauge function $\varphi$ dictates whether we make a deterministic
or random step. If we
do $\alpha_k$ deterministic steps for $i< k$, then $\card \Fk_k =
4^{\alpha_k}$, and  we make sure that $4^{-\alpha_k} \asymp \varphi(2^{-k})$.
Once $\Fk_k$ and $\Rk_k$ are defined, we consider $\Rk = \bigcap_{k=1}^\infty
\Rk_k$, which satisfies $0 < \Hau^\varphi(\Rk) < \infty$.

We do not know whether the resulting random set has a.e.\
projection of zero length (almost surely),
so we modify the construction slightly,
removing certain squares at specific levels chosen in advance. This
yields a new random set $\Rk'$ which has all the desired properties.

\section{{\bf Construction in detail}}

Consider a gauge function $\varphi(r)$ as in Theorem~\ref{th-main}.
Without loss of generality, we can assume that $\varphi(1)=1$.
Let
$$
\alpha_n = \lfloor \log(\varphi(2^{-n}))/\log(1/4) \rfloor\ \ \ \mbox{for}\ \
n\ge 0,
$$
so that $\alpha_0=0$ and
\begin{equation} \label{defal}
\varphi(2^{-n}) \le 4^{-\alpha_n} \le 4\cdot \varphi(2^{-n}).
\end{equation}
Since $\varphi(r)$ is increasing, we have $\alpha_n \ge \alpha_{n-1}$, and since
$\varphi(r)/r^2$ is decreasing, we have $\alpha_n \le \alpha_{n-1}+1$.

Now,
$$
\infty = \int_0^1\frac{\varphi(r)}{r^2}dr = \sum_{n=0}^\infty
\int_{2^{-n-1}}^{2^{-n}} \frac{\varphi(r)}{r^2}dr  \le \sum_{n=0}^\infty
2^{n+1} \varphi(2^{-n}) \le \sum_{n=0}^\infty 2^{n+1} 4^{-\alpha_n},
$$
so
\begin{equation} \label{series}
\sum_{n=0}^\infty \lam_n = \infty,\ \ \ \mbox{where}\ \ \lam_n:=
2^n 4^{-\alpha_n}.
\end{equation}

Now we specify when we perform deterministic steps. The transition from
$\Fk_{n-1}$ to $\Fk_n$ is deterministic if $\alpha_n>\alpha_{n-1}$,
that is, $\alpha_n = \alpha_{n-1}+1$. Otherwise, the
transition is random, as described in Section 2. This is well-defined,
and we make $\alpha_n$ deterministic steps on levels $i<n$, so
$\# \Fk_n = 4^{\alpha_n}$. Recall that $\Rk_n = \bigcup_{B\in \Fk_n} B$
and $\Rk = \bigcap_{n=1}^\infty \Rk_n$.
Consider the probability measure $\mu$ on $\Rk$ defined as the weak$^*$
limit of uniform measures on $\Rk_n$. We claim that there exists $C_1>0$
such that
\begin{equation} \label{est1}
\mu(B_r) \le C_1\cdot \varphi(r)
\end{equation}
for any ball of radius $r$. Indeed,
up to a multiplicative constant, we can replace the ball $B_r$ in
(\ref{est1}) by a square $B \in \Fk_{n}$ where $2^{-n} \asymp r$.
We have $\mu(B) = 1/\#\Fk_n = 4^{-\alpha_n} \asymp \varphi(2^{-n})$,
and the desired estimate follows. Thus, $\Hau^\varphi(\Rk)>0$ by the Mass
Distribution Principle. We can assume that $\lam_n$ are bounded,
\begin{equation} \label{supcon}
\lam_n < C_2,
\end{equation}
for all $n$.
Indeed, the set $\Rk_n$ consists of $4^{\alpha_n}$ squares of side length
$2^{-n}$. If (\ref{supcon}) does not hold, then  $\inf (4^{\alpha_n} 2^{-n})
=0$, hence $\Hau^1(\Rk)=0$, and then all projections of $\Rk$ have
zero length.

\medskip

Next we specify how we modify the set $\Rk$ by removing certain squares
at  prescribed levels.
We label the four dyadic subsquares of a square as in Figure 1.

\begin{figure}
\centerline{\epsfxsize=1.in \epsffile{fig3.ps}}
\caption{Labeling subsquares.}
\end{figure}

This labeling induces a natural addressing scheme for each dyadic
square $B\in \Gk_k$. The address has length $k$ and the symbols
are from $\{0,1,2,3\}$; we write it as $\om(B) = \{\om_i(B)\}_{i=1}^{k}$.

For $B \in \Gk_{n}$ and $1 \le i \le n$
define
$$
Y_i(B) =
\left\{ \begin{array}{ll} \lam_i,
                                      & \mbox{if}\ \om_{i}(B)\in \{0,2\}; \\
                                   0, & \mbox{if}\ \om_{i}(B)\in \{1,3\}.
\end{array} \right.
$$
We can consider $\Gk_{n}$ as a discrete probability space with the
uniform measure. Then $\{Y_i\}_{i=1}^{n}$ are independent random variables,
such that $\Pro(Y_i = \lam_i) = \Pro(Y_i = 0) = \half$.
Let
$$
S_n(B) = \sum_{i=1}^{n} Y_{k_i}(B),
$$
where $k_i$ is the $i$-th deterministic step, so that $\alpha_{k_i-1}=i-1$ and
$\alpha_{k_i}=i$. Denote
$$
\gam_i := \lam_{k_i} = 2^{k_i} 4^{-\alpha_{k_i}} = 2^{k_i} 4^{-i};
$$
then
$\Exp (S_n) = \half \sum_{i=1}^{n} \gam_{i}$ and
$\Var (S_n) =  \frac{1}{4} \sum_{i=1}^{n} \gam_i^2$.
Observe that $S_n(B)$ depends only on the digits of $B$ corresponding to
deterministic steps, hence it is independent of the event  $B\in \Fk_{k_n}$
which can be considered on the same probability space $\Gk_{k_n}$.

We say that $B\in \Gk_{k_n}$ is {\it deviant\/} if
$$
|S_n(B) - \Exp(S_n)| >  \Exp(S_n)/2.
$$
Using Chebyshev's inequality we obtain
$$
\Pro(B \mbox{\ is  deviant}) \le \frac{\Var(S_n)}{(\half \Exp(S_n))^2}
= \frac{\sum_{i=1}^{n} \gam_{i}^2}{(\sum_{i=1}^{n} \gam_{i})^2} \le
C_2\left(\sum_{i=1}^{n} \gam_{i}\right)^{-1},
$$
since $\gam_{i}^2 \le C_2 \gam_{i}$ by (\ref{supcon}).
Denoting by $\Dk_{n}$ the collection of deviant squares in $\Gk_{k_n}$
and using independence of $B$ being deviant from the event
$(B\in \Fk_{k_n})$,
we obtain that
$$
\# (\Fk_{k_n} \cap \Dk_{n})
\le C_2 \cdot \# \Fk_{k_n} \left(\sum_{i=1}^{n} \gam_{i}\right)^{-1}\,.
$$
By the definition of measure $\mu$ we have
\begin{equation} \label{dop2}
\mu\left(\Rk \setminus \bigcup_{B \in \Dk_{n}} B \right) \ge 1 -
C_2 \left(\sum_{i=1}^{n} \gam_{i}\right)^{-1}\,.
\end{equation}

Note that $\alpha_i = j$ for $k_j \le i < k_{j+1}$, hence denoting $k_0=0$
we have
$$
\sum_{i=0}^{k_n-1} \lam_i =
\sum_{j=0}^{n-1} \sum_{i=k_j}^{k_{j+1}-1} \lam_i
= \sum_{j=0}^{n-1} 4^{-j} \sum_{i=k_j}^{k_{j+1}-1} 2^i <
\sum_{j=0}^{n-1} 4^{-j} 2^{k_{j+1}} = 4 \sum_{j=1}^n \gam_j.
$$
Observe that $k_n\to \infty$, otherwise $\Rk$ is a finite set.
Thus, $\sum_{i=1}^{n} \gam_{i}\to \infty$ as $n\to\infty$ by
(\ref{series}), and
we can find $n(j) \in \Nat$, $j\ge 1$,  such that
\begin{equation} \label{dop3}
C_2^{-1}\sum_{i=1}^{n(j)} \gam_{i} > 2^{j+1}.
\end{equation}
Let
$$
\Rk' := \Rk \setminus \bigcup_{j=1}^\infty \bigcup_{B\in \Dk_{n(j)}} B.
$$
We have
$\mu(\Rk') \ge 1 - \half \sum_{j=1}^\infty 2^{-j} = \half$ by (\ref{dop2})
and (\ref{dop3}).  Thus, $\mu|_{\Rk'}$ is a positive measure, and
$\mu(B_r \cap \Rk') \le \mu(B_r) \le C_1\cdot \varphi(r)$ for any ball $B_r$ of
radius $r$, hence $\Hau^\varphi(\Rk') > 0$ by the Mass Distribution Principle.

\section{{\bf Proof of Proposition \ref{prop-main} and  Theorem~\ref{th-main} }}

Denote
$$
\Rk_{k_n}' = \Rk_{k_n} \setminus \bigcup_{B\in \Dk_n} B.
$$

\begin{lemma} \label{lem-main}
Let $\ell$ be a line intersecting $[0,1]^2$
that does not hit any vertices of the squares in $\Gk_{k_n}$. Then
$$
\Pro(\Rk'_{k_n} \cap \ell \ne \es) \le 64\cdot
\left(\sum_{i=1}^{n} \gam_i\right)^{-1}.
$$
\end{lemma}

\noindent {\sc Proof of Theorem~\ref{th-main} assuming
Lemma~\ref{lem-main}.}
Observe that $\Rk' = \bigcap_{j=1}^\infty \Rk_{k_{n(j)}}'$, hence
Lemma~\ref{lem-main} implies that the probability of $\ell$ hitting
$\Rk'$ equals zero.
Let $\Th \in [0,\pi]$ be such that the
line $y\cos\Th = x\sin\Th$ is orthogonal to $\ell$, and let $\bn$
be the unit normal vector for $\ell$.
Then by Fubini's Theorem,
$$
\Exp\left[ \Leb^1(p_\theta(\Rk')) \right] = \int_{\R} \Pro(\Rk' \cap
(\ell + t \bn) \ne \es)\,dt =0 \,.
$$
The proposition, and hence  the theorem, follow  by integrating over $\Th$. \qed

\medskip

\noindent {\sc Proof of Lemma~\ref{lem-main}.}
Let $\alpha$ be the angle that $\ell$ forms with the horizontal.
First we assume that $\alpha \in [0,\pi/2]$ and
then indicate how to consider the case $\alpha \in (\pi/2,\pi]$.
Let
$$
A_n(\ell) = \card\{B\in \Gk_{k_n}:\ B \cap \ell \ne \emptyset\}.
$$
Observe that
\begin{equation} \label{eq-estan} A_{n}(\ell) \le  2^{k_n+1}.
\end{equation}
To verify this  we may assume, using symmetry, that $\ell$ forms
an angle in $[0,\pi/4]$ with the horizontal. Then $\ell$
intersects at most two squares in each of the $2^{k_n}$ columns of
$\Gk_{k_n}$, and (\ref{eq-estan}) follows.

Recall that none of the squares in $\Rk_{k_n}'$ are
deviant, i.e.
$$
B \in \Rk_{k_n}' \ \Rightarrow\ (1/4) \sum_{i=1}^{n} \gam_i \le S_n(B) \le
(3/4) \sum_{i=1}^{n} \gam_i,
$$
where $S_n(B) = \sum \{\gam_i:\ i\le n,\,\om_{k_i}(B)\in \{0,2\}\}$.

(i) Say that $B \in \Gk_{2n}$ is {\bf $0$-rich} if
$$
\sum\{\gam_i:\ i \le n,\,\om_{k_i}(B) = 0\} \ge (1/8)
\sum_{i=1}^{n} \gam_i.
$$

(ii) Say that $B \in \Gk_{2n}$ is {\bf $2$-rich} if
$$
\sum\{\gam_i:\ i \le n,\,\om_{k_i}(B) = 2\} \ge (1/8)
\sum_{i=1}^{n} \gam_i.
$$

\noindent By the definition of deviant
squares, every non-deviant square is
either  $0$-rich or  $2$-rich  (or both).
Consider the events
$$
Z_i = \Big\{ \exists\, B\in \Fk_{k_n}:\ B\ \mbox{is $i$-rich}\ \&\ B\cap
\ell \ne \emptyset \Big\} \ \ \ \mbox{for}\ \ i=0,2.
$$
Since $\Fk_{k_n}'$ contains only  non-deviant
squares from $\Fk_{k_n}$, we have
$$
\Pro(\Rk_{k_n}' \cap \ell \ne \es) \le \Pro(Z_0)+\Pro(Z_2).
$$
Let us estimate $\Pro(Z_0)$. We have
\begin{equation} \label{eq-condex}
\E[\card\{B\in \Fk_{k_n}:\ B\cap \ell \ne \es\}\,|\, Z_0] \le
\frac{\E[\card\{B\in \Fk_{k_n}:\ B\cap \ell \ne \es\}]}{\Pro(Z_0)}\,.
\end{equation}
Observe that
\begin{equation} \label{eq-dop1}
\Pro(B\in \Fk_{k_n}) = \#\Fk_{k_n}/\#\Gk_{k_n} = 4^{n}/2^{2k_n}=
4^{n-k_n}
\end{equation}
for any $B\in \Gk_{k_n}$.
Writing
$$
\card\{B\in \Fk_{k_n}:\ B\cap \ell \ne \es\} =
\sum_{B\in \Gk_{k_n}} {\bf 1}_{\{B\in \Fk_{k_n}:\ B\cap \ell \ne \es\}}
$$
and using (\ref{eq-dop1}) we obtain by (\ref{eq-estan}) that
\begin{equation} \label{eq-ex}
\E\Big[\card\{B\in \Fk_{k_n}:\ B\cap \ell \ne \es\}\Big] =
A_n(\ell)\cdot 4^{n-k_n} \le 2\cdot 2^{2n-k_n}.
\end{equation}

Thus by (\ref{eq-condex}),
\begin{equation} \label{eq-condez}
{\Pro(Z_0)} \le
\frac{2^{2n-k_n+1}} {\E[\card\{B\in \Fk_{k_n}:\ B\cap \ell \ne \es\}\,|\, Z_0]}
\, .
\end{equation}
 It remains to estimate the denominator in (\ref{eq-condez}) from below. Let
$$
\Psi_0 := \{Q\in \Gk_{k_n}:\ Q\ \mbox{is $0$-rich}\ \&\ Q\cap \ell \ne \es\}.
$$
Order the squares in $\Gk_{k_n}$ hit by $\ell$ from left to right and from
bottom to top. This is a total order by the assumption on slope of the
line $\ell$. For $Q\in \Psi_0$ consider the event
$$
Y_Q = \Big\{ \mbox{ $Q$\ \  is the first square in $\Psi_0$ such that
 $Q\in \Fk_{k_n}$} \Big\}.
$$
Then $Z_0 = \bigcup_{Q\in \Psi_0} Y_Q$ is a disjoint union, and
so, for any random variable $f$,
\begin{equation} \label{eq-conv}
\E[f\,|\,Z_0] = \sum_{Q\in \Psi_0} \frac{\Pro(Y_Q)}{\Pro(Z_0)}
\E[f\,|\,Y_Q] \ge \min_{Q \in \Psi_0}\E[f\,|\,Y_Q].
\end{equation}
Fix $Q\in \Psi_0$. We have
\begin{equation} \label{eq-condex2}
\E[\card\{B\in \Fk_{k_n}:\ B\cap\ell\ne\es\}\,|\,Y_Q] =
\sum_{B\in \Gk_{k_n}:\ B\cap\ell\ne\es} \Pro(B\in \Fk_{k_n}\,|\,Y_Q).
\end{equation}
Fix $i$ such that $\om_{k_i}(Q)=0$, and denote by
$\widetilde{Q}$ the dyadic square in $\Gk_{k_i-1}$ having the
address $\om(\widetilde{Q}) = \om_1(Q)\ldots w_{k_i-1}(Q)$. The fact
that $Q\in \Fk_{k_n}$ implies that

$\widetilde{Q}\in \Fk_{k_i-1}$.
(Note that by definition, $[0,1]^2
\supset \Rk_{k_1} \supset \ldots \supset \Rk_{k_n}$.)
Recall that the step $k_i$ is deterministic, so all four offspring of
$\widetilde{Q}$ are in $\Fk_{k_i}$.
Since the slope of $\ell$ is positive, $\ell$
intersects at least $\half 2^{k_n-k_i}$ squares $B\in \Gk_{k_n}$ whose
addresses start with $\om(\widetilde{Q})k$, for $k\in \{1,2,3\}$
(see Figure 1). For each of these squares we have (using the
independence of $Y_Q$ from the random choices involving the
descendants of $\om(\widetilde{Q})k$ with  $k\in \{1,2,3\}$), that
$$
\Pro(B\in \Fk_{k_n}\,|\,Y_Q) = \Pro(B\in \Fk_{k_n}\,|\,
\widetilde{Q}\in \Fk_{k_i-1}) = 4^{k_i-k_n+n-i}.
$$
This is because we made $(n-i)$ deterministic steps between
stages $k_i$ and $k_n$, and hence $(k_n-k_i)-(n-i)$ random steps.
Therefore, the sum of $\Pro(B\in \Fk_{k_n}\,|\,Y_Q)$ over the set
of squares
$$
\Uk_i=\Big\{ B\in \Gk_{k_n}, \, : \ B\cap\ell\ne\es,\ \
\{\om_j(B)\}_1^{k_i-1} = \{\om_j(Q)\}_1^{k_i-1},\ \ \om_{k_i}(B) \in
\{1,2,3\} \Big\},
$$

is at least $\half 2^{k_n-k_i}\cdot 4^{k_i-k_n+n-i}= \half 2^{2n-k_n}\gam_i$.
Notice that the sets $\Uk_i$

are disjoint for distinct $i$ with
$\om_{k_i}(Q)=0$. Thus, by the definition of $0$-rich squares,
the right-hand side of (\ref{eq-condex2}) is at least
$$
(1/2) 2^{2n-k_n} \cdot \sum \{\gam_i:\ i\le n,\ \om_{k_i}(B)=0 \} \ge (1/16)
2^{2n-k_n} \cdot \sum_{i=1}^{n} \gam_i \,.
$$
Therefore by (\ref{eq-conv}),
$$
\E[\card\{B\in \Fk_{k_n}:\ B\cap \ell \ne \es\}\,|\, Z_0] \ge 2^{2n-k_n-4}
\cdot \sum_{i=1}^{n} \gam_i \,.
$$
Together with (\ref{eq-condez}), this
yields that
$$\Pro(Z_0) \le  32 \cdot \left({\sum_{i=1}^{n} \gam_i}\right)^{-1}.
$$

The estimate $P(Z_2) \le 32 \cdot \left({\sum_{i=1}^{n} \gam_i}\right)^{-1}$
is proved similarly. We consider the set $\Psi_2$ of squares in $\Gk_{k_n}$
hit by $\ell$ that are $2$-rich and condition on the
{\em last} square in $\Psi_2$ which belongs to $\Fk_{k_n}$.

This concludes the proof of
the lemma when the angle $\alpha$ is in $[0,\pi/2]$. In the case when
$\alpha \in [\pi/2,\pi]$ we interchange the roles of
the subsquares  $\{0,2\}$ and $\{1,3\}$ and use that for a  non-deviant
square $B$ we have
$$
\sum \{\gam_i:\ i \le n,\ \om_{k_i}(B) \in \{1,3\}\} \ge
(1/4) \sum_{i=1}^{n} \gam_i.
$$
\qed

\section{{\bf Proof of Lemma~\ref{lem-gauge}}}

Let
$$
\varphi_1(s) = \inf_r \left\{(s/r)^2 \varphi(r):\ r\le s \right\}\,.
$$
Then $\varphi_1(s) \le \varphi(s)$, so the left-hand side in (\ref{eq-gauge}) is clear.
For the right-hand side consider a cover of a set $A$ for $\Hau^{\varphi_1}$.
Let $S$ be a set in this cover where $S$ has diameter $s$. Then we can find
$r\le s$ with $(s/r)^2 \varphi(r) < 2 \varphi_1(s)$. Let $v<r$ be the largest number
of the form $s/2^k$. Then $(s/v)^2 \varphi(v) < 8 \varphi_1(s)$. Now $S$ may be
covered by a square of side $s$, hence by $4^{k+1} = 4(s/v)^2$ squares of
side $v/2$. Replacing $S$ by these squares and using them in the Hausdorff
sum for $\Hau^{\varphi}$ we conclude that $\Hau^\varphi(A) \le 32 \Hau^{\varphi_1}(A)$.

The condition $\varphi_1(s)/s^2 \downarrow$ is immediate, since
$\varphi_1(s)/s^2 = \inf \{\varphi(r)/r^2:\ r\le s\}$.
It remains to check that $\varphi_1(s) \uparrow$. Suppose $s< t$. Then
$\varphi_1(t) =\min\{A_1, A_2\}$, where
$$ A_1= \inf_{r\le s}\left\{(t^2/r^2)
\cdot \varphi(r)\right\} \mbox{ \rm and }  A_2=
\inf_{r\in [s,t]}   \left\{(t^2/r^2)\cdot \varphi(r)\right\} \,.
$$
It is clear that $A_1 \ge \varphi_1(s)$, and
$A_2 \ge (t^2/r^2)\cdot
\varphi(s) \ge \varphi(s) \ge \varphi_1(s)$ since $\varphi(r)$ is increasing.
The lemma is proved. \qed

\section{{\bf Proof of Corollary~\ref{cor-int} }}
We only need to prove that (iii)$\Rightarrow$(i). We prove the contrapositive,
i.e., given $\psi$ satisfying the regularity conditions,

we show that $\Hau^\psi$ is not absolutely continuous to  $\I^m$
as Borel measures on $\R^n$.

Let $\varphi(r)=\psi(r)/r^{m-1}$. Then Proposition \ref{prop-main}
yields a  Borel set $A\subset \R^2$ such that $\Hau^\varphi(A) > 0$
and $\I^1(A)=0$. Define $B \subset \R^n$ as the Cartesian product
$B=A \times [0,1]^{m-1} \times \{(0,\ldots,0)\}$
(appending $n-m-1$ zeros). Then it is standard to verify that
$\Hau^\psi(B)>0$ yet $\I^m(B)=0$. For the latter, observe that for a
generic $m$-dimensional subspace $V$ and $a\in V$,
the $(n-m)$-dimensional subspace
$p_V^{-1}(a)$ intersected with $\R^{m+1}\times \{(0,\ldots,0)\}$,
is a line which hits $B$ if and only if its projection on the first
two coordinates hits $A$. Thus the inner integral in (\ref{integg}) is
zero for a generic $V$.
\qed

\bibliographystyle{amsplain}

config: arxiv


TITLE: lift starts at floor 1 and goes up to floor 6. find number ways in which people can leave lift
QUESTION [1 upvotes]: A lift can record how many people leave lift at each floor. It starts at floor 1 and goes up to floor 6.
(A) How many different records are possible of people leaving the lift?
(B) what if 8 people consist of 5 men and 3 women and lift can distinguish a man from a woman?
I tried solving part A by doing $C^{n+5-1}_{5-1}$ where $n$ is total number of people in lift
And for part B, I was unable to do anything

REPLY [0 votes]: Since $n$ people each have $5$ different options to decide which floor to leave the lift, and each person's decision on which floor to leave is independent, we get the answer to [A] as $5^n$.


TITLE: Least value of $|z-w|$
QUESTION [2 upvotes]: On an argand diagram, sketch the locus representing complex numbers $z$ satisfying $|z+i|=1$ and the locus representing complex numbers $w$ satisfying arg$(w-2)=\frac{3}{4}\pi$ find the least value of $|z-w|$ for points on these loci.
I have sketched the loci. And can anyone teach me how to find the least value with a diagram? Thanks

REPLY [1 votes]: $\qquad\qquad\qquad$
See the above diagram. Here, note that $\triangle{ABC},\triangle{AOD}$ and $\triangle{DEF}$ are triangles with $45^\circ,45^\circ,90^\circ$ and with $|AC|=|AO|=|OD|=1,|ED|=|EO|-|OD|=2-1=1$.
Thus,
$$\begin{align}\text{the least value of $|z-w|$}&=|CF|\\&=|CD|+|DF|\\&=|AD|-|AC|+\frac{1}{\sqrt 2}|DE|\\&=\color{red}{\frac{3}{2}\sqrt 2-1}\end{align}$$


TITLE: How to show discontinuity of a complex function at given points?
QUESTION [0 upvotes]: Let $f:\mathbb C \to \mathbb C$ be defined by $f(z)=\sqrt{z}=\sqrt{r}e^{i arg_\alpha(z)/2}$ where $arg_{\alpha}(z)$ is the angle in $(\alpha,\alpha+2\pi]$ and $A=\{z\in \mathbb C: z=te^{i\alpha}, t\ge 0\}$, how to show the discontinuity of $f$ at $z\in A$? Any help to understand the problem?
Is the set $A$ a bunch cut or a set of bunch points?

REPLY [0 votes]: If you define $\sqrt {re^{it}}$ as $re^{it/2}$  when $0<t\leq 2\pi$ and take $\alpha =0$ then the statement is correct and it can be proved by showing that square root of $1+\frac i n$ tends to $1$ whereas square root of $1-\frac i n$ tends to $-1$. For other values of $\alpha$ you have to take $\alpha <t \leq 2\pi +\alpha $ instead of $0<\alpha \leq 2\pi$. In other words you haev to rotate $1\pm \frac i n$ by angle $\alpha$.


TITLE: What is the minimum possible exterior surface area of an open glassed-top aquarium?
QUESTION [0 upvotes]: This is a question from Khan Academy I do not understand.
An open-topped glass aquarium with a square base is designed to hold $32$ space cubic feet of water. What is the minimum possible exterior surface area of the aquarium?
This is what I realize:
Surface area = $x^2+4xy$
Volume = $32=x^2\cdot y$
How do I use this to solve the problem?

REPLY [1 votes]: As well as Cameron Buie's suggestion to express $y$ in terms of $x$, and then use calculus to find the minimum area, there is a neat trick that gives the answer more directly:
Instead of making one such aquarium, make two. Then invert the second one, and place it on top of the first one. Now you have a cuboid with volume $64$.
The minimum surface area of a cuboid with given volume is attained by a cube, which in this case has sides of length $\sqrt[3]{64}=4$. So each of the two original aquaria has a $4\times 4$ base and a height of $y=2$.
You can use this method for other problems. For instance, what is the shape of a soup bowl of maximum volume, with a given surface area? Just make two bowls, and put the second one upside-down on top of the first! We see immediately that each bowl must be hemispherical, because a sphere is the maximum volume enclosed by a given surface area.


text:
\begin{document}

\maketitle

\section{Introduction}
Consider the polynomial ring $\mathbb{C}[x_1,\ldots,x_n]$ in $n$ commuting indeterminates $x_1,\ldots, x_n$.
Let $X=\{x_1,\ldots,x_n\}$ and $[n]=\{1,\ldots,n\}$.
For a nonempty subset $S$ of $[n]$, define the linear form $X_S$ in
$\mathbb{C}[x_1,\ldots,x_n]$ by the sum
\[
X_S\coloneqq\sum_{i\in S}x_i.
\]
For integers $k,n$ satisfying $1\leq k\leq n$, define homogeneous
polynomials $B_{n,k}(X)$ of degree $\binom{n}{k}$ and $B_{n}(X)$ of
degree $2^n - 1$ by the products
\begin{equation}\label{eq:b.nk}
  B_{n,k}(X)\coloneqq\prod_{S\subseteq [n], |S|=k} X_S \hspace{.1in}
  \text{ and } \hspace{.1in}
  B_n(X) \coloneqq\prod_{k=1}^n B_{n,k}(X).
\end{equation}

For brevity, let $B_{n,k}\coloneqq B_{n,k}(X)$ and
$B_n\coloneqq B_n(X)$ when the alphabet is understood.  We refer to
$B_{n,k}$ as the \emph{$(n,k)$-th Boolean product polynomial}
and to $B_n$ as the \emph{$n$-th total Boolean product polynomial}.

Observe that the polynomials $B_{n}(X)$ and $B_{n,k}(X)$ are symmetric
under permutations of the variables.  In fact, one might consider the
Boolean product polynomials to be among the most ``natural symmetric
polynomials'' along with elementary, homogeneous, and monomial
symmetric polynomials, and Stanley's chromatic symmetric polynomials.

The main focus of this article is the Schur-positivity of all Boolean product polynomials.

\begin{theorem}\label{thm:Billey-Billera} For any  positive integers
  $k\leq n$, the $(n,k)$-th Boolean product polynomial $B_{n,k}(X)$ is
  Schur-positive. That is, there exist nonnegative integers
  $\kappa_{\lambda}^{(n,k)}$ such that
 \[
B_{n,k}(X)=\sum_{\lambda} \kappa_{\lambda}^{(n,k)}s_{\lambda}(X).
\]
Furthermore, the $n$-th total Boolean product polynomial is
  Schur-positive.
\end{theorem}

Our motivation for studying the Boolean product polynomials stems from three
long-standing open problems.  One  comes from matroid
theory/hyperplane arrangements, one comes from economics/game
theory/physics, and one comes from Hadamard's maximal determinant problem (see Section~\ref{s:remarks}).

The first interesting open problem is to find the characteristic
polynomial $\chi_{n}(t)$ for the real matroid $M_n$ spanned by 0--1
vectors in $\mathbb{R}^n$.  The same polynomial $\chi_{n}(t)$ is the
characteristic polynomial of the hyperplane arrangement corresponding
to hyperplanes given by the vanishing of $X_S$ for all nonempty
$S \subseteq [n]$.  This arrangement has also appeared in the work of
\cite{CJM11}, where it is called the \emph{resonance arrangement}. One
of the main results in \cite{CJM11} is that the regions of the
resonance arrangement are in fact the chambers of polynomiality of the
genus $g$ double Hurwitz numbers.  One approach to computing
$\chi_{n}(t)$ is the \emph{finite field method} due to Athanasiadis
\cite{Athanasiadis96}, which asserts that, for large enough primes $p$,
$\chi_{n}(p)$ is the number of points in $\mathbb{F}_p^n$ in the
complement of the resonance arrangement.  The finite field method is
also described nicely in \cite{Stanley07} and \cite[Section
3.11.4]{Stanley-EC1}.  This approach was used in \cite[Lemma
5.3]{Terao11} to compute $\chi_{n}$ for small $n$.

As a variety, the resonance  arrangement is the zero locus
of the total Boolean product polynomial $B_{n}(X)$ as defined in
\eqref{eq:b.nk}.  In addition, the zero locus of $B_{n,k}(X)$ in
$\mathbb{R}^n$ is a central hyperplane arrangement in $\mathbb{R}^n$.
Klivans and Reiner \cite{Klivans-Reiner06} study the zonotope dual to
this subarrangement of the resonance  arrangement in the
context of degree sequences of hypergraphs.
 This zonotope, called the
 polytope of degree sequences in \cite{Klivans-Reiner06} is the
 Minkowski sum of the line segments $[0,\mathbf{e}_{S}]$ where $S$
 ranges over all $k$-subsets of $[n]$,\
 $\mathbf{e}_{S}\coloneqq \sum_{i\in S}\mathbf{e}_{i}$,\ and
 $\mathbf{e}_{i}$ is the i$^{th}$ unit vector in $\mathbb{R}^{n}$.

Our second motivating problem has roots in the work of Shapley
\cite{Shapley67} in his study of \emph{economic equilibria}, \emph{i.e.}
the core of an $n$-person cooperative game.  We start by considering
collections of subsets of the finite set $[n]$ as sets of vertices of
the $n$-cube $[0,1]^{n}$.
Let $2^{[n]}$ denote the set of all subsets of $[n]$, and let
$2^{2^{[n]}}$ denote the Boolean algebra on $2^{[n]}$.  A collection
$\mathcal{C} \subseteq 2^{[n]}$ is \emph{balanced} if the convex hull of
the set $\{\mathbf{e}_{S} : S \in \mathcal{C} \}$ meets the main
diagonal in $[0,1]^{n}$, {\it i.e.}, the line between
$\mathbf{e}_{\emptyset}$ and $\mathbf{e}_{[n]}$.  A collection is
\emph{unbalanced} otherwise.  The set of unbalanced collections is an
order ideal in $2^{2^{[n]}}$, while the complementary set of balanced
collections is an order filter.  Thus, it makes sense to consider
\emph{minimal balanced} collections and \emph{maximal unbalanced}
collections.  The former were first considered by Shapley, while the
latter have arisen more recently in two independent studies by
Billera-Moore-Moraites-Wang-Williams \cite{BMMWW12} and Bj\"orner
\cite{Bjorner15}.

We are interested in the enumerative problem of counting the maximal
unbalanced collections for a given $n$.  These collections of subsets
of $[n+1]$ are in bijection with the regions in the
resonance arrangement \cite{BMMWW12}.  In addition, the regions of the
resonance arrangement are said to count so-called ``generalized
retarded functions'' of quantum field theory \cite{Evans95}, while in
\cite{Terao11}, where the arrangement is called the \emph{all-subsets
  arrangement}, its regions are shown to correspond to certain
preference rankings of interest in psychology and economics.

From work of Zuev \cite{Zuev89}, it is known that the number of maximal
unbalanced collections for a given $n$ is asymptotically on the order
of $2^{n^{2}}$, while specific upper and lower bounds were derived in
\cite{BMMWW12}.  One way of obtaining the number exactly is to compute
the characteristic polynomial $\chi_{n}(t)$ of the resonance
arrangement as described above and to apply the theorem of Zaslavsky
relating regions of hyperplane arrangements to the characteristic
polynomial \cite{Zaslavsky75}, see also \cite[Thm
3.11.7]{Stanley-EC1}.  See \cite[A034997]{oeis}
for additional known results on this integer sequence.

The outline of this extended abstract is as follows.  In
Section~\ref{s:notation}, we review our notation and key theorems by
Lascoux and Pragacz.  In Section~\ref{s:main}, we prove our main
results.  As a consequence, we consider the special case of
$B_{n,n-1}(X)$ where we can give the explicit Schur expansion using
work of D\'esarm\'enien and Wachs on descent sets of derangements.  We
also generalize the Boolean product polynomials to multiple alphabets, and
give additional positivity results.  In Section~\ref{s:remarks}, we
state some additional open problems related to Boolean product
polynomials.

\smallskip

\noindent\textbf{Acknowledgments.} We thank Patricia Hersh, Steve Mitchell, and Jair
Taylor for helpful discussions.  We thank BIRS for the opportunity to
begin this collaboration at the Algebraic Combinatorics Workshop in
August 2015.

\section{Notation and Background}\label{s:notation}
Throughout this extended abstract, we fix a positive integer $n$ and
an alphabet $X=\{x_1,\ldots, x_n\}$. Denote the symmetric group on $n$
letters by $\mathfrak{S}_n$.  We refer the reader to Fulton-Harris
\cite{FultonHarris91}, Macdonald \cite{Macdonald95} or Stanley
\cite[Chapter 7]{Stanley-EC2} for a detailed treatment of the
combinatorics of symmetric polynomials and its relation to the
representation theory of both the symmetric group and the general
linear group.

\subsection{Partitions, tableaux and symmetric polynomials}
A \emph{partition} $\lambda$ of a positive integer $m$ is a finite ordered list of positive integers $(\lambda_1,\ldots,\lambda_k)$ such that $\sum_{i=1}^k\lambda_i=m$.
We call the $\lambda_i$ the \emph{parts} of $\lambda$ and denote the number of parts by $\ell(\lambda)$. We denote the sum of the parts of $\lambda$ by $|\lambda|$.
If $\lambda$ is a partition of $m$, we denote this by $\lambda\vdash m$.
Pictorially we depict $\lambda=(\lambda_1,\cdots,\lambda_k)\vdash m$
via its Young diagram drawn in French notation, which is a left-justified array of $m$ boxes with $\lambda_i$ boxes in row $i$ from the bottom.
Finally, we denote the unique partition of $0$ by $\varnothing$.

A \emph{semistandard Young tableau} $T$ of shape $\lambda$ is a filling of the boxes of its Young diagram with positive integers such that the entries in each row increase weakly when read from left to right, whereas the entries in each column increase strictly when read from bottom to top.
For any positive integer $m$ and partition $\lambda$, we denote by $\ssyt{\lambda,m}$ the set of semistandard Young tableaux $T$ of shape $\lambda$ satisfying the condition that their entries do not exceed $m$.
A semistandard Young tableau $T$ of shape $\lambda$ with distinct
entries drawn from the set $[|\lambda|]$ is said to be \emph{standard}.
An entry $i$ in a standard Young tableau (abbreviated SYT) $T$ is a \emph{descent} if $i+1$ belongs to a row strictly above that occupied by $i$. Otherwise, it is an \emph{ascent}.

The symmetric group $\mathfrak{S}_n$ acts on
$\mathbb{C}[x_1,\ldots,x_n]$ by permuting variables. The resulting
ring of invariants, denoted by $\Lambda_n$, is the well-known ring of
\emph{symmetric polynomials} in $n$ variables. It is a polynomial
algebra generated by the $e_p(X)$ for
$1\leq p\leq n$ defined by
\begin{align}\label{eqn:definition of elementary}
e_p(X)=\sum_{1\leq j_1<\cdots <j_p\leq n} x_{j_1}\cdots x_{j_p}.
\end{align}
We refer to $e_p(X)$ as the \emph{$p$-th elementary symmetric polynomial}. Given a partition $\lambda=(\lambda_1,\ldots, \lambda_k)$, define $e_{\lambda}(X)$ multiplicatively by setting $e_{\lambda}(X)=e_{\lambda_1}(X)\cdots e_{\lambda_k}(X)$. Furthermore, set $e_{\varnothing}(X)=1$.
The ring of symmetric polynomials is a graded ring with the grading given by setting $\deg(e_p(X))=p$. The $d$-th degree graded piece, denote by $\Lambda_n^{d}$, is the $\mathbb{C}$-linear span of the $e_{\lambda}(X)$ where $\lambda\vdash d$ and $\ell(\lambda)\leq n$.
The ring of symmetric polynomials is endowed with a distinguished involution $\omega$ that maps $e_{\lambda}(X)$ to the \emph{complete homogeneous symmetric polynomial} $h_{\lambda}(X)$.

From \eqref{eqn:definition of elementary}, it is clear how to define $e_p(Y)$ for any finite alphabet $Y$.
Note further that $e_p(Y)$ is $0$ if $p>|Y|$.
Given a positive integer $1\leq k\leq n$, define the following new alphabet
$$\kboolean{k}\coloneqq \left\lbrace\displaystyle X_S= \sum_{i\in S}x_i :  S\subseteq [n], |S|=k\right\rbrace.$$
The $(n,k)$-th Boolean product polynomial can alternatively be written as $e_{\binom{n}{k}}(\kboolean{k})$.

The most important linear basis of $\Lambda_n$ is given by the Schur functions $s_{\lambda}(X)$ for all partitions $\lambda$. Consider $T\in \ssyt{\lambda,n}$ and let $\cont{T}=(\alpha_1,\ldots,\alpha_n)$ be the ordered sequence of nonnegative integers where $\alpha_i$ is the number of instances of $i$ in $T$, for $1\leq i\leq n$.
Let $X^{\cont{T}}\coloneqq \prod_{i=1}^n x_i^{\alpha_i}$.
The Schur function $s_{\lambda}(X)$ is defined as follows.
\begin{align}
s_{\lambda}(X)=\sum_{T\in \ssyt{\lambda,n}} X^{\cont{T}}.
\end{align}
Since we also work with alphabets other than $X$, we remark here that
to define $s_{\lambda}(Y)$ for any finite alphabet $Y$, the sole
change required, other than changing $X$ to $Y$,  is to replace $n$ by the cardinality of $Y$ throughout.

\begin{example}\label{ex:(3,2)-boolean sum}
If $n=3$ and $k=2$, the alphabet $\kboolean{2}= \{x_1+x_2,x_1+x_3,x_2+x_3\}$.  We have

\begin{align*}
e_1(\kboolean{2})=&(x_1+x_2)+(x_1+x_3)+(x_2+x_3)= 2s_{(1)}(x_1,x_2,x_3),\\
e_2(\kboolean{2})= &(x_1+x_2)(x_1+x_3)+(x_1+x_2)(x_2+x_3)+(x_1+x_3)(x_2+x_3)\\ =&2s_{(11)}(x_1,x_2,x_3)+s_{(2)}(x_1,x_2,x_3),\\
e_3(\kboolean{2})= &(x_1+x_2)(x_1+x_3)(x_2+x_3)=s_{(21)}(x_1,x_2,x_3).
\end{align*}
As a more involved example, consider $n=5$, $k=3$ and $X=\{x_1,\ldots,x_5\}$.
The reader may verify that $e_{10}(X^{(3)})$ equals the expression below, where the commas and parenthesis in our notation for partitions and the alphabet $X$ have all been omitted:
\begin{align*}
 &6s_{32221} + 9s_{3 3 2 1 1} + 3s_{3 3 2 2} + 3s_{3 3 3 1} + 9s_{4 2 2 1 1} + 3s_{4 2 2 2} + 6s_{4 3 1 1 1} + 9s_{4 3 2 1} + 3s_{4 3 3} + 3s_{4 4 1 1} \\ &+ 3s_{4 4 2} + 4s_{5 2 1 1 1} + 4s_{5 2 2 1} + 4s_{5 3 1 1} + 4s_{5 3 2} + 2s_{5 4 1} + s_{6 1 1 1 1} + s_{6 2 1 1} + s_{6 2 2} + s_{6 3 1}.
\end{align*}
\end{example}
Example~\ref{ex:(3,2)-boolean sum} suggests that the $e_{p}(\kboolean{k})$ expand positively in terms of Schur functions.
This is indeed true and to establish this fact, we need a geometric perspective on obtaining the alphabet $\kboolean{k}$ starting from $X$.

\subsection{Schur functors and Chern classes of vector bundles}

We briefly discuss  some representation theory of the general linear group $\mathrm{GL}_n(\mathbb{C})$ and the symmetric group  $\mathfrak{S}_n$.
The reader is referred to \cite[Lecture 6]{FultonHarris91} for more details.
Consider a  vector space $V$ of dimension $n$ over $\mathbb{C}$.
 We denote the irreducible polynomial representation of $\mathrm{GL}_{n}(\mathbb{C})$ corresponding to $\lambda\vdash m$ by $\schurfunc{\lambda}(V)$, obtained by acting with the Young symmetrizer $c_{\lambda}$ on $V^{\otimes m}$.  We assume here that $\ell(\lambda)\leq n$. The association $V\mapsto \schurfunc{\lambda}(V)$ is a functor in the category of finite dimensional vector spaces and is called the \emph{Schur functor}.
In particular, $\schurfunc{(1^k)}(V)$ corresponds to the exterior power $\Exterior^{k}V$, whereas $\schurfunc{(k)}(V)$ corresponds to the symmetric power $\mathrm{Sym}^k V$.
The connection to the ring of symmetric polynomials is made explicit by the \emph{character} map  $\mathsf{Ch}$ defined by
\[
\mathsf{Ch}(\schurfunc{\lambda}(V))=s_{\lambda}(X).
\]
Just as partitions index the irreducible polynomial representations of $\mathrm{GL}_n(\mathbb{C})$, they index the irreducible representations of  $\mathfrak{S}_n$.
The link to Schur polynomials is made manifest by the map $\mathsf{Frob}$, referred to as the \emph{Frobenius characteristic}, that sends the irreducible representation of $\mathfrak{S}_n$ indexed by $\lambda\vdash n$ to $s_{\lambda}(X)$.

We turn our attention to Chern classes of vector bundles over a smooth projective variety $V$. The reader is referred to \cite{Fulton84} for further details. We  merely collect facts that  allow us to cast the question of the Schur-positivity of the Boolean product polynomials as one involving Chern roots.
Let $\mathcal{E}$ be a vector bundle of rank $r$ over  $V$. The \emph{total Chern class} $c(\mathcal{E})$ is the sum of the individual Chern classes
\[
c(\mathcal{E})=1+c_1(\mathcal{E})+\cdots +c_r(\mathcal{E}).
\]
Note that $c_i(\mathcal{E})=0$ for all $i>r$ \cite[Theorem 3.2a]{Fulton84}.
If one assumes temporarily that $\mathcal{E}$ is the direct sum of line bundles $\mathcal{L}_1,\ldots, \mathcal{L}_r$, then the Whitney-sum property \cite[Theorem 3.2e]{Fulton84} implies that
\[
c(\mathcal{E})=\prod_{i=1}^{r}(1+c_1(\mathcal{L}_i)).
\]
If $E$ is not a direct sum, the splitting principle \cite[Remark 3.2.3]{Fulton84} says that by constructing an appropriate filtration of $\mathcal{E}$ where the successive quotients are line bundles, one may still factor the total Chern class of $\mathcal{E}$ formally as
$
c(\mathcal{E})=\prod_{i=1}^{r}(1+\alpha_i).
$
The $\alpha_i$ for $1\leq i\leq r$ are said to be the \emph{Chern roots} of $\mathcal{E}$.
We treat  Chern roots as formal variables.
The observation \cite[Remark 3.2.3c]{Fulton84} that is key for us is that the Chern roots of $\schurfunc{(1^k)}(\mathcal{E})=\Exterior^{k}\mathcal{E}$ for any positive integer $1\leq k\leq r$ are given by
\[
\left\lbrace\sum_{i\in S}\alpha_i: S\subseteq [r], |S|=k\right\rbrace.
\]
This should remind the reader of the construction of the alphabet $X^{(k)}$  from $X$.

From this point onwards, fix a complex vector bundle $\mathcal{E}$ of
rank $n$.

Given a positive integer $k$, let $\delta_k$ be the partition of
staircase shape $(k,k-1,\ldots,1)$.  We have the following influential
theorem due to Lascoux.

\begin{theorem}\cite{Lascoux78}\label{thm:Lascoux}
The total Chern class of $\Exterior^{2}\mathcal{E}$ and
$\mathrm{Sym}^{2}\mathcal{E}$ is Schur-positive in terms of the Chern
roots $x_1,\ldots, x_n$ of $\mathcal{E}$.  Specifically, there exist
integers $d_{\lambda,\mu}\geq 0$  for $\mu \subseteq \lambda$
such that
\begin{align*}
&c(\Exterior^2\mathcal{E})=\prod_{1\leq i <j\leq n}(1+x_i+x_j)=2^{-\binom{n}{2}}\sum_{\mu\subseteq \delta_{n-1}}d_{\delta_{n-1},\mu}2^{|\mu|}s_{\mu}(X),\\
&c(\mathrm{Sym}^2\mathcal{E})=\prod_{1\leq i \leq j\leq n}(1+x_i+x_j)=2^{-\binom{n}{2}}\sum_{\mu\subseteq \delta_n}d_{\delta_n,\mu}2^{|\mu|}s_{\mu}(X).
\end{align*}
\end{theorem}
The $d_{\lambda,\mu}$ appearing in Theorem~\ref{thm:Lascoux} are defined as follows: pad  $\lambda$ and $\mu$ with $0$s so that the resulting sequences have length $n$ each. Say we obtain $(\lambda_1,\ldots,\lambda_n)$ and $(\mu_1,\ldots, \mu_n)$ from $\lambda$ and $\mu$ respectively.
Then, assuming $\mu\subseteq \lambda$,
\[
d_{\lambda,\mu}=\det\left( \binom{\lambda_i+n-i}{\mu_j+n-j}\right)_{1\leq i,j\leq n}.
\]
Determinants such as the one above are called \emph{binomial
  determinants}.  It is not immediate that these determinants are
positive.  Lascoux \cite{Lascoux78} appeals to geometric
considerations to establish positivity.  Establishing the positivity
combinatorially is the primary motivation of the seminal work of
Gessel-Viennot \cite{Gessel-Viennot85} who identify $d_{\lambda,\mu}$
as counting certain non-intersecting lattice paths in the
plane.
This combinatorial interpretation implies that the coefficients in the
expansion in Theorem~\ref{thm:Lascoux} are positive rational numbers.
To prove integrality, observe that the products yielding
$c(\Exterior^{2}\mathcal{E})$ and $c(\mathrm{Sym}^2\mathcal{E})$
expand integrally in the basis of monomial symmetric polynomials.  The
inverse of the Kostka matrix, whose entries are integral, allows us to
obtain an integral expansion in terms of Schur polynomials.  To the
best of our knowledge, there is no known combinatorial proof
establishing that $2^{\binom{n}{2}}$ divides
$d_{\delta_{n-1},\mu}2^{|\mu|}$.

Given partitions $\lambda$ and $\mu$ not necessarily comparable by containment, denote by $s_{\lambda}(\schurfunc{\mu}(\mathcal{E}))$ the Schur polynomial $s_{\lambda}$ evaluated at the alphabet comprising the Chern roots of $\schurfunc{\mu}(\mathcal{E})$.
\begin{example}\label{ex:Pragacz's function}
Let $n=3$ and $X=\{x_1,x_2,x_3\}$ consist of the Chern roots of some vector bundle $\mathcal{E}$ of rank $3$. Then we have
\begin{align*}
s_{(21)}(\schurfunc{(1^2)}(\mathcal{E}))=s_{(21)}(x_1+x_2,x_1+x_3,x_2+x_3)
= 2s_{(3)}(X)+5s_{(21)}(X)+4s_{(111)}(X).
\end{align*}
\end{example}

\begin{remark}\label{rem:Not plethysm}
The reader should not confuse the earlier operation of substituting the alphabet corresponding to the Chern roots of $\schurfunc{\mu}(\mathcal{E})$ into $s_{\lambda}$ for plethysm, which corresponds to taking the character of $\schurfunc{\lambda}(\schurfunc{\mu}(\mathcal{E}))$.
\end{remark}

We recall a theorem due to Pragacz which generalizes Lascoux's result above.
The gist of the statement is also present in \cite[Page
34]{Pragacz04}.
\begin{theorem}\label{thm:Pragacz} \cite[Corollary 7.2]{Pragacz96}
 Let $\mathcal{E}_1, \ldots, \mathcal{E}_k$ be vector bundles, and let $Y_1,\ldots, Y_k$ be the alphabets consisting of their Chern roots respectively. For partitions $\lambda, \mu^{(1)},\ldots, \mu^{(k)}$, there exists nonnegative integers $c_{(\nu^{(1)},\ldots,\nu^{(k)})}^{\lambda, (\mu^{(1)},\ldots,\mu^{(k)})}$ such that
 \[
 s_{\lambda}(\schurfunc{\mu^{(1)}}(\mathcal{E}_1)\otimes \cdots \otimes \schurfunc{\mu^{(k)}}(\mathcal{E}_k))=\sum_{\nu_1,\ldots, \nu_k} c_{(\nu^{(1)},\ldots,\nu^{(k)})}^{\lambda,(\mu^{(1)},\ldots,\mu^{(k)})} s_{\nu_1}(Y_1)\cdots s_{\nu_k}(Y_k).
 \]

\end{theorem}
\noindent Pragacz's proof of Theorem~\ref{thm:Pragacz} relies on deep work of
Fulton-Lazarsfeld \cite{Fulton-Lazarsfeld83} in the context of
numerical positivity. The Hard Lefschetz theorem is a key component in
the aforementioned work.

\section{Schur-positivity using Chern roots}\label{s:main}

In this section, we establish the Schur-positivity of $B_{n,k}(X)$ and
$B_n(X)$.    Additional consequences are then described below.

\begin{proof}[Proof of Theorem~\ref{thm:Billey-Billera}]
Let $\mathcal{E}$ be a complex vector bundle of rank $n$.  Observe
that $c(\Exterior^{2}\mathcal{E})=\sum_p e_p(X^{(2)})$ provided $X$ is
the alphabet of Chern roots of $\mathcal{E}$.  On comparing
homogeneous summands on the right hand side of the preceding equality
with those in Theorem~\ref{thm:Lascoux}, we see that Lascoux's result
yields the Schur-positivity of $e_{p}(X^{(2)})$ for each $p\geq 0$. In
particular, $B_{n,2}(X)$ is Schur-positive.  To establish positivity in the
general case, we follow the route laid out by Lascoux.

For a positive integer $k$, recall from Section~\ref{s:notation} that
the Chern roots of
$\schurfunc{(1^k)}(\mathcal{E})=\Exterior^{k}\mathcal{E}$ are given by
the elements in the alphabet $X^{(k)}$.  We have
\[
c(\Exterior^{k}\mathcal{E})=\prod_{S\subseteq [n], |S|=k} \left(1+\sum_{i\in S}x_i\right)=\sum_{p\geq 0}e_{p}(X^{(k)}).
\]
We will show that each $e_{p}(X^{(k)})$ is Schur-positive in terms of the Chern roots of $\mathcal{E}$.

From Theorem~\ref{thm:Pragacz}, we infer
that  the structure coefficients $c^{\lambda,\mu}_{\nu}$  in the
following expansion are all nonnegative,
\begin{align}\label{eqn: Pragacz}
s_{\lambda}(\mathcal{S}^{\mu}(\mathcal{E}))=\displaystyle\sum_{\nu}c^{\lambda,\mu}_{\nu}s_{\nu}(X).
\end{align}
In the case where $\mu=(1^k)$ for some positive integer $k$ and
$\lambda=(1^p)$ for some nonnegative integer
$0\leq p\leq \binom{n}{k}$, the left hand side of \eqref{eqn: Pragacz}
equals
$e_p(X^{(k)})$. This establishes the Schur-positivity of
$e_p(X^{(k)})$.
The Schur-positivity of $B_{n,k}(X)$ is the special case $p=\binom{n}{k}$.
Finally,  the Schur-positivity of  $B_n(X)=\prod_{k=1}^{n}B_{n,k}(X)$ follows from the
Littlewood-Richardson rule, which is an explicit positive combinatorial
rule to multiply Schur polynomials.
\end{proof}

\begin{remark}
One can show $B_{n,2}(X)=e_{\binom{n}{2}}(X^{(2)})=s_{\delta_{n-1}}(X)$ as in \cite[\S 3, Example 7]{Macdonald95}.
However, we do not know a combinatorial proof for the positivity of $e_p(X^{(2)})$ for $1<p<\binom{n}{2}$, or of $e_{p}(X^{(k)})$ for higher $k$ in general.
\end{remark}

\subsection{The $(n,n-1)$-Boolean product polynomial: A special case}
Let $q$ be an indeterminate and consider the $q$-deformation of $B_{n,n-1}(X)$ defined as
\begin{align}\label{eqn:q-deformation}
B_{n,n-1}(X;q)\coloneqq\prod_{i=1}^{n}(h_1(X)+qx_i).
\end{align}
To motivate the above deformation, we consider the special cases $q=0$ and $q=-1$.
Observe that $B_{n,n-1}(X;0)=h_{(1^n)}(X)$.
Let $V$ be an $n$-dimensional vector space over $\mathbb{C}$. If $\rho$ is the polynomial representation of $\mathrm{GL}_n(\mathbb{C})$ obtained by its action on $V^{\otimes n}$, then its character $\mathsf{Ch}(\rho)$ is equal to $h_{(1^n)}(X)$.
Furthermore, recall that $h_{(1^n)}(X)$ is also the Frobenius characteristic of the regular representation of $\mathfrak{S}_n$. Thus, in the $q=0$ case, we recover well-known representations.

The case $q=-1$ is more interesting.
 Clearly, $B_{n,n-1}(X;-1)$ equals $B_{n,n-1}(X)$.  On expanding  the product in \eqref{eqn:q-deformation}, we obtain
 \begin{align}
 B_{n,n-1}(X)=\sum_{j=0}^{n}(-1)^{j}e_{j}(X)h_{(1^{n-j})}(X).
 \end{align}
 Comparing the expression on the right hand side with the equality in \cite[Theorem 8.1]{GesselReutenauer93}, we conclude that $B_{n,n-1}(X)$ equals the  symmetric function denoted by $D_n$ therein.
 The symmetric function $D_n$ was introduced by Desarmenien and Wachs \cite{DesarmenienWachs88} in the context of descents sets of derangements.
Given the expansion of $D_n$ in the basis of fundamental quasisymmetric functions, we obtain the following result.
\begin{theorem}

   For $n\geq 2$, we have the Schur-positive expansion $\displaystyle
   B_{n,n-1}(X)=\sum_{\lambda\vdash n}a_\lambda s_{\lambda}(X) $ where
   $a_\lambda$ is the number of $T\in SYT(\lambda)$ with smallest
   ascent given by an even number.
 \end{theorem}

 \iffalse
 \begin{example}
 Let $n=4$ and $X=\{x_1,x_2,x_3,x_4\}$. Then
 \begin{align*}
 B_{4,3}(X)=&(x_1+x_2+x_3)(x_1+x_2+x_4)(x_1+x_3+x_4)(x_2+x_3+x_4)\\
 =& s_{(1,1,1,1)}(X) + s_{(2,1,1)}(X)+ s_{(2, 2)}(X) + s_{(3, 1)}(X).
 \end{align*}
 Here are the four SYTs  with $4$ boxes each whose smallest ascent (in the shaded box) is even.
 \[
 \ytableausetup{mathmode,boxsize=0.9em}
 \begin{ytableau}
 *(blue!40)4\\3\\2\\1
 \end{ytableau}
 \hspace{10mm}
 \begin{ytableau}
 4\\*(blue!40)2\\1 &3
 \end{ytableau}
 \hspace{10mm}
 \begin{ytableau}
 *(blue!40)2 & 4\\1 &3
 \end{ytableau}
 \hspace{10mm}
 \begin{ytableau}
 *(blue!40)2\\1 & 3 & 4
 \end{ytableau}
 \hspace{10mm}
 \]
 \end{example}
 \fi
Now consider the case where $q$ is a positive integer. We have
 \begin{align}\label{eqn:generic q}
 B_{n,n-1}(X;q)=\sum_{j=0}^{n}q^{j}e_{j}(X)h_{(1^{n-j})}(X).
 \end{align}
 From \eqref{eqn:generic q}  it is clear, for instance by the Pieri rule, that $B_{n,n-1}(X;q)$ is Schur-positive.
 We briefly remark on how to construct (ungraded) $\mathfrak{S}_n$-modules whose Frobenius characteristic is $B_{n,n-1}(X;q)$.
 Let $\mathbf{1}$ denote the trivial character of the Young subgroup $\mathfrak{S}_1^{j}\times \mathfrak{S}_{n-j}$ of $\mathfrak{S}_n$.
 Then the Frobenius characteristic of the induced character $\mathbf{1}\!\!\uparrow_{\mathfrak{S}_1^{j}\times \mathfrak{S}_{n-j}}^{\mathfrak{S}_n}$ is equal to $h_{n-j}(X)h_{(1^j)}(X)$.
 On taking a direct sum of $q^j$ copies of this induced character for $0\leq j\leq n$ and subsequently tensoring with the sign representation of $\mathfrak{S}_n$, we obtain a character with Frobenius characteristic $B_{n,n-1}(X;q)$.
 This construction is mildly unsatisfactory and one would ideally want a more `natural' graded representation where $q$ records the grading.

 We conclude with a curious observation when $q=1$. In this case, the dimension of a $\mathbb{C}\mathfrak{S}_n$-module with Frobenius characteristic $B_{n,n-1}(X;1)$ is equal to $\displaystyle\sum_{k=0}^n \frac{n!}{k!}$.
 By \cite[Theorem 10.4]{ArdilaRinconWilliams16}, this is also the number of positroids on
 $[n]$. In ongoing work, we are investigating a natural $\mathfrak{S}_n$-action on the distinguished
 indexing set for positroids given by decorated permutations.
 See \cite[A000522]{oeis} for many further interpretations of this sequence of dimensions.

\subsection{The case of two alphabets}
We investigate the case where we have two different alphabets.
The result that follows allows for further generalization to the case of multiple alphabets.

Let $X=\{x_1,\ldots, x_n\}$ and $Y= \{y_1,\ldots, y_m\}$ be two distinct alphabets.
Note that the cardinalities of $X$ and $Y$ can be distinct.
Assume further that $X$ and $Y$ consist of Chern roots of vector bundles $\mathcal{E}$ and $\mathcal{F}$ of ranks $n$ and $m$ respectively.
 Given nonempty subsets $S\subseteq [n]$ and $T\subseteq [m]$, define the subset sums
\begin{align*}
X_S\coloneqq\sum_{i\in S}x_i \text{ and }
Y_T\coloneqq\sum_{i\in T}y_i.
\end{align*}
Fix positive integers $j$ and $k$. Consider the following product that naturally generalizes the $(n,k)$-th Boolean product polynomial.
\begin{align*}
\mathcal{P}_{j,k}(X,Y)\coloneqq\prod_{\substack{S\subseteq [n]\\|S|=j}}\prod_{\substack{T\subseteq [m]\\|T|=k}} (X_S+Y_T).
\end{align*}
This expression is clearly symmetric in the $X$ variables and $Y$ variables. Note further that $\mathcal{P}_{j,k}(X,Y)$ is equal to $e_p(\Exterior^{j}\mathcal{E}\otimes \Exterior^{k}\mathcal{F})$ where $p=\binom{n}{j}\binom{m}{k}$.
 Therefore, by invoking Theorem~\ref{thm:Pragacz} again, we obtain the following extension.
 \begin{theorem}\label{cor:multiple alphabets}
The bivariate polynomial $\mathcal{P}_{j,k}(X,Y)$ is Schur-positive. That is, there exist nonnegative integers $a_{\lambda\mu}$ such that
\begin{align*}\label{eqn: dual Cauchy generalization}
\mathcal{P}_{j,k}(X,Y)
=\sum_{\lambda,\mu}a_{\lambda\mu}s_{\lambda}(X)s_{\mu}(Y).
\end{align*}
\end{theorem}
\noindent Observe that the Schur-positivity in Theorem~\ref{cor:multiple alphabets} subsumes that in the statement of Theorem~\ref{thm:Billey-Billera} if we pick exactly one of $j$ or $k$ to equal $0$.

 \section{Further remarks}\label{s:remarks}
\begin{enumerate}
\item Recall that part of motivation for studying Boolean product polynomials was to understand the matroid $M_n$ spanned by nonzero $n$-vectors with components $0$ or $1$.  We should note here that understanding the real linear algebra of these vectors can go much deeper than knowledge of the matroid $M_{n}$, which effectively only needs to know for each $n \times n$ matrix $A$ of $0's$ and $1's$ whether $\det A$ is zero or not.  For example, to ``know'' the \emph{arithmetic matroid}  of all 0-1 vectors, one needs to know, in addition, the absolute value $|\det A|$ of each such matrix.  Now to really know the possible determinants of all 0-1 matrices would include the solution of the problem of Hadamard, \emph{i.e.}, whether there is an $n \times n$ Hadamard matrix whenever $n=4k$.  The reason for this is that for each 0-1 $n \times n$ matrix $A$, $\det A \le (n+1)^{(n+1)/2}/2^{n}$ with equality if and only if there is a Hadamard matrix of order $n+1$.

\item From the work of Gessel-Reutenauer \cite[Theorem 3.6]{GesselReutenauer93}, it  follows that $B_{n,n-1}(X)$ is also related to representations of the free Lie algebra. Let $V$ be an $n$-dimensional vector space over $\mathbb{C}$.
Then we have
\[
B_{n,n-1}(X)=\sum_{\substack{\lambda\vdash n\\\lambda \text{ does not have parts equaling } 1}}\mathsf{Ch}(\mathrm{Lie}_{\lambda}(V)),
\]
where $\mathrm{Lie}_{\lambda}(V)$ are certain $\mathrm{GL}(V)$-modules known as \emph{higher Lie modules} \cite{Reutenauer93}.
This suggests that there might be a link between Boolean product polynomials and representations of the free Lie algebras, and we intend to explore this in the future.

\item Observe that
$
\mathcal{P}_{1,1}(X,Y)=\prod_{i=1}^{n}\prod_{j=1}^{m}(x_i+y_j)
$
and by the dual Cauchy identity, this product is well-known to have a nice Schur function expansion.
Thus, a natural question is to find an appropriate analogue to the Robinson-Schensted insertion algorithm that allows us to establish the Schur-positivity in Corollary~\ref{cor:multiple alphabets} combinatorially.

\end{enumerate}

\printbibliography


TITLE: How prove this integral $ \int\limits_0^1 \int\limits_0^1 \ln\Gamma(x+y^3)\,dx\,dy =-\frac 7 {16}+\frac{1}{2}\ln 2\pi$
QUESTION [12 upvotes]: show that

$$ I=\int_0^1 \int_0^1 \ln\Gamma(x+y^3) \, dx \, dy =-\frac 7 {16} + \frac 1 2 \ln 2\pi$$

where $$\Gamma(a)=\int_0^\infty x^{a-1}e^{-x} \, dx$$
then
$$I=\int_0^1 \int_0^1 \ln\left(\int_0^\infty t^{x+y^3-1}e^{-t} \, dt\right) \, dx \, dy$$
Then I can't works,Thank you

REPLY [7 votes]: Using
$$
\int_0^1 \ln\Gamma\left(x+\alpha\right)\ dx
=\frac{1}{2}\ln2\pi+\alpha \log \alpha -\alpha\quad;\quad \alpha \geq 0,
$$
and
$$
\int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\  n=0,1,2,\ldots
$$
then
$$
\int_0^1 \ln\Gamma\left(x+y^3\right)\ dx=\frac{1}{2}\ln2\pi+3y^3 \ln y -y^3
$$
and
\begin{align}
\int_0^1\int_0^1 \ln\Gamma\left(x+y^3\right)\ dx\ dy&=\int_0^1 \left(\frac{1}{2}\ln2\pi+3y^3 \ln y -y^3\right)\ dy\\
&=\frac{1}{2}\ln2\pi-\frac{3}{4^2}-\frac14\\
&=\large\color{blue}{\frac{1}{2}\ln2\pi-\frac{7}{16}}.\tag{Q.E.D.}
\end{align}


TITLE: Find the number of pair (A,B)
QUESTION [0 upvotes]: For a pair (A,B) of subsets of the set X ={1,2,3,..,100} let A∆B denote
the set of all elements of X which belong to exactly one of A or B. The
number of pairs (A,B) of subsets of X such that A∆B = {2,4,6,8,...,100}
is
(A) 2^151 (B) 2^102 (C) 2^101 (D) 2^100.
My approach- all even nummber should be taken by both A and B. No even number is common between them. Odd number taken by A should taken by B also. But I can't count how much pair should be there.
Please help me to find out. Thanks for your help.

REPLY [1 votes]: Every odd number is either in both of $A$ and $B$, or neither of $A$ and $B$.
Every even number is either in $A$ and not in $B$, or not in $A$ and in $B$.
Therefore, for each number (even or odd), you have two choices to make, so there are $2^{100}$ such pairs.