TITLE: Purity of a quantum state in the Heisenberg picture
QUESTION [4 upvotes]: The purity of a quantum state is defined as
$$
\gamma = \mathrm{Tr}(\rho^2)
$$
In the Schrödinger picture, it's easy to use this formula to see how the purity of the system changes as the system evolves. In the Heisenberg picture, the density matrix $\rho$ is constant. How does one calculate the purity?

REPLY [4 votes]: For a closed system, whose time-evolution is unitary and given by the Schrodinger equation, any quantity $S$ quantifying a state's "mixed-ness" (or entanglement entropy, depending on your perspective) that takes the form $S = \text{Tr}\left[ f(\rho) \right]$ for some analytic function $f$ has a value that is identical in the Schrodinger and Heisenberg pictures. You can see this by the cyclic property of the trace and the equation $\rho_\text{Schro}(t) = U(t) \rho_\text{Heis} U^\dagger(t)$. $S$ could be the purity, the von Neumann entropy, or the Renyi entropy. Moreover, $S$ is constant in either picture because the Schmidt weights (the eigenvalue spectrum of $\rho$) do not change over time in either picture.
For an open system, which is directly coupled to its environment, $\rho$ is not constant in either picture. It doesn't evolve unitarily but instead via Kraus maps, which are basically what you get when you take Schrodinger time evolution and trace out part of the Hilbert space.


TITLE: Terminology about the tensorial represention of a group
QUESTION [0 upvotes]: This could be extremely trivial but i need to be sure I am not wrong.
I am encountering many times statements where the author says

"Tensors are examples of representations for the Lorentz group".
   (A modern introduction to quantum field theory, Michele Maggiore. Page 20).

It is clear to me that if I consider for example $SO(3)$, or $SO(3,1)$ I can think to act its transformations on scalars, vectors or event tensors and find the related representations.
So the previous sentence isn't an abuse of notation? I think I should say a tensorial representation of the Lorentz group is a representation that acts on tensors, is this correct?
I found a similar problem in a more physical context. With the statement

"$d$ degenerate eigenstates furnish a d-dimensional irreducible representation for the group $G$".
  (Group theory in a nutshelli for physicists, A. Zee. Page 163).

Here is the context. Given a hamiltonian $H$ and its symmetry group $G$, then $H$ is invariant under the transoformations $T$ of $G$
$$H=T^{-1}HT\rightarrow HT=TH$$
considering the eigenvalue problem $$H\psi=E\psi$$ $$H(T\psi)=TH\psi=E(T\psi)$$
so we can see that $T$ doesn't mix states of differents eigenspaces, and the eigenspaces are like invariant subspaces of the Hilbert space from the point of view of $T$. Using a suitable basis for the Hilbert space I can show $T$ has a block diagonal form and hence is reducible. But considering $H$ just on a specific eigenspace $V_E$ then the hamiltonian has the form $H=EI$ and following from Schur's lemma I can say that the specific block associated to $V_E$ of the reducible representation of T is an irreducible representation. Is this what it is meant with the previous statment? That a correct choiche of the basis $\psi$ in the eigenspace showes me what is the irreducible representation?

REPLY [1 votes]: A representation of a group $G$ is a pair $(\varphi,V)$ where $V$ is a vector space and $\varphi:G\to\mathrm{GL}(V)$ is a group homomorphism between the group $G$ and the set of invertible linear transformations over $V$. Sometimes authors call $\varphi$ the representations and sometimes authors call $V$ the representation.
You can say that the first way is fine since $\varphi$ carries information about the space it acts on. On the other hand the second one is, as you point out, an abuse of terminology. However, often (not always!) once $V$ is fixed, there is only one choice for $\varphi$, so it's fine.
For tensors, $V$ is the tensor product of $\mathbb{R}^n$ $\mathrm{rank}$ times and $\varphi$ is the unique homomorphism that makes $G$ acts on tensors in the usual way that we know.
For the Hamiltonian example you presented, $V$ is the eigenspace of energy $E$ and $\varphi$ is some unitary operator restricted to $V$, $U|_V$, that acts on the states and commutes with the Hamiltonian. Precisely because it commutes with it, we never go out of the eigenspace, so the operator $U$ can be restricted to $\mathrm{GL}(V)$.
By the way, you are using Schur's lemma backwards. It says that if $[H,U|_V]=0$ and the representation is irreducible, then $H \propto \mathbb{1}$ (which is true). Not the other way around. That means that $(U|_V,V)$ can still be a reducible representation if there are invariant subspaces. For example, suppose your Hamiltonian has degenerate energies for states with the same total spin $J = S+L$. Then if we look at energy eigenspaces and only care about spin $S$, and not $J$, we may find a direct sum of spin representations there.


TITLE: Lower bound for constrained ordered partitions (i.e., compositions)?
QUESTION [1 upvotes]: An ordered $m$-partition (also called a composition) of an integer $n$ is an ordered sequence of positive integers $a_1, \ldots, a_m$ such that $\sum_i a_i = n$. Such a partition is $N$-constrained if additionally $a_i \leq N$ for all $i$.
Let $p(N, m, n)$ denote the number of $N$-constrained, ordered $m$-partitions of $n$. Bounds on $p(N, m, n)$ are known (see Andrews, "The Theory of Partitions" or Ratsaby, "Estimate of the Number of Restricted Integer Partitions"), however the ones I have seen are given either in terms of generating functions or ugly summations.
I am looking for a closed-form lower bound on $p(N, m, n)$. Is it known that for $n$ in a certain range (depending on $N, m$) it holds that $p(N, m, n) \geq c \cdot N^m$ for some constant $c$?

REPLY [1 votes]: For my answer, I assume that compositions are on non-negative numbers (allowing only positive numbers does not change my statements, qualitatively).
First, observe that $p(N,m,n)=\binom{m}{n}_{N+1}$, where $\binom{m}{n}_{N+1}$ is a generalized, or extended, binomial coefficient (so there you have an exact formula, not only a lower bound). Since $\binom{m}{n}_A\ge \binom{m}{n}_B$ if and only if $A\ge B$, a (very weak) lower bound for $p(N,m,n)$ is $\binom{m}{n}$, the ordinary binomial coefficient.
However, much better lower bounds can be obtained by noting that $\binom{m}{n}_{N+1}$ arises as the distribution of the sum of $m$ iid random variables, uniformly distributed on $\{0,\ldots,N\}$. Then apply the central limit theorem (or a variant of the DeMoivre-Laplace theorem) to get nice bounds ...


TITLE: Why does the antisymmetrization map factor through $n$-forms?
QUESTION [5 upvotes]: Consider a $k$-algebra $A$ and a bimodule $M$. One can construct two complexes, the Hochschild complex $C_n(A,M)$ and the Chevalley-Eilenberg complex $C'_n(A,M)=M\otimes \Lambda^n(A)$. Given an element $m\otimes a_1\otimes \cdots\otimes a_n$ and a permutation $\sigma\in S_n$, there is an action $$\sigma(m\otimes a_1\otimes\cdots\otimes a_n)=m\otimes a_{\sigma^{-1}1}\otimes\cdots\otimes a_{\sigma^{-1}n}$$
of $k[S_n]$ on $C_n(A,M)$. The element $\varepsilon_n = \sum_{\sigma\in S_n} (-1)^\sigma \sigma$ induces well defined maps $C_n'(A,M)\to C_n(A,M)$ which turn out to give a chain map $C'(A,M)\to C(A,M)$. When $A$ is commutative and $M$ is symmetric the differential of $C'$ is trivial and taking homology there is a map $\varepsilon : M\otimes \Lambda^n(A)\to H_n(A,M)$.
If $\Omega^1(A)$ is the $A$-module of Kähler differentials of $k\to A$, one defines $\Omega^n(A) = \Lambda_A^n(\Omega^1(A))$ (where the exterior product is taken over $A$), so it is spanned by the elements $a_0da_1\cdots da_n$. There is a canonical $k$-map $d:A\to \Omega^1(A)$ that sends $a\mapsto da$. In his book Cyclic Homology, Loday claims that $\varepsilon$ factors through $M\otimes_A \Omega^n(A)$, and to check this is suffices to check that any element of the form $$\varepsilon(mx,y,a_3,a_4,\cdots,a_{n+1})+\varepsilon(my,x,a_3,\cdots,a_{n+1})-\varepsilon(m,xy,a_3,\cdots,a_{n+1})$$
is a boundary.
I have two problems:

First, I assume the map $\varepsilon$ wants to factor through is the exterior power of $d:A\to \Omega^1(A)$ over $k$ followed by the projection to the exterior power over $A$.
Assuming this, I really cannot see how checking the above suffices. Perhaps he wants to use that $d$ is a universal derivation? For example, one needs to check that one can replace $\otimes=\otimes_k$ with $\otimes_A$, and for this one needs to show an element of the form

$$\varepsilon(my,x,a_3,\cdots,a_{n+1})-\varepsilon(m,xy,a_3,\cdots,a_{n+1})$$
also goes to a boundary. I am not sure how this is included in Loday's verification.

REPLY [3 votes]: The $n$-th exterior power $\Omega^nA=\wedge_A^n(\Omega^1_A)$ represents the functor $$N\mapsto\left\{ d : A^n\to N\ |\ d\text{ antisymmetric, }k\text{-multilinear, derivation in all arguments}\right\}\\\ \ \ \ \ \ \ \ =\left\{ d : A^n\to N\ |\ d\text{ antisymmetric, }k\text{-multilinear, derivation in first argument}\right\}.$$
In particular, to construct the desired map
$$\tilde{\varepsilon}: M\otimes_A \Omega^n A\to H_n(A,M)$$ it suffices to construct a map $$\hat{\varepsilon}: A^n\to\text{Hom}_A(M,H_n(A,M))$$ which is antisymmetric, $k$-multilinear and a derivation in the first component; then $\tilde{\varepsilon}$ is the unique morphism with $$\tilde{\varepsilon}(m\otimes\text{d}a_1\wedge\ldots\wedge\text{d}a_n)=\hat{\varepsilon}(a_1,\ldots,a_n)(m).$$ Here, for $\hat{\varepsilon}$ you take the adjoint of $\varepsilon$, which we already know to be $k$-multilinear and antisymmetric, and the Leibniz rule for the first argument is what Loday claims needs to be checked.


TITLE: How to write join of multiple graphs
QUESTION [2 upvotes]: Let $\overline{K}_{a_i}$ denote a null graph on $a_i$ vertices where $1\le i\le n$.
I want to do join of   $\overline{K}_{a_i}$.
I am not able to  understand how to write join of the graphs $\overline{K}_{a_i}$ mathematically.
Can someone please help me to write it down formally?

REPLY [1 votes]: In general, you might denote graph join with $+$ (in which case you'd write $\overline{K}_{a_1} + \overline{K}_{a_2} + \dots + \overline{K}_{a_n}$) or with $\vee$ (in which case we have an additional option: $\overline{K}_{a_1} \vee \overline{K}_{a_2} \vee \dots \vee \overline{K}_{a_n}$ can also be $\bigvee_{i=1}^n \overline{K}_{a_i}$).
In this particular case, the join of empty graphs with $a_1, a_2, \dots, a_n$ vertices happens to be the complete multipartite graph $K_{a_1, a_2, \dots, a_n}$ which has its own notation. (Provided that $n>1$.)


TITLE: Is one tuple set transitive relation?
QUESTION [1 upvotes]: I am wondering if a relationship with only one tuple e.g. {$(a,b)$} is transitive. If so, what about empty relation {$\emptyset$}.
We know that if a relation is transitive if $(a,b)\in R$ and $(b,c)\in R$, then $(a,c) \in R$. I don't know how to use the definition to interpret my question. Please show me your thread.
We got only one tuple, and it seems does not fit in the definition. Appreciate for any help.
Updated notes: *Not sure about the notation of empty relation should be either $\emptyset$ or {$\emptyset$}. Undoubtedly, $\emptyset$ is an element of {$\emptyset$}, if  $\emptyset$  still be a subset of every set in this case, e.g {$\emptyset$} *

REPLY [2 votes]: Here's Yves Stalder's proof, written out in more detail, since you found his argument hard to follow. Suppose the relation $R=\{(a,b)\}$ contains two pairs $(u,v)$ and $(v,w)$; I need to show that it also contains $(u,w)$.
The fact that $(u,v)\in R=\{(a,b)\}$means that $(u,v)=(a,b)$, i.e., that $u=a$ and $v=b$. Similarly, from $(v,w)\in R$, we get that $v=a$ and $w=b$. In particular, $u=a$ and $w=b$, which means that $(u,w)=(a,b)$ and therefore $(u,w)\in R$.
(The hypotheses also imply that $a=b$ because $v$ equals both $a$ and $b$, but that information isn't needed for this proof.)


TITLE: Prove there is a point where $f''=0$ when f is three times differentiable.
QUESTION [0 upvotes]: Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a three times differentiable function. Suppose that $f(0)=3,f(3)=6,f(6)=9$. Prove that there exists $x_0\in \mathbb{R}$ such that $f(x_0)=0$.
I have concluded that I should use Mean Value Theorem on some finite interval for $f'(x)$ so then I would get $f''(x)=0$ but I am not sure what this exact interval would be. Since $f$ is three times differentiable, I already know that $f'(x)$ exists on the interval but how could I use the three given points to show that there exists a local maximum or minimum?

REPLY [2 votes]: We have some $\alpha\in(0,3)$ and $\beta\in(3,6)$ such that $1=\dfrac{f(3)-f(0)}{3-0}=f'(\alpha)$ and $1=\dfrac{f(6)-f(3)}{6-3}=f'(\beta)$, so there exists some $\gamma\in(\alpha,\beta)$ such that $f''(\gamma)=0$.


TITLE: Fourier series for $e^{-|x|}$
QUESTION [1 upvotes]: So I'm trying to find the Fourier series of the $PC_{2\pi}$ function $e^{-|x|}$ for $x\in[-\pi,\pi]$ using the formula
$$c_0+\sum_{k=-\infty}^\infty c_k e^{-ikx}$$
where
$$
c_k(f)=\frac{1}{2\pi} \int_{-\pi}^\pi f(x)e^{-ikx} dx
$$
I've reached an answer (below), but I think that the result is wrong.  Is my answer correct? If not, then what is the correct answer?

In this case that would be finding
$$c_k(f)=\frac{1}{2\pi} \int_{-\pi}^\pi e^{-|x|}e^{-ikx} dx$$
which I've separated into
$$=\frac{1}{2\pi} \int_{-\pi}^0 e^{x-ikx} dx + \int_0^\pi e^{-x-ikx}dx $$
For each of them I found
\begin{align}
\int_{-\pi}^{0} e^{x-i k x} d x &= \int e^{t} \cdot \frac{1}{1-i k} d t \\
&= \int \frac{e^{t}(1+i k)}{1+k^{2}} d t \\
&= \int \frac{e^{t}}{1+k^{2}}+i \frac{k e^{t}}{1+k^{2}} d t \\
&= \int_{-\pi}^{0} \frac{e^{x-i k x}}{1+k^{2}} d t+\int_{-\pi}^{0} i \frac{k e^{x-i k x}}{1+k^{2}} d t \\
&=\frac{1-e^{-\pi+i \pi k}}{1+k^{2}}+\frac{i k}{1+k^{2}}\left(1-e^{-\pi+i \pi k}\right)
\end{align}
and
\begin{align}
\int_{0}^{\pi} e^{-x-i k x} d x &= \int e^{t} \cdot \frac{1}{-1-i k} d t \\
&= \int \frac{e^{t}(-1+i k)}{1+k^{2}} d t \\
&= \int-\frac{e^{t}}{1+k^{2}}+i \frac{k e^{t}}{1+k^{2}} d t \\
&= \int_{0}^{\pi}-\frac{e^{-x-i k x}}{1+k^{2}} d t+\int_{-\pi}^{0} i \frac{k e^{-x-i k x}}{1+k^{2}} d t \\
&=\frac{e^{-\pi-i \pi k}-1}{1+k^{2}}+i \frac{k}{1+k^{2}}\left(e^{-\pi-i \pi k}-1\right)
\end{align}
Hence my final result:
$$ c_k(f)= \frac{e^{i\pi k}+e^{-\pi-i\pi k}+ike^{-\pi-i\pi k}-ike^{i\pi k-\pi}}{2\pi(1+k^2)}$$

REPLY [1 votes]: The Fourier cosine series for $e^{-\left|x\right|}$ is as follows.
(1) $\quad e^{-\left|x\right|}=\frac{1+\sinh (\pi )-\cosh(\pi)}{\pi}+\frac{2}{\pi}\underset{K\to\infty}{\text{lim}}\ \sum\limits_{k=1}^K \frac{\left(1-(-1)^k e^{-\pi }\right)}{k^2+1}\cos(k\ x)$

Figure (1) below illustrates formula (1) above for $e^{-\left|x\right|}$ evaluated at $K=100$ in orange overlaid on the blue reference function.

Figure (1): Illustration of formula (1) for $e^{-\left|x\right|}$

The following links provide information on the derivation of Fourier cosine series for even functions.

Wikipedia: Cosine series
Weisstein, Eric W. "Fourier Cosine Series." From MathWorld--A Wolfram Web Resource.


TITLE: about the series $\frac{1}{\sinh ^2(s)}$
QUESTION [0 upvotes]: I been reading the paper Novel integral representations of the Riemann zeta-function and Dirichlet eta-function, close expressions for Laurent series expansions of powers of trigonometric functions and digamma function, and summation rules of Sergey K. Sekatskii and he gives
$$\frac1{\sinh^2 x}=\frac1{x^2}-\frac13+\frac2{\pi^2}\sum_\limits{n=1}^\infty(-1)^{n+1}(2n+1)\left(\frac{x}{\pi}\right)^{2n}\zeta(2n+2)$$
I have recalculated the series and gives
$$\frac{1}{\sinh ^2(x)}=\sum _{k=1}^{\infty } x^k \left(-2 (k+1) \pi ^{-k-2} \zeta (k+2) \cos \left(\frac{\pi  k}{2}\right)\right)+\frac{1}{x^2}-\frac{1}{3}$$
What do you think?

REPLY [2 votes]: By applying $-\frac{d^2}{dx^2}\log(\cdot)$ to both sides of
$$ \sinh(x) = x \prod_{n\geq 1}\left(1+\frac{x^2}{\pi^2 n^2}\right) \tag{1}$$
we get
$$ \frac{1}{\sinh^2 x}=\frac{1}{x^2}-2\sum_{n\geq 1}\frac{\pi^2 n^2-x^2}{(\pi^2 n^2+x^2)^2}=\frac{1}{x^2}-2\sum_{n\geq 1}\frac{1}{\pi^2 n^2}\sum_{m\geq 0}(-1)^m(2m+1)\left(\frac{x}{\pi n}\right)^{2m}\tag{2} $$
and
$$ \frac{1}{\sinh^2 x}=\frac{1}{x^2}-2\sum_{m\geq 0}\frac{(-1)^m(2m+1)\zeta(2m+2)}{\pi^{2m+2}}\,x^{2m}\tag{3}$$
so Sergey K. Sekatskii is definitely right. On the other hand $\cos\frac{\pi k}{2}$ is zero for odd values of $k$, hence the two representations are equivalent.


TITLE: Summation over set notation
QUESTION [1 upvotes]: I am reading some lecture notes on quantum computing and ran into the following expression that I cannot seem to understand:

For example, if we apply the Hadamard gate $H$ to each bit in a register of $n$ zeroes, we obtain $\frac{1}{\sqrt{2}} \sum_{j \in \{0,1\}^n} | j \rangle$, which is a superposition of all $n$-bit strings.
  More generally, if we apply $H^{\otimes n}$ to an initial state $| i \rangle$, with $i \in \{0,1\}^n$, we obtain
  \begin{equation}
\tag{2.1}
    H^{\otimes n} | i \rangle
  = \frac{1}{\sqrt{2^n}}
    \sum_{j \in \{0,1\}^n}
    (-1)^{i \cdot j} | j \rangle,
\end{equation}
  where $i \cdot j = \sum_{k=1}^n i_k j_k$ denotes the inner product of the $n$-bit strings $i, j \in \{0, 1\}^n$.
  For example:
  $$
    H^{\otimes 2} | 01 \rangle
  =         \frac{1}{\sqrt{2}} ( | 0 \rangle + | 1 \rangle )
    \otimes \frac{1}{\sqrt{2}} ( | 0 \rangle - | 1 \rangle )
  = \frac{1}{2} \sum_{j \in \{0,1\}^2} (-1)^{01 \cdot j} | j \rangle.
$$
(Original image here.)

In both of the summations, a set is implied but I do not see how it would work if it were expanded. Any help would be appreciated. Thank you.

REPLY [3 votes]: $\{0,1\}^n$ denotes the set of all bit strings of length $n$.  So, for instance, with $n=3$ and $i = 011$, we have
$$
\begin{align*}
H ^{\otimes 3}|i \rangle &= \frac{1}{\sqrt{2^3}} \sum_{j \in \{0,1\}^3} (-1)^{i \cdot j}
\\ & = \frac{1}{\sqrt{2^3}}\left[(-1)^{(000)\cdot i}|000 \rangle + (-1)^{(001) \cdot i}|001 \rangle  + \cdots + (-1)^{(111) \cdot i}|111 \rangle\right]
\\ & =
\frac{1}{\sqrt{2^3}}\left[|000 \rangle -|001 \rangle  -|010 \rangle + |011 \rangle + |100 \rangle - |101 \rangle - |110 \rangle + |111 \rangle\right]
\end{align*}
$$

REPLY [3 votes]: The sum in (2.1) is over all length-$n$ vectors with each entry $0$ or $1$. The symbol $i$ is also such a vector, so the dot product $i\cdot j$ in $(-1)^{i\cdot j}$ is the sum of values of $k$ for which $i_k=j_k=1$. In the unnumbered display-line equation below, the same thing happens with the special case $n=2,\,i_1=0,\,i_2=1$.


TITLE: Equation for position of mass suspended by a spring on an incline
QUESTION [0 upvotes]: I'm interested an application of hooke's spring equations to solve for the position of an object with respect to time in the following scenario:

Where $T_2$ is a spring with coefficient $k$. I assume the incline is infinitely long to the right and am interested in the distance moved along a coordinate system rotated by $\theta$, i.e. $x(t)$ along the incline itself. This simplifies it by making it a one-dimensional problem, whose solution can be generalized.
Because I've been accosted in the past for asking homework questions, when in fact the question cam from my own mind, I will note on the onset that this diagram was borrowed from an unrelated problem in the textbook – which is why $T_2$ is drawn as a rope not a spring.
What I have tried
I'm looking for $x(t)= d(t)+x_0$, the position with respect to time, where $d(t)$ is the displacement with respect to time. Setting distance at which the spring is at rest as the origin, we have $x(t)=d(t) \implies x=d$
From the diagram, the leftwards force of the spring is $F_s=kd$ and the rightwards force of gravity is $mg\sin(\theta)$.
Thus the net force:
$$F_{net} = ma = kd-mg\sin(\theta)$$
Solving for $a$, we have:  $$a=\left(\frac{k}{m}\right)d-g\sin(\theta)$$
My simple knowledge of physics led me first to using this equation (it didn't occur to me at the time that the situation is more nuanced that I thought, i.e. non constant acceleration):
$$x-x_0 = v_0t+\frac{1}{2}at^2$$
Plugging in $a$, we can solve for d:
$$d = \frac{mg\sin(\theta)t^2}{1-kt^2}$$
and plug this back into the equation for $a$, yielding
$$a=\frac{gsin(\theta)}{kt^2-1}$$
Using a CAS to double integrate, as I'm just taking differential calculus now, we can get an equation for position.
$$x(t)=d(t)+0=g \left(\frac{t \sqrt{\frac{1}{k}}}{2} \log{\left (t - \sqrt{\frac{1}{k}} \right )} - \frac{t \sqrt{\frac{1}{k}}}{2} \log{\left (t + \sqrt{\frac{1}{k}} \right )} - \frac{1}{2 k} \log{\left (t - \sqrt{\frac{1}{k}} \right )} - \frac{1}{2 k} \log{\left (t + \sqrt{\frac{1}{k}} \right )}\right) \sin{\left (O \right )}$$
Why it is Wrong
I know my equation for position was derived assuming a constant acceleration – and here acceleration is not constant.
What I am Looking For and Why Help is Necessary
After seeing the answers to this question I realize I'll need more involved calculus to solve my question of interest. However, knowing only the rudimentary rules of integration without having a solid theoretical basis precludes me from solving more involved integrals, such as the circular integral with acceleration and position we have here. I'm looking for a step by step derivation of an $x(t)$ equation. Thanks so much.
Edit
From the comments I learned I am dealing with a differential equation of the form
$$x''(t)=\frac{k}{m}x(t)-g\sin(\theta)$$
Which I solved with wolfram alpha, yielding:
$$x\left(t\right)=e^{\left(\frac{\sqrt{k}t}{\sqrt{\left(m\right)}}\right)}+e^{\left(-\frac{\sqrt{k}t}{m}\right)}+\frac{gm\sin \left(\theta _1\right)}{k}$$
This seems to be correct, but I need confirmation.

REPLY [2 votes]: You can't really solve this by simple integration because the equation of motion is a 2nd order differential equation.
Imagine an $x$-axis parallel to the incline and pointing right (and somewhat down).
The Newtonian equation of motion is now:
$$ma=-kx+mg\sin \theta \tag{1}$$
Where $\theta$ is the angle between horizontal and incline. With $a=\frac{d^2x}{dt^2}=\ddot{x}$, $(1)$ becomes:
$$m\ddot{x}+kx-mg\sin \theta=0\tag{2}$$
$(2)$ would be a simple 2nd order differential equation for which standard solutions exist, if it wasn't for that niggling $mg\sin \theta$ tail, so first we get rid of that with a little trick.
Divide both sides by $m$ and define:
$$\omega^2=\frac{k}{m}$$
This gives:
$$\ddot{x}+\omega^2x-g\sin \theta=0\tag{3}$$
Now substitute:
$$u=\omega^2x-g\sin \theta\tag{4}$$
$$\implies \dot{u}=\omega^2\dot{x}$$
$$\implies \ddot{u}=\omega^2\ddot{x}\implies \ddot{x}=\frac{1}{\omega^2}\ddot{u}$$
Into $(3)$ we get:
$$\frac{1}{\omega^2}\ddot{u}+u=0$$
Multiply both sides by $\omega^2$ to get:
$$\ddot{u}+\omega^2u=0\tag{5}$$
$(4)$ is the classic 2nd order linear homogeneous (ordinary) differential equation and it has the general solution:
$$u(t)=c_1\cos \omega t+ c_2\sin \omega t$$
To determine the integration constants $c_1,c_2$ we need initial conditions:
$x(0)=x_0$, which is the displacement at $t=0$.
We also assume that at $t=0$, $v=\dot{x}=0$.
Use $(4)$ to define $u(0)$, then $\dot{u}(0)$ and use these expressions to work out $c_1$ and $c_2$.
This gives you a full expression for $u(t)$ and with $(4)$:
$$u(t)=\omega^2x(t)-g\sin \theta$$
Extract $x(t)$ from there, for a full description of displacement in time. It's the kinematic equation of a simple harmonic oscillator with angular velocity $\omega=\sqrt{\frac{k}{m}}$.
Alternative substitution to $(3)$ (starting from $(2)$):
$$u= kx-mg\sin \theta$$
This also works and leads to the same end result.


TITLE: Uniform distribution with probability density function. Find the value of $k$.
QUESTION [7 upvotes]: For a random sample $X_1,X_2,...X_n$ from a uniform $[0,\Theta]$ distribution, with probability density function
$$f(x;\Theta) = \left\{ \begin{array} \ \frac{1}{\Theta} & 0\le x \le\Theta,\\ 0 & \text{otherwise}.\end{array}\right.$$
What is the value of $k$ such that $\hat{\Theta}=k\bar{X}$ is an unbiased estimator of $\Theta$?
I've done some questions similar to this but I'm not sure how to go about this one. I have a test in 3 hours so help is really appreciated!

REPLY [10 votes]: $\overline{X}= \frac1n\sum X_i$  $\mathrm{E}(X_i)=\theta/2 =\mathrm{E}(\overline{X})$. So $\theta =\mathrm{E}(2\overline{X})$. Hence $k=2$. Joriki gave the correct answer but for the OP to understand an explanation seems in order.  I hope this is in time for your test.


TITLE: Higher energy of outer orbitals around nucleus
QUESTION [1 upvotes]: I remember reading that the outer orbitals of atom have higher energy. If an electron is given enough energy it goes to higher energy orbitals from $n=1$ to $n=2,3,4...$
My question is that why do the outer orbitals have higher energy? Is is because of charges?

REPLY [0 votes]: Although this is quantum mechanical...If I explain in classical way you can think electrostatic force between nucleus and electron is directing inwards and due to this effect electron rotating around nuclear. Then due to rotation there appears a centrifugal force $\frac{m_ev^2}{r}$ directing outward. now when it getting more energy its kinetic energy $\frac{mv^2}{2}$ increased I.e velocity $v$ increased i.e more centrifugal force $r$ also increased .In this way more energy absorbing by electron and due to more outward centrifugal force increases its orbit radius $r$.I know this is classical but I just try to make it simplify.


TITLE: Questions on measurable sets
QUESTION [1 upvotes]: I'm learning about measure theory, specifically measurable sets, and need help with the following exercises:

$(1)$ Find the measure of the set $E_1 = \mathbb{Z} \cup \mathbb{Q} \cup (\mathbb{R} \setminus \mathbb{N})$;
$(2)$ Find the measure of the set $E_2 = \bigcap_{n=1}^{\infty}\big[1 - {1 \over 2^n}, 3 + {1 \over 3^n}\big]$;
$(3)$ Let $\{a_n\}$ be an increasing sequence of positive terms. If $E_n = (-a_n, a_n]$, find $m(\bigcup_n E_n)$;
$(4)$ (True or False) Let $A \subset \mathbb{R}$. Then $m(A) = 0$ iff all the subsets of $A$ are measurable.

Note : $m$ is the Lebesgue measure.

Since I'm having difficulties for question $(4)$ I am going to share my thoughts on $(1)$, $(2)$ and $(3)$.
$(1)$ Since $\mathbb{N} \subsetneq \mathbb{Q}$, then $(\mathbb{R} \setminus \mathbb{N}) \cup \mathbb{Q} = \mathbb{R}$. But $\mathbb{R} \cup \mathbb{Z}$ is just $\mathbb{R}$. Therefore, since the Lebesgue measure can reach infinity,
$$m(E_1) = m(\mathbb{R}) = \infty.$$
$(2)$ Calculating the first two terms: for $n = 1$ we get the interval $[{1 \over 2}, {10 \over 3}]$. For $n  = 2$ we have the interval $[{3 \over 4}, {28 \over 9}]$. So as $n$ increases the interval increases from the left and decreases from the right. As $n$ tends to infinity we get the interval $[1, 3]$. Therefore,
$$m(E_2) = m\big(\bigcap_{n=1}^{\infty}\big[1 - {1 \over 2^n}, 3 + {1 \over 3^n}\big]\big) = m([1, 3]) = 2.$$
$(3)$ Since $E_n = (-a_n, a_n]$ where $\{a_n\}$ is an increasing sequence of positive terms, then $\bigcup_n E_n = \mathbb{R}$. It follows that
$$m(\bigcup_n E_n) = m(\mathbb{R}) = \infty.$$

Is my work correct for $(1)$, $(2)$ and $(3)$? How do I prove or disprove $(4)$?

REPLY [3 votes]: Your work is correct for $1$ and $2$. For $3$, how about the sequence: $a_n = 1 - \frac1n$? It is increasing, though:
$$\bigcup_n E_n = (-1,1)$$
We have to separate two cases. We know that either $(a_n)$ converges or goes to $+ \infty$. If it converges to $a$, $0 < a < \infty$, then we have:
$$\bigcup_n E_n= (-a,a)$$
Otherwise, the union is $(-\infty, \infty)$.
For $4$, recall that the Lebesgue measure is complete, which means that every subset of a null set is measurable. If $m(A) = 0$, then $A$ is a null set and every subset of $A$ is measurable. Conversely, it is known that every set of positive Lebesgue measure contains a nonmeasurable subset (for a proof, see e.g. this)


TITLE: Is there any order behind the chaos of multiplication modulo $p$?
QUESTION [2 upvotes]: Let's lift the finite group $\mathbb{Z}_p$ to $\mathbb{Z}$ when we're taking magnitudes so the usual ordering we're using.  So we say $a < b$ in $\mathbb{Z}_p$ iff $a < b$ under inclusion into $\mathbb{Z}$.  That is all unimportant with respect to my question but in case anyone was wondering what ordering we refer to.
So we know that, modulo $p$, $a \lt b$ does not imply $ca \lt cb$, but has anyone studied what actually happens to the direction of $\lt$ under multiplication?
It would be of use to me for googling if I knew what to call a group together with an ordering on the group since google confuses this order with "order of an element".

REPLY [0 votes]: The quick and short aswer is that there is no useful ordering in $\Bbb{Z}_p$. A fundamental difficulty is that the "lift" is not unique. Modulo $11$ do you want to lift $10$ as $10$ or $-1$, or $-12\ldots$? Yes, we can make lifting well-defined by insisting
that the lifts are in the interval $[0,p-1]$ or some fixed interval of $p$ integers. Another popular choice is $[-(p-1)/2,(p-1)/2]$. Defining the lifting is not the problem. Lack of an order relation compatible with arithmetic is.
This means that people who are used to solving equations by methods that depend on having the order relation around (such as finding a solution of an equation by dissecting smaller and smaller intervals) need to learn other methods. Lack of ordering means lack of concept of a "small error", so a method based on improving an approximate solution loses its basis.
Yet, ordering the lifts is occasionally useful. Gauss' Lemma comes to mind. There we are to count how many of the lifts of $a,2a,\ldots,\dfrac{(p-1)}2a$ are larger than $p/2$. That leads eventually to the law of quadratic reciprocity. An essential tool in deciding whether quadratic equations have a solution modulo $p$ or not (among other things).
Lack of an ordering occasionally gives a headache to people working in continuous math side also. Angles between two lines have a similar ambiguity coming from the fact an angle is often defined only up to an integer multiple of $2\pi$. Usually the difficulties there are not severe, because confining ourselves to an interval of length $2\pi$ allows us to isolate any eventual discontinuities, and we can e.g. integrate using polar coordinates quite well. The type of problems that linger are of the type: what's the average of two angles? Is the average of $+1^\circ$ and $-1^\circ$ zero? Why does it change, when we use $359^\circ$ in place of $-1^\circ$? Of course, in a sense, all of the calculus of residues in complex analysis rides on the non-uniqueness of the lift of a compass direction to a real number :-)


text:
\begin{document}

\title{Ehrhart Theory for Lawrence Polytopes and Orbifold Cohomology of Hypertoric Varieties}

\author{Alan Stapledon}
\address{Department of Mathematics, University of Michigan, Ann Arbor, MI 48109, U.S.A.}
\email{astapldn@umich.edu}

\thanks{The author would like to thank Nicholas Proudfoot

for some useful comments.

}

\begin{abstract}
We establish a connection between the orbifold cohomology of hypertoric varieties and the Ehrhart theory of Lawrence polytopes.
More specifically, we show that the dimensions of the orbifold cohomology groups of a hypertoric variety are equal to the coefficients of the Ehrhart $\delta$-polynomial of the associated Lawrence polytope. As a consequence, we deduce a formula for the Ehrhart $\delta$-polynomial of a Lawrence polytope and use the injective part of the Hard Lefschetz Theorem for hypertoric varieties to deduce some inequalities between the coefficients of the $\delta$-polynomial.
\end{abstract}

\maketitle

\section{Introduction}

Hypertoric varieties were introduced by Biewlawski and Dancer \cite{BDGeometry} using a hyperk\"ahler analogue of the construction of toric varieties
by K\"ahler quotients. Hausel and Sturmfels \cite{HauQuaternionic,HSToric} showed that the cohomology of hypertoric varieties is intimately related to the combinatorics of matroids and hyperplane arrangements. Recently, Jiang and Tseng \cite{JTOrbifold} and Goldin and Harada \cite{GHOrbifold} independently gave explicit descriptions of the orbifold cohomology rings of hypertoric varieties. We will establish a relationship between the orbifold cohomology of  hypertoric varieties and the lattice point enumeration of Lawrence polytopes.

Let $N$ be a lattice of rank $d$ and let $\mathcal{B} = \{ b_{1}, \ldots, b_{n} \}$ be a configuration of $n$ non-zero lattice points in $N$  that  generate $N$ as an abelian group.

Let $M = \Hom_{\mathbb{Z}}(N,\mathbb{Z})$ denote the dual lattice to $N$ and let $\mathcal{R} = \{r_{1}, \ldots, r_{n} \}$ be a collection of real numbers.
Consider the \emph{weighted, co-oriented hyperplane arrangement} $\mathcal{H}$ in $M_{\mathbb{R}} = M \otimes_{\mathbb{Z}} \mathbb{R}$ defined by

the hyperplanes
$H_{i} = \{ u \in M_{\mathbb{R}} \mid \langle u, b_{i} \rangle = r_{i}     \}$,
for $i = 1, \ldots, n$, and the co-normal vectors $\mathcal{B}$. We may and will choose $\mathcal{R}$ so that the hyperplane arrangement $\mathcal{H}$ is \emph{simple}. That is, the intersection of any $l$ hyperplanes in $\mathcal{H}$ is either empty or has codimension $l$.
The \emph{Lawrence toric variety} $X_{\mathcal{H}}$
associated to $\mathcal{H}$ is a $(d + n)$-dimensional simplicial toric variety,

 and the corresponding hypertoric variety $\mathcal{M}_{\mathcal{H}}$
 is a $2d$-dimensional complete intersection in $X_{\mathcal{H}}$, with a canonical orbifold structure \cite[Section 6]{HSToric}. In Section \ref{construction}, we will give a concrete local construction of $X_{\mathcal{H}}$ and $\mathcal{M}_{\mathcal{H}}$ in terms of $\mathcal{H}$, expanding on the descriptions in \cite{HSToric} and \cite{JTOrbifold}. We refer the reader to \cite{HSToric} and \cite{ProSurvey} for an algebraic description of hypertoric varieties using Gale diagrams and for the construction of hypertoric varieties as hyperk\"ahler quotients.

 The theory of orbifold cohomology, developed by Chen and Ruan \cite{CRNew, CROrbifold},
associates to an orbifold $Y$ a finite-dimensional $\mathbb{Q}$-algebra
$H_{\orb}^{*}(Y, \mathbb{Q})$, graded by $\mathbb{Q}$. In our case, the orbifold cohomology groups of $X_{\mathcal{H}}$ and $\mathcal{M}_{\mathcal{H}}$ are graded over $\mathbb{Z}$ and depend only on $\mathcal{B}$ and not on $\mathcal{H}$ \cite[Theorem 1.1]{JTOrbifold}.

\begin{theorem}\cite[Theorem 3.10]{JTNote}\label{lemma1}
The orbifold cohomology groups $H_{\orb}^{*}(X_{\mathcal{H}}, \mathbb{Q})$ and
$H_{\orb}^{*}(\mathcal{M}_{\mathcal{H}}, \mathbb{Q})$
are isomorphic as graded $\mathbb{Q}$-algebras and vanish in all degrees not in $2\mathbb{Z}$.
\end{theorem}

If $e_{1}, \ldots, e_{n}$ denote the standard generators of
$\mathbb{Z}^{n}$, then the \emph{Lawrence polytope} $P_{\mathcal{B}}$ associated to $\mathcal{B}$ is the
$(d + n - 1)$-dimensional lattice polytope in $N \times \mathbb{Z}^{n}$ with
vertices
$\{ (b_{i},e_{i}),(0,e_{i})\mid  i = 1, \ldots, n \}$.

Lawrence polytopes have been crucial in the construction of non-rational polytopes in the work of  Perles \cite{GruConvex}, Mn\"ev \cite{MnevUniversality} and
Ziegler \cite{ZieNon, ZieLectures}, and their combinatorial properties have been studied by Bayer and Sturmfels \cite{BSLawrence} and Santos \cite{SanTriangulations}. Recently, they appeared in Batyrev and Nill's classification of degree $1$ lattice polytopes \cite{BNMultiples} and in
the construction  of hypertoric varieties \cite{HSToric}.

For every positive integer $m$, let $f_{\mathcal{B}}(m) := \# \left( mP_{\mathcal{B}} \cap (N \times \mathbb{Z}^{n}) \right)$ denote the
number of lattice points in the $m$'th dilate of $P_{\mathcal{B}}$. A famous theorem of Ehrhart \cite{ehrhartpolynomial} asserts that
$f_{\mathcal{B}}(m)$ is a polynomial in $m$ of degree $\dim P_{\mathcal{B}} = n + d - 1$, called the
\emph{Ehrhart polynomial} of $P_{\mathcal{B}}$.
Equivalently,
the generating series of $f_{\mathcal{B}}(m)$ can be written in the form
\begin{equation*}
\sum_{m \geq 0} f_{\mathcal{B}}(m) \, t^{m} = \frac{ \delta_{\mathcal{B}}(t) }{ (1 - t)^{n + d} } \, ,
\end{equation*}
where $\delta_{\mathcal{B}}(t) = \delta_{0} + \delta_{1}t + \cdots +  \delta_{n + d - 1}t^{n + d - 1}$ is a polynomial of degree at most $d$ with
integer coefficients,
called the \emph{Ehrhart $\delta$-polynomial} of $P_{\mathcal{B}}$.

Ehrhart $\delta$-polynomials of lattice polytopes have been studied extensively over the last forty years by many authors including Stanley \cite{StaHilbert2, StaDecompositions, StaHilbert1,  StaMonotonicity} and Hibi \cite{HibSome, HibEhrhart,  HibLower, HibStar}.

In a recent paper \cite{YoWeightI}, the author expressed the coefficients of  the Ehrhart $\delta$-polynomial of a lattice polytope as sums of dimensions of orbifold cohomology groups of a toric orbifold. We will use the following result.

\begin{theorem}\cite[Theorem 4.6]{YoWeightI}\label{oldie}
Let $P$ be an $m$-dimensional lattice polytope in a lattice $N'$ and let $\mathcal{T}$ be a lattice triangulation of $P$.
If $\sigma$ denotes the cone over $P \times \{ 1 \}$ in $(N' \times \mathbb{Z})_{\mathbb{R}}$ and if $\triangle$ denotes the simplicial fan refinement of $\sigma$ induced by $\mathcal{T}$, then we may consider the corresponding toric variety $Y = Y(\triangle)$, with its canonical orbifold structure.
The Ehrhart $\delta$-polynomial of $P$ has the form
\[
\delta_{P}(t) = \sum_{i = 0}^{m} \dim_{\mathbb{Q}} H_{\orb}^{2i}(Y, \mathbb{Q}) t^{i}.
\]

\end{theorem}

We deduce the following geometric interpretation of $\delta_{\mathcal{B}}(t)$.

\begin{theorem}\label{punter}
The Ehrhart $\delta$-polynomial of the Lawrence polytope $P_{\mathcal{B}}$ has the form
\[
\delta_{\mathcal{B}}(t) = \sum_{i = 0}^{n + d - 1} \dim_{\mathbb{Q}} H_{\orb}^{2i}(X_{\mathcal{H}}, \mathbb{Q}) t^{i}
= \sum_{i = 0}^{n + d - 1} \dim_{\mathbb{Q}} H_{\orb}^{2i}(\mathcal{M}_{\mathcal{H}}, \mathbb{Q}) t^{i}.
\]
\end{theorem}
\begin{proof}
If $\psi: N \times \mathbb{Z}^{n} \rightarrow \mathbb{Z}$ is given by $\psi(\lambda + \sum_{i = 1}^{n} \mu_{i}e_{i} ) = \sum_{i = 1}^{n} \mu_{i}$, for $\lambda \in N$ and integers $\mu_{i}$,  then we may choose an isomorphism $N \cong \psi^{-1}(0) \times \mathbb{Z}$ such that $P_{\mathcal{B}}$
is a $(n + d - 1)$-dimensional lattice polytope in $\psi^{-1}(0) \times \{1\}$.
If $\Sigma_{\mathcal{H}}$ denotes the fan associated to the toric variety $X_{\mathcal{H}}$, then
the hyperplane arrangement $\mathcal{H}$ induces a regular lattice triangulation $\mathcal{T}$ of $P_{\mathcal{B}}$ such that the cones of $\Sigma_{\mathcal{H}}$ are given by the cones over the faces of $\mathcal{T}$ \cite[Proposition 4.2]{HSToric}.
The result now follows from
Theorem \ref{lemma1} and Theorem \ref{oldie}.
\end{proof}

Theorem \ref{punter} and computations of the orbifold cohomology ring of $\mathcal{M}_{\mathcal{H}}$ by Jiang and Tseng \cite{JTNote,JTOrbifold}  give a combinatorial formula for $\delta_{\mathcal{B}}(t)$. In order to state the result,
first recall that  the
\emph{matroid} $M_{\mathcal{B}}$ associated to  $\mathcal{B}$ is the collection of all linearly independent subsets of $\mathcal{B}$, as well as the origin $\{ 0 \}$.
The dimension of an element $F$ in $M_{\mathcal{B}}$ is equal to the dimension of $\Span F$, the
span
of the elements of $F$ in $N_{\mathbb{R}}$.
For any $F$ in $M_{\mathcal{B}}$, consider the projection
$\phi_{F}: N \rightarrow N/ (N \cap \Span F)$ and the induced configuration
$\mathcal{B}_{F} = \{ \phi_{F}(b_{i}) \mid b_{i} \notin \Span F\}$ in $N/ (N \cap \Span F)$, with associated matroid $M_{F}$. The hyperplane arrangement $\mathcal{H}$ restricts to a simple hyperplane arrangement $\mathcal{H}^{F}$ on the affine space $\cap_{b_{i} \in F} H_{i}$,
which we identify with $M_{F,\mathbb{R}} = \Hom_{\mathbb{R}}(N_{\mathbb{R}}/\Span F,\mathbb{R})$ after translation.

The $h$-vector of $M_{F}$ is defined by
\[
h_{F}(t) = \sum_{i = 0}^{\codim F} f_{i} t^{i} (1 - t)^{\codim F - i},
\]
where $f_{i}$ equals the number of elements in $M_{F}$ of dimension  $i$.
The hyperplane arrangement $\mathcal{H}^{F}$ divides $M_{F,\mathbb{R}}$ into locally closed cells (see Section \ref{construction})
and we will consider the $h$-vector associated to the complex of bounded cells,
\[
h^{\bd}_{F}(t) = \sum_{i = 0}^{\codim F} f^{\bd}_{i}(t - 1)^{i},
\]
where $f^{\bd}_{i}$ equals the number of bounded cells in $\mathcal{H}^{F}$ of dimension $i$.
Since $\mathcal{H}^{F}$ is simple, it follows from results of Zaslavsky that $h_{F}(t) = h^{\bd}_{F}(t)$ \cite{ZasFacing}.

The following geometric interpretation of $h_{F}(t) = h^{\bd}_{F}(t)$
was observed by Hausel and Sturmfels, after Konno had computed the cohomology ring of a hypertoric variety in \cite{KonCohomology}.

\begin{theorem}\cite[Theorem 1.1,Corollary 6.6]{HSToric}\label{hussey}
With the notation above, for any element $F$ in $M_{\mathcal{B}}$, the varieties $X_{\mathcal{H}^{F}}$ and $\mathcal{M}_{\mathcal{H}^{F}}$ have no odd cohomology and
\[ h_{F}(t) = h^{\bd}_{F}(t) = \sum_{i = 0}^{\codim F} \dim_{\mathbb{Q}} H^{2i}(X_{\mathcal{H}^{F}}, \mathbb{Q}) t^{i}
= \sum_{i = 0}^{\codim F} \dim_{\mathbb{Q}} H^{2i}(\mathcal{M}_{\mathcal{H}^{F}}, \mathbb{Q}) t^{i}. \]
\end{theorem}

If $F = \{ b_{k_{1}},  \ldots, b_{k_{r}} \}$ lies in $M_{\mathcal{B}}$, then let $\BOX(F) = \{ v \in N_{\mathbb{R}} \mid v = \sum_{j = 1}^{r} \alpha_{j} b_{k_{j}}, 0 < \alpha_{j} < 1 \}$ and let $\BOX( \{ 0 \} )  = \{ 0 \}$. The following result follows from Theorem \ref{hussey} and the work of Jiang and Tseng.

\begin{theorem}\cite[Proposition 4.7]{JTOrbifold}\label{bussey}
With the notation above,

\[
\sum_{i = 0}^{n + d - 1} \dim_{\mathbb{Q}} H_{\orb}^{2i}(X_{\mathcal{H}}, \mathbb{Q}) t^{i}
= \sum_{i = 0}^{d + n - 1} \dim_{\mathbb{Q}} H_{\orb}^{2i}(\mathcal{M}_{\mathcal{H}}, \mathbb{Q}) t^{i} = \sum_{F \in M} \# \left( \BOX(F) \cap N \right) t^{\dim F} h_{F}(t).
\]
\end{theorem}

We deduce the following result and refer the reader

to Section \ref{combinatorics} for a combinatorial proof.

\begin{theorem}\label{newbe}
The Ehrhart $\delta$-polynomial of the Lawrence polytope $P_{\mathcal{B}}$ is equal to
\[
\delta_{\mathcal{B}}(t) = \sum_{F \in M} \# \left( \BOX(F) \cap N \right) t^{\dim F} h_{F}(t) =
\sum_{F \in M} \# \left( \BOX(F) \cap N \right) t^{\dim F} h^{bd}_{F}(t).
\]
In particular, if we write $b_{i} = a_{i}v_{i}$, for some primitive integer vector $v_{i}$ and some positive integer $a_{i}$,
and set $R^{bd}_{F}$ to be the number of bounded regions of $M_{F, \mathbb{R}} \smallsetminus \mathcal{H}^{F}$,
then
$\delta_{0} = 1$, $\delta_{1} = \sum_{i = 1}^{n}a_{i} - d$, $\delta_{d} = \sum_{F \in M} \# \left( \BOX(F) \cap N \right) R^{bd}_{F}$ and
$\delta_{d + 1} = \ldots = \delta_{d + n - 1} = 0$. If $V_{F}$ denotes the number of vertices in $\mathcal{H}^{F}$, which equals the number of maximal elements of $M_{F}$, then $(d + n - 1)!$ times the volume of $P_{\mathcal{B}}$ is equal to $\sum_{F \in M} \# \left( \BOX(F) \cap N \right) V_{F}$.
\end{theorem}

\begin{rem}
If we change the co-orientation of $\mathcal{B}$, i.e. replace some of the $b_{i}$ by $-b_{i}$, then we obtain an isomorphic hypertoric variety $\mathcal{M}_{\mathcal{H}}$ \cite[Theorem 2.2]{HPProperties}, and hence Theorem \ref{punter} implies that $\delta_{\mathcal{B}}(t)$ remains unchanged.

This can also be deduced using the formula for $\delta_{\mathcal{B}}(t)$ in Theorem \ref{newbe}.

\end{rem}

The matroid $M_{\mathcal{B}}$ is called \emph{coloop free} if there does not exist an element $b_{i}$ which lies in every maximal linearly independent subset of $\mathcal{B}$. One verifies that this condition holds if and only if $M_{\mathbb{R}} \smallsetminus \mathcal{H}$ contains a bounded region.

Hausel and Sturmfels showed that if $M_{\mathcal{B}}$ is coloop free then the
injective part of the Hard Lefschetz theorem holds for the hypertoric variety $\mathcal{M}_{\mathcal{H}}$ and hence the $h$-vector of $M_{\mathcal{B}}$ is a
\emph{MacCauley vector} \cite[Theorem 7.4]{HSToric} (c.f. \cite{SwaElements}). That is, if we write $h_{\{0\}}(t) = \sum_{i = 0}^{d} h_{i} t^{i}$, then there exists a graded $\mathbb{Q}$-algebra $R = R_{0} \oplus R_{1} \oplus \cdots
\oplus R_{\lfloor \frac{d}{2} \rfloor}$ generated by $R_{1}$ and with $g_{i} = h_{i} - h_{i -1} = \dim_{\mathbb{Q}} R_{i}$. We will use the following corollary and refer the reader to \cite{HauQuaternionic} for the definition of the $g$-inequalities.

\begin{theorem}\cite[Corollary 4.2]{HauQuaternionic}\label{rubbish}
If $M_{\mathcal{B}}$ is coloop free then its $h$-vector satisfies
$h_{i} \leq h_{j}$,
for $i \leq j \leq d - i$, and satisfies the $g$-inequalities.
\end{theorem}

We deduce the following inequalities between the coefficients of the Ehrhart $\delta$-polynomial of $P_{\mathcal{B}}$.

\begin{theorem}
If $M_{\mathcal{B}}$ is coloop free, then the Ehrhart $\delta$-polynomial of $P_{\mathcal{B}}$ satisfies
$\delta_{i} \leq \delta_{j}$, for $i \leq j \leq d - i$.
\end{theorem}
\begin{proof}
One verifies that if $M_{\mathcal{B}}$ is coloop free then $M_{F}$ is coloop free for any element $F$ in $M_{\mathcal{B}}$.
By Theorem \ref{newbe}, $\delta_{\mathcal{B}}(t) = \sum_{F \in M} \# \left( \BOX(F) \cap N \right) t^{\dim F} h_{F}(t)$.
If we write $h_{F}(t) = \sum_{i = 0}^{\codim F} h_{F,i} t^{i}$ then Theorem \ref{rubbish} implies that
$h_{F, i - \dim F} \leq h_{F, j - \dim F}$ for $i \leq j \leq d + \dim F - i$ and the result follows.
\end{proof}

\begin{rem}
The theorem above fails if $M_{\mathcal{B}}$ is not coloop free. For example, if $N = \mathbb{Z}$ and $\mathcal{B} = \{ 1 \}$, then $P$ is an interval of length $1$

and $\delta_{0} = 1 > \delta_{1} = 0$.
\end{rem}

\begin{exa}\label{visit}
Let $N = \mathbb{Z}^{2}$ and set $\mathcal{B} = \{ b_{1}, b_{2}, b_{3}, b_{4} \} = \{ (1,0), (0,1), (-2,0), (2,-1)  \}$ and
$\{ r_{1}, r_{2}, r_{3}, r_{4} \} = \{ 0, 0, 2, -1 \}$. The matroid $M_{\mathcal{B}}$ is given by $\{ 0, b_{1}, b_{2}, b_{3}, b_{4}, b_{1}b_{2}, b_{1}b_{4},
b_{2}b_{3}, b_{2}b_{4}, b_{3}b_{4} \}$ and $h_{\{ 0\}}(t) = 1 + 2t + 2t^{2}$. We have $\BOX(\{b_{2}b_{4}\}) \cap N = \{(1,0)\}$, $M_{\{b_{2}b_{4}\}} =
\{ 0 \}$ and $h_{\{b_{2}b_{4}\}}(t) = 1$. Also $\BOX(\{b_{3}\}) \cap N = \{(-1,0)\}$, $M_{\{b_{3}\}} =
\{ 0, b_{2}, b_{4} \}$ and $h_{\{b_{3}\}}(t) = 1 + t$. We conclude that $P_{\mathcal{B}}$ is a $5$-dimensional lattice polytope with $8$ vertices and
$\delta_{P}(t) = (1 + 2t + 2t^{2}) + t^{2} + t(1 + t) = 1 + 3t + 4t^{2}$.

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\psline[linewidth=0.04cm](10.780937,2.0)(10.780937,-2.0)
\psdots[dotsize=0.12](10.780937,0.0)
\psdots[dotsize=0.12](10.780937,-1.0)
\usefont{T1}{ptm}{m}{n}
\rput(11.562344,0.81){$\mathcal{H}^{\{b_{3}\}}$}
\end{pspicture}
}

\end{exa}

\begin{rem}
We have the following toric interpretation of the Ehrhart polynomial $f_{\mathcal{B}}(m)$ (see, for example, \cite{HNPCayley}).
If $P_{i}$ is the convex hull of the origin and $b_{i}$ in $N_{\mathbb{R}}$, then one can consider the Minkowski sum $Q = P_{1} + \cdots + P_{n}$ in $N_{\mathbb{R}}$.
Each $P_{i}$ determines a line bundle $L_{i}$ on the $d$-dimensional toric variety $Y$ corresponding to the polytope $Q$ and the toric variety corresponding to the polytope $P_{\mathcal{B}}$ is the $(d + n - 1)$-dimensional projective bundle $\mathbb{P}_{Y}( L_{1} \oplus \cdots \oplus L_{n} )$, with corresponding line bundle $\mathcal{O}(1)$. The Ehrhart polynomial is given by $f_{\mathcal{B}}(m) = \dim H^{0}(\mathbb{P}( L_{1} \oplus \cdots \oplus L_{n} ), \mathcal{O}(1)^{\otimes m})$, for positive integers $m$ \cite{FulIntroduction}.
\end{rem}

We conclude the introduction with a brief outline of the contents of the paper and note that all varieties and orbifolds will be over $\mathbb{C}$.
In Section \ref{construction}, we present a local construction of hypertoric varieties and use it to explain Theorem \ref{bussey}. In Section \ref{combinatorics}, we give a combinatorial proof of Theorem \ref{newbe}.

\section{Orbifold Cohomology of Hypertoric Varieties}\label{construction}

The goal of this section is to explain Theorem \ref{bussey}, which describes the dimensions of the orbifold cohomology groups of a hypertoric variety.
We will first review the construction of Lawrence toric varieties and refer the reader to Section 4 in \cite{HSToric}
for an equivalent presentation using Gale duality.
We continue with the notation of the introduction and let
$H_{i, +} = \{ u \in M_{\mathbb{R}} \mid \langle u, b_{i} \rangle \geq r_{i}     \}$ and $H_{i, -} = \{ u \in M_{\mathbb{R}} \mid \langle u, b_{i} \rangle \leq r_{i}     \}$,
for $i = 1, \ldots, n$.
The hyperplane arrangement $\mathcal{H}$ divides $M_{\mathbb{R}}$ into locally closed \emph{cells} as follows: an element $u$ in $M_{\mathbb{R}}$ lies in the cell
given by the intersection of $\cap_{u \in H_{i,+}} H_{i,+} $ and $\cap_{u \in H_{i,-}} H_{i,-}$.
If $\sigma$ denotes the cone over the Lawrence polytope $P_{\mathcal{B}}$ in $N_{\mathbb{R}} \times \mathbb{R}^{n}$, then
$\mathcal{H}$ determines a simplicial fan refinement $\Sigma_{\mathcal{H}}$ of $\sigma$ \cite{HSToric}.
 If $C$ is a cell of $\mathcal{H}$, then there is a corresponding simplicial cone $\sigma_{C}$ in $\Sigma_{\mathcal{H}}$ with codimension equal to $\dim C$ and rays passing through the primitive integer vectors
$\{ (b_{i}, e_{i}) \mid C \subseteq H_{i,-} \}$ and $\{ (0, e_{i}) \mid C \subseteq H_{i,+} \}$.
 The cones in $\Sigma_{\mathcal{H}}$ not contained in the boundary of $\sigma$ are in bijection with the bounded cells of $\mathcal{H}$
 \cite[Theorem 4.7]{HSToric}.

In particular, the vertices of $\mathcal{H}$ correspond to the maximal cones of $\Sigma_{\mathcal{H}}$. The simplicial toric variety $X_{\mathcal{H}} = X(\Sigma_{\mathcal{H}})$ is the \emph{Lawrence toric variety} associated to $\mathcal{H}$.
The proper torus-invariant subvarietes $V_{C}$ of $X_{\mathcal{H}}$ are in bijective correspondence with the bounded cells $C$ of $\mathcal{H}$.
The union of the proper torus-invariant subvarieties of $X_{\mathcal{H}}$ is called the \emph{core} of $X_{\mathcal{H}}$.

Fix a vertex $\gamma$ of $\mathcal{H}$ corresponding to a maximal element $F$ in $M_{\mathcal{B}}$ and to the

maximal cone $\sigma_{\gamma}$ in $\Sigma_{\mathcal{H}}$.
We will describe the corresponding open toric subvariety $U_{\gamma}$ of $X_{\mathcal{H}}$.

If $N_{F}$ denotes the sublattice of $N$ generated by $\{ b_{i} \mid b_{i} \in F \}$, then the elements of the finite group $N(F) = N/N_{F}$ are in bijective correspondence with  the elements of $\prod_{G \subseteq F} \BOX(G)\cap N$.
Consider $\mathbb{A}^{d + n}$ with co-ordinates $\{ z_{i} \mid \gamma \in H_{i, -} \}$ and $\{ w_{i} \mid \gamma \in H_{i, +} \}$ and consider the embedding
$\iota: N(F) \hookrightarrow (\mathbb{C}^{*})^{d + n}$ such that if $g$ in $N(F)$ corresponds to
$v = \sum_{b_{i} \in G} \alpha_{i} b_{i} \in \BOX(G) \cap N$ with  $0 < \alpha_{i} < 1$, then,
for every
$b_{i}$ in $G$, $\iota(g)$  equals $e^{2 \pi i \alpha_{i}}$ in the co-ordinate corresponding to $z_{i}$ and equals $e^{2 \pi i (1 - \alpha_{i})}$ in the co-ordinate corresponding to $w_{i}$, and $\iota(g)$ equals $1$ in all other co-ordinates.
The action of $(\mathbb{C}^{*})^{d + n}$ on $\mathbb{A}^{d + n}$ induces an

action of $N(F)$ on $\mathbb{A}^{d + n}$ such that the age of $g$ (see Subsection 7.1 \cite{AGVAlgebraic}) is given by
$\age(g) = \sum_{b_{i} \in G} \alpha_{i} + (1 - \alpha_{i}) = \dim G$. The corresponding open toric subvariety of $X_{\mathcal{H}}$ is given by  $U_{\gamma} = \mathbb{A}^{d + n}/ N(F)$ and the orbifold structure of $X_{\mathcal{H}}$ is locally induced by this quotient.
If $C$ is a bounded cell of $\mathcal{H}$, then the corresponding proper torus-invariant subvariety $V_{C}$ of $X_{\mathcal{H}}$ has $\dim V_{C} = \dim C$.
If $\gamma$ is not in the closure of $C$, then $U_{\gamma} \cap V_{C} = \emptyset$. If $\gamma$ lies in the closure of $C$, then $U_{\gamma} \cap V_{C}$ is defined by setting the co-ordinates $\{ z_{i} \mid C \subseteq H_{i,-} \}$ and $\{ w_{i} \mid C \subseteq H_{i, +} \}$ equal to zero.

The hypertoric variety $\mathcal{M}_{\mathcal{H}}$ associated to $\mathcal{H}$ is a $2d$-dimensional complete intersection in $X_{\mathcal{H}}$ with a canonical orbifold structure. We will describe $\mathcal{M}_{\mathcal{H}} \cap U_{\gamma}$ (c.f. \cite[Proposition 4.3]{JTOrbifold}).

If $b_{i} \notin F$, then we may write $b_{i} = \sum_{b_{j} \in F} a_{i,j} b_{j}$, for unique rational numbers $a_{i,j}$.
The ideal $I$ generated by $\langle z_{i} = \sum_{b_{j} \in F} a_{i,j}z_{j}w_{j} \mid
u \in H_{i, -} \smallsetminus H_{i, +}  \rangle$ and $\langle w_{i} = \sum_{b_{j} \in F} a_{i,j}z_{j}w_{j} \mid
u \in H_{i, +} \smallsetminus H_{i, -}  \rangle$,  defines a subvariety $V(I) \subseteq \mathbb{A}^{d + n}$ which is invariant under the action of
$N(F)$. The induced subvariety $V(I)/N(F)$ of $U_{\gamma}$ is equal to $\mathcal{M}_{\mathcal{H}} \cap U_{\gamma}$ and the orbifold structure of $\mathcal{M}_{\mathcal{H}}$ is locally induced by this quotient. It can be seen from this description that the core of $X_{\mathcal{H}}$ is contained in $\mathcal{M}_{\mathcal{H}}$.
If $\mathbb{A}^{2d}$ has co-ordinates $\{ z_{i} \mid b_{i} \in F \}$ and $\{ w_{i} \mid b_{i} \in F \}$, then the corresponding projection $\pi: \mathbb{A}^{d + n} \rightarrow \mathbb{A}^{2d}$ induces an isomorphism
$\pi: V(I) \rightarrow \mathbb{A}^{2d}$. The embedding $\iota: N(F) \hookrightarrow (\mathbb{C}^{*})^{d + n}$ factors as an embedding $\nu: N(F) \hookrightarrow (\mathbb{C}^{*})^{2d}$ followed by the inclusion of $(\mathbb{C}^{*})^{2d}$ into $(\mathbb{C}^{*})^{d + n}$ by adding $1$'s in the extra co-ordinates. The embedding $\nu$ induces an action of $N(F)$ on $\mathbb{A}^{2d}$ such that the projection $\pi$ is $N(F)$-equivariant and induces an isomorphism of orbifolds $\mathcal{M}_{\mathcal{H}} \cap U_{\gamma} \cong \mathbb{A}^{2d}/N(F)$.

We have seen in the introduction that any $G$ in $M_{\mathcal{B}}$ determines a hyperplane arrangement $\mathcal{H}^{G}$ with co-normal vectors $\mathcal{B}_{G}$. The corresponding Lawrence toric variety $X_{G}$ and hypertoric variety $\mathcal{M}_{G}$ may be regarded as subvarieties of $X_{\mathcal{H}}$ and $\mathcal{M}_{\mathcal{H}}$ respectively. For $g$ in $N(F)$
corresponding to
$v \in \BOX(G) \cap N$,
one verifies that,  as varieties,  $X_{G} \cap U_{\gamma} = (\mathbb{A}^{d + n})^{g}/N(F)$
and $\mathcal{M}_{G} \cap U_{\gamma} \cong (\mathbb{A}^{2d})^{g}/N(F)$, where $(\mathbb{A}^{d + n})^{g}$ and  $(\mathbb{A}^{2d})^{g}$ are the subvarieties of $(\mathbb{A}^{d + n})$ and $(\mathbb{A}^{2d})$ respectively that are invariant under the action of $g$.

\begin{exa}
Consider the notation of Example \ref{visit} and fix the vertex $\gamma = (-1/2, 0)$ of $\mathcal{H}$ corresponding to $F = \{ b_{2}b_{4} \}$ in $M_{\mathcal{B}}$.
The open toric subvariety $U_{\gamma}$ of $X_{\mathcal{H}}$ is the quotient of $\mathbb{A}^{6}$ with co-ordinates $\{z_{1},z_{2},z_{3},
z_{4},w_{2},w_{4}\}$ by the action of $N(F) = \mathbb{Z}/2\mathbb{Z}$, acting via multiplication by $(1,-1,1,-1,-1,-1)$.
The open subvariety $\mathcal{M}_{\mathcal{H}} \cap U_{\gamma}$ of the hypertoric variety $\mathcal{M}_{\mathcal{H}}$ is the complete intersection defined by $z_{1} = \frac{1}{2} z_{2}w_{2} + \frac{1}{2} z_{4}w_{4}$ and $z_{3} = -z_{2}w_{2} - z_{4}w_{4}$.
The core of $X_{\mathcal{H}}$ consists of two weighted projective spaces $\mathbb{P}(1,1,2)$ that intersect along their unique singular points at $V_{\gamma} = X_{F} = \mathcal{M}_{F}$.
\end{exa}

\begin{rem}
Hausel and Sturmfels proved that the embedding of the core of $X_{\mathcal{H}}$ into either $X_{\mathcal{H}}$ or $\mathcal{M}_{\mathcal{H}}$ induces an isomorphism of cohomology rings over $\mathbb{Z}$ \cite[Lemma 6.5]{HSToric} and used this to prove Theorem \ref{hussey} and Theorem \ref{rubbish}.
\end{rem}

Orbifold cohomology was introduced by Chen and Ruan in \cite{CRNew} and associates to an orbifold $Y$ a graded $\mathbb{Q}$-algebra $H^{*}_{\orb}(Y, \mathbb{Q})$. More specifically, one associates to $Y$ an orbifold $I(Y)$, called the \emph{inertia orbifold} of $Y$, such that,

if $Y = X/G$ for some smooth variety $X$ and finite abelian group $G$, then $I(Y) = \coprod_{g \in G} X^{g}/G$, where $X^{g}$ is the subvariety of $X$ fixed by $g$. If $Y_{1}, \ldots, Y_{r}$ denote the connected components of $I(Y)$, then,
for any $i \in \mathbb{Q}$, Chen and Ruan defined the $i$'th
orbifold cohomology group of $Y$ by
\begin{equation*}
H_{\orb}^{i}(Y, \mathbb{Q}) = \bigoplus_{j = 1}^{r} H^{i - 2\age(Y_{j})}(Y_{j}, \mathbb{Q}),
\end{equation*}
where $\age(Y_{j})$ is the \emph{age} of $Y_{j}$.

We have $I(U_{\gamma}) = \coprod_{G \subseteq F} \coprod_{  v \in \BOX(G) \cap N } (\mathbb{A}^{d + n})^{g}/N(F)$, $I(\mathcal{M}_{\mathcal{H}} \cap U_{\gamma}) = \coprod_{G \subseteq F} \coprod_{  v \in \BOX(G) \cap N } (\mathbb{A}^{2d})^{g}/N(F)$, and, in both cases, the age of the connected component corresponding to a pair $(v,G)$ is equal to $\dim G$.
In fact,  we have isomorphisms of varieties, $I(X_{\mathcal{H}}) \cong \coprod_{G \in M} \coprod_{  v \in \BOX(G) \cap N } X_{G}$ and
$I(\mathcal{M}_{\mathcal{H}}) \cong \coprod_{G \in M} \coprod_{  v \in \BOX(G) \cap N } \mathcal{M}_{G}$ \cite{JTNote}. We conclude that Theorem \ref{hussey} and the above computation of inertia orbifolds implies Theorem \ref{bussey}, which computes the orbifold cohomology of $X_{\mathcal{H}}$ and $\mathcal{M}_{\mathcal{H}}$.

\section{Ehrhart Theory for Lawrence Polytopes}\label{combinatorics}

The goal of this section is to give a combinatorial proof of Theorem \ref{newbe}. We will first show that
\begin{equation}\label{done}
\delta_{\mathcal{B}}(t) =
\sum_{F \in M} \# \left( \BOX(F) \cap N \right) t^{\dim F} h^{bd}_{F}(t).
\end{equation}

The proof should be compared with the combinatorial proof of Theorem \ref{oldie} (see \cite{YoWeightI}) and the proof of \cite[Theorem 1.1]{PayEhrhart}.  Recall that $\sigma$ denotes the cone over $P_{\mathcal{B}}$ in $N_{\mathbb{R}} \times \mathbb{R}^{n}$ and that $\Sigma_{\mathcal{H}}$ is a fan refining $\sigma$, such that  the cones $\sigma_{C}$ in $\Sigma_{\mathcal{H}}$ that do not lie in the boundary of $\sigma$ are in bijection with the bounded cells $C$ of $\mathcal{H}$. The cone $\sigma_{C}$ has codimension equal to $\dim C$ and its rays correspond to the primitive integer vectors
$\{ (b_{i}, e_{i}) \mid C \subseteq H_{i,-} \}$ and $\{ (0, e_{i}) \mid C \subseteq H_{i,+} \}$. If $F$ is an element of $M_{\mathcal{B}}$ and $w = \sum_{b_{i} \in F} a_{i}b_{i} \in \BOX(F) \cap N$, for some $0< a_{i} < 1$, then let
$l(w) = \sum_{b_{i} \in F} (a_{i}b_{i},e_{i}) \in \sigma \cap (N \times \mathbb{Z}^{n})$.
If $v$ is a lattice point in the interior of $\sigma$ and $C$ is the largest bounded cell in $\mathcal{H}$ such that $v \in \sigma_{C}$, then we have a unique expression
\[
v = \: l(w) + \sum_{b_{i} \notin F, C \subseteq H_{i,-}} (b_{i}, e_{i}) + \sum_{b_{i} \notin F, C \subseteq H_{i,+} } (0, e_{i})
+ \sum_{C \subseteq H_{i,-}} \alpha_{i} (b_{i}, e_{i}) + \sum_{C \subseteq H_{i,+}} \beta_{i}(0, e_{i}),
\]
where $w \in \BOX(F) \cap N$, for some $F$ in $M_{\mathcal{B}}$ such that $C$ is a bounded cell in $\mathcal{H}^{F}$, and $\alpha_{i}, \beta_{i}$ are non-negative integers. Conversely, given $w \in \BOX(F) \cap N$, a bounded cell $C$ in $\mathcal{H}^{F}$
and non-negative integers $\{ \alpha_{i} \mid C \subseteq H_{i,-} \}$ and $\{ \beta_{i} \mid C \subseteq H_{i,+} \}$,
the lattice point
$v = l(w) + \sum_{b_{i} \notin F, C \subseteq H_{i,-}} (b_{i}, e_{i}) + \sum_{b_{i} \notin F, C \subseteq H_{i,+} } (0, e_{i})
+ \sum_{C \subseteq H_{i,-}} \alpha_{i} (b_{i}, e_{i}) + \sum_{C \subseteq H_{i,+}} \beta_{i}(0, e_{i})$, lies in the interior of $\sigma$ and $C$ is the largest bounded cell in $\mathcal{H}$ such that $v \in \sigma_{C}$. Recall from the proof of Theorem \ref{punter} that
$\psi: N \times \mathbb{Z}^{n} \rightarrow \mathbb{Z}$ is given by $\psi(\lambda + \sum_{i = 1}^{n} \mu_{i}e_{i} ) = \sum_{i = 1}^{n} \mu_{i}$, for $\lambda \in N$ and integers $\mu_{i}$, and that $\psi^{-1}(m) \cap \sigma = mP_{\mathcal{B}}$ and $\psi(l(w)) = \dim F$. It is a standard result of Ehrhart theory (see, for example, \cite{BRComputing}) that
\[ \sum_{m \geq 1} \# \left( \Int(mP_{\mathcal{B}}) \cap (N \times \mathbb{Z}^{n}) \right)t^{m} = \frac{t^{d + n} \delta_{\mathcal{B}}(t^{-1})}{(1 - t)^{d + n}}. \]
Using the above facts and setting $\mathcal{H}_{\bd}^{F}$ to be the collection of bounded cells in $\mathcal{H}^{F}$, we calculate
\begin{align*}
t^{d + n} \delta_{\mathcal{B}}(t^{-1}) &= (1 - t)^{d + n} \sum_{ v \in \Int(\sigma) \cap (N \times \mathbb{Z}^{n})} t^{\psi(v)} \\
&= (1 - t)^{d + n} \sum_{F \in M} \# \left( \BOX(F) \cap N \right) t^{\dim F} \sum_{C \in \mathcal{H}_{\bd}^{F}} \frac{t^{d + n - \dim C -2\dim F}}{( 1- t)^{d + n - \dim C}} \\
&= t^{d + n} \sum_{F \in M} \# \left( \BOX(F) \cap N \right) t^{-\dim F} \sum_{C \in \mathcal{H}_{\bd}^{F}} (t^{-1} - 1)^{\dim C},
\end{align*}
and (\ref{done}) follows immediately as desired. We present below the remainder of the proof of Theorem \ref{newbe}.

\begin{proof}

Observe that the constant coefficient in $h_{F}(t)$ is $1$ and the coefficient of $t$ in $h_{\{ 0 \}}(t)$ is
$n - d$. The elements of $M_{\mathcal{B}}$ of dimension $1$ correspond to the elements $b_{i}$ in $\mathcal{B}$ and $\BOX(\{b_{i}\} ) \cap N$ consists of $a_{i} - 1$ lattice points. We deduce that $\delta_{1} = n - d + \sum_{i = 1}^{n} (a_{i} - 1) = \sum_{i = 1}^{n}a_{i} - d$. The leading term of
$h^{bd}_{F}(t)$ is $R^{bd}_{F} t^{\codim F}$ and hence $\delta_{d} = \sum_{F \in M} \# \left( \BOX(F) \cap N \right) R^{bd}_{F}$ and
$\delta_{d + 1} = \ldots = \delta_{d + n - 1} = 0$. It is a standard fact of Ehrhart theory (see, for example, \cite{BRComputing}) that the normalised volume of $P_{B}$ is equal to $\frac{1}{(d + n - 1)!} \delta_{\mathcal{B}}(1)$. The last statement follows from the observation that
$h_{F}(1) = h^{\bd}_{F}(1) = V_{F}$.
\end{proof}

\begin{rem}\label{smashing}
If $P_{1}, \ldots, P_{n}$ are lattice polytopes in $N_{\mathbb{R}}$ and $\{e_{1}, \ldots, e_{n} \}$ is the standard basis of $\mathbb{R}^{n}$, then the \emph{Cayley sum} $P = P_{1} \ast \cdots \ast P_{n}$ is the convex hull of
$P_{1} \times \{e_{1}\}$, \ldots, $P_{n} \times \{e_{n}\}$ in $N \times \mathbb{R}^{n}$. If the affine span of the union of the $P_{i}$ is $N_{\mathbb{R}}$, then $P$ is a $(d + n - 1)$-dimensional lattice polytope. Setting $P_{i}$ to be the convex hull of the origin and $b_{i}$, we
recover the Lawrence polytope $P_{\mathcal{B}}$.
The degree $s$ of a lattice polytope $Q$ is the degree of its Ehrhart $\delta$-polynomial and it is a standard fact that  $(\dim Q + 1 - s)Q$ is the smallest dilate of $Q$
that contains a lattice point in its relative interior (see, for example, \cite{BRComputing}). Cayley sums provide examples of lattice polytopes with small degree relative to their dimension since the degree of $P$ is at most $d$ \cite[Proposition]{BNMultiples}. In \cite{BNMultiples}, Batyrev and Nill
used Lawrence polytopes in the classification of lattice polytopes of degree $1$.
\end{rem}

\begin{rem}
It is standard result of Ehrhart theory (see, for example, \cite{BRComputing}) that $\delta_{1} = f_{\mathcal{B}}(1) - \dim P_{\mathcal{B}} - 1$

and hence Theorem \ref{newbe} implies that $P_{\mathcal{B}}$ contains $\sum_{i = 1}^{n} (a_{i} + 1)$ lattice points. More specifically, the lattice points in $P_{\mathcal{B}}$ are
$\{ (\lambda_{i}v_{i}, e_{i}) \mid 0 \leq \lambda_{i} \leq a_{i}, 1 \leq i \leq n \}$. We noted in Remark \ref{smashing} that $mP_{\mathcal{B}}$ contains no interior lattice points for $1 \leq m \leq n - 1$. It is a standard fact that $\delta_{d} = \# ( \Int(nP_{\mathcal{B}}) \cap (N \times \mathbb{Z}^{n}) )$ (see, for example, \cite{BRComputing}),
and hence Theorem \ref{newbe} implies that $nP$ contains $\sum_{F \in M} \# \left( \BOX(F) \cap N \right) R^{bd}_{F}$ interior lattice points. More specifically, if $w$ is a lattice point in $\BOX(F)$ and $R$ is a bounded region in $\mathcal{H}^{F}$, then the corresponding lattice point in the relative interior of $nP_{\mathcal{B}}$ is $l(w) + \sum_{b_{i} \notin F, R \subseteq H_{i,-}} (b_{i}, e_{i}) + \sum_{b_{i} \notin F,  R \subseteq H_{i,+}} (0, e_{i}) \in \sigma_{R}$.
\end{rem}

\begin{rem}
We present an alternative combinatorial proof of the formula
\begin{equation*}
\delta_{\mathcal{B}}(t) =
\sum_{F \in M} \# \left( \BOX(F) \cap N \right) t^{\dim F} h_{F}(t).
\end{equation*}
If $v$ is a lattice point in $\sigma$, then $v$ lies in some maximal cone $\sigma_{\gamma}$ of $\Sigma_{\mathcal{H}}$, and we may write
$v = \sum_{\gamma \in H_{i,-}} \alpha_{i} (b_{i}, e_{i}) + \sum_{\gamma \in H_{i,+}} \beta_{i} (0, e_{i})$, for unique non-negative integers $\alpha_{i}$ and $\beta_{i}$. The element $G$ of $M_{\mathcal{B}}$ generated by $\{ b_{i} \mid \alpha_{i}, \beta_{i} \neq 0 \}$ and the sets
 $I_{-} = \{ i \mid \alpha_{i} \neq 0, b_{i} \notin \Span G  \}$ and
$I_{+} = \{ i \mid \beta_{i} \neq 0, b_{i} \notin \Span G  \}$
are independent of the choice of maximal cone $\sigma_{\gamma}$.

Note that $b_{i} \in \Span G$ if and only if  $M_{G,\mathbb{R}} = \cap_{b_{i} \in G} H_{i}$ is contained in either  $H_{i,-}$ or $H_{i,+}$.
If we consider the projection $\phi_{G}: N \rightarrow  N/ (N \cap \Span G)$ and the hyperplane arrangement $\{ \phi_{G}(H_{i}) \mid i \in I = I_{-} \cup I_{+}  \}$, then the intersection of the half spaces $\{ \phi_{G}(H_{i, -}) \mid  i \in I_{-} \}$ and $\{ \phi_{G}(H_{i, +}) \mid  i \in I_{+} \}$ is a region in the hyperplane arrangement and, moreover, every region in the hyperplane arrangement has this form.
We deduce that $v$ has a unique expression
\[
v = \: l(w) + \sum_{b_{i} \in G \smallsetminus F} ((b_{i}, e_{i}) + (0, e_{i}))
+ \sum_{M_{G,\mathbb{R}} \subseteq H_{i,-}} \mu_{i} (b_{i}, e_{i}) + \sum_{M_{G,\mathbb{R}} \subseteq H_{i,+}} \mu_{i}'(0, e_{i})
+ \sum_{i \in I_{-}} \nu_{i}(b_{i}, e_{i}) + \sum_{i \in I_{+}} \nu_{i}'(0, e_{i}),
\]
where $w \in \BOX(F) \cap N$, for some $F \subseteq G$, $\mu_{i}, \mu_{i}'$ are non-negative integers and $\nu$, $\nu'$ are positive integers.

Conversely, consider elements $F \subseteq G$ in $M_{\mathcal{B}}$, a subset $I$ in $\{ i \mid b_{i} \notin \Span G \}$ and a region $R$ in the hyperplane arrangement $\{ \phi_{G}(H_{i}) \mid i \in I  \}$. If we set $I_{-} = \{ i \mid R \subseteq \phi_{G}(H_{i, -}) \}$ and $I_{+} = \{ i \mid R \subseteq \phi_{G}(H_{i, +}) \}$, and consider some $w \in \BOX(F) \cap N$, some non-negative integers $\mu_{i}, \mu_{i}'$ and some positive integers $\nu$, $\nu'$, then the right hand side of the above expression gives a lattice point $v$ in $\sigma$.

By a theorem of Zaslavsky \cite{ZasFacing}, the number of regions $r_{G,I}$ in the hyperplane arrangement $\{ \phi_{G}(H_{i}) \mid i \in I  \}$
is equal to the number of elements $G'$ in $M_{\mathcal{B}}$ satisfying $G \subseteq G'\subseteq G \cup \{ b_{i} \mid i \in I \}$.
Setting
$I_{G} = \{ i \mid b_{i} \notin \Span G \}$, we compute

\begin{align*}
\delta_{\mathcal{B}}(t) &= (1 - t)^{d + n} \sum_{ v \in \sigma \cap (N \times \mathbb{Z}^{n})} t^{\psi(v)} \\
&= (1 - t)^{d + n} \sum_{F \in M} \# \left( \BOX(F) \cap N \right) t^{\dim F} \sum_{F \subseteq G}  \frac{t^{2(\dim G - \dim F)}}{(1 - t)^{\dim G + n - |I_{G}|}} \sum_{I \subseteq I_{G} } r_{G,I} \frac{t^{|I|}}{(1 - t)^{|I| }} \\
&= \sum_{F \in M} \# \left( \BOX(F) \cap N \right) t^{-\dim F} \sum_{F \subseteq G} t^{2\dim G} (1 - t)^{d - \dim G + |I_{G}|} \sum_{G \subseteq G'} \frac{t^{\dim G' - \dim G}}{(1 - t)^{|I_{G}|}} \\
&= \sum_{F \in M} \# \left( \BOX(F) \cap N \right) t^{\dim F }  \sum_{F \subseteq G'} t^{\dim G' - \dim F }(1 - t)^{d - \dim G'} \sum_{F \subseteq G \subseteq G'} t^{\dim G - \dim F} (1 - t)^{\dim G' - \dim G} \\
&= \sum_{F \in M} \# \left( \BOX(F) \cap N \right) t^{\dim F} h_{F}(t).
\end{align*}

\end{rem}

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TITLE: Recovering information about braids from their decomposition into positive and negative braids
QUESTION [8 upvotes]: Suppose that $b$ is a braid. Then $b$ can be uniquely written as
$D_{RL}(b)^{-1}N_{RL}(b)$ where $D_{RL}(b),N_{RL}(b)$ are the unique positive braids such that $b=D_{RL}(b)^{-1}N_{RL}(b)$ and where
$D_{RL}(b)^{-1}\wedge_{L}N_{RL}(b)=e$ where $r\wedge_{L}s$ denotes the left gcd of the positive braids $r$ and $s$. See Chapter 2 of Dehornoy's book Braids and Self-Distributivity for more information about this decomposition.
I wonder what information about the braid $b$ can be recovered from the positive braids $D_{RL}(b),N_{RL}(b)$.
Suppose that $b$ is a braid. Then if Karen has access to both $D_{RL}(b),N_{RL}(b)$, then Karen is able to recover the entire braid $b$. What information about the braid $b$ can Karen recover if she has access to only $D_{RL}(b)$ or only $N_{RL}(b)$? I am mainly interested in orderly braids $b$ where if $w=\sigma_{\alpha_{1}}^{e_{1}}...\sigma_{\alpha_{k}}^{e_{k}}$ is a braid word for $b$ with a minimized length, then there will likely be many $i$'s where $\alpha_{i}$ is very close to $\alpha_{i+1}$. I am also interested in what properties of the braid $b$ make it likely or unlikely for one to recover information about $b$ when one knows $D_{RL}(b)$ or $N_{RL}(b)$.

REPLY [0 votes]: I claim that $D_{RL}(b)$ in practice contains most of the information about the left half of the braid $b$ and $N_{RL}(b)$ contains most information about the right half of the braid $b$. Furthermore, I claim that whenever $b=uv$ where $u$ is positive and $v$ is negative, then one can recover much information about the braid $b$ from either $u$ or $v$.
If $b$ is a highly structured braid, then it is usually very difficult to hide any information about $b$. After all, all the current braid based cryptographic algorithms have been broken since braids do not do a very good job at concealing information. In particular, if $b$ is a braid, then Dehornoy's handle reduction could be used to find a braid word that represents $b$ and which reveals much of the information about the braid $b$. It seems like handle reduction is
the best technique for revealing information about a braid. In fact, one can do slightly better by using Algorithm 1 in this paper to minimize the length of a braid and also reveal information about the braid. One could also improve Algorithm 1
by sorting and other techniques.
Since positive braids are already handle reduced, handle reduction cannot be used to transform a positive braid word or a negative braid word because positive
braids word and negative braid words are already reduced. We however can use the following technique that allows us to use handle reduction to reveal information even about positive braids.
If $b$ is a positive braid, then one could find a canonical positive braid $c$ where $b$ and $c$ have about the same length and then use handle reduction to find a reduced word that represents
$bc^{-1}$ and then the reduced word representing $bc^{-1}$ would reveal all the information that one was attempting to conceal using the braid $b$.
If you know anything about braids, you would know that the canonical choice
of the braid $c$ is either a power of $\Delta_{n}$ or a power of $\delta_{n}$ where we define $\delta_{n}=\sigma_{n-1}\dots\sigma_{1}$ and we define
$\Delta_{n}=\delta_{1}\dots\delta_{n}$. It is not too difficult to show that $\delta_{n}^{n}=\Delta_{n}^{2}$ (so there is not much difference in choosing
$\delta_{n}$ or $\Delta_{n}$) and that
$\Delta_{n}^{-1}\sigma_{i}\Delta_{n}^{-1}=\sigma_{n-i}$ for $1\leq i<n$. The braids $\delta_{n},\Delta_{n}$ are essential in the Garside normal form and the
Birman-Ko-Lee normal forms of the braid groups. Furthermore, for each positive braid $b$, there is some $k$ and positive braid $c$ where
$bc=\Delta_{n}^{k}$.
Suppose that $b=uv$ where $u$ is positive and $v$ is negative. Let $L:B_{n}\rightarrow\mathbb{Z}$ be the homomorphism where
$L(\sigma_{i})=1$ for all $i$. Let $r=\lfloor\frac{2\cdot L(u)}{n(n-1)}\rfloor$ and $s=-\lfloor\frac{2\cdot L(v)}{n(n-1)}\rfloor$.
Now, $$b=uv=u\Delta_{n}^{-r}\Delta_{n}^{r-s}\Delta_{n}^{s}v.$$
An entity with access to $u$ can extract information about the braid $b$ by applying handle reduction and other techniques on the braid $u\Delta_{n}^{-r}$ and an entity with access to $v$ can extract information about $b$ from $\Delta_{n}^{s}v$.
In practice, handle reduction a very fast algorithm and handle reduction is conjectured to always terminate within quadratically many steps.
Example
Suppose that $b$ is the following braid.
[ 1, 1, -2, -2, -3, -3, -4, -4, 5, 5, -6, -6, 7, 7, 8, 8, -9, -9, -10, -10, 1, 1, 2, 2, 3, 3, -4, -4, -5, -5, -6, -6, 7, 7,
  8, 8, -9, -9, 10, 10, -1, -1, 2, 2, -3, -3, 4, 4, 5, 5, -6, -6, -7, -7, -8, -8, 9, 9, -10, -10, 1, 1, -2, -2, 3, 3, 4, 4, 5, 5, -6, -6, 7, 7, 8, 8, -9, -9, 10, 10, 1, 1, 2, 2, 3, 3,
  4, 4, -5, -5, 6, 6, 7, 7, 8, 8, 9, 9, 10, 10, 1, 1, -2, -2, -3, -3, 4, 4, 5, 5, 6, 6, -7, -7, -8, -8, -9, -9, 10, 10, 1,
  1, -2, -2, 3, 3, 4, 4, -5, -5, -6, -6, -7, -7, -8, -8, 9, 9, -10, -10, 1, 1, -2, -2, -3, -3, 4, 4, -5, -5, -6, -6, -7, -7, 8, 8, -9, -9, 10, 10, -1, -1, 2, 2, 3, 3, 4, 4, 5, 5, -6, -6,
  7, 7, 8, 8, 9, 9, -10, -10, 1, 1, 2, 2, 3, 3, 4, 4, 5, 5, 6, 6, -7, -7, -8, -8, -9, -9, -10, -10, 1, 1, -2, -2, -3, -3,
  -4, -4, -5, -5, 6, 6, 7, 7, 8, 8, -9, -9, -10, -10, -1, -1, -2, -2, -3, -3, 4, 4, 5, 5, -6, -6, -7, -7, -8, -8, -9, -9, -10, -10, 1, 1, -2, -2, 3, 3, -4, -4, 5, 5, -6, -6, 7, 7, 8, 8, 9,
  9, -10, -10, 1, 1, 2, 2, -3, -3, 4, 4, -5, -5, -6, -6, -7, -7, 8, 8, 9, 9, -10, -10, 1, 1, -2, -2, -3, -3, 4, 4, -5, -5, 6,
  6, -7, -7, 8, 8, 9, 9, -10, -10, -1, -1, 2, 2, -3, -3, -4, -4, 5, 5, -6, -6, -7, -7, 8, 8, 9, 9, 10, 10, 1, 1, 2, 2, -3, -3, 4, 4, 5, 5, 6, 6, 7, 7, 8, 8, -9, -9, -10, -10, 1, 1,
  -2, -2, 3, 3, -4, -4, -5, -5, -6, -6, -7, -7, 8, 8, 9, 9, 10, 10, -1, -1, 2, 2, 3, 3, 4, 4, -5, -5, -6, -6, 7, 7, 8, 8, 9, 9,
  10, 10, 1, 1, 2, 2, -3, -3, 4, 4, -5, -5, -6, -6, 7, 7, -8, -8, 9, 9, -10, -10 ]
Write $b=D_{RL}(b)^{-1}N_{RL}(b)$. Then the following braid word obtained through techniques listed in this answer represents a braid of the form $\Delta_{n}^{r}N_{RL}(b)$. On the other hand, the following braid gives most of the information about the right half of the braid $b$.
[ 9, 8, -9, -10, 7, 8, 9, -10, 9, 10, 10, 9, 10, 8, 9, 6, 7, 8, 9, 10, 5, 6, 7, -8, 9, 10, 10, -8, -9, -9, 10, 10, -4, -5,
  -3, -4, -5, -5, -1, 2, -3, 4, 4, 5, 2, -3, 4, 5, 2, 3, 4, 5, -6, -6, 7, 7, 8, 8, 9, 9, 10, 10, 5, 5, 6, 6, -7, -7, -8, -8, -9, -9, -10, -10, -1, 2, -3, 2, -3, -3, -1, 2, 3, 4, 4, -5, -5,
  6, -7, 6, 6, 7, -8, -8, 9, 9, 10, 10, -1, 2, 3, 3, -4, -4, -5, -6, -6, -6, -5, 6, 7, 7, -8, -8, 9, 9, 10, 10, -1, -2, -2,
  -3, -3, -4, 5, -4, 5, -6, -6, -7, -7, -8, -8, -9, -9, -10, -10, -1, -1, -2, -2, -3, 4, -5, -6, -7, -8, -9, -10, -3, 4, -5, -6, -7, -8, -9, 4, -5, -6, -7, -8, 9, 10, 4, -5, -6, -7, 8, 9,
  10, 4, -5, -6, 4, -5, 6, 7, 8, 9, 10, 4, 5, 6, 7, 8, 9, 10, -3, -4, -5, 6, -2, -3, -1, -2, -3, -3, -4, 5, -6, -7, -8, -9,
  -10, 5, -6, -7, -8, -9, 10, -4, -5, -5, 6, -7, -8, 9, 10, -5, -6, -7, 8, -4, -5, -6, 7, -8, 9, -3, -4, -5, -6, -6, -7, 8, -9, -10, 8, -9, 10, -7, 8, 9, 10, -7, 8, 9, -6, -7, -7, 8, -9,
  -10, 8, -9, 10, 8, 9, 10, -7, -8, -8, 9, -10, 9, -8, -9, 10, -9, -9, -10, -10, -2, 3, -4, 5, 3, -2, -3, -4, 5, -4, -5, -6, -6, -7,
  -7, 8, 8, 9, 9, 10, 10, -5, -5, -6, -6, 7, 7, 8, 8, 9, 9, 10, 10, -3, -3, 4, 4, -5, -5, -6, -6, 7, 7, -8, -8, 9, 9, -10, -10 ]


TITLE: Finding the limit of a series
QUESTION [1 upvotes]: I got the n'th term but clueless about what to do next

REPLY [0 votes]: We have
$$t_r = \dfrac{r(r+1)(2r+1)/6}{r^2(r+1)^2/4} = \dfrac23 \dfrac{2r+1}{r(r+1)} = \dfrac23 \left(\dfrac1r + \dfrac1{r+1}\right)$$
Hence,
\begin{align}
S_n & = \sum_{r=1}^n (-1)^r t_r = \sum_{r=1}^n (-1)^r \dfrac23 \left(\dfrac1r + \dfrac1{r+1}\right) = \dfrac23 \left(\sum_{r=1}^n \dfrac{(-1)^r}r + \sum_{r=1}^n \dfrac{(-1)^r}{r+1} \right)\\
& = \dfrac23 \left(-\dfrac11 + \sum_{r=2}^n \dfrac{(-1)^r}r + \sum_{r=1}^{n-1} \dfrac{(-1)^r}{r+1} + \dfrac{(-1)^n}{n+1} \right)\\
& = \dfrac23 \left(-1 + \sum_{r=1}^{n-1} \dfrac{(-1)^{r+1}}{r+1} + \sum_{r=1}^{n-1} \dfrac{(-1)^r}{r+1} + \dfrac{(-1)^n}{n+1}\right)\\
& = -\dfrac23 \left(1+ \dfrac{(-1)^{n+1}}{n+1}\right)
\end{align}
Now conclude the limit of $S_n$.