TITLE: Lebesgue-integrability and almost everywhere equal functions
QUESTION [1 upvotes]: Claim: If two functions $f,g: \Omega \to \overline{\mathbb{R}}$ are equal almost everywhere in $E$, i.e. $f=g$ on $E \setminus N$ with $\mu(N)=0$, and $f \in \mathcal{L}(E)$, then $g \in \mathcal{L}(E)$ and the integrals have the same value.
My definitions are as follows:

Definition: Let $(\Omega,\Sigma,\mu)$ be a measure space and $E \in \Sigma$. Then the integral of a non-negative simple function $s: \Omega \to [0,\infty]$ with representation $s(x) = \sum \limits_{j=1}^{k} \alpha_j \chi_{A_j(x)}$ on $E$ is
$$\int_E s d\mu := \sum \limits_{j=1}^{k} \alpha_j \mu(A_j \cap E).$$

Note that it can be shown that the integral does not depend on the representation of $s$.

Definition: Let $(\Omega,\Sigma,\mu)$ be a measure space and $E \in \Sigma$, then the integral of a non-negative function $f: \Omega \to [0,\infty]$ is
$\int_E f d\mu := sup \{\int_E s d\mu: 0 \leq s \leq f \text{ on E, s simple}\}.$

Note that $f$ is not required to be measurable, but integral functions in the end will be required to be measurable.

Definition: Let $(\Omega,\Sigma,\mu)$ be a measure space, $E \in \Sigma$ and $f: \Omega \to \overline{\mathbb{R}}$.
(i) If at least one of the integrals of $\int_E f_+ d\mu$, $\int_E f_- d\mu$ is finite, then the integral of $f$ is
$$\int_E f d\mu := \int_E f_+ d\mu$ - \int_E f_- d\mu.$$
(ii) If $f$ is measurable and both integrals are finite, i.e. $\int_E f d\mu \in \mathbb{R}$, then $f$ is Lebesgue-integrable ($f \in \mathcal{L}(E)$).

Now my idea to prove the claim is as follows:
There is a lemma in my lecture notes that states:

Lemma: Let $(\Omega,\Sigma,\mu)$ be a measure space, $E \in \Sigma$ and $f,g: \Omega \to \overline{\mathbb{R}}$.
$A \subset E, A \in \Sigma$ with $\mu(E \setminus A)=0$ and $f \in \mathcal{L}(E) \implies f \in \mathcal{L}(A)$ and $\int_A f d\mu = \int_E f d\mu.$

The proof proceeds step-wise for simple, non-negative and then general functions.
It is pretty easy to see that the integral equality holds for step functions and then follows for non-negative functions:

A simple function with $0 \leq s \leq f$ on E, also satisfies this condition on $A$, so $\int_A f d\mu \geq \int_E f d\mu$.

A simple function with $0 \leq s \leq f$ on A can be extended to $E$ with $s(x)=0$ $\forall x \in E \setminus A$, so $\int_A f d\mu \leq \int_E f d\mu.$

Since $f \in \mathcal{L}(E)$, both positive and negative part have finite integrals, so $f \in \mathcal{L}(A)$ and the integral equality follows from the one for positive and negative parts.
However, the proof shows that the lemma also holds for non-negative functions without the need to assume that $f \in \mathcal{L}(E)$ and I think we can use this to show the claim if we already know that $g$ is measurable:
$f=g$ on $E \setminus N$ implies $f_{\pm}=g_{\pm}$ on $E \setminus N$.
So $\int_E f_{\pm} = \int_{E \setminus N} f_{\pm} = \int_{E \setminus N} g_{\pm} = \int_E g_{\pm}$, which proves that $g \in \mathcal{L}(E)$ and that the integrals have the same value. For the last step we obviously need $f \in \mathcal{L}(E)$, but the lemma can be used even before we know that $g \in \mathcal{L}(E)$.
Now it remains to show that $g$ is in fact measurable if $f \in \mathcal{L}(E)$. If the measure space is complete, then the measurability of $g$ follows from the measurability of $f$ and we are done (see here).
Now I have two questions:

Is the completeness of the space necessary for the result to be correct?
Am I correct that we only need completeness to show that $g$ is measurable, i.e. is my proof correct?

Thanks a lot!

REPLY [1 votes]: You need to know that $g$ is measurable in order to integrate $g$. But yes, if you add the hypothesis of measurability of $g$, then $\int f = \int g$.

Yes you need completeness to establish measurability of $g$. $g$ can fail to be measurable if the space is not complete.


TITLE: Are there open questions in Markov chains?
QUESTION [3 upvotes]: I would be curious to know if there's still open question about discrete markovian chains

REPLY [5 votes]: Chapter 23 of Markov Chains and Mixing Times by Levin, Peres, and Wilmer is called "Open Problems". The book is available online here.


TITLE: For which topological spaces $X$ is $X^n$ homeomorphic to the space of unordered n tuples of points in $X$
QUESTION [6 upvotes]: Consider a topological space $X$. Consider the spaces $A_n=X^n$ and $B_n=X^n/q_n$, the space of where $q_n$ is the equivalence relation where two points in $X^n$ are equivalent if one can be constructed from another via a permutation of the points.
For which topological spaces $X$ are $A_n$ and $B_n$ homeomorphic for all natural numbers $n$? For example, when $X=\mathbb{R}$, $A_n$ and $B_n$ are not homeomorphic in general, consider $n=2$ in particular.
Another example, when $X=\mathbb{R}^2$, $A_n,B_n$ are homeomorphic for each natural $n$. This can be proven by thinking of $X$ as the complex plane, $A_n$ as the space of monic polynomials of degree $n+1$ over $\mathbb{C}$, and thinking of $B_n$ as the space of unordered $n$ tuples of roots. It is a theorem that the map which takes the coefficients of a polynomial to its roots is bijective and bicontinuous, thus these spaces are homeomorphic.
Some more details on this proof:
1: Each list of coefficients gives a unique set of roots and vice versa
2: the coefficient list as a function of the roots is continuous because this function is a polynomial
3: the roots as a function of the coefficients is continuous in the quotient space $B_n$ (for a proof, see Bhatia matrix analysis, beginning of chapter 6).
I'd like to know for which other spaces this works.

REPLY [0 votes]: Via this post, we have the following characterization:

Let $X$ be a topological space. Then, $X^n$ is homeomorphic to $X^n / q_n$ if and only if there exists a quotient map $f : X^n \to X^n$ such that $f$ induces the equivalence relation $q_n$ on $X^n$.

So, this gives a (presumably difficult) way to test if $X$ satisfies the property you like.

It suffices to find a surjection $f : X^n \to X^n$ such that

(quotient map definition) Given $U \subseteq X^n$, $U$ is open in $X^n$ if and only f $f^{-1}(U)$ is open in $X^n$.
(induced equivalence relation) Given $x = (x_1,\ldots,x_n),y=(y_1,\ldots,y_n) \in X^n$, $x$ is equivalent to $y$ under $q_n$ if and only if $f(x)=f(y)$. In other words, the coordinates $y_1,\ldots,y_n$ are a permutation of the coordinates $x_1,\ldots,x_n$ if and only if $f(x)=f(y)$.

You can replace 1. by any of the other equivalent ways of defining quotient maps, as outlined again here.
This is perhaps not a satisfying answer, but at least it gives a basic characterization. Maybe others will be able to develop this thought process further. Feel free to comment ideas below. I was thinking maybe 1. and 2. can be shown to prove that $f$ must satisfy a stronger property, since we know what $q_n$ is, but I couldn't work that out.


TITLE: example of a commutative ring without zero divisor that is not an integral domain
QUESTION [2 upvotes]: I'm not sure if I understand this question.
An integral domain is a commutative ring (with unity) without zero-divisors.
The question ask for an integral domain that is not an integral domain?
Can someone shed some understanding?
Thanks in advance

REPLY [3 votes]: Essentially, you want a commutative ring without zero-divisors and without unity. So take $2\Bbb Z$ for instance


TITLE: Prove that, for any sentence A of the predicate calculus with identity, at least one of spectrum(A) and spectrum(¬ A) is cofinite
QUESTION [2 upvotes]: I got the following exercise:

Prove that, for any sentence A of the predicate calculus with
  identity, at least one of spectrum(A) and spectrum(¬ A) is cofinite.

I already tried to prove this theorem by a proof by contradiction. So suppose spectrum(A) and spectrum(¬ A) are both not cofinite. Then clearly, spectrum(A) and spectrum(¬ A) are infinite. Thus A and ¬ A have both infinite models. But I don't no how to go on ...
Could you please help me, dear stackexchange?
We got the hint that the following lemma is helpful: For any firstorder sentence s, if spectrum(s) is infinite, then s has an infinite model.

REPLY [1 votes]: Ok well I just looked it up. Take the theory that says there are at least $n$ elements for each $n$, and in which all constants are equal and all relations are satisfied by all tuples. There is essentially only one model of this theory and if that model satisfies $\sigma$ then some finite set of the axioms will imply $\sigma$ and so $\sigma$ has cofinite spectrum.


TITLE: Explanation of $\sum_n\langle\psi_n(x)|(O\psi_n)(x)\rangle=:(\mathrm{tr}\,O)(x)=\mathrm{tr}\int\frac{\mathrm{d}k}{(2\pi)^4}e^{ikx}Oe^{-ikx}$
QUESTION [2 upvotes]: Let $D$ be the Dirac operator, $O_N:=e^{-(D/N)^2}$ for $N\in\mathbf{N}$ and $\{\psi_n\}$ a complete set of eigenfunctions of $D$.
On page $69$ and $78$ of Path Integrals and
Quantum Anomalies and in the paper Path Integral for Gauge Theories with Fermions, Fujikawa uses the equation
\begin{equation}\tag{1}
(\mathrm{tr}\,O_N)(x):=\sum_{n=1}^{\infty}\big\langle\psi_n(x)|(O_N\psi_n)(x)\big\rangle=\mathrm{tr}\int\frac{\mathrm{d}k}{(2\pi)^4}e^{ikx}O_Ne^{-ikx}
\end{equation}
to derive
\begin{equation}
\lim_{N\to\infty}(\mathrm{tr}\,O_N)(x)=-\frac{1}{8\pi^2}\mathrm{tr}(F_{ij}F_{kl})(x)\epsilon^{ijkl}.
\end{equation}
Unfortunately, I don't understand how the right hand side of $(1)$ is defined - however, he uses the relation
\begin{equation}
\nabla_\mu\mathrm{e}^{ikx}=\mathrm{e}^{ikx}(\nabla_\mu+ik_\mu)
\end{equation}
later on - since $\nabla_\mu(u\cdot\psi)=u\cdot(\nabla_\mu+ik_\mu)\psi$ if $u(x):=e^{ikx}$, this suggests that $Ae^{-ikx}$ is actually the operator $\psi\mapsto A(u\cdot\psi)$.
He also claims that $(1)$ is a "unitary transformation" from the basis $\{\psi_n\}$ to plane waves $\{e^{ikx}\}$, but this doesn't make sense to me - I think $\psi_n(x)\in\mathbf{C}^N\otimes \mathfrak{g}$, whereas $e^{ikx}\in\mathbf{C}$.
If you've seen similar equations somewhere else, please also let me know.

REPLY [0 votes]: Let's assume $O$ acts on complex valued functions. We can improve the notation by setting $\langle x|\psi\rangle:=\psi(x)$, $\langle\psi|x\rangle:=\overline{\psi(x)}$ (complex conjugate) and $|n\rangle:=\psi_n$. We can obtain the result from the title by introducing the plane wave basis defined by $\langle x|k\rangle=\mathrm{e}^{\mathrm{i}kx}$ and using the Fourier inversion theorem:
\begin{align}
\sum_n \langle \psi_n(x)|(O\psi_n)(x)\rangle=\sum_n \overline{\psi_n(x)}(O\psi_n)(x)=\sum_n\langle n|x\rangle \langle x|O|n\rangle=\sum_n\int\frac{\mathrm{d}k}{(2\pi)^d}\langle n|k\rangle\langle k|x\rangle\langle x|O|n\rangle\\
    =\int\frac{\mathrm{d}k}{(2\pi)^d}\langle k|x\rangle\langle x|O\sum_n\langle n|k\rangle|n\rangle=\int\frac{\mathrm{d}k}{(2\pi)^d}\langle k|x\rangle\langle x|O|k\rangle=\int\frac{\mathrm{d}k}{(2\pi)^d}\mathrm{e}^{-\mathrm{i}kx}\langle x|O|k\rangle
\end{align}
Of course, writing $\langle x|O|k\rangle=O\langle x|k\rangle$ is simply wrong.


TITLE: Modal set-theory
QUESTION [3 upvotes]: In his “The Potential Hierarchy of Sets”, Review of Symbolic Logic 6:2 (2013), 205-28 Øystein Linnebo has proposed a modal set-theory. I was wondering what kind of utility can such a theory have for a mathematician and if there have been other similar attempts to develop theories of that kind.

REPLY [3 votes]: Not an answer, but too long for a comment:
Something you may be interested in is the modal logic approach to forcing, which is probably the most important and ubiquitous technique in modern set theory, as developed in the book by Fitting and Smullyan (see the review http://www.jstor.org/stable/2586777?seq=2#page_scan_tab_contents, which includes a critique of the approach).
NOTE: by "modal logic approach to forcing" I mean the use of modal logic in providing a framework for forcing constructions, not the modal logics which arise in studying forcing extensions - as in http://arxiv.org/abs/math/0509616 - which is really cool but not the same thing.)
Of course, a caveat is in order here: how is the modal approach to forcing useful? Philosophically I am told it is appealing and intuitive, at least to some; mathematically, I don't know of any time it has been used to achieve new results. Much more useful is the approach via Boolean-valued models (see "A proof of the independence of the continuum hypothesis" http://link.springer.com/article/10.1007%2FBF01705520 by Scott for a gentle example, or the section in Jech's book for a more thorough approach) - this is similar in spirit, in that it uses non-classical logic and structures to achieve a purely syntactic approach to forcing. The advantage that Boolean-valued models have in forcing seems to be that the relevant combinatorial properties of Boolean algebras are well-understood and easy to work with; there are many results about forcing which are proved straightforwardly via Boolean-valued models, but whose proofs in the usual poset formulation are quite messy.


text:
\subsection{Klein type and Grushin type metrics}\label{Klein-and-Grushin}

In this section, we consider geodesics in discontinuous metrics of the following two types:
\begin{equation}
ds^2= \frac{v(y)\,dx^2 +  w(y)\,dy^2}{y^2},  \ \quad   ds^2= \frac{v(y)}{y^2}\,dx^2 + w(y)\,dy^2 ,
\label{X1}
\end{equation}
where $v(y),w(y)$ are smooth positive functions.
The metrics \eqref{X1} are natural generalizations of the Klein metric \eqref{21} and the Grushin metric, respectively. Grushin type metrics have various applications in control theory and mechanics, see e.g.~\cite{BC}.

Both metrics \eqref{X1} have discontinuity on the line $\AA \colon y=0$ that lead to the same phenomenon as in the case of pseudo-Riemannian metrics: geodesics cannot pass through a point $q_0 \in \AA$ in arbitrary tangential directions, but only in the admissible direction $p=\infty$, see \cite{GR, Rem-Klein}.
Without loss of generality assume that $q_0=0$ (the origin) and $v(0)=w(0)=1$.
As proved in \cite{GR, Rem-Klein}, geodesics of the family $\Gamo$ have the form
\begin{equation}
x = \alf y^2 + o(y^2), \quad x = \alf y^3 + o(y^3), \quad \alf \in \bR,
\label{X11}
\end{equation}
for Klein type and Grushin type metrics, respectively.\footnote{
One of the differences among Klein type and Grushin type metrics can be found in the different asymptotic characters of the natural parametrization of geodesics tending to a point $q_0 \in \AA$. Naturally parametrized geodesics in Klein type metrics reach $q_0$ in infinite time, while in Grushin type metrics they reach it in finite time.
It follows from \eqref{3} similarly to what was done earlier for pseudo-Riemannian metrics.
}

\medskip

Unparametrized geodesics in the metrics \eqref{X1} are solutions of the Euler--Lagrange equation \eqref{6} with the Lagrangians $\LL = \sqrt{v+wp^2}/y$, $\LL = \sqrt{v+w(yp)^2}/y$, respectively. The corresponding the energy integrals \eqref{20} read
\begin{equation}
H(y,p) = \frac{v}{\sqrt{y^2(v + wp^2)}},  \ \quad
H(y,p) = \frac{v}{\sqrt{y^2(v + w(yp)^2)}}.
\label{X12}
\end{equation}
To establish the set $\HH$ of all possible constants $\ha^2$, one can substitute \eqref{X11} and the corresponding expressions $p^{-1} = 2\alf y + o(y)$, $p^{-1} = 3\alf y^2 + o(y^2)$ in the functions $H^2(y,p)$ from \eqref{X12} and pass to the limit $y \to 0$. Since $v(0)=w(0)=1$, for Klein type metrics this yields
\begin{equation}
\ha^2 = \lim_{y \to 0} H^2(y,p) =
\lim_{y \to 0} \frac{v^2(y)}{y^2v(y) + y^2w(y) (2\alf y + o(y))^{-2}} = 4\alf^2,
\label{X2}
\end{equation}
hence $\HH = [0, +\infty)$.
Analogous reasonings give the same result for Grushin type metrics.

\medskip

To define the behavior of geodesics $\gamapm \in \Gamo$, it is sufficient to consider the equation $v(y)/y^2 = \ha^2$ for all possible values $\ha^2 \ge 0$. This gives the following conclusions:

\begin{itemize}
\item
Geodesics $\gam^{\pm}_{\alf}$ and $\gam^{\pm}_{-\alf}$ with the same superscript return (or do not return)
to the initial line $y=0$ simultaneously.

\item
The family $\Gamo$ contains at least one couple $\gamapm$ that do not return to the initial line $y=0$:
$\gam_0^{\pm}$ with $\ha^2=0$ are the halves of the $y$-axis.

\item
The family $\Gamo$ contains an infinite number of geodesics that return to the initial line $y=0$.
Moreover, for any $\eps>0$ there exists $A>0$ such that all $\gamapm \in \Gamo$ with $|\alf|>A$
do not leave the strip $|y| < \eps$ and return to $y=0$.

\item
Geodesics $\gamap \in \Gamo$ (respectively, $\gamam \in \Gamo$) with all $\alf \neq 0$ return to the line $y=0$
if and only if $v(y)/y^2 \to 0$ as $y \to +\infty$ (respectively, $y \to -\infty$) and the condition $yv'(y) - 2v(y) \neq 0$ holds for all $y>0$ (respectively, $y<0$).
\end{itemize}

{\bf Example 3.3.}
For $v(y) = 1$, we have $v(y)/y^2 \to 0$ as $y \to \pm \infty$ and $yv'(y) - 2v(y) \neq 0$ for all $y$.
Hence $\gam_0^{\pm} \in \Gamo$ are only two geodesics that do not return to the line $y=0$,
while all $\gamapm \in \Gamo$ with $\alf \neq 0$ return.
In the case of the Klein metric \eqref{21} this statement is obvious: $\gamapm \in \Gamo$ with $\alf \neq 0$ are the halves of the circles $(x-x_*)^2 + y^2 = h_{\alf}^{-2}$.
\hfill\ebox

{\bf Example 3.4.}

Let $v(y) = 1 + y^4$.
The graph of the function $a(y) = v(y)/y^2$ is presented on Fig.~\ref{Fig11} (left).
It has two global minimums at $y = \pm 1$, hence there exist two horizontal geodesics $y = \pm 1$.

Further without loss of generality consider geodesics in the upper semiplane $y>0$.
Moreover, it is sufficient to analyze the family $\Gamo$ for $q_0 \in \AA$ ($y_0=0$) and for three arbitrary non-sungular points $q_0$ with $y_0>1$, $y_0=1$, $0<y_0<1$. In the last three cases, one can use results obtained in Section~\ref{Riemannian} for Riemannian metrics. The families $\Gamo$ for $q_0$ with $y_0>1$, $y_0=1$, $0<y_0<1$ are presented on Fig.~\ref{Fig12}.

Consider the family $\Gamo$ for $q_0 \in \AA$. There are three different classes of geodesics $\gamap \in \Gamo$ (and the same for $\gamam \in \Gamo$), presented on Fig.~\ref{Fig11} (right).
Firstly, $\gamap \in \Gamo$ with $\ha^2 > 2$, which do not leave the strip $0<y<1$ and return to the initial line $y=0$. Secondly, two geodesics $\gamap \in \Gamo$ with $\ha^2 = 2$ ($\alf = \pm \frac{1}{\sqrt{2}}$),
which do not leave the strip $0<y<1$, do not return to $y=0$, and asymptotically tend to the horizontal geodesic $y=1$.
Finally, $\gamap \in \Gamo$ with $0 \le \ha^2 < 2$, which do not return to $y=0$ (the $y$-coordinate along them monotonically increases from $0$ to $+\infty$).
\hfill\ebox

\begin{figure}[h!]
\begin{center}
\includegraphics{Fig11.jpg}
\caption{
Illustration for Example~3.4.
On the left: the graph of the function $a(y) = v(y)/y^2 = y^2 + y^{-2}$.
On the right: geodesics $\gamap \in \Gamo$ for a point $q_0 \in \AA$.
}
\label{Fig11}
\end{center}
\end{figure}

\begin{figure}[h!]
\begin{center}
\includegraphics{Fig12.jpg}
\caption{
Illustration for Example~3.4.
Geodesics $\gamapm \in \Gamo$ for a point $q_0$ with $y_0>1$, $y_0=1$, $0<y_0<1$ (from left to right).
}
\label{Fig12}
\end{center}
\end{figure}

config: arxiv


TITLE: Different ways of evaluating $\int_{0}^{\pi/2}\frac{\cos(x)}{\sin(x)+\cos(x)}dx$
QUESTION [3 upvotes]: My friend showed me this integral (and a neat way he evaluated it) and I am interested in seeing a few ways of evaluating it, especially if they are "often" used tricks.
I can't quite recall his way, but it had something to do with an identity for phase shifting sine or cosine, like noting that $\cos(x+\pi/2)=-\sin(x)$ we get:
$$ I=\int_{0}^{\pi/2}\frac{\cos(x)}{\sin(x)+\cos(x)}dx=\int_{\pi/2}^{\pi}\frac{-\sin(x)}{-\sin(x)+\cos(x)}dx\\
$$
Except for as I have tried, my signs don't work out well. The end result was finding
$$ 2I=\int_{0}^{\pi/2}dx\Rightarrow I=\pi/4
$$
Any help is appreciated! Thanks.

REPLY [3 votes]: Let
\begin{equation}
I=\int_0^{\pi/2}\frac{\cos x}{\sin x+\cos x}\ dx
\end{equation}
and
\begin{equation}
J=\int_0^{\pi/2}\frac{\sin x}{\sin x+\cos x}\ dx
\end{equation}
then
\begin{equation}
I+J=\frac{\pi}{2}\tag1
\end{equation}
and
\begin{align}
I-J&=\int_0^{\pi/2}\frac{\cos x-\sin x}{\sin x+\cos x}\ dx\\[10pt]
&=\int_0^{\pi/2}\frac{1}{\sin x+\cos x}\ d(\sin x+\cos x)\\[10pt]
&=0\tag2
\end{align}
Hence, $$I=J=\frac\pi4$$by linear combinations $(1)$ and $(2)$.


text:
\begin{document}

\title[Generalized parking functions and descent numbers]
{Generalized parking functions, descent numbers,
and chain polytopes of ribbon posets}

\author{Denis Chebikin and Alexander Postnikov}

\keywords{Labeled trees, tree inversions, parking functions, descent sets
of permutations, chain polytopes, ribbon posets}

\address{Department of Mathematics, Massachusetts Institute of
  Technology,
Cambridge, MA 02139}

\email{chebikin@gmail.com, apost@math.mit.edu}

\begin{abstract}
We consider the inversion enumerator $I_n(q)$, which
counts labeled trees or, equivalently, parking functions.
This polynomial has a natural extension to generalized parking functions.
Substituting $q=-1$ into this generalized polynomial produces the number
of permutations with a certain descent set. In the classical case,
this result implies the formula $I_n(-1) = E_n$, the number of
alternating permutations.
We give a combinatorial proof of these formulas based on the
involution principle.
We also give a geometric interpretation of these identities in terms
of volumes of
generalized chain polytopes of ribbon posets.  The volume of such a
polytope is given by
a sum over generalized parking functions, which is similar to an expression
for the volume of the parking function polytope of Pitman and Stanley.

\end{abstract}

\maketitle

\section{introduction}\label{intro}

Let $\mathcal{T}_n$ be the set of all trees on vertices labeled
$0$, $1$, $2$, \dots, $n$ rooted at $0$. For $T\in\mathcal{T}_n$, let
$\inv(T)$ be the number of pairs $i>j$ such that $j$ is a descendant of $i$
in $T$. Define the $n$-th \emph{inversion enumerator} to be the polynomial
$$
I_n(q):=\sum_{T\in\mathcal{T}_n} q^{\inv(T)}.
$$
Another way to define this polynomial
is via parking functions. A sequence $(b_1,b_2,\dots,b_n)$ of
positive integers is a \emph{parking function of length $n$}
if for all $1\leq j \leq n$,
at least $j$ of the $b_i$'s do not exceed $j$. A classical bijection of
Kreweras \cite{Kreweras}
establishes a correspondence between trees in $\mathcal{T}_n$ with
$k$ inversions and parking functions of length $n$ whose components add up to
${n+1\choose 2} - k$. Hence we can write
$$
I_n(q) = \sum_{(b_1,\dots,b_n)\in\mathcal{P}_n}
q^{{n+1\choose 2} - b_1-b_2-\dots-b_n},
$$
or
$$
\sum_{(b_1,\dots,b_n)\in\mathcal{P}_n} q^{b_1+b_2+\dots+b_n-n}=
q^{n\choose 2}\cdot I_n(q^{-1}),
$$
where $\mathcal{P}_n$ is the set of all parking functions of length $n$.
Cayley's formula states that $|\mathcal{T}_n|=|\mathcal{P}_n|=(n+1)^{n-1}$,
hence $I_n(1) = (n+1)^{n-1}$.

Here we focus on the formula
\begin{equation}\label{In_minus_one}
I_n(-1) = E_n,
\end{equation}
where $E_n$ is the
$n$-th \emph{Euler number}, most commonly defined as the
number of permutations $\sigma_1\sigma_2\dots\sigma_n$ of $[n]=
\{1,2,\dots,n\}$ such that $\sigma_1 < \sigma_2 > \sigma_3 < \sigma_4 >\dots$,
called \emph{alternating} permutations. This formula can be obtained
by deriving
a closed form expression for
the generating function
$\sum_{n\geq 0} I_n(q) x^n/n!$ and showing that setting $q=-1$ yields
$\tan x + \sec x = \sum_{n\geq 0} E_n x^n/n!$
(see the paper \cite{Gessel} by Gessel or Exercises
3.3.48--49 in \cite{GouldenJackson}).
A~direct combinatorial proof was given
by
Pansiot \cite{Pansiot}.
In this paper we give two
other ways to prove this fact, one of which,
presented in Section \ref{parking_functions},
is an involution argument on the set of all
but $E_n$ members of $\mathcal{P}_n$. This involution
is a special case of a more general argument
valid for a broader version of parking functions, which we now describe.

Let $\va=(a_1,a_2,\dots,a_n)$ be a \emph{non-decreasing} sequence
of positive integers.
Let us call
a sequence $(b_1,b_2,\dots,b_n)$ of positive integers an
\emph{$\va$-parking function} if the increasing rearrangement
$b_1' \leq b_2' \leq \dots \leq b_n'$ of this sequence satisfies
$b_i' \leq a_i$ for all $i$. Note that
$(1,2,\ldots,n)$-parking functions
are the regular parking functions of length $n$.
These $\va$-parking functions are
$(a_1, a_2 - a_1, a_3-a_2,\dots)$-parking functions in the original notation
of Yan \cite{Yan}, but the present definition is consistent with later
literature, such as the paper \cite{KungYan} of Kung and Yan.
Let $\mathcal{P}_{\va}$
be the set of all $\va$-parking functions, and define
\begin{equation}\label{Igamma}
I_{\va}(q) := \sum_{(b_1,\dots,b_n)\in\mathcal{P}_{\va}}
q^{b_1+b_2+\dots+b_n-n}
\end{equation}
(this is the \emph{sum enumerator} studied in \cite{KungYan}).
For a subset $S\subseteq [n-1]$, let $\beta_n(S)$ be the number of permutations
of size $n$ with descent set $S$.
In Section~\ref{parking_functions}
(Theorem~\ref{Igamma_minus_one_thm}) we prove the following generalization
of (\ref{In_minus_one}):
\begin{equation}\label{Igamma_minus_one}
|I_{\va}(-1)| = \left\{
\begin{array}{ll}
0,& \mbox{if $a_1$ is even;}\\
\beta_n(S),& \mbox{if $a_1$ is odd,}
\end{array}
\right.
\end{equation}
where
\begin{equation}\label{Igamma_S}
S = \Bigl\{i\in [n-1]\ \Bigl|\ a_{i+1}\mbox{ is odd}\Bigr\}.
\end{equation}
Indeed, for $\va = (1,2,\ldots,n)$ we have
$S = \{2,4,6,\dots\}\cap [n-1]$, so that $\beta_n(S)$ counts alternating
permutations of size $n$. The formula (\ref{Igamma_minus_one}) arises in
a more sophisticated algebraic context in the paper \cite{PakPostnikov}
of Pak and Postnikov.

In Section \ref{chain_polytopes}
we obtain a geometric
interpretation of these results by considering generalized
chain polytopes of ribbon posets.
Given a subset $S \subseteq \{2,3,\dots,n-1\}$,
define $u_S = u_1u_2\dots u_{n-1}$ to be the monomial in non-commuting
formal variables $\aaa$ and $\bbb$ with $u_i = \aaa$ if $i\notin S$ and
$u_i = \bbb$ if $i\in S$. Let $\comp(S)$ be the

composition
$(1,\delta_1,\delta_2,\dots,\delta_{k-1})$ of $n$,

where the $\delta_i$'s are defined by
$u_S = \aaa^{\delta_1} \bbb^{\delta_2} \aaa^{\delta_3} \bbb^{\delta_4}
\dots$.
For example, for $n=7$ and $S = \{2,3,4\}$ we have
$u_S = \aaa\bbb\bbb\bbb\aaa\aaa = \aaa\bbb^3\aaa^2$, so
$\comp(S) = (1,1,3,2)$.
Now define the polytope $\mathcal{Z}_S(d_1,d_2,\dots,
d_k)$, where $0<d_1\leq d_2 \leq \cdots \leq d_k$ are real numbers,
to be the set of
all points $(x_1,x_2,\dots,x_n)$ satisfying
the inequalities $x_j \geq 0$ for $j\in [n]$,
$x_1 \leq d_1$, and
$$
x_{\delta_1+\delta_2+\dots+\delta_{i-1}+1}
+ x_{\delta_1+\delta_2+\dots+\delta_{i-1}+2}
+ \dots + x_{\delta_1+\delta_2+\dots+\delta_i + 1} \leq d_{i+1}
$$
for $1\leq i \leq k-1$. Thus to the above example corresponds the polytope
$\mathcal{Z}_S(d_1,d_2,d_3,d_4)$ in $\mathbb{R}_{\geq 0}^7$ defined by
$$
x_1 \leq d_1;
$$
$$
x_1+x_2 \leq d_2;
$$
$$
x_2+x_3+x_4+x_5 \leq d_3;
$$
$$
x_5+x_6+x_7 \leq d_4.
$$
We require $1\notin S$ here to ensure that $\delta_1\neq 0$, but
there is no essential loss of generality because the chain polytope
of the poset $Z_S$ is defined by the same relations as $Z_{[n-1]-S}$.

For a poset $P$ on $n$ elements, the
\emph{chain polytope} $\mathcal{C}(P)$ is the set of points
$(x_1,x_2,\dots,x_n)$ of the unit hypercube $[0,1]^n$ satisfying the
inequalities $x_{p_1} + x_{p_2} + \dots + x_{p_\ell} \leq 1$ for every
chain $p_1 < p_2 < \dots < p_\ell$ in $P$; see \cite{StanleyPosetPolytopes}.
Hence
$\mathcal{Z}_S(1,1,\dots)$ is the chain polytope of the
\emph{ribbon poset} $Z_S$, which is the poset on $\{z_1,z_2,\dots,z_n\}$
generated by the cover relations $z_i \covers z_{i+1}$ if $i\in S$ and
$z_i \covered z_{i+1}$ if $i\notin S$. The volume of $\mathcal{C}(P)$ equals
$1/n!$ times the number of linear extensions of $P$, which in the case
$P = Z_S$ naturally correspond to permutations of size $n$ with descent set
$S$. Our main result concerning the polytope $\mathcal{Z}_S$ is a formula
for its volume. For a composition $\gamma$ of $n$,
let $\Kappa_\gamma$ denote the set of \emph{weak} compositions $\alpha
= (\alpha_1, \alpha_2, \dots)$ of $n$,
meaning that $\alpha$ can have parts equal to $0$, such that
$\alpha_1  + \alpha_2 +\dots + \alpha_i \geq \gamma_1 + \gamma_2 + \dots
+ \gamma_i$ for all $i$.
Define $\va(\gamma)$ to be the sequence consisting of
$\gamma_1$ $1$'s, followed by $\gamma_2$ $2$'s, followed by
$\gamma_3$ $3$'s, and so on.
Then $\alpha$ is in $\Kappa_\gamma$ if and only
if
$\alpha$ is the content of an $\va(\gamma)$-parking function.
(The \emph{content} of a parking function
is the composition whose $i$-th part is the number of
components of the parking function equal to $i$.)
In Section~\ref{chain_polytopes}
(Theorem~\ref{vol_ZS_thm}) we show that
\begin{equation}\label{vol_ZS}
n!\cdot \Vol\bigl(\mathcal{Z}_S(d_1,d_2,\dots,d_k)\bigr)=
\left|
\sum_{(b_1,\dots,b_n)\in\mathcal{P}_{\va(\comp(S))}} \prod_{i=1}^n
(-1)^{b_i} d_{b_i}
\right|
=
\end{equation}
$$
=
\left|
\sum_{\alpha \in \Kappa_{\comp(S)}}
{n\choose \alpha}\cdot (-1)^{\alpha_1+\alpha_3+\alpha_5+\dots}\cdot
d_1^{\alpha_1} d_2^{\alpha_2} \dots d_k^{\alpha_k}
\right|,
$$
where ${n\choose \alpha} = {n!\over \alpha_1!\alpha_2!\dots\alpha_k!}$
and $k = \ell(\comp(S))$ is the number of parts of
$\comp(S)$. For
example, for $n=5$ and $S=\{4\}$,
we have
$$
\Kappa_{\comp(S)} = \Kappa_{(1,3,1)} =
\{(1,3,1),(1,4,0),(2,2,1),(2,3,0),(3,1,1),(3,2,0),
$$
$$
(4,0,1), (4,1,0), (5,0,0)\},
$$
so
we get from (\ref{vol_ZS}) that
$$
5!\cdot\Vol\bigl(\mathcal{Z}_{\{1\}}(d_1,d_2,d_3)\bigr)
=  20 d_1   d_2^3 d_3
-  5  d_1   d_2^4
-  30 d_1^2 d_2^2 d_3
+  10 d_1^2 d_2^3
$$
$$
+ 20  d_1^3 d_2   d_3
- 10  d_1^3 d_2^2
- 5   d_1^4       d_3
+ 5   d_1^4 d_2
-     d_1^5.
$$

Setting $d_i = q^{i-1}$ in (\ref{vol_ZS}), where we take
$q\geq 1$ so that the sequence $d_1$, $d_2$, \ldots\ is non-decreasing,
and recalling~(\ref{Igamma}) gives
$$
n!\cdot \Vol(1,q,q^2,\dots) =
\left|

\sum_{(b_1,\ldots,b_n)\in\mathcal{P}_{\va(\comp(S))}}
(-q)^{b_1+b_2+\cdots+b_n-n}
\right|
= \left|
I_{\va(\comp(S))}(-q)
\right|.
$$
Specializing further by setting $q=1$ yields the identity
$$
\left|I_{\va(\comp(S))}(-1)\right| = \beta_n(S).
$$
Observe that this identity is consistent with (\ref{Igamma_minus_one}).
Indeed, the first part of $\comp(S)$ is positive, and thus the first
element of $\va(\comp(S))=(a_1,a_2,\dots,a_n)$
is $1$, i.e.\ an odd number. Comparing the sequence $(a_1,a_2,\dots,a_n)$ with
the letters of the word $\bbb u_S$ we see that $a_{i+1}
= a_i+1$ if the corresponding letters of $u_S$ are different, and
$a_{i+1}=a_i$ otherwise; in other words, changes of parity between
consecutive elements of $(a_1,a_2,\dots,a_n)$ correspond to letter changes
in the word $\bbb u_S$. (The extra $\bbb$ in front corresponds to the first
part $1$ of $\comp(S)$.)
For example, for $n=7$ and $S=\{2,3,4\}$, we have
$\comp(S) = (1,1,3,2)$, $\bbb u_S = \bbb\aaa\bbb^3\aaa^2$, and
$\va(\comp(S)) = (1,2,3,3,3,4,4)$.
It follows that the subset constructed from
$\va(\comp(S))$ according to the rule (\ref{Igamma_S}) of an earlier result
is $S$, so
the results agree.

Considering once more the case $S=\{2,4,6,\dots\}\cap[n-1]$, let us point
out the similarity between the formula (\ref{vol_ZS}) and the expression
that Pitman and Stanley \cite{PitmanStanley} derive for the volume of their
\emph{parking function polytope}. This polytope,
which we denote by $\Pi_n(c_1,c_2,\dots,c_n)$,
is defined by the inequalities $x_i \geq 0$ and
$$
x_1+x_2+\dots + x_i \leq c_1+c_2+\dots+c_i
$$
for all $i\in [n]$. The volume-preserving change of coordinates
$y_i = c_n+c_{n-1} + \dots + c_{n+1-i} - (x_1+x_2+\dots+x_i)$ transforms the
defining relations above into $y_i \geq 0$ for $i\in [n]$, $y_1 \leq c_1$, and
$y_i - y_{i+1} \geq c_i$ for $i\in [n-1]$, and these new relations look much
like the ones defining $\mathcal{Z}_{\{2,4,6,\dots\}}(c_1,c_2,\dots,c_n)$:
in essence we have here a difference instead of a sum. This similarity
somewhat explains the close resemblance of the volume formulas for the two
polytopes,
as for $\Pi_n(c_1,c_2,\dots,c_n)$ we have
$$
n!\cdot \Vol(\Pi_n(c_1,c_2,\dots,c_n))
= \sum_{(b_1,\dots,b_n)\in \mathcal{P}_n} \prod_{i=1}^n c_{b_i}
= \sum_{\alpha \in \Kappa_{1^n}} \prod_{i=1}^n {n\choose \alpha}c_i^{\alpha_i}.
$$

\section*{Acknowledgments}

We would like to thank Ira Gessel for providing useful references.

\section{An involution on $\va$-parking functions}\label{parking_functions}

The idea of the
combinatorial argument presented in this section was first discovered by
the second author and Igor Pak during their work on \cite{PakPostnikov}.

Let $\va=(a_1,a_2,
\dots,a_n)$ be a non-decreasing sequence of positive integers.
As a first step in the construction of our involution on
$\va$-parking functions, let $Y_{\va}$ be the Young diagram whose
column lengths from left to right are $a_n$, $a_{n-1}$, \dots, $a_1$.
Define a \emph{horizontal strip} $H$ inside $Y_{\va}$ to be a set of cells
of $Y_{\va}$ satisfying the following conditions:

(i) for every $i\in [n]$, the set
$H$ contains exactly one cell $\sigma_i$ from column $i$
(we number the columns $1$, $2$, \dots, $n$
from left to right);

(ii) for $i < j$, the cell $\sigma_i$ is in the same or in a lower row than
the cell $\sigma_j$.

For a horizontal strip $H$, let us call a filling of the cells of $H$ with
numbers $1$, $2$, \dots, $n$ \emph{proper} if the numbers in row $i$ are in
increasing order if $i$ is odd, or in decreasing order if $i$ is even (we
number the rows $1$, $2$, \dots, from top to bottom). Let $\mathcal{H}_{\va}$
denote the set of all properly filled horizontal strips inside $Y_{\va}$.

For an $\va$-parking function $\vb = (b_1,b_2,\dots,b_n)$, define
$H(\vb)\in\mathcal{H}_{\va}$ in the following way.
Let $I_j \subseteq [n]$ be the set
of indices $i$ such that $b_i=j$. Construct
the filled horizontal strip $H(\vb)$ by first writing the elements
of $I_1$ in \emph{increasing} order in the $|I_1|$ \emph{rightmost} columns
in row $1$ of $Y_{\va}$, then writing the elements of $I_2$ in
\emph{decreasing} order in the next $|I_2|$ columns from the right in
row $2$,
then writing the elements of $I_3$ in \emph{increasing} order
in the next $|I_3|$ columns from the right in row $3$, and so on, alternating
between increasing and decreasing order.

\begin{lemma}\label{Hb_fits_into_Y}
A sequence
$\vb$ is an $\va$-parking function if and only if the horizontal strip
$H(\vb)$ produced in the above construction fits into $Y_{\va}$.
\end{lemma}

\begin{proof}
Let $(b_1',b_2',\ldots,b_n')$ be the increasing rearrangement of $\vb$. Then
the cell of $H(\vb)$ in the $i$-th column from the right belongs to row
$b_i'$. Thus the condition of the lemma is equivalent to $b_i' \leq a_i$ for
all $i$.
\end{proof}

Clearly, the filling of $H(\vb)$ from the above procedure is proper, and
hence $\vb \leftrightarrow H(\vb)$ is a bijection between $\mathcal{P}_{\va}$
and $\mathcal{H}_{\va}$. We will describe our involution on
$\va$-parking functions in terms of the corresponding filled
horizontal strips.

Let $H$ be a properly filled horizontal strip in $\mathcal{H}_{\va}$.
In what follows we use $\sigma_i$ to refer
to both the cell of $H$ in column $i$ and to the number written in it.
Let $r(\sigma_i)$ be the number of the row containing $\sigma_i$.
We begin by defining the \emph{assigned direction} for each of
the $\sigma_i$'s.
For the purpose of this definition it is convenient to imagine that
$H$ contains a cell labeled $\sigma_{n+1}=n+1$ in row $1$ and column $n+1$,
that is, just outside the first row $Y_{\va}$ on the right.

Let
$$
\epsilon_i = \sgn(\sigma_i-\sigma_{i+1})(-1)^{r(\sigma_i)},
$$
where $\sgn(x)$ equals $1$ if $x>0$, or $-1$ if $x<0$.
Define the assigned direction for $\sigma_i$ to be
\emph{up} if $\epsilon_i = -1$ and \emph{down} if $\epsilon_i=1$.

Let us say that
$\sigma_i$ is \emph{moveable down} if the assigned direction for $\sigma_i$
is down, $\sigma_i$ is not the bottom cell of column $i$, and moving
$\sigma_i$ to the cell immediately below it would not violate the rules of
a properly filled horizontal strip. The latter condition prohibits moving
$\sigma_i$ down if there is another cell of $H$ immediately to the left of it,
or if moving $\sigma_i$ down
by one row would violate the rule for the relative order
of the numbers in row $r(\sigma_i)+1$.
Let us say that $\sigma_i$ is \emph{moveable up} if the assigned direction
for $\sigma_i$ is up. Note that we do not need any complicated conditions
in this case: if $\sigma_i$ has another cell of $H$ immediately to its right,
or if $\sigma_i$ is in the top row, then the assigned direction for
$\sigma_i$ is down.

It is a good time to consider an example. Figure \ref{hor_strip_example}
shows the diagram $Y_{\va}$ and a
properly filled horizontal strip $H\in\mathcal{H}_{\va}$ for
$\va=(3,3,6,7,7,7,8)$.
The horizontal strip shown is $H(\vb)$, where $\vb = (5,7,2,5,1,5,2) \in
\mathcal{P}_{\va}$. Moveable cells are equipped with arrows pointing in
their assigned directions. Note that the assigned direction for
$\sigma_7 = 5$ is down
because of the ``imaginary''
$\sigma_8=8$ next to it, but it is not moveable down because the
numbers $7$, $3$, $5$ in row $2$ would not be ordered properly. The assigned
direction for $\sigma_3=4$ and $\sigma_4=6$ is also down, but these cells
are not moveable down because moving them down would not produce a horizontal
strip.

\begin{figure}[h!]
\begin{center}
\input{example1.pstex_t}
\end{center}
\caption{A properly filled horizontal strip with assigned directions for
its cells}
\label{hor_strip_example}
\end{figure}

The validity of the involution we are about to define depends on the following
simple but crucial fact.

\begin{lemma}\label{move_back_forth}
Let $\sigma_i$ be a moveable cell of $H\in\mathcal{H}_{\va}$, and let
$H'\in\mathcal{H}_{\va}$ be the horizontal strip obtained from $H$ by
moving $\sigma_i$ by one row in its assigned direction.
Then

(a) $\sigma_i$
is moveable in the opposite direction in $H'$;

(b) if $j \neq i$ and $\sigma_j$ is moveable in $H$, then $\sigma_j$
is moveable in
the same direction in $H'$;

(c) if $\sigma_j$ is not moveable in $H$, then it is not
moveable in $H'$.
\end{lemma}

\begin{proof}
Follows by inspection of the moving rules.
\end{proof}

Let $\tilde{\mathcal{H}}_{\va}$
be the set of all $H\in\mathcal{H}_{\va}$ such that
$H$ has at least one moveable cell (up or down). Define the map
$\psi : \tilde{\mathcal{H}}_{\va} \rightarrow \tilde{\mathcal{H}}_{\va}$
as follows:
given $H\in\tilde{\mathcal{H}}_{\va}$, let $\psi(H)$ be the horizontal
strip obtained from $H$ by choosing the \emph{rightmost} moveable cell
of $H$ and moving it by one row in its assigned direction. In view of
Lemma \ref{move_back_forth}, $\psi$ is an involution.

For $\vb = (b_1,b_2,\dots,b_n)\in \mathcal{P}_{\va}$
and $H = H(\vb)$, define
$s(\vb) = s(H) := b_1 + b_2+\dots+b_n - n$. Observe that $s(\vb)$ is the number
of cells of $Y_{\va}$ that lie above one of the cells of $H(\vb)$. In the
example in Figure \ref{hor_strip_example}, we have $s(\vb)=s(H)=
6+4+4+4+1+1+0=20$. Clearly, $s(H) = s(\psi(H)) \pm 1$ for $H\in
\tilde{\mathcal{H}}_{\va}$. Since $\psi$ is fixed-point free,
it follows that
$$
\sum_{H\in\tilde{\mathcal{H}}_{\va}} (-1)^{s(H)} = 0,
$$
and that
\begin{equation}\label{Igamma_minus_one_remainder}
I_{\va}(-1) = \sum_{H\in\tilde{\mathcal{H}}_{\va}-\mathcal{H}_{\va}}
(-1)^{s(H)}
\end{equation}
(cf.\ (\ref{Igamma})). It remains to examine the members of
$\mathcal{H}_{\va}-\tilde{\mathcal{H}}_{\va}$ in order to evaluate the
right hand side of (\ref{Igamma_minus_one_remainder}).

\begin{lemma}\label{fixed_points}
If $a_1$ is even, then $\mathcal{H}_{\va}-\tilde{\mathcal{H}}_{\va}
= \varnothing$. If $a_1$ is odd, then
$\mathcal{H}_{\va}-\tilde{\mathcal{H}}_{\va}$ consists of all filled
horizontal strips $H$ in $Y_{\va}$ such that
the cell $\sigma_i$ of $H$ is at the bottom of column $i$ for all
$i\in [n]$, and
the permutation
$\sigma_n \dots \sigma_2\sigma_1$ has descent set
$$
S = \Bigl\{i\in [n-1]\ \Bigl|\ a_{i+1}\mbox{ is odd}\Bigr\}
$$
(cf.\ (\ref{Igamma_S})).
\end{lemma}

\begin{proof}
Let $H\in\mathcal{H}_{\va}$, and consider the cell
$\sigma_n$ in $H$.
Since $\sigma_n < \sigma_{n+1}=n+1$, it follows that $\sigma_n$ is moveable
up if $r(\sigma_n)$ is even, or moveable down if $r(\sigma_n)$ is odd, unless
in the latter case $\sigma_n$ is at the bottom of column $n$. Thus if
$a_1$ is even, that is, if the rightmost column of $Y_{\va}$ has even
height, $\sigma_n$ is always moveable and
$\mathcal{H}_{\va}=\tilde{\mathcal{H}}_{\va}$.

Suppose that $a_1$ is odd, and let
$H\in\mathcal{H}_{\va}-\tilde{\mathcal{H}}_{\va}$.
Then no cells of $H$ are moveable, and hence the assigned direction for
every cell is down.

First, let us show that all cells of $H$ are at the bottom of their respective
columns. Suppose it is not so, and choose the \emph{leftmost}
cell $\sigma_i$ of $H$ such that
the cell immediately below it is in $Y_{\va}$. Our choice guarantees that
$\sigma_i$ does not have another cell of $H$ immediately to its left, so the
only way $\sigma_i$ can be not moveable down is if $\sigma_{i-1}$ is in
column $i-1$ one row below $\sigma_i$ and $\sgn(\sigma_{i-1}-\sigma_i)
= -(-1)^{r(\sigma_{i-1})}$.

But in this case
the assigned direction for $\sigma_{i-1}$ is up --- a contradiction.

Now let us compute the descent set of $\sigma_n\sigma_{n-1}\dots \sigma_1$.
We just proved that $r(\sigma_{n+1-i}) = a_{i}$ for all $i\in [n]$.
For $i\in [n-1]$, we have
$$
1 = \epsilon_{n-i} =
\sgn(\sigma_{n-i} - \sigma_{n+1-i})(-1)^{r(\sigma_{n-i})},
$$
and hence $\sigma_{n+1-i} > \sigma_{n-i}$, that is,
$\sigma_n\sigma_{n-1}\cdots\sigma_1$ has a descent in position $i$,
if and only if $r(\sigma_{n-i}) = a_{i+1}$ is odd.
\end{proof}

Note that from Lemma \ref{fixed_points} it follows that for all
$\mathcal{H}_{\va}-\tilde{\mathcal{H}}_{\va}$, the value
of $s(H)$ is the same, namely,
$a_1 + a_2 + \cdots + a_n - n$.
Combining with (\ref{Igamma_minus_one_remainder}),
we obtain the following theorem (cf.\ (\ref{Igamma_minus_one})).

\begin{theorem}[cf.\ \cite{PakPostnikov}]\label{Igamma_minus_one_thm}
For a non-decreasing
sequence $\va = (a_1,a_2,\dots,a_n)$ of positive integers, we have
$$
I_{\va}(-1) = \left\{
\begin{array}{ll}
0,& \mbox{if $a_1$ is even;}\\
(-1)^{a_1+\cdots+a_n-n}\cdot\beta_n(S),& \mbox{if $a_1$ is odd,}
\end{array}
\right.
$$
where $S = \Bigl\{i\in [n-1]\ \Bigl|\ a_{i+1}\mbox{ is odd}\Bigr\}$,
and $\beta_n(S)$ is the number of permutations of $[n]$ with descent set $S$.
\end{theorem}

\section{Generalized chain polytopes of ribbon posets}
\label{chain_polytopes}

In this section we prove the formula (\ref{vol_ZS}) of Section \ref{intro}:

\begin{theorem}\label{vol_ZS_thm}
For a positive integer $n$, a subset $S\subseteq [n-1]$ such that
$$\comp(S) = (1,\delta_1, \delta_2, \ldots, \delta_{k-1}),$$
and a sequence $0 < d_1 \leq d_2 \leq \cdots \leq d_k$ of real numbers,
we have
\begin{eqnarray}\label{vol_ZS_2}
\nonumber
n!\cdot \Vol\bigl(\mathcal{Z}_S(d_1,d_2,\dots,d_k)\bigr)&=&
(-1)^{1+\delta_2+\delta_4+\cdots}
\sum_{(b_1,\dots,b_n)\in\mathcal{P}_{\va(\comp(S))}} \prod_{i=1}^n
(-1)^{b_i} d_{b_i}
\\
&=&
(-1)^{1+\delta_2+\delta_4+\cdots}
\sum_{\alpha \in \Kappa_{\comp(S)}}
{n\choose \alpha}\cdot (-1)^{\alpha_1+\alpha_3+\alpha_5+\dots}\cdot
d_1^{\alpha_1}

\cdots d_k^{\alpha_k}.
\end{eqnarray}
\end{theorem}

\begin{proof}
First, note that the
expression in the right hand side of (\ref{vol_ZS_2}) is obtained
from the middle one by grouping together the terms corresponding
to all ${n\choose \alpha}$ $\va$-parking functions of content $\alpha$;
each of these terms equals
$$
(-1)^{\alpha_1+2\alpha_2+3\alpha_3+\cdots}\cdot d_1^{\alpha_1}
\cdots d_k^{\alpha_k}
= (-1)^{\alpha_1+\alpha_3+\alpha_5+\cdots}\cdot d_1^{\alpha_1}
\cdots d_k^{\alpha_k}.
$$

In what follows we prove the equality between the left and the right hand sides of
(\ref{vol_ZS_2}).
For $i\in [k]$, let $\rho_i = 1+\delta_1+\delta_2+\cdots+\delta_{i-1}$.
The volume of $\mathcal{Z}_S(d_1,d_2,\ldots,d_k)$ can be expressed as
the following iterated integral:
\begin{equation}\label{integrals}
\int_0^{d_1}
\int_0^{d_2 - x_{1}}
\int_0^{d_2 - x_{1} - x_{2}}
\dots
\int_0^{d_2 - x_{1} - x_{2} - \cdots - x_{\rho_2-1}}
\end{equation}
$$
\int_0^{d_{3}-x_{\rho_{2}}}
\int_0^{d_{3}-x_{\rho_{2}}-x_{\rho_{2} + 1}}
\dots
\int_0^{d_{3}-x_{\rho_{2}}-x_{\rho_{2} + 1}-\cdots-x_{\rho_{3}-1}}
$$
$$
\dots
$$
$$
\int_0^{d_{k}-x_{\rho_{k-1}}}
\int_0^{d_{k}-x_{\rho_{k-1}}-x_{\rho_{k-1}+1}}
\dots
\int_0^{d_{k}-x_{\rho_{k-1}}-x_{\rho_{k-1}+1}-\cdots-x_{\rho_k-1}}
dx_n\ dx_{n-1}\ \cdots\ dx_1
$$
(Similar integral formulas appear in \cite{KungYan}
and in \cite[Sec.\ 18]{PostnikovStanley}.)
Note that the assumption $d_1\leq d_2 \leq \cdots \leq d_k$ validates
the upper limits of those integrals taken with respect to
variables $x_2$,
$x_{\rho_2+1}$, $x_{\rho_3+1}$, \dots, $x_{\rho_{k-1}+1}$:
for $2\leq i\leq k-1$, the condition
$$
x_{\rho_i+1} \leq d_{i} - x_{\rho_{i-1}+1} - x_{\rho_{i-1}+2} -
\cdots - x_{\rho_i}
$$
implies that
$$
d_{i+1} - x_{\rho_i+1} \geq 0,
$$
and $x_1 \leq d_1$ implies $d_2 - x_1 \geq 0$.

For $\ell\in [n]$, let $J_\ell$ denote the evaluation of the
$n+1-\ell$ inside integrals of (\ref{integrals}), that is, the
integrals with respect to the variables $x_n$, $x_{n-1}$, \ldots,
$x_{\ell}$.

\begin{lemma}\label{inside_integrals}
For $i\in [k]$, we have
\begin{eqnarray}\label{int_lemma}
\nonumber
J_{\rho_i+1}
&=&
(-1)^{\delta_{i+1}+\delta_{i+3}+\delta_{i+5}+\cdots}
\\
&&
\ \ \ \ \ \ \ \cdot\sum_{\alpha\in\Kappa_{(0,\delta_i,\delta_{i+1},\ldots,\delta_{k-1})}}
(-1)^{\alpha_1+\alpha_3+\alpha_5+\cdots}
\cdot{1\over \alpha_1!\ \alpha_2!\ \cdots}\cdot
x_{\rho_i}^{\alpha_1} d_{i+1}^{\alpha_2} d_{i+2}^{\alpha_3} \cdots.
\end{eqnarray}
\end{lemma}

\begin{proof}
We prove the lemma by induction on $i$, starting with the trivial base
case of $i=k$, in which we have $J_{\rho_k+1} = J_{n+1} = 1$. Now suppose
the claim is true for some $i$. By straightforward iterated integration one
can show that for non-negative integers $r$ and $s$,
\begin{equation}\label{int_identity}
\int_0^a \int_0^{a-y_{r}} \int_0^{a-y_{r}-y_{r-1}} \dots
\int_0^{a-y_{r}-y_{r-1}-\cdots-y_2}
y_1^s\ dy_1\ \cdots\ d_{y_{r-1}}\ d_{y_{r}}
= {s!\ a^{r+s}\over (r+s)!}.
\end{equation}
Using (\ref{int_identity}) to
integrate the term of
(\ref{int_lemma})
corresponding to a particular $\alpha \in
\Kappa_{(0,\delta_i,\delta_{i+1},\ldots,\delta_{k-1})}$, we get
\begin{eqnarray}\label{int_one_term}
\nonumber
&&
\int_0^{d_i-x_{\rho_{i-1}}}
\int_0^{d_i-x_{\rho_{i-1}}-x_{\rho_{i-1}+1}}
\dots
\int_0^{d_i-x_{\rho_{i-1}}-x_{\rho_{i-1}+1}-\cdots-x_{\rho_i-1}}\\
\nonumber
&&\ \ \ \
(-1)^{\alpha_1+\alpha_3+\cdots}
\cdot {1\over \alpha_1!\ \alpha_2!\ \cdots}\cdot
x_{\rho_i}^{\alpha_1} d_{i+1}^{\alpha_2} d_{i+2}^{\alpha_3} \cdots\
dx_{\rho_i}\ \cdots\ dx_{\rho_{i-1}+1}\\[6pt]
\nonumber
&&
= (-1)^{\alpha_1+\alpha_3+\cdots}
\cdot {1\over \alpha_1!\ \alpha_2!\ \cdots} \cdot
{\alpha_1!\ (d_i-x_{\rho_{i-1}})^{\delta_{i-1}+\alpha_1}\over
(\delta_{i-1}+\alpha_1)!} \cdot
d_{i+1}^{\alpha_2} d_{i+2}^{\alpha_3} \cdots\\[6pt]
\nonumber
&&
= (-1)^{\alpha_1+\alpha_3+\cdots}\cdot
\sum_{j,m\geq 0\ :\atop j+m=\delta_{i-1}+\alpha_1}
{1\over \alpha_2!\ \alpha_3!\cdots}
\cdot
(-1)^j \cdot
{x_{\rho_{i-1}}^j d_i^{m}\over
 j!\ m!}\cdot
d_{i+1}^{\alpha_2} d_{i+2}^{\alpha_3} \cdots\\[6pt]
&&
=
\sum_{{{j,m\ :\ j+m=\delta_{i-1}+\alpha_1,\atop
{\ \ \ \ \ \ (j,m,\alpha_2,\alpha_3,\ldots)\in\atop
\ \ \ \ \ \ \ \ \ \Kappa_{(0,\delta_{i-1},\delta_i,\ldots)}}}}}
(-1)^{j+\alpha_1+\alpha_3+\cdots}\cdot
{1\over j!\ m!\ \alpha_2!\ \alpha_3!\ \cdots} \cdot
x_{\rho_{i-1}}^j d_i^m d_{i+1}^{\alpha_2} d_{i+2}^{\alpha_3} \cdots
\end{eqnarray}
Observe that $(j,m,\alpha_2,\alpha_3,\ldots)\in
\Kappa_{(0,\delta_{i-1},\delta_i,\ldots)}$ if and only if
$(\alpha_1,\alpha_2,\ldots)\in\Kappa_{(0,\delta_i,\delta_{i+1},\ldots)}$,
where $\alpha_1 = j+m-\delta_{i-1}$. Hence summing the above equation
over all $\alpha\in\Kappa_{(0,\delta_i,\delta_{i+1},\ldots)}$, we get
\begin{eqnarray*}\label{int_sum}
&&
J_{\rho_{i-1}+1} \\[6pt]
&&
=
\int_0^{d_i-x_{\rho_{i-1}}}
\int_0^{d_i-x_{\rho_{i-1}}-x_{\rho_{i-1}+1}}
\dots
\int_0^{d_i-x_{\rho_{i-1}}-x_{\rho_{i-1}+1}-\cdots-x_{\rho_i-1}}
J_{\rho_i+1}\
dx_{\rho_i}\ \cdots\ dx_{\rho_{i-1}+1}
\\[6pt]
&&
= (-1)^{\delta_{i}+\delta_{i+2}+\delta_{i+4}+\cdots}\\
&&
\ \ \ \ \cdot
\sum_{(j,m,\alpha_2,\alpha_3,\ldots)\atop
\ \ \ \ \in\Kappa_{(0,\delta_{i-1},\delta_i,\ldots)}}
(-1)^{j+\alpha_2+\alpha_4+\cdots}\cdot
{1\over j!\ m!\ \alpha_2!\ \alpha_3!\ \cdots} \cdot
x_{\rho_{i-1}}^j d_i^m d_{i+1}^{\alpha_2} d_{i+2}^{\alpha_3} \cdots .
\end{eqnarray*}
Note that the signs are consistent: taking into account the factor
$(-1)^{\delta_{i+1}+\delta_{i+3}+\delta_{i+5}+\cdots}$ omitted from
(\ref{int_one_term}), the total sign of a term of
(\ref{int_one_term}) is
$$
(-1)^{\delta_{i+1}+\delta_{i+3}+\delta_{i+5}+\cdots}\cdot
(-1)^{j+\alpha_1+\alpha_3+\cdots}\\
=
(-1)^{\delta_{i}+\delta_{i+2}+\delta_{i+4}+\cdots}\cdot
(-1)^{j+\alpha_2+\alpha_4+\cdots},
$$
which is true because
$$
\alpha_1+\alpha_2+\cdots = \delta_i+\delta_{i+1}+\cdots = n-\rho_i,
$$
and hence
all the exponents on both sides add up to
$2(n-\rho_{i})+2j$, i.e.\ an even number.
\end{proof}

To finish the proof of Theorem~\ref{vol_ZS_thm}, set $i=1$ in
Lemma~\ref{inside_integrals} and integrate with respect to $x_n$:
\begin{eqnarray*}
&&
n!\cdot J_n\\[6pt]
&&
= n!\int_0^{d_1} J_1\ dx_1\\[6pt]
&&
= (-1)^{\delta_1+\delta_3+\delta_5+\cdots}\ n!\\
&&\ \ \ \
\cdot
\sum_{\alpha\in\Kappa_{(0,\delta_1,\delta_2,\ldots)}}
(-1)^{\alpha_2+\alpha_4+\alpha_6+\cdots}\cdot
{1\over (\alpha_1+1)!\ \alpha_2!\ \alpha_3!\ \cdots}\cdot
d_1^{\alpha_1+1} d_2^{\alpha_2} d_3^{\alpha_3} \cdots \\[6pt]
&&=
(-1)^{\delta_1+\delta_3+\cdots}
\sum_{(\alpha_1+1,\alpha_2,\alpha_3,\ldots)\atop
\ \ \ \ \in\Kappa_{(1,\delta_1,\delta_2,\ldots)}}
(-1)^{\alpha_2+\alpha_4+\cdots}
{n\choose \alpha_1+1,\ \alpha_2,\ \alpha_3,\ldots}\cdot
d_1^{\alpha_1+1}d_2^{\alpha_2}d_3^{\alpha_3}\cdots,
\end{eqnarray*}
and it is clear that $(\alpha_1+1,\alpha_2,\alpha_3,\ldots)
\in \Kappa_{(1,\delta_1,\delta_2,\ldots)}$ if and only if
$\alpha \in \Kappa_{(0,\delta_1,\delta_2,\ldots)}$.
\end{proof}