# The Hilbert-Kunz density functions of quadric hypersurfaces

Vijaylaxmi Trivedi  School of Mathematics, Tata Institute of Fundamental
Research, Homi Bhabha Road, Mumbai-40005, India<EMAIL_ADDRESS>
###### Abstract.

We show that the Hilbert-Kunz density function of a quadric hypersurface of
Krull dimension $n+1$ is a piecewise polynomial on a subset of $[0,n]$, whose
complement in $[0,n]$ has measure zero. Our explicit description of the
Hilbert-Kunz density function confirms a conjecture of Watanabe-Yoshida on the
lower bound of the Hilbert-Kunz multiplicity of the quadric of dimension
$n+1$, provided the characteristic is at least $n-1$. We also show that the
Hilbert-Kunz multiplicity of a quadric of fixed dimension is an eventually
strictly decreasing function of the characteristic confirming a conjecture of
Yoshida.

The main input comes from the classification of Arithmetically Cohen-Macaulay
bundles on the projective variety defined by the quadric via matrix
factorizations.

###### Key words and phrases:

Hilbert-Kunz multiplicity, spinor bundles, smooth quadric, ACM bundles

##  1\. Introduction

Let $R$ be a Noetherian ring containing a field of characteristic $p>0$ and
$I$ be an ideal of finite colength in $R$. For such a pair P. Monsky (in [M])
had introduced a characteristic $p$ invariant known as the Hilbert-Kunz (HK)
multiplicity $e_{HK}(R,I)$. This is a positive real number given by

$e_{HK}(R,I)=\lim_{n\to\infty}\frac{\ell(R/I^{[q]})}{q^{\dim R}}\geq 1.$

If $(R,{\bf m},k)$ is a formally unmixed Noetherian local ring then it was
proved by Watanabe-Yoshida (Theorem 1.5 in [WY1]) that $e_{HK}(R,{\bf m})=1$
if and only if $R$ is regular. For the next best class of rings, namely the
quadric hypersurfaces, they made the following conjecture in 2005:

Conjecture (1) (Conjecture 4.2 in [WY2]). Let $p>2$ be prime and
$K={\bf{\bar{F}_{p}}}$ and let

$R_{p,n+1}=K[x_{0},\ldots,x_{n+1}]/(x_{0}^{2}+\cdots+x_{n+1}^{2})$

denote the quadric hypersurface of dimension $n+1$. Then for any formally
unmixed non regular local ring $(A,{\bf m}_{A},K)$ of dimension $n+1$ we have

$e_{HK}(A,{\bf m}_{A})\geq e_{HK}(R_{p,n+1},{\bf m})\geq 1+m_{n+1},$

where $m_{n+1}$ are the constants occurring as the coefficients of the
following expression

$\mbox{sec}(x)+\mbox{tan}(x)=1+\sum_{n=0}^{\infty}m_{n+1}x^{n+1},\quad\mbox{for}\quad|x|<\pi/2.$

In the same paper they showed that the conjecture holds for $n\leq 3$. The
second inequality of the conjecture for $n\leq 5$ was proved by Yoshida in
[Y]. Later the conjecture up to $n\leq 5$ was proved by Aberbach-Enescu in
[AE2].

Enescu and Shimomoto in [ES] have proved the first inequality $e_{HK}(A)\geq
e_{HK}(R_{p,n+1})$, where $A$ belongs to the class of complete intersection
local rings.

On the other hand, around 2010, Gessel-Monsky in [GM], proved the following
result:

$\lim_{p\to\infty}e_{HK}(R_{p,n+1},{\bf m})=1+m_{n+1}.$

Progresses on Conjecture (1) is discussed in Huneke’s survey article [Hu].

Various people ([AE1], [AE2], Celikbas-Dao-Huneke-Zhang in [CDHZ]) have given
a lower bound of the form $e_{HK}(R,{\bf m})\geq C(d)$, where $C(d)$ is an
explicit constant and $(R,{\bf m})$ is an arbitrary formally unmixed
nonregular local ring of dimension $d$.

However the above result of Gessel-Monsky implies that the lower bounds such
as $C(d)$ are weaker than the bound given in the above conjecture: Recall so
far the best constant $C(d)=1+\frac{1}{d!d^{d}}$ (see Section 8, Theorem 13 in
[Hu]), whereas, as noted in [AE2],

$1+m_{3}=1+\frac{1}{3!},\quad 1+m_{4}=1+\frac{5}{4!},\quad
1+m_{5}=1+\frac{16}{5!},\quad 1+m_{6}=1+\frac{61}{6!}\quad\mbox{etc}.$

Regarding the HK multiplicities of the quadric hypersurfaces, another
conjecture was made by Yoshida ([Y]):

Conjecture (2). $e_{HK}(R_{p,n+1},{\bf m})$ is a decreasing function in $p$
for a fixed $n$.

Note that it is not true in general that $e_{HK}(R_{p})$ is a monotonic
function of $p$. The first example which is given by Han-Monsky ([HM]) is the
ring $R_{p}=k[x,y,z]/(x^{4}+y^{4}+z^{4})$, where $k$ is a field of
characteristic $p>2$ and $e_{HK}(R_{p},(x,y,z)))$ is a function of the
congruence class of $p~{}~{}(\mbox{mod}~{}~{}8)$ and not monotonic.

Conjecture (1) and Conjecture (2) have been revisited in the recent paper
[JNSWY].

Hilbert-Kunz density functions, which were introduced in [T1], generalize
Hilbert-Kunz multiplicities in the graded setting. In this paper, our key new
idea is to describe the Hilbert-Kunz density functions of quadric
hypersurfaces.

As applications of our description in Section 4, we obtain upper and lower
bounds of the values of the respective HK multiplicities of the quadric
hypersurfaces.

In particular we prove the second inequality of the Conjecture (1) by showing

Theorem (A) (Theorem 4.3). Let $p\neq 2$ and let $p>n-2$. Then, for $n\geq 3$,

$1+m_{n+1}+\tfrac{2n-4}{p}\geq e_{HK}(R_{p,n+1},{\bf m})\geq 1+m_{n+1}.$

In fact the second inequality is strict if $p\geq 3n-4$.

In Section 8, we obtain a more refined description of the Hilbert-Kunz density
function than in Section 4. As an application, we show that

Theorem (B) (Theorem 8.8 and Theorem 8.10). Given $n\geq 4$, there exist
polynomials $p_{n}(t),q_{n}(t)\in{\mathbb{Q}}[t]$ of degrees
$\leq(n+1)^{n+3}$, such that for all $p>2^{\lfloor n/2\rfloor}(n-2)$

$e_{HK}(R_{p,n+1})=1+m_{n+1}+\frac{p_{n}(t)}{q_{n}(t)}\Bigr{\rvert}_{t={1/p}}.$

The same assertion holds if $n=3$ and $p\geq 5$.

Given $n$, the polynomials $p_{n}(t)$ and $q_{n}(t)$ can be computed
explicitly by Theorem 8.8 and Theorem 8.10. For a prime $p>2$ and $n=3,4,5$,
the respective Hilbert-Kunz multiplicity is also computed in [WY2] and [Y].

As an application of our computation, we provide the following confirmation of
Conjecture (2).

Theorem (C) (Theorem 8.6). Given any $n\geq 3$, there exists $\epsilon>0$ such
that $e_{HK}(R_{p,n+1})$ is a strictly decreasing function of $p$ for $p\geq
1/\epsilon$.

A value of $\epsilon$ in Theorem (C) is explicitly determined in Section 8;
see Theorem 8.6.

Now we outline our methods of obtaining the Hilbert-Kunz density function and
provide descriptions of the Hilbert-Kunz density functions.

Recall that for a pair $(R,I)$, where $I$ is a finite colength homogeneous
ideal in a standard graded ring of dimension at least two, the HK density
function $f_{R,I}:{\mathbb{R}}\longrightarrow[0,\infty)$ is a compactly
supported continuous function given by

$f_{R,I}(x)=\lim_{s\to\infty}\frac{\ell(R/I^{[q]})_{\lfloor
xq\rfloor}}{q^{\dim R-1}},\quad\mbox{where}\quad q=p^{s}.$

The HK density function recovers the HK multiplicity since

$e_{HK}(R,I)=\int_{0}^{\infty}f_{R,I}(x)dx.$

For further details we refer to Section 2.

The theory of Hilbert-Kunz density function was initiated in [T1] and
subsequently developed in [TW2] (see Theorem 1.1 and the preceding definition
in [T1] and [TW2]), and [T2] (see Proposition 2.1).

The HK density function of $(R,I)$ tracks the length of the graded pieces of
$R/I^{[p^{n}]}$ separately. This very feature of the HK density functions
allows us to use the machinery of sheaf cohomology and other algebro-geometric
gadgets on $\mbox{Proj}(R)$ to describe the HK density functions; and thus in
turn the HK multiplicity. These advantages of using HK density functions over
the mere HK multiplicities are subsequently apparent.

Before going further we remark that for $n=1$ and $n=2$, the ring $R_{p,n+1}$
is the homogeneous coordinate ring of ${\mathbb{P}}^{1}_{k}$ and
${\mathbb{P}}_{k}^{1}\times{\mathbb{P}}_{k}^{1}$ respectively. In both the
cases the invariants $e_{HK}(R_{p,n+1})$ and $f_{R_{p,n+1},{\bf m}}$ are
independent of the characteristic (see Eto-Yoshida [EY] and [T1]).

Therefore in the rest of the paper we consider

$R_{p,n+1}=K[x_{0},\ldots,x_{n+1}]/(x_{0}^{2}+\cdots+x_{n+1}^{2}),\quad\mbox{where}\quad
n\geq 3,$

where $K$ is a perfect field of characteristic $p>0$; $R_{p,n+1}$ is equipped
with the standard graded structure. Set $Q_{n}=\mbox{Proj}(R_{p,n+1})$.

One key input that facilitates calculation in the quadric hypersurface case is
the complete classification of arithmetically Cohen-Macaulay bundles
(abbreviated as ACM bundles) on $Q_{n}$ as a direct sum of line bundles and
twisted spinor bundles.

This classification follows from the classification of maximal Cohen-Macaulay
modules on $R_{p,n+1}$ due to Knorrer [K] and Buchweitz-Eisenbud-Herzog [BEH];
we recall the relevant details in Section 2. Since
$F^{s}_{*}({\mathcal{O}}(a))$ is an ACM bundle on $Q_{n}$, for every $s^{th}$
iterated Frobenius map $F^{s}:Q_{n}\longrightarrow Q_{n}$ we have

(1.1)
$F_{*}^{s}({\mathcal{O}}(a))=\oplus_{t\in{\mathbb{Z}}}{\mathcal{O}}(t)^{\nu^{s}_{t}(a)}\oplus\oplus_{t\in{\mathbb{Z}}}{\mathcal{S}}(t)^{\mu^{s}_{t}(a)},$

where ${\mathcal{O}}(t)={\mathcal{O}}_{Q_{n}}(t)$ and ${\mathcal{S}}(t)$ is a
twisted spinor bundle if $n$ is odd, and a twisted direct sum of the two
spinor bundles if $n$ is even.

Later Achinger in [A] showed that the multiplicities of the bundles
${\mathcal{O}}(t)$ and ${\mathcal{S}}(t)$ occurring in (1.1) are related to
the lengths of graded components of the ring $R_{p,n+1}/{\bf m}^{[q]}$, with
$q=p^{s}$, by the formula

(1.2) $\ell(R_{p,n+1}/{\bf
m}^{[q]})_{a}=\nu^{s}_{0}(a)+2\lambda_{0}\mu^{s}_{1}(a),$

where ${\bf m}=(x_{0},\ldots,x_{n+1})$ and $\lambda_{0}=2^{\lfloor
n/2\rfloor}$.

Here we crucially use this, and another result, namely Theorem 2 from [A]
which determines the integers $t_{1}$ and $t_{2}$, in terms of $q=p^{s}$ and
$a$, for which the bundles ${\mathcal{O}}(t_{1})$ and ${\mathcal{S}}(t_{2})$
occur with non-zero multiplicity in the decomposition of
$F^{s}_{*}({\mathcal{O}}(a))$ (or of $F^{s}_{*}({\mathcal{S}}(a))$). However
[A] does not determine this multiplicity.

Using (1.2) and the theory of HK density functions we show that the functions
${\boldsymbol{\nu}}_{i}^{(p)}:{\mathbb{R}}\longrightarrow{\mathbb{R}}$,
${\boldsymbol{\mu}}_{i-1}^{(p)}:{\mathbb{R}}\longrightarrow{\mathbb{R}}$ given
by

(1.3)
${\boldsymbol{\nu}}_{i}^{(p)}(x)=\lim_{s\to\infty}\frac{\nu^{s}_{-i}(\lfloor
xq\rfloor)}{q^{n}},\quad{\boldsymbol{\mu}}_{i-1}^{(p)}(x)=\lim_{s\to\infty}\frac{2\lambda_{0}\mu^{s}_{-i+1}(\lfloor
xq\rfloor)}{q^{n}}$

are well defined functions continuous functions and for any nonnegative
integer $i$

$f_{R_{p,n+1}}(x+i)={\boldsymbol{\nu}}_{i}^{(p)}(x)+{\boldsymbol{\mu}}_{i-1}^{(p)}(x),\quad\mbox{for}~{}~{}x\in[0,1).$

Next we analyze the functions ${\boldsymbol{\nu}}_{i}^{(p)}$,
${\boldsymbol{\mu}}_{i-1}^{(p)}$.

We show that there exists a subset of $[0,1]$ which we call the difficult
range in the interval $[0,1)$, provided $p>n-2$, given as follows:

  1. (1)

$[\frac{1}{2}-\frac{n-2}{2p},\quad\frac{1}{2}+\frac{n-2}{2p})$, if $n\geq 3$
is an odd integer and

  2. (2)

$[0,~{}~{}~{}\frac{n-2}{2p})\cup[1-\frac{n-2}{2p},~{}~{}~{}1)$, if $n\geq 4$
is an even integer.

We show (see Proposition 5.3) that the rank functions
${\boldsymbol{\nu}}_{i}^{(p)}$, ${\boldsymbol{\mu}}_{j}^{(p)}$, when
restricted to the complement of the difficult range in $[0,1]$, are piecewise
polynomials. Moreover we see that these polynomials are independent of the
characteristic $p$. Such a description springs from the fact that the bundle
$F^{s}_{*}({\mathcal{O}}(\lfloor xq\rfloor))$ has atmost one twist of spinor
bundle(s) if $x$ is outside the difficult range.

The dependence of the density function of $R_{p,n+1}$ on the characteristic
$p>0$ is only via its description on the difficult range. We note that the
length of the difficult range goes to zero as $p\to\infty$. and in this range
characteristic $p$ shows up in the guise of functions
${\boldsymbol{\mu}}^{(p)}_{i}$, which is a bounded function (see Lemma 3.8).
More precisely we have:

### The first formulation of $f_{R_{p,n+1}}$

Here we express $f_{R_{p,n+1}}$, piecewise, as a sum of a polynomial in
${\mathbb{Q}}[x]$ (and therefore characteristic free) and another nonnegative
continuous function which is exactly the “pure characteristic $p$”
contribution. We prove Proposition 1.1 and Corollary 1.2 in Section 4.

######  Proposition 1.1.

Let $p\neq 2$ be a prime such that $p>n-2$ and $n_{0}=\lceil{n}/{2}\rceil-1$.
There exists an explicit set $\\{{\bf Z}_{0}(x),\ldots,{\bf
Z}_{n_{0}+1}(x),{\bf Y}_{n_{0}+2}(x),\ldots,{\bf Y}_{n-1}(x)\\}$ of degree $n$
polynomials in ${\mathbb{Q}}[x]$ which are independent of the characteristic
$p$, and a set
$\\{{\boldsymbol{\mu}}^{(p)}_{n_{0}-1},{\boldsymbol{\mu}}^{(p)}_{n_{0}},{\boldsymbol{\mu}}^{(p)}_{n_{0}+1}\\}$
of bounded continuous functions from $[0,1)\longrightarrow{\mathbb{R}}_{\geq
0}$ such that

  1. (1)

if $n\geq 4$ is an even number then

$f_{R_{p,n+1}}(x)=\begin{cases}{\bf Z}_{i}(x),\quad\mbox{if}\quad i\leq
x<i+1\quad\mbox{and}\quad 0\leq i\leq n_{0}\\\ \\\ {\bf
Z}_{n_{0}+1}(x),\quad\mbox{if}\quad(n_{0}+1)\leq x<(n_{0}+2)-\frac{n-2}{2p}\\\
\\\ {\bf
Z}_{n_{0}+1}(x)+{{\boldsymbol{\mu}}}^{(p)}_{n_{0}-1}(x-n_{0}-1),\quad\mbox{if}\quad
1-\tfrac{n-2}{2p}\leq x-(n_{0}+1)<1\\\ \\\ {\bf
Y}_{n_{0}+2}(x)+{{\boldsymbol{\mu}}}^{(p)}_{n_{0}+1}(x-n_{0}-2),\quad\mbox{if\quad$0\leq
x-(n_{0}+2)<\small{\frac{n-2}{2p}}$}\\\ \\\ {\bf
Y}_{n_{0}+2}(x),\quad\mbox{if}\quad(n_{0}+2)+\frac{n-2}{2p}\leq x<(n_{0}+3)\\\
\\\ {\bf Y}_{i}(x),\quad\mbox{if}\quad i\leq x<i+1\quad\mbox{and}\quad
n_{0}+3\leq i<n\end{cases}$

and $f_{R_{p},n+1}(x)=0$ elsewhere.

  2. (2)

If $n\geq 3$ is an odd number then

$f_{R_{p},n+1}(x)=\begin{cases}{\bf Z}_{i}(x)\quad\mbox{if}\quad i\leq
x<i+1\quad\mbox{and}\quad 0\leq i\leq n_{0}\\\ \\\ {\bf
Z}_{n_{0}+1}(x),\quad\mbox{if}\quad(n_{0}+1)\leq
x<(n_{0}+\frac{3}{2})-\frac{n-2}{2p}\\\ \\\ {\bf
Y}_{n_{0}+1}(x)+{\boldsymbol{\mu}}^{(p)}_{n_{0}}(x-n_{0}-1),\quad\mbox{if}\quad\tfrac{1}{2}-\tfrac{n-2}{2p}\leq
x-(n_{0}+1)<\tfrac{1}{2}+\tfrac{n-2}{2p}\\\ \\\ {\bf
Y}_{n_{0}+1}(x),\quad\mbox{if}\quad(n_{0}+1)+\frac{1}{2}+\frac{n-2}{2p}\leq
x<(n_{0}+2)\\\ \\\ {\bf Y}_{i}(x),\quad\mbox{if}\quad i\leq
x<i+1\quad\mbox{and}\quad n_{0}+2\leq i<n\end{cases}$

and $f_{R_{p},n+1}(x)=0$ elsewhere. Here we can also write

$f_{R_{p},n+1}(x)={\bf
Z}_{n_{0}+1}(x)+{{\boldsymbol{\mu}}^{(p)}_{n_{0}-1}}(x-n_{0}-1),\quad\mbox{if}\quad\tfrac{1}{2}-\tfrac{n-2}{2p}\leq
x-(n_{0}+1)<\tfrac{1}{2}.$

Moreover in both the cases the function $f_{R_{p},n+1}$ is partially symmetric
and the symmetry is given by

$f_{R_{p},n+1}(x)=f_{R_{p},n+1}(n-x)\quad\mbox{for}\quad 0\leq
x\leq\tfrac{n-2}{2}\left(1-\tfrac{1}{p}\right).$

We can look at this result as follows

######  Corollary 1.2.

If $p>n-2$ and $p\neq 2$ then

$\begin{array}[]{lcll}f_{R_{p,n+1},{\bf
m}}(x)&=&f_{R^{\infty}_{n+1}}(x)&x\in[0,\quad\frac{n+2}{2}-\frac{n-2}{2p}]\\\
\\\
&=&f_{R^{\infty}_{n+1}}(x)+{\boldsymbol{\mu}}^{(p)}_{n_{0}-1}(x-n_{0}-1)&x\in[\frac{n+2}{2}-\frac{n-2}{2p},\quad\frac{n+2}{2})\\\
\\\
&=&f_{R^{\infty}_{n+1}}(x)+{\boldsymbol{\mu}}^{(p)}_{i}(x-i-1)&x\in[\frac{n+2}{2},\quad\frac{n+2}{2}+\frac{n-2}{2p}]\\\
\\\
&=&f_{R^{\infty}_{n+1}}(x)&x\in[\frac{n+2}{2}+\frac{n-2}{2p},\quad\infty),\end{array}$

where $i=n_{0}+1$ if $n$ is even and $i=n_{0}$ if $n$ is odd. Moreover the
function $f_{R^{\infty}_{n+1}}:[0,\infty)\longrightarrow[0,\infty)$ given by

$f_{R^{\infty}_{n+1}}(x):=\lim_{p\to\infty}f_{R_{p,n+1},{\bf m}}(x)$

is a well defined continuous function.

Theorem (A) follows from Proposition 1.1 and Corollary 1.2 as detailed in
Theorem 4.3.

Further, this explicit description of support of the $f_{R_{p},n+1}$ along
with the result of [TW1] gives the following result (in Corollary 4.4)

Corollary Let $p>2$ be a prime number such that $p>n-2$ then the $F$-threshold
of the ring $R_{p,n+1}$ is $c^{\bf m}({\bf m})=n$.

### The second formulation of $f_{R_{p,n+1}}$

In Section 7, we further analyse the rank functions
${\boldsymbol{\nu}}_{i}^{(p)}$, ${\boldsymbol{\mu}}^{(p)}_{i}$ on the
difficult range.

  * •

We find a countable collection of mutually disjoint semi open intervals, i.e.,
interval of the form $[a,b)$, which are contained in the difficult range. The
complement of the union of these intervals in the difficult range has measure
zero. Our indexing of the members in this countable collection is independent
of the underlying characteristic. See Lemma 6.5, Lemma 6.7.

  * •

In Proposition 7.2 and Proposition 7.4, we show that there is a finite set of
matrices with entries in ${\mathbb{Q}}[t]$, ${\mathbb{Q}}[x]$ such that on a
given semi open subinterval the rank functions
${\boldsymbol{\nu}}_{i}^{(p)}(x)$, ${\boldsymbol{\mu}}^{(p)}_{i}(x)$ are
products of matrices from the above collection, evaluated at $t=1/{p}$. Given
the semi open subinterval the choices of matrices in the product only depends
on the chosen indexing of the interval.

The description of ${\boldsymbol{\mu}}_{i}^{(p)}$ in Proposition 7.2 and
Proposition 7.4 gives us a ground where we can compare the integrals of
${\boldsymbol{\mu}}_{i}^{(p)}$ as $p$ varies; the characteristic free indexing
of the semi open intervals facilitates the comparison. Further, using detailed
analysis, we choose one subinterval on which the integral of
${\boldsymbol{\mu}}_{i}^{(p)}$ is a strictly decreasing function of $p$ and on
all other subintervals it is a decreasing function of $p$.

Moreover the integrals of ${\boldsymbol{\mu}}_{i}^{(p)}$ have nice features
over such subintervals: For example, for given integer $l\geq 1$,
$\int{\boldsymbol{\mu}}_{i}^{(p)}$ over the union of subintervals which are
indexed by tuples of length $l$, will correspond to ${\mathbb{A}}^{l}$ where
${\mathbb{A}}$ is the sum of the above mentioned finite set of matrices in
${\mathbb{Q}}[t]$. In particular $\int{\boldsymbol{\mu}}_{i}^{(p)}$ over the
difficult range involves a power series expression of the matrix. Now arguing
that ${\mathbb{A}}$ has no eigenvalue $\geq 1$ leads to the polynomials as
given in Theorem (B).

Using Theorem (B) and the result of [ES] one can answer the Conjecture (1) for
the class of complete local rings in the following way:

Let $n\geq 4$ and let $p>2^{\lfloor n/2\rfloor}(n-2)$. Let $(A,{\bf m}_{A},K)$
be a complete intersection but nonregular local ring of dimension $n+1$. Then

$e_{HK}(A,{\bf m}_{A})\geq e_{HK}(R_{p,n+1},{\bf
m})=1+m_{n+1}+\frac{p_{n}(t)}{q_{n}(t)}\bigr{\rvert}_{t=1/p}>1+m_{n+1},$

where for given $n$, the polynomials $p_{n}(t)$ and $q_{n}(t)$ can be computed
explicitly by Theorem 8.8 and Theorem 8.10.

In particular we have a sharp lower bound on the class of complete
intersection rings.

In the end we explicitly write down the HK density function for $n=3$ case in
Theorem 9.2. Note that $n=3$ is the first $n$ where the $e_{HK}(R_{p,n+1})$
involves the characteristic $p$. Here the HK density function, which is a
compactly supported continuous function, is non smooth at infinitely many
points, however, intriguingly, it is a ${\mathcal{C}}^{2}({\mathbb{R}})$
function. This is different (see Remark 9.3) from the previous known cases
such as two dimensional graded rings (see Example 3.3 in [T1]) or projective
toric varieties (see [MT] Theorem 3.4 and the proof of Theorem 1.1, or Theorem
4.1.18 of [Mo]) where the set of non smooth points is a finite set.

Question 1: If $R$ is a standard graded ring over a perfect field and is of
dimension $d\geq 2$ and $I\subset R$ is a graded ideal of finite colength,
then when does the HK density function $f_{R,I}$ belong to
${\mathcal{C}}^{d-2}({\mathbb{R}})$?

An approach to address this question is mentioned in [Muk], where it is
claimed that an affirmative answer to one of the questions in Question 5.6
shows that the HK density function is in ${\mathcal{C}}^{d-2}({\mathbb{R}})$.

Based on the examples of quadric hypersurfaces and also other above mentioned
examples we pose a variant (suggested by the referee) of a question due to
K.I. Watanabe (private communication).

Question 2: Let $(R,I)$ be a standard graded pair of dimension $d\geq 2$. Let
$[0,\alpha]$ be the support of the the HK density function $f_{R,I}$ of
$(R,I)$. Does there exist a subset $S$ of $[0,\alpha]$ of Lebesgue measure
zero such that

  1. (1)

There is a countable collection of mutually disjoint semi open intervals
$\\{[a_{n},b_{n})\\}_{n\in{\mathbb{N}}}$ such that

$[0,\alpha]\setminus S=\cup_{n\in{\mathbb{N}}}[a_{n},b_{n});$

  2. (2)

and for each $n$, there is a polynomial $P_{n}\in{\mathbb{R}}[x]$ such that on
$[a_{n},b_{n})$ the density function is given by $P_{n}$?

Moreover, can $S$ be taken to a countable set?

Looking further, the methods used in this paper suggest possible computations
for the HK density and related invariants in other situations, where we have
information on ACM bundles using matrix factorizations.

We would like to thank the referee for a careful and thorough reading of the
paper, and for detailed suggestions to improve the exposition.

######  Notations 1.3.

In the rest of the paper

$R_{p,n+1}=\frac{k[x_{0},\ldots,x_{n+1}]}{(x_{0}^{2}+\cdots+x_{n+1}^{2}]}\quad\mbox{and}\quad
Q_{n}=\mbox{Proj}~{}R_{p,n+1},$

where $n\geq 3$ and $k$ is a perfect field of characteristic $p>2$.

##  2\. preliminaries

In this section we recall the relevant results which are known in the
literature.

First we recall the following notion of HK density function for $(R,I)$, where
$R$ is a standard graded ring and $I$ is a homogeneous ideal in $R$ of finite
colength, which was introduced by the author ([T1]) for standard graded rings
and later this notion was generalized by the author and Watanabe ([TW2]) for
${\mathbb{N}}$-graded rings.

######  Definition 2.1.

Let $R$ be a Noetherian standard graded ring of dimension $d\geq 2$ over a
perfect field of characteristic $p>0$, and let $I\subset R$ be a homogeneous
ideal such that $\ell(R/I)<\infty$. For $s\in{\mathbb{N}}$ and $q=p^{s}$, let
$f_{s}:[0,\infty)\longrightarrow[0,\infty)$ be defined as

$f_{s}(R,I)(x)=\frac{\ell(R/I^{[q]})_{\lfloor xq\rfloor}}{q^{\dim
R-1}},\quad\mbox{where}\quad q=p^{s}.$

Theorem (Theorem 1.1 of [T1]). The sequence
$\\{f_{s}(R,I)\\}_{s\in{\mathbb{N}}}$ converges uniformly to a compactly
supported continuous function $f_{R,I}:[0,\infty)\longrightarrow[0,\infty)$,
where

$f_{R,I}(x)=\lim_{s\to\infty}f_{s}(R,I)(x).\quad\mbox{Moreover}\quad
e_{HK}(R,I)=\int_{0}^{\infty}f_{R,I}(x)dx.$

We call $f_{R,I}$ to be the HK density function of $R$ with respect to the
ideal $I$.

######  Definition 2.2.

A vector bundle $E$ on a smooth $n$-dimensional hypersurface
$X=\mbox{Proj}~{}S/(f)$, where $S=k[x_{0},\ldots,x_{n+1}]$ is called
arithmetically Cohen-Macaulay (ACM) if $H^{i}(X,E(m))=0$, for $0<i<n$ and for
all $m$.

It is easy to check that a vector bundle $E$ on $X$ is ACM if and only if the
corresponding graded $S/(f)$-module $\oplus_{j\in{\mathbb{N}}}H^{0}(X,E(j))$
is maximal Cohen-Macaulay (MCM).

Let $Q_{n}=\mbox{Proj}~{}S/(f)$ be the quadric given by the hypersurface
$f=x_{0}^{2}+\cdots+x_{n+1}^{2}=0$ in
${\mathbb{P}}^{n+1}_{k}=\mbox{Proj}~{}S$, where $n\geq 3$. Let $k$ be an
algebraically closed field. Henceforth we assume $n>2$.

By B-E-H classification ([BEH]) of indecomposable graded MCM modules over
quadric we have: Other than free modules on $S/(f)$, there is (up to shift)
only one indecomposable module $M$ (which corresponds to the single spinor
bundle $\Sigma$ on $Q_{n}$) if $n$ is odd and there are only two of them
$M_{+}$ and $M_{-}$ (which correspond to the two spinor bundles $\Sigma_{+}$
and $\Sigma_{-}$ on $Q_{n}$) if $n$ is even.

Moreover an MCM module over $S/(f)$ corresponds to a matrix factorization of
the polynomial $f$ (such an equivalence is given by Eisenbud in [E], for more
general hypersurfaces $(f)$), which is a pair $(\phi,\psi)$ of square matrices
of polynomials, of the same size, such that $\phi\cdot\psi=f\cdot
id=\psi\cdot\phi$ and the MCM module is the cokernel of $\phi$.

Now the matrix factorization $(\phi_{n},\psi_{n})$ for indecomposable bundles
on $Q_{n}$ (see Langer [L], Section 2.2) gives an exact sequence of locally
free sheaves on ${\mathbb{P}}_{k}^{n+1}$, where
$i:Q_{n}\longrightarrow{\mathbb{P}}_{k}^{n+1}$ is the natural inclusion.

(2.1) $0\longrightarrow{\mathcal{O}}_{{\mathbb{P}}_{k}^{n+1}}(-2)^{2^{\lfloor
n/2\rfloor+1}}\stackrel{{\scriptstyle\Phi_{n}}}{{\longrightarrow}}{\mathcal{O}}_{{\mathbb{P}}_{k}^{n+1}}(-1)^{2^{\lfloor
n/2\rfloor+1}}\longrightarrow i_{*}{\mathcal{S}}\longrightarrow 0,$

${\mathcal{S}}=\Sigma$ and $\Phi_{n}=\phi_{n}=\psi_{n}$ for $n$ odd and
${\mathcal{S}}=\Sigma_{+}\oplus\Sigma_{-}$ and
$\Phi_{n}=\phi_{n}\oplus\psi_{n}$ for $n$ even. Moreover we have (see (2.4),
(2.5) and (2.6) in [L]) the short exact sequences of vector bundles on
$Q_{n}$: If $n$ is odd then

$0\longrightarrow{\mathcal{S}}\longrightarrow{\mathcal{O}}_{Q_{n}}^{2^{\lfloor
n/2\rfloor+1}}\longrightarrow{\mathcal{S}}(1)\longrightarrow 0.$

If $n$ is even then

$0\longrightarrow\Sigma_{-}\longrightarrow{\mathcal{O}}_{Q_{n}}^{2^{\lfloor
n/2\rfloor}}\longrightarrow\Sigma_{+}(1)\longrightarrow 0$

and

$0\longrightarrow\Sigma_{+}\longrightarrow{\mathcal{O}}_{Q_{n}}^{2^{\lfloor
n/2\rfloor}}\longrightarrow\Sigma_{-}(1)\longrightarrow 0.$

We also have the natural exact sequence

(2.2)
$0\longrightarrow{\mathcal{O}}_{{\mathbb{P}}_{k}^{n+1}}(-2)\longrightarrow{\mathcal{O}}_{{\mathbb{P}}_{k}^{n+1}}\longrightarrow{\mathcal{O}}_{Q_{n}}\longrightarrow
0.$

We denote ${\mathcal{O}}_{Q_{n}}(m)={\mathcal{O}}(m)$ and

$R_{p,n+1}=\frac{k[x_{0},\ldots,x_{n+1}]}{(x_{0}^{2}+\cdots+x_{n+1}^{2}]}=\oplus_{m\geq
0}H^{0}(Q_{n},{\mathcal{O}}(m))\quad\mbox{and}\quad n\geq 3,$

where $k$ is a field of characteristic $p>2$. In particular the $m^{th}$
graded component of $R_{p,n+1}$ is $H^{0}(Q_{n},{\mathcal{O}}(m))$. We will be
using the following set of equalities in our forthcoming computations.

(2.3)
$\left\\{\begin{matrix}h^{0}(Q_{n},{\mathcal{O}}(m))=h^{0}({\mathbb{P}}_{k}^{n+1},{\mathcal{O}}_{{\mathbb{P}}_{k}^{n+1}}(m))-h^{0}({\mathbb{P}}_{k}^{n+1},{\mathcal{O}}_{{\mathbb{P}}_{k}^{n+1}}(m-2))\\\
\
h^{0}(Q_{n},{\mathcal{S}}(m))=2\lambda_{0}[h^{0}({\mathbb{P}}_{k}^{n+1},{\mathcal{O}}_{{\mathbb{P}}_{k}^{n+1}}(m-1))-h^{0}({\mathbb{P}}_{k}^{n+1},{\mathcal{O}}_{{\mathbb{P}}_{k}^{n+1}}(m-2))],\end{matrix}\right.$

where $2\lambda_{0}=2^{\lfloor n/2\rfloor+1}$.

We note that $\omega_{Q_{n}}={\mathcal{O}}(-n)$ and
${\mathcal{S}}^{\vee}={\mathcal{S}}(1)$, where $\omega_{Q_{n}}$ denotes the
canonical bundle of $Q_{n}$. Therefore by Serre duality

(2.4)
$\left\\{\begin{matrix}h^{n}(Q_{n},{\mathcal{O}}(m))=h^{0}(Q_{n},{\mathcal{O}}(-m-n))\\\
\
h^{n}(Q_{n},{\mathcal{S}}(m))=h^{0}(Q_{n},{\mathcal{S}}(1-m-n)).\end{matrix}\right.$

The rank of $Q_{n}$-bundle ${\mathcal{S}}=\lambda_{0}=2^{\lfloor n/2\rfloor}$.

The statement and the proof of the following lemma is contained in [A].

######  Lemma 2.3.

For given integer $a$ and $q=p^{s}$, if the nonnegative integers
$\nu^{s}_{t}(a)$, $\mu^{s}_{t}(a)$ are the integers occurring in the
decomposition

$F_{*}^{s}({\mathcal{O}}(a))=\oplus_{t\in{\mathbb{Z}}}{\mathcal{O}}(t)^{{\nu}^{s}_{t}(a)}\oplus\oplus_{t\in{\mathbb{Z}}}{\mathcal{S}}(t)^{{\mu}^{s}_{t}(a)}$

then

$\ell(R_{p,n+1}/{\bf
m}^{[q]})_{a}=\nu^{s}_{0}(a)+2\lambda_{0}\mu^{s}_{1}(a),\quad\mbox{where}\quad{\bf
m}=(x_{0},\ldots,x_{n+1}).$

###### Proof.

Since ${\mathcal{O}}(a)$ and ${\mathcal{S}}(a)$ are ACM bundles (follows from
(2.1) and (2.2)), the projection formula implies that
$F_{*}^{s}({\mathcal{O}}(a))$ is an ACM bundle on $Q_{n}$. Thus for $q=p^{s}$
and $a\in{\mathbb{Z}}$, we indeed have a decomposition of the form

(2.5)
$F_{*}^{s}({\mathcal{O}}(a))=\oplus_{t\in{\mathbb{Z}}}{\mathcal{O}}(t)^{\nu^{s}_{t}(a)}\oplus\oplus_{t\in{\mathbb{Z}}}{\mathcal{S}}(t)^{\mu^{s}_{t}(a)}.$

Restricting the Euler sequence in ${\mathbb{P}}_{k}^{n+1}$ to $Q_{n}$ we get
the short exact sequence

$0\longrightarrow\Omega^{1}_{{\mathbb{P}}^{n+1}_{k}}(1)\mid_{Q_{n}}\longrightarrow\oplus^{n+2}{\mathcal{O}}\longrightarrow{\mathcal{O}}(1)\longrightarrow
0$

of sheaves of ${\mathcal{O}}$-modules, where the second map is given by
$(a_{0},\ldots,a_{n+1})\to\sum a_{i}x_{i}$. This gives the long exact sequence

$0\longrightarrow
H^{0}(Q_{n},F^{s*}\Omega^{1}_{{\mathbb{P}}^{n+1}_{k}}(1)\mid_{Q_{n}}\otimes{\mathcal{O}}(a))\longrightarrow
H^{0}(Q_{n},\oplus^{n+2}F^{s*}{\mathcal{O}}(a))\stackrel{{\scriptstyle\Psi_{a+q}}}{{\longrightarrow}}H^{0}(Q_{n},{\mathcal{O}}(a+q))\\\
\longrightarrow
H^{1}(Q_{n},F^{s*}\Omega^{1}_{{\mathbb{P}}^{n+1}_{k}}(1)\mid_{Q_{n}}\otimes{\mathcal{O}}(a))\longrightarrow
0.$

Therefore

$\ell\Bigl{(}\frac{R_{p,n+1}}{{\bf m}^{[q]}}\Bigr{)}_{a+q}=\ell({\rm
coker}\,~{}\Psi_{a+q})\\\
=h^{1}\bigl{(}Q_{n},F^{s*}\Omega^{1}_{{\mathbb{P}}^{n+1}_{k}}(1)\mid_{Q_{n}}\otimes{\mathcal{O}}(a)\bigr{)}=h^{1}\bigl{(}Q_{n},\Omega^{1}_{{\mathbb{P}}^{n+1}_{k}}\mid_{Q_{n}}\otimes
F^{s}_{*}{\mathcal{O}}(a+q)\bigr{)}.$

Now by Lemma 1.2 in [A] we have

$h^{1}\bigl{(}Q_{n},\Omega^{1}_{{\mathbb{P}}^{n+1}_{k}}(t)\mid_{Q_{n}}\bigr{)}=\delta_{t,0}\quad
and\quad
h^{1}\bigl{(}Q_{n},{\mathcal{S}}\otimes\Omega^{1}_{{\mathbb{P}}^{n+1}_{k}}(t)\mid_{Q_{n}}\bigr{)}=2^{\lfloor
n/2\rfloor+1}\delta_{t,1}.$

Hence

$\ell(R_{p,n+1}/{\bf m}^{[q]})_{a}={\rm
coker}\,~{}{\Psi_{a}}=\nu^{s}_{0}(a)+2\lambda_{0}\mu^{s}_{1}(a).$

∎

######  Lemma 2.4.

Let $i\geq 0$ be an integer and $x\in[i,i+1)$ then

$\nu^{s}_{0}(\lfloor xq\rfloor)=\nu^{s}_{-i}(\lfloor xq\rfloor-
iq)\quad\mbox{and}\quad\mu^{s}_{1}(\lfloor xq\rfloor)=\mu^{s}_{-i+1}(\lfloor
xq\rfloor-iq).$

In particular

$f_{R_{p,n+1},{\bf
m}}(x)={\boldsymbol{\nu}}_{i}^{(p)}(x-i)+{\boldsymbol{\mu}}_{i-1}^{(p)}(x-i)\quad\mbox{for}\quad
x\in[i,i+1),$

where ${\boldsymbol{\nu}}_{i}^{(p)}:{\mathbb{R}}\longrightarrow{\mathbb{R}}$
and ${\boldsymbol{\mu}}_{i}^{(p)}:{\mathbb{R}}\longrightarrow{\mathbb{R}}$ are
the functions as defined in (1.3).

###### Proof.

Note that for any integer $m\geq 0$, there is an integer $i\geq 0$ such that
$iq\leq m<(i+1)q$. Hence by the projection formula

(2.6) $F_{*}^{s}({\mathcal{O}}(m))\simeq
F^{s}_{*}\bigl{(}{\mathcal{O}}(m-iq)\otimes
F^{s*}({\mathcal{O}}(i))\bigr{)}\simeq
F^{s}_{*}\bigl{(}{\mathcal{O}}(m-iq)\bigr{)}\otimes{\mathcal{O}}(i).$

In particular

(2.7) $\ell(R_{p,n+1}/{\bf
m}^{[q]})_{m}=\nu^{s}_{0}(m)+2\lambda_{0}\mu^{s}_{1}(m)=\nu^{s}_{-i}(m-iq)+2\lambda_{0}\mu^{s}_{-i+1}(m-iq).$

Therefore for $x\in[i,i+1)$

$f_{R_{p,n+1},{\bf m}}(x)=\lim_{s\to\infty}\frac{\ell(R/{\bf
m}^{[q]})_{\lfloor xq\rfloor}}{q^{n}}\\\
=\lim_{s\to\infty}\frac{\nu^{s}_{-i}(\lfloor xq\rfloor-
iq)+2\lambda_{0}\mu^{s}_{-i+1}(\lfloor xq\rfloor-
iq)}{q^{n}}=({\boldsymbol{\nu}}_{i}^{(p)}+{\boldsymbol{\mu}}_{i-1}^{(p)})(x-i).$

∎

We also use the following result of Achinger (Theorem 2 in [A]) which
determines, in terms of $s$, $a$ and $n$, when the numbers $\nu^{s}_{i}(a)$
and $\mu^{s}_{i}(a)$ are nonzero in the decomposition of
$F^{s}_{*}({\mathcal{O}}(a))$. Langer in [L] has given such formula for the
occurance of line bundles in the Frobenius direct image.

Theorem [A]. Let $p\neq 2$, $s\geq 1$ and $n\geq 3$ and

$F_{*}^{s}({\mathcal{O}}(a))=\oplus_{t\in{\mathbb{Z}}}{\mathcal{O}}(t)^{\nu^{s}_{t}(a)}\oplus\oplus_{t\in{\mathbb{Z}}}{\mathcal{S}}(t)^{\mu^{s}_{t}(a)}.$
$F_{*}^{s}({\mathcal{S}}(a))=\oplus_{t\in{\mathbb{Z}}}{\mathcal{O}}(t)^{{\mathchoice{\widetilde{\nu}}{\widetilde{\nu}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{\nu}$}}}$}}{\tilde{\nu}}}^{s}_{t}(a)}\oplus\oplus_{t\in{\mathbb{Z}}}{\mathcal{S}}(t)^{{{\mathchoice{\widetilde{\mu}}{\widetilde{\mu}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{\mu}$}}}$}}{\tilde{\mu}}}^{s}}_{t}(a)}.$

Then

  1. (1)

$F_{*}^{s}({\mathcal{O}}(a))$ contains ${\mathcal{O}}(t)$ if and only if
$0\leq a-tq\leq n(q-1)$.

  2. (2)

$F_{*}^{s}({\mathcal{O}}(a))$ contains ${\mathcal{S}}(t)$ if and only if

$\left(\tfrac{(n-2)(p-1)}{2}\right)\tfrac{q}{p}\leq
a-tq\leq\left(\tfrac{(n-2)(p-1)}{2}+n-2+p\right)\tfrac{q}{p}-n.$

  3. (3)

$F_{*}^{s}({\mathcal{S}}(a))$ contains ${\mathcal{O}}(t)$ if and only if
$1\leq a-tq\leq n(q-1)$.

  4. (4)

$F_{*}^{s}({\mathcal{S}}(a))$ contains ${\mathcal{S}}(t)$ if and only if

$\left(\tfrac{(n-2)(p-1)}{2}\right)\tfrac{q}{p}+1-\delta_{s,1}\leq
a-tq\leq\left(\tfrac{(n-2)(p-1)}{2}+n-2+p\right)\tfrac{q}{p}-n+\delta_{s,1}.$

##  3\. Formula for the rank functions ${\boldsymbol{\nu}}_{i}^{(p)}$ and
${\boldsymbol{\mu}}_{i-1}^{(p)}$

Let the rank functions ${\boldsymbol{\nu}}_{i}^{(p)}$
${\boldsymbol{\mu}}_{i-1}^{(p)}$ be as given in (1.3). As noted earlier to
analyze the rank functions it is enough to study on the interval $[0,1)$.

One of the goal of this section is to determine, for a given rank function a
subinterval in $[0,1)$ such that the function is polynomial, that is there is
a polynomial $p(x)\in{\mathbb{Q}}[x]$ such that the rank function at $x$ is
equal to $p(x)$ for $x$ in that subinterval. We give the description of such
subintervals in Proposition 3.10.

Since this section is technical, the reader can choose to skip the proof of
Lemmas 3.2, 3.5, 3.7 and 3.8 for the moment and move to the next section.

######  Notations 3.1.

  1. (1)

For given integer $a$ and $q=p^{s}$, the nonnegative integers
$\nu^{s}_{t}(a)$, $\mu^{s}_{t}(a)$ are the integers occurring in the
decomposition

$F_{*}^{s}({\mathcal{O}}(a))=\oplus_{t\in{\mathbb{Z}}}{\mathcal{O}}(t)^{{\nu}^{s}_{t}(a)}\oplus\oplus_{t\in{\mathbb{Z}}}{\mathcal{S}}(t)^{{\mu}^{s}_{t}(a)}.$

  2. (2)

For the sake of abbreviation henceforth we will denote

    1. (a)

${\mathcal{O}}_{Q_{n}}(m))={\mathcal{O}}(m)$,

    2. (b)

$h^{0}(Q_{n},{\mathcal{O}}(m))=L_{m}$ and

    3. (c)

$2\lambda_{0}=2^{\lfloor n/2\rfloor+1}$.

  3. (3)

Let $n_{0}$ and $\Delta$ be given as

$n_{0}=\lceil\tfrac{(n-2)(p-1)}{2p}\rceil\quad\mbox{and}\quad
n_{0}-\Delta=\tfrac{(n-2)(p-1)}{2p}.$

Hence for $p>n-2\implies n_{0}=\lceil n/2\rceil-1$

  4. (4)

A bundle is a spinor bundle of type $t$ if it is isomorphic to
${\mathcal{S}}(t)$, where ${\mathcal{S}}=\Sigma$ if $n$ is odd and
${\mathcal{S}}=\Sigma_{+}+\Sigma_{-}$ if $n$ is even. We say two spinor
bundles ${\mathcal{S}}(t)$ and ${\mathcal{S}}(t^{\prime})$ are of the same
type if $t=t^{\prime}$.

First we make the observation that there can occur at most
$\lceil(n-2)/p\rceil+2$ twists of spinor bundles in the decomposition of
$F^{s}_{*}({\mathcal{O}}(a))$, hence for $p>n-2$ the number reduces to three.
The next result shows for $p>n-2$ and any $s$ and $a$, there can be at most
two twists of Spinor bundles appearing in the decomposition of
$F_{*}^{s}({\mathcal{O}}(a))$.

######  Lemma 3.2.

If $1-n\leq a<q=p^{s}$ and $p>2$ then

$F^{s}_{*}({\mathcal{O}}(a))=\oplus_{t=0}^{n-1}{\mathcal{O}}(-t)^{{\nu}^{s}_{-t}(a)}\oplus\oplus_{i=-1}^{\lceil\tfrac{n-2}{p}\rceil}{\mathcal{S}}(-n_{0}-i)^{\mu^{s}_{-n_{0}-i}(a)}.$

  1. (1)

If $n\geq 4$ is an even number and $n-2<p$ then $n_{0}=({n-2})/{2}$ and

(3.1)
$F_{*}^{s}({\mathcal{O}}(a))=\oplus_{t=0}^{n-1}{\mathcal{O}}(-t)^{{\nu}^{s}_{-t}(a)}\oplus\oplus_{i=n_{0}-1}^{n_{0}+1}{\mathcal{S}}(-i)^{\mu^{s}_{-i}(a)}.$

Moreover

    1. (a)

$\mu^{s}_{-n_{0}+1}(a)\neq 0\implies a/q\in[1-\frac{n-2}{2p},~{}~{}1)$

    2. (b)

$\mu^{s}_{-n_{0}}(a)\neq 0\implies a/q\in[0,~{}~{}1)$

    3. (c)

$\mu^{s}_{-n_{0}-1}(a)\neq 0\implies a/q\in[0,~{}~{}\frac{n-2}{2p})$.

  2. (2)

If $n\geq 3$ is an odd number and $n-2<p$ then $n_{0}=({n-1})/{2}$ and

(3.2)
$F_{*}^{s}({\mathcal{O}}(a))=\oplus_{t=0}^{n-1}{\mathcal{O}}(-t)^{{\nu}^{s}_{-t}(a)}\oplus{\mathcal{S}}(-n_{0}+1)^{{\mu}^{s}_{-n_{0}+1}(a)}\oplus{\mathcal{S}}(-n_{0})^{{\mu}^{s}_{-n_{0}}(a)}.$

Moreover

    1. (a)

$\mu^{s}_{-n_{0}+1}(a)\neq 0\implies
a/q\in[\frac{1}{2}-\frac{n-2}{2p},~{}~{}1)$

    2. (b)

$\mu^{s}_{-n_{0}}(a)\neq 0\implies
a/q\in[0,~{}~{}\frac{1}{2}+\frac{n-2}{2p})$.

###### Proof.

The formula for the decomposition for $F^{s}_{*}({\mathcal{O}}(a))$ for
$1-n\leq a<q$ follows from the assertion (1) of Theorem [A].

By the assertion (2) of [A], if ${\mathcal{S}}(t)$ occurs in the decomposition
of $F^{s}_{*}({\mathcal{O}}(a))$ then

$(n_{0}-\Delta)q\leq a-tq\leq(n_{0}-\Delta)q+\tfrac{(n-2)q}{p}+q-n$
$\implies(n_{0}-\Delta)\leq\tfrac{a}{q}-t\leq(n_{0}-\Delta)+\tfrac{(n-2)}{p}+1-\tfrac{n}{q}.$

Hence

(3.3) $0\leq\tfrac{a}{q}+\Delta-t-n_{0}\leq\tfrac{n-2}{p}+1-\tfrac{n}{q}.$

We note that $0\leq a/q+\Delta<2$ and $t$ and $n_{0}$ are integers. Therefore

$n_{0}\leq a/q+\Delta-t<2-t\implies n_{0}-1\leq-t$

and

$-t-n_{0}-1\leq a/q+\Delta-t-
n_{0}-1\leq\tfrac{n-2}{p}-\tfrac{n}{q}<\lceil\tfrac{n-2}{p}\rceil.$

This proves $-n_{0}-t\in\\{-1,0,\cdots,\lceil{(n-2)}/{p}\rceil\\}$.

In particular if $n-2<p$ then $-n_{0}-t\in\\{-1,0,1\\}$.

  1. Let $n$ be even and $n-2<p$ then $n=2m$ for some integer $m\geq 2$. Then

$n_{0}=\lceil\tfrac{(n-2)(p-1)}{2p}\rceil=\lceil\tfrac{(m-1)(p-1)}{p}\rceil=\lceil(m-1)-\tfrac{m-1}{p}\rceil=m-1=\tfrac{n-2}{2}.$

This implies $\Delta=(n-2)/2p$.

  2. Let $n$ be odd and $n-2<p$ then $n=2m+1$ for some integer $m\geq 1$. Now

$n_{0}=\lceil\tfrac{(2m-1)(p-1)}{2p}\rceil=\lceil\tfrac{2m(p-1)}{2p}-\tfrac{p-1}{2p}\rceil=\lceil
m-\tfrac{2m+p-1}{2p}\rceil=m=\tfrac{n-1}{2},$

where the second last equality follows as $0<(2m+p-1)/2p<1$. This implies
$\Delta=(n-2)/2p+1/2$.

Putting the three possible values of $t$ in (3.3) we get the following three
inequalities

  1. (1)

If $-t=n_{0}-1$ then we have $0\leq\tfrac{a}{q}+\Delta-1\leq
1+\frac{n-2}{p}-\frac{n}{q}<1+\frac{n-2}{p}$.

  2. (2)

If $-t=n_{0}$ then we have $0\leq\tfrac{a}{q}+\Delta\leq
1+\frac{n-2}{p}-\frac{n}{q}<1+\frac{n-2}{p}$.

  3. (3)

If $-t=n_{0}+1$ then we have
$0\leq\tfrac{a}{q}+\Delta\leq\frac{n-2}{p}-\frac{n}{q}<\frac{n-2}{p}$.

From these three equalities it is easy to obtain the assertions (1) (a), (1)
(b), (1) (c), (2) (a) and (2) (b). ∎

######  Notations 3.3.

Fix an integer $n\geq 3$. Then for an integer $a\geq 1-n$

$L_{a}=(2a+n)\frac{(a+n-1)\cdots(a+1)}{n!}=\frac{2a^{n}}{n!}+O(a^{n-1}).$

We note that $L_{a}=0$, for $1-n\leq a\leq-1$. Let $q=p^{s}$, where $s\geq 1$.
For $0\leq i\leq n_{0}+1$, we define iteratively the integers $Z_{-i}(a,q)$ as
follows:

$Z_{0}(a,q)=Z_{0}(a)=L_{a},\quad\quad Z_{-1}(a,q)=Z_{0}(a+q)-L_{1}Z_{0}(a)$

and in general

$Z_{-i}(a,q)=Z_{0}(a+iq)-\left[L_{1}Z_{-i+1}(a,q)+L_{2}Z_{-i+2}(a,q)+\cdots+L_{i}Z_{0}(a)\right].$

Similarly, for $n_{0}+1\leq i\leq n-1$, we define iteratively another set of
integers $Y_{-i}(a,q)$ as follows:

$Y_{-n+1}(a,q)=L_{q-a-n}$

and for $n_{0}+1\leq i<n-1$

$Y_{-i}(a,q)=L_{(n-i)q-a-n}-\left[L_{n-i-1}Y_{-n+1}(a,q)+\cdots+L_{1}Y_{-i-1}(a,q)\right].$

######  Remark 3.4.

By construction it follows that there exists unique rational numbers
$\\{r_{ij},s_{ik}\\}_{j,k}$ such that for all $a\in{\mathbb{Z}}$ and $q=p^{s}$
we have $Z_{0}(a)=L_{a}$ and for $0\leq i\leq n_{0}+1$

$Z_{-i}(a,q)=r_{i0}L_{a}+r_{i1}L_{a+q}+\cdots+r_{i(i-1)}L_{a+(i-1)q}+L_{a+iq}$

and

$Y_{-i}(a,q)=s_{i0}L_{q-a-n}+s_{i1}L_{2q-a-n}+\cdots+s_{i(n-i-2)}L_{(n-i-1)q-a-n}+L_{(n-i)q-a-n}.$

Now let $\\{{\bf Z}_{i}(x)\\}_{0\leq i\leq n_{0}+1}$ and $\\{{\bf
Y}_{i}(x)\\}_{n_{0}+1\leq i<n}$ denote the two sets of polynomials, where

${\bf
Z}_{i}(x)=\frac{2}{n!}\left[r_{i0}(x-i)^{n}+r_{i1}(x-i+1)^{n}+\cdots+r_{i,i-1}(x-1)^{n}+(x)^{n}\right]$

and

${\bf
Y}_{i}(x)=\frac{2}{n!}\left[s_{i0}(i+1-x)^{n}+s_{i1}(i+2-x)^{n}+\cdots+s_{i,n-i-2}(n-1-x)^{n}+(n-x)^{n}\right].$

Since the set ${\mathbb{Z}}[1/p]$ is a dense subset of ${\mathbb{R}}$ we have
$\lim_{q\to\infty}{\lfloor xq\rfloor}/{q}=x$ for $x\geq 0$. This implies

${\bf Z}_{i}(x+i)=\lim_{q\to\infty}\frac{Z_{-i}(\lfloor
xq\rfloor,q)}{q^{n}}\quad\mbox{and}\quad{\bf
Y}_{i}(x+i)=\lim_{q\to\infty}\frac{Y_{-i}(\lfloor xq\rfloor,q)}{q^{n}}$

for $x\in[0,1)$.

######  Lemma 3.5.

Let $p$ be an odd prime such that $p>n-2$. Then for given integer $1-n\leq
a<q=p^{s}$

  1. (1)

$\nu^{s}_{-i}(a)+2\lambda_{0}\mu^{s}_{-i+1}(a)=Z_{-i}(a,p^{s})$, if $0\leq
i\leq n_{0}$.

  2. (2)

$\nu^{s}_{-n_{0}-1}(a)+2\lambda_{0}\mu^{s}_{-n_{0}}(a)=Z_{-n_{0}-1}(a,p^{s})+2\lambda_{0}\mu^{s}_{-n_{0}+1}(a)$.

  3. (3)

$\nu^{s}_{-i}(a)+2\lambda_{0}\mu^{s}_{-i+1}(a)=Y_{-i}(a,p^{s})$, if
$n_{0}+3\leq i\leq n-1$.

  4. (4)

$\nu^{s}_{-n_{0}-2}(a)=Y_{-n_{0}-2}(a,p^{s})$.

  5. (5)

$\nu^{s}_{-n_{0}-1}(a)=Y_{-n_{0}-1}(a,p^{s})-2\lambda_{0}\mu^{s}_{-n_{0}-1}(a)$.

  6. (6)

$\nu^{s}_{-i}(a)=0$, for $i\geq n$ and $\mu^{s}_{-j}(a)=0$ if
$j\not\in\\{n_{0}+1,n_{0},n_{0}-1\\}$.

###### Proof.

We fix $1-n\leq a<q=p^{s}$.

Note that the assertion (6) is just the assertions (1) and (2) of Lemma 3.2.

Now, by Lemma 3.2

$F^{s}_{*}({\mathcal{O}}(a))={\mathcal{O}}(-n+1)^{\nu^{s}_{-n+1}(a)}\oplus\cdots\oplus{\mathcal{O}}(-1)^{\nu^{s}_{-1}(a)}\oplus{\mathcal{O}}^{\nu^{s}_{0}(a)}\\\
\oplus{\mathcal{S}}(-n_{0}+1)^{\mu^{s}_{-n_{0}+1}(a)}\oplus{\mathcal{S}}(-n_{0})^{\mu^{s}_{-n_{0}}(a)}\oplus{\mathcal{S}}(-n_{0}-1)^{\mu^{s}_{-n_{0}-1}(a)}.$

Tensoring the above equation by ${\mathcal{O}}(i)$ and by the projection
formula, we get

$F^{s}_{*}({\mathcal{O}}(a+iq))={\mathcal{O}}(i-n+1)^{\nu^{s}_{-n+1}(a)}\oplus\cdots\oplus{\mathcal{O}}(i-1)^{\nu^{s}_{-1}(a)}\oplus{\mathcal{O}}(i)^{\nu^{s}_{0}(a)}\\\
\
\oplus{\mathcal{S}}(i-n_{0}+1)^{\mu^{s}_{-n_{0}+1}(a)}\oplus{\mathcal{S}}(i-n_{0})^{\mu^{s}_{-n_{0}}(a)}\oplus{\mathcal{S}}(i-n_{0}-1)^{\mu^{s}_{-n_{0}-1}(a)}.$

By (2.3), $h^{0}(Q,{\mathcal{O}}(m))=0$ for $m\leq-1$ and
$h^{0}(Q,{\mathcal{S}}(m))=0$ for $m\leq 0$. Also
$h^{0}(Q,{\mathcal{O}}(m))=L_{m}$ for all $m\geq 1-n$. Hence applying the
functor $H^{0}(Q,-)$ to the above decomposition we get

(3.4) $\begin{split}\nu^{s}_{0}(a)=&L_{a}=Z_{0}(a,p^{s})\quad\mbox{and}\\\
\nu^{s}_{-1}(a)=&L_{a+q}-L_{1}L_{a}=Z_{-1}(a,p^{s}).\\\ \end{split}$

In general, for $i\leq n_{0}-1$,

$L_{a+iq}=L_{0}\nu^{s}_{-i}(a)+L_{1}\nu^{s}_{-i+1}(a)+\cdots+L_{i}\nu^{s}_{0}(a)$

which implies

$\nu^{s}_{-i}(a)=\nu^{s}_{-i}(a)+2\lambda_{0}\mu^{s}_{-i+1}(a)=Z_{-i}(a,p^{s}).$

Let $i=n_{0}$, Since $h^{0}(Q,{\mathcal{S}}(1))=2\lambda_{0}$ we have

$\begin{array}[]{lcl}L_{a+n_{0}q}&=&L_{0}\nu^{s}_{-n_{0}}(a)+L_{1}\nu^{s}_{-n_{0}+1}(a)+\cdots+L_{n_{0}}\nu^{s}_{0}(a)+2\lambda_{0}\mu^{s}_{-n_{0}+1}(a)\\\
\\\
&=&\nu^{s}_{-n_{0}}(a)+L_{1}Z_{-n_{0}+1}(a,p^{s})+\cdots+L_{n_{0}}Z_{0}(a,p^{s})+2\lambda_{0}\mu^{s}_{-n_{0}+1}(a)\end{array}$

which implies

(3.5)
$\nu^{s}_{-n_{0}}(a)+2\lambda_{0}\mu^{s}_{-n_{0}+1}(a)=Z_{-n_{0}}(a,p^{s}).$

This proves assertion (1).

Let $i=n_{0}+1$. Since $h^{0}(Q,{\mathcal{S}}(2))=2\lambda_{0}(L_{1}-L_{0})$,
we have

$L_{a+(n_{0}+1)q}=L_{0}\nu^{s}_{-n_{0}-1}(a)+L_{1}\nu^{s}_{-n_{0}}(a)+\cdots+L_{n_{0}+1}\nu^{s}_{0}(a)\\\
+2\lambda_{0}\left[(L_{1}-L_{0})\mu^{s}_{-n_{0}+1}(a)+\mu^{s}_{-n_{0}}(a)\right]\\\
=(\nu^{s}_{-n_{0}-1}(a)+2\lambda_{0}\mu^{s}_{-n_{0}}(a))+\left[L_{1}(\nu^{s}_{-n_{0}}(a)+2\lambda_{0}\mu^{s}_{-n_{0}-1}(a))\right.\\\
\left.+L_{2}\nu^{s}_{-n_{0}+1}(a)+\cdots+L_{n_{0}+1}\nu^{s}_{0}(a)\right]-2\lambda_{0}\mu^{s}_{-n_{0}+1}(a)$

which implies, by assertion (1) of the lemma,

(3.6)
$\nu^{s}_{-n_{0}-1}(a)+2\lambda_{0}\mu^{s}_{-n_{0}}(a)=Z_{-n_{0}-1}(a,p^{s})+2\lambda_{0}\mu^{s}_{-n_{0}+1}.$

This proves assertion (2).

Now if we tensor the decomposition of $F^{s}_{*}({\mathcal{O}}(a))$ by
${\mathcal{O}}(-j)$, where $1\leq j\leq n-n_{0}-1$, then we get

$F^{s}_{*}({\mathcal{O}}(a-jq))={\mathcal{O}}(-j-n+1)^{\nu^{s}_{-n+1}(a)}\oplus\cdots\oplus{\mathcal{O}}(-j-1)^{\nu^{s}_{-1}(a)}\oplus{\mathcal{O}}(-j)^{\nu^{s}_{0}(a)}\\\
\oplus{\mathcal{S}}(-j-n_{0}+1)^{\mu^{s}_{-n_{0}+1}(a)}\oplus{\mathcal{S}}(-j-n_{0})^{\mu^{s}_{-n_{0}}(a)}\oplus{\mathcal{S}}(-j-n_{0}-1)^{\mu^{s}_{-n_{0}-1}(a)}.$

Now applying the functor $H^{n}(Q_{n},-)$ and then using the Serre duality

$h^{n}(Q,{\mathcal{O}}_{Q}(m))=h^{0}(Q,{\mathcal{O}}_{Q}(-m-n))=L_{-m-n}$

and $h^{n}(Q,{\mathcal{S}}(m))=h^{0}(Q,{\mathcal{S}}(1-m-n))$ (here $jq-
a-n\geq 1-n$) we get

$L_{jq-
a-n}=L_{j-1}\nu_{-n+1}^{s}(a)+L_{j-2}\nu_{-n+2}^{s}(a)+\cdots+L_{0}{\nu^{s}}_{-n+j}(a)\\\
+\mu^{s}_{-n_{0}+1}(a)h^{0}(Q_{n},S(n_{0}+j-n))+\mu^{s}_{-n_{0}}(a)h^{0}(Q_{n},S(n_{0}+j+1-n))\\\
+\mu^{s}_{-n_{0}-1}(a)h^{0}(Q_{n},S(n_{0}+j+2-n)).$

Hence $1\leq j\leq n-n_{0}-2$, we get

$L_{jq-
a-n}=L_{j-1}\nu_{-n+1}^{s}(a)+L_{j-2}\nu_{-n+2}^{s}(a)+\cdots+L_{0}\nu_{-n+j}^{s}(a).$

Therefore

$\nu^{s}_{-n+j}(a)=L_{jq-
a-n}-\left[L_{j-1}\nu^{s}_{-n+1}(a)+\cdots+L_{2}\nu^{s}_{-n+j-2}(a)+L_{1}\nu^{s}_{-n+j-1}(a)\right]\\\
=Y_{-n+j}(a,p^{s}).$

(3.7) $\begin{split}\mbox{For}\quad j=1,\quad\nu^{s}_{-n+1}(a)=&L_{q-a-n},\\\
\mbox{For}\quad
j=2,\quad\nu^{s}_{-n+2}(a)=&L_{2q-a-n}-L_{1}\nu^{s}_{-n+1}(a).\end{split}$

In general

$\begin{array}[]{lcl}\nu^{s}_{-i}(a)&=&\nu^{s}_{-i}(a)+2\lambda_{0}\mu^{s}_{-i+1}(a)=Y_{-i}(a,p^{s}),\quad{for}\quad
n_{0}+3\leq i\leq n-1\\\ \
\nu^{s}_{-n_{0}-2}(a)&=&Y_{-n_{0}-2}(a,p^{s}).\end{array}$

This proves assertions (3) and (4).

For $j=n-n_{0}-1$ we get

$L_{(n-n_{0}-1)q-a-n}=L_{n-n_{0}-2}\nu^{s}_{-n+1}(a)+\cdots+L_{0}\nu^{s}_{-n_{0}-1}(a)+2\lambda_{0}\mu^{s}_{-n_{0}-1}(a).$

(3.8)
$\nu^{s}_{-n_{0}-1}(a)=Y_{-n_{0}-1}(a,p^{s})-2\lambda_{0}\mu^{s}_{-n_{0}-1}(a).$

This proves assertion (5). ∎

######  Remark 3.6.

By Lemma 3.5 it follows that if there is an interval $I_{1}\subset[0,1)$ with
the property that $a/p^{s}\in I_{1}$ implies there is at the most one type of
spinor bundle in the decomposition of $F^{s}_{*}({\mathcal{O}}(a))$, then all
the functions $\nu^{s}_{-i}(a)$ and $\mu^{s}_{-i+1}(a)$ have polynomial
expressions for $a/q\in I_{1}$. More precisely we have the following.

######  Lemma 3.7.

Let $0\leq a<q=p^{s}$, where $p\neq 2$ is a prime such that $p>n-2$ and $n\geq
3$ is an integer. Let $n_{0}=\lceil\tfrac{n}{2}\rceil-1$. Then we have

$F_{*}^{s}({\mathcal{O}}(a))=\bigoplus_{t=0}^{n-1}{\mathcal{O}}(-t)^{\nu^{s}_{-t}(a)}\oplus\bigoplus_{t=n_{0}-1}^{n_{0}+1}{\mathcal{S}}(-t)^{\mu^{s}_{-t}(a)}.$

Case (1). Let $n\geq 3$ be odd then

  1. $\begin{array}[]{lcl}0\leq a/q<1&\implies&\nu^{s}_{-i}(a)=\begin{cases}Z_{-i}(a,p^{s})&\quad{for}\quad 0\leq i\leq n_{0}-1\\\ \ Y_{-i}(a,p^{s})&\quad{for}\quad n_{0}+1\leq i<n\end{cases}\\\ &&\mu^{s}_{-i}(a)=0\quad{for}\quad i\neq n_{0}-1,n_{0},\end{array}$

  2. $\begin{array}[]{lcl}a/q\in[0,\frac{1}{2}-\frac{n-2}{2p}]\implies\begin{cases}\nu^{s}_{-n_{0}}(a)=Z_{-n_{0}}(a,p^{s})\\\ \mu^{s}_{-n_{0}+1}(a)=0\\\ \mu^{s}_{-n_{0}}(a)=\frac{1}{2\lambda_{0}}\left[Z_{-n_{0}-1}(a,p^{s})-Y_{-n_{0}-1}(a,p^{s})\right],\end{cases}\end{array}$

  3. $\begin{array}[]{lcl}a/q\in[\frac{1}{2}+\frac{n-2}{2p},1)&\implies&\begin{cases}\nu^{s}_{-n_{0}}(a)=Z_{-n_{0}}(a,p^{s})-Y_{-n_{0}-1}(a,p^{s})+Z_{-n_{0}-1}(a,p^{s})\\\ \mu^{s}_{-n_{0}+1}(a)=\frac{1}{2\lambda_{0}}\left[Y_{-n_{0}-1}(a,p^{s})-Z_{-n_{0}-1}(a,p^{s})\right]\\\ \mu^{s}_{-n_{0}}(a)=0.\end{cases}\end{array}$

Case (2). If $n\geq 4$ is even then

  1. $\begin{array}[]{lcl}0\leq a/q<1&\implies&\nu^{s}_{-i}(a)=\begin{cases}Z_{-i}(a,p^{s})&\quad{for}\quad 0\leq i\leq n_{0}-1\\\ \ Y_{-i}(a,p^{s})&\quad{for}\quad n_{0}+2\leq i<n\end{cases}\\\ &&\mu^{s}_{-i}(a)=0\quad{for}\quad i\neq n_{0}-1,n_{0},n_{0}+1,\end{array}$

  2. $\begin{array}[]{lcl}a/q\in[\tfrac{n-2}{p},\quad 1-\tfrac{n-2}{p})\implies\begin{cases}\nu^{s}_{-n_{0}}(a)=Z_{-n_{0}}(a,p^{s})\\\ \nu^{s}_{-n_{0}-1}(a)=Y_{-n_{0}-1}(a,p^{s})\\\ \mu^{s}_{-n_{0}+1}(a)=0\\\ \mu^{s}_{-n_{0}}(a)=\frac{1}{2\lambda_{0}}\bigl{[}Z_{-n_{0}-1}(a,p^{s})-Y_{-n_{0}-1}(a,p^{s})\bigr{]},\end{cases}\end{array}$

  3. $a/q\in[\tfrac{n-2}{p},\quad 1)\implies\mu^{s}_{-n_{0}-1}(a)=0.$

###### Proof.

We prove the lemma for the case when $n$ is odd, as the proof is similar when
$n$ is even.

Now by Lemma 3.2 (2) we have $\mu^{s}_{-j}(a)=0$ if
$j\not\in\\{n_{0}-1,n_{0}\\}$. Therefore by Lemma 3.5 (1) for $0\leq i\leq
n_{0}-1$

$\nu^{s}_{-i}(a)=\nu^{s}_{-i}(a)+2\lambda_{0}\mu^{s}_{-i+1}(a)=Z_{-i}(a,p^{s}).$

and for $n_{0}+2\leq i<n$, by Lemma 3.5 (3) and (4),

$\nu^{s}_{-i}(a)=Y_{-i}(a,p^{s}).$

If $i=n_{0}+1$ then by Lemma 3.5 (5)

$\nu^{s}_{-n_{0}-1}(a)=Y_{-n_{0}-1}(a,p^{s})-2\lambda_{0}\mu^{s}_{-n_{0}-1}(a)=Y_{-n_{0}-1}(a,p^{s}).$

Assertion (1). If $a/q\in[0,\tfrac{1}{2}-\tfrac{n-2}{2p}]$ then
$\mu_{-n_{0}+1}^{s}(a)=0$, by Lemma 3.2 (2)(b).

This gives, by Lemma 3.5 (1)

$\nu^{s}_{-n_{0}}(a)=\nu^{s}_{-n_{0}}(a)+\mu^{s}_{-n_{0}+1}(a)=Z_{-n_{0}}(a,p^{s})$

and by Lemma 3.5 (2)

$2\lambda_{0}\mu^{s}_{-n_{0}}(a)=Z_{-n_{0}-1}(a,p^{s})-\nu^{s}_{-n_{0}-1}(a),$

whereas by Lemma 3.5 (5), $\nu^{s}_{-n_{0}-1}(a)=Y_{-n_{0}-1}(a,p^{s})$.

Assertion (2). If $a/q\in[\tfrac{1}{2}+\tfrac{n-2}{2p},~{}~{}1]$ then
$\mu^{s}_{-n_{0}}(a)=0$, by Lemma 3.2 (2)(a). By Lemma 3.5 (2) and (5),

$Y_{-n_{0}-1}(a,p^{s})=\nu^{s}_{-n_{0}-1}(a)=Z_{-n_{0}-1}(a,p^{s})+2\lambda_{0}\mu^{s}_{-n_{0}+1}(a)$

which implies
$\mu^{s}_{-n_{0}+1}(a)=\frac{1}{2\lambda_{0}}\left[L^{s}_{-n_{0}-1}(a)-Z^{s}_{-n_{0}-1}(a)\right]$
and

$\begin{array}[]{lcl}\nu^{s}_{-n_{0}}(a)&=&Z_{-n_{0}}(a,p^{s})-2\lambda_{0}\mu^{s}_{-n_{0}+1}(a)\\\
&=&Z_{-n_{0}}(a,p^{s})-Y_{-n_{0}-1}(a,p^{s})+Z_{-n_{0}-1}(a,p^{s}).\end{array}$

∎

Now using the fact that the HK density function $f_{R_{p,n+1}}$ is continuous
and is the limit function of the sequence of functions
$\\{f_{s}(R,I)\\}_{s\in{\mathbb{N}}}$, we deduce that the rank functions are
well defined and continuous.

######  Lemma 3.8.

The functions
${\boldsymbol{\nu}}_{i}^{(p)}:{\mathbb{R}}\longrightarrow{\mathbb{R}}$ and
${\boldsymbol{\mu}}_{i}^{(p)}:{\mathbb{R}}\longrightarrow{\mathbb{R}}$ given
by

${\boldsymbol{\nu}}_{i}^{(p)}(x)=\lim_{s\to\infty}\frac{\nu_{-i}^{s}(\lfloor
xq\rfloor)}{q^{n}}\quad\mbox{and}\quad{\boldsymbol{\mu}}_{i}^{(p)}(x)=\lim_{s\to\infty}2\lambda_{0}\frac{\mu_{-i}^{s}(\lfloor
xq\rfloor)}{q^{n}}$

are well defined continuous functions.

Moreover, for each $i$ we have $0\leq{\boldsymbol{\nu}}_{i}^{(p)}(x)\leq 1$
and $0\leq{\boldsymbol{\mu}}_{i}^{(p)}(x)\leq 2$.

###### Proof.

By (2.6), it is enough to prove the assertion for all
${\boldsymbol{\nu}}_{i}^{(p)}$ and ${\boldsymbol{\mu}}_{i}^{(p)}$ restricted
to the interval $[0,1)$.

By Theorem 1.1 of [T1] (see Definition 2.1) the sequence
$\\{f_{s}(R_{p,n+1},{\bf m})\\}_{s\in{\mathbb{N}}}$ converges uniformly to the
HK density function $f_{R_{p,n+1}}$. Therefore

$f_{R_{p,n+1}}(x)=\lim_{s\to\infty}f_{s}(R_{p,n+1},{\bf
m})(x)=\lim_{s\to\infty}\ell(R_{p,n+1}/{\bf m}^{[q]})_{\lfloor
xq\rfloor}/q^{n}.$

On the other hand if $x\in[0,1)$ then by Lemma 2.3 and Lemma 2.4

(3.9) $\ell(R_{p,n+1}/{\bf
m}^{[q]})_{\lfloor(x+i)q\rfloor}/q^{n}=\frac{1}{q^{n}}\left(\nu^{s}_{-i}(\lfloor
xq\rfloor)+2\lambda_{0}\mu^{s}_{-i+1}(\lfloor xq\rfloor)\right).$

By Lemma 3.7, $\mu^{s}_{-i+1}(\lfloor xq\rfloor)=0$ if
$i\not\in\\{n_{0},n_{0}+1,n_{0}+2\\}$. Hence

${\boldsymbol{\nu}}_{i}^{(p)}(x)=\lim_{s\to\infty}\frac{\nu^{s}_{-i}(\lfloor
xq\rfloor)}{q^{n}}=f_{R_{p,n+1}}(x+i)$

is a well defined continuous function. We can argue for other rank functions
as follows. By Lemma 3.5 (4)

${\boldsymbol{\nu}}_{n_{0}+2}^{(p)}(x)=\lim_{s\to\infty}\frac{\nu_{-n_{0}-2}^{s}(\lfloor
xq\rfloor)}{q^{n}}=\lim_{s\to\infty}\frac{Y_{-n_{0}-2}(\lfloor
xq\rfloor,q)}{q^{n}}={\bf Y}_{n_{0}+2}(x+n_{0}+2).$

By (3.9)

$\frac{\mu^{s}_{-n_{0}-1}(\lfloor xq\rfloor)}{q^{n}}=\frac{\ell(R_{p,n+1}/{\bf
m}^{[q]})_{\lfloor(x+n_{0})q\rfloor}}{q^{n}}-\frac{\nu^{s}_{-n_{0}-2}(\lfloor
xq\rfloor)}{q^{n}}.$

Since both the terms on the right hand side has limit as $s\to\infty$ we get

(3.10) ${\boldsymbol{\mu}}_{n_{0}+1}^{(p)}(x)=f_{R_{p,n+1}}(x+n_{0}+2)-{\bf
Y}_{n_{0}+2}(x+n_{0}+2)$

is a well defined and continuous function.

Using similar arguments and Lemma 3.5, we can show that the functions
${\boldsymbol{\nu}}_{n_{0}+1}^{(p)}$, ${\boldsymbol{\mu}}_{n_{0}}^{(p)}$,
${\boldsymbol{\mu}}_{n_{0}-1}^{(p)}$ then ${\boldsymbol{\nu}}_{n_{0}}^{(p)}$
are well defined and continuous:

${\boldsymbol{\nu}}_{n_{0}+1}^{(p)}(x)={\bf
Y}_{n_{0}+1}(x+n_{0}+1)-{\boldsymbol{\mu}}_{n_{0}+1}^{(p)}(x),\quad\mbox{by
Lemma~{}\ref{l2}~{}(5)},$

${\boldsymbol{\mu}}_{n_{0}}^{(p)}(x)=f_{R_{p,n+1}}(x+n_{0}+1)-{\boldsymbol{\nu}}_{n_{0}+1}^{(p)}(x),\quad\mbox{by}~{}(\ref{conte})$

(3.11)
${\begin{array}[]{lcl}{\boldsymbol{\mu}}_{n_{0}-1}^{(p)}(x)&=&{\boldsymbol{\nu}}_{n_{0}+1}^{(p)}(x)+{\boldsymbol{\mu}}_{n_{0}}^{(p)}(x)-{\bf
Z}_{n_{0}+1}(x+n_{0}+1),\quad\mbox{by Lemma~{}\ref{l2}~{}(2)}\\\
&=&f_{R_{p,n+1}}(x+n_{0}+1)-{\bf Z}_{n_{0}+1}(x+n_{0}+1).\end{array}}$

and

${\boldsymbol{\nu}}_{n_{0}}^{(p)}(x)=f_{R_{p,n+1}}(x+n_{0})-{\boldsymbol{\mu}}_{n_{0}-1}^{(p)}(x).$

Hence all the ranks functions are well defined continuous functions.

To prove the second assertion of the lemma. we compute the ranks of the
bundles on both the sides of (3.1) and (3.2), and use the fact that the spinor
bundle ${\mathcal{S}}$ has rank $\lambda_{0}$. This gives

$p^{sn}=\nu_{0}^{s}(a)+\cdots+\nu_{-n+1}^{s}(a)+\lambda_{0}\bigl{[}\mu_{-n_{0}+1}^{s}(a)+\mu_{-n_{0}}^{s}(a)+\mu_{-n_{0}-1}^{s}(a)\bigr{]}.$

In particular, we have
$0\leq\nu^{s}_{-i}(a),~{}~{}\lambda_{0}\mu^{s}_{-i}(a)\leq q^{n}=p^{sn}$.
Therefore

${{\boldsymbol{\mu}}}_{i}^{(p)}(x)=2\lambda_{0}\lim_{s\to\infty}{\mu^{s}_{-i}(\lfloor
xq\rfloor)}/{q^{n}}\leq
2\quad\mbox{and}\quad{{\boldsymbol{\nu}}}_{i}^{(p)}(x)=\lim_{s\to\infty}{\nu^{s}_{-i}(\lfloor
xq\rfloor)}/{q^{n}}\leq 1.$

∎

######  Definition 3.9.

We call ${\boldsymbol{\nu}}_{i}^{(p)}$ and ${\boldsymbol{\mu}}_{i}^{(p)}$ the
rank functions of $Q_{n}$ or of $R_{p,n+1}$.

We say ${\boldsymbol{\nu}}_{i}^{(p)}$ is polynomial in the interval
$I_{1}\subset[0,1)$ if there exists a polynomial $F_{i}(X)\in{\mathbb{Q}}[x]$
such that ${\boldsymbol{\nu}}_{i}^{(p)}(x)=F_{i}(x)$, for all $x\in I_{1}$. We
can define the similar notion for ${\boldsymbol{\mu}}^{(p)}_{i}$.

In the following Proposition, which is one of the key steps in the paper, we
describe the range where these functions are in fact polynomial functions and
hence are independent of $p$ though the range itself might depend on $p$.

######  Proposition 3.10.

Let $p>2$ such that $p>n-2$. If $n\geq 4$ is even then for $x\in[0,1)$ and for
$i\not\in\\{n_{0}-1,n_{0},n_{0}+1\\}$ the function
${\boldsymbol{\mu}}_{i}^{(p)}(x)=0$, and

  1. (1)

${\boldsymbol{\nu}}_{i}^{(p)}(x)+{\boldsymbol{\mu}}_{i-1}^{(p)}(x)={\bf
Z}_{i}(x+i)$, if $x\in[0,1)$ and $0\leq i\leq n_{0}$.

  2. (2)

${\boldsymbol{\nu}}_{n_{0}+1}^{(p)}(x)+{\boldsymbol{\mu}}_{n_{0}}^{(p)}(x)={\bf
Z}_{n_{0}+1}(x+n_{0}+1)$, if $x\in[0,1-\tfrac{n-2}{2p})$.

  3. (3)

${\boldsymbol{\nu}}^{(p)}_{n_{0}+2}(x)={\bf Y}_{n_{0}+2}(x+n_{0}+2)$, if
$x\in[0,1)$.

  4. (4)

${\boldsymbol{\nu}}_{n_{0}+2}^{(p)}(x)+{\boldsymbol{\mu}}_{n_{0}+1}^{(p)}(x)={\bf
Y}_{n_{0}+2}(x+n_{0}+2)$, if $x\in[\tfrac{n-2}{2p},1))$.

  5. (5)

${\boldsymbol{\nu}}_{i}^{(p)}(x)+{\boldsymbol{\mu}}_{i-1}^{(p)}(x)={\boldsymbol{\nu}}_{i}^{(p)}(x)={\bf
Y}_{i}(x+i)$, if $x\in[0,1)$ and $n_{0}+3\leq i<n$.

If $n\geq 3$ is odd then for $x\in[0,1)$ and for $i\not\in\\{n_{0}-1,n_{0}\\}$
the function ${\boldsymbol{\mu}}_{i}^{(p)}(x)=0$, and

  1. (1)

${\boldsymbol{\nu}}_{i}^{(p)}(x)+{\boldsymbol{\mu}}_{i-1}^{(p)}(x)={\bf
Z}_{i}(x+i)$, if $x\in[0,1)$ and $0\leq i\leq n_{0}$.

  2. (2)

${\boldsymbol{\nu}}_{n_{0}+1}^{(p)}(x)+{\boldsymbol{\mu}}_{n_{0}}^{(p)}(x)={\bf
Z}_{n_{0}+1}(x+n_{0}+1)$, if $x\in[0,\frac{1}{2}-\tfrac{n-2}{2p})$.

  3. (3)

${\boldsymbol{\nu}}_{n_{0}+1}^{(p)}(x)+{\boldsymbol{\mu}}_{n_{0}}^{(p)}(x)={\bf
Y}_{n_{0}+1}(x+n_{0}+1)$, if $x\in[\tfrac{1}{2}+\tfrac{n-2}{2p},1))$.

  4. (4)

${\boldsymbol{\nu}}^{(p)}_{n_{0}+1}(x)={\bf Y}_{n_{0}+1}(x+n_{0}+1)$, if
$x\in[0,1)$.

  5. (5)

${\boldsymbol{\nu}}^{(p)}_{n_{0}+2}(x)={\bf Y}_{n_{0}+2}(x+n_{0}+2)$, if
$x\in[0,1)$.

  6. (6)

${\boldsymbol{\nu}}_{i}^{(p)}(x)+{\boldsymbol{\mu}}_{i-1}^{(p)}(x)={\boldsymbol{\nu}}_{i}^{(p)}(x)={\bf
Y}_{i}(x+i)$, if $x\in[0,1)$ and $n_{0}+2\leq i<n$.

Moreover for both even and odd $n$,
${\boldsymbol{\nu}}_{i}^{(p)}(x)={\boldsymbol{\mu}}_{i-1}^{(p)}(x)=0$, for all
$x\in[0,1)$ and $i\geq n$.

###### Proof.

For both parities of $n$ and for $i=n_{0}$ the assertion (1) follows from
(3.5). Lemma 3.7 gives the vanishing of $\mu^{s}_{-n_{0}+1}(a)$ for $a$ in the
certain range. This applied to (3.11) gives assertion (2). The rest of the
assertions follow from Lemma 3.7, Lemma 3.5 along with Remark 3.4.∎

##  4\. The HK density function $f_{R_{p,n+1}}$ and $f_{R^{\infty}_{n+1}}$

Now we are ready to give the first formulation of the HK density function
$f_{R_{p,n+1}}$ provided $p>n-2$.

We follow the Notations 3.1. By Lemma 3.2, for $q=p^{s}$ and an integer $0\leq
a<q$, there exist integers $\nu^{s}_{t}(a)$ and $\mu^{s}_{t}(a)$ given by the
decomposition

$F_{*}^{s}({\mathcal{O}}(a))=\oplus_{t=0}^{n-1}{\mathcal{O}}(-t)^{\nu^{s}_{-t}(a)}\oplus\oplus_{t=-n_{0}+1}^{-n_{0}-1}{\mathcal{S}}(t)^{\mu^{s}_{t}(a)}.$

Proof of Proposition 1.1. If $x\in[i,i+1)$ then by Lemma 2.4

$f_{R_{p,n+1}}(x)={\boldsymbol{\nu}}_{i}^{(p)}(x-i)+{\boldsymbol{\mu}}_{i-1}^{(p)}(x-i).$

Case (A). Suppose $n\geq 4$ is an even number.

Case (a). Let $i\in\\{0,1,\ldots,n_{0}\\}$. Then by Proposition 3.10

$f_{R_{p,n+1}}(x)={\boldsymbol{\nu}}_{i}^{(p)}(x-i)+{\boldsymbol{\mu}}_{i-1}^{(p)}(x-i)={\bf
Z}_{i}(x).$

Case (b). Let $i=n_{0}+1$ and

Sub case (b) (1). If $x\in[n_{0}+1,n_{0}+2-\tfrac{n-2}{2p})$, then by
Proposition 3.10 (2)

$f_{R_{p,n+1}}(x)={\boldsymbol{\nu}}_{n_{0}+1}^{(p)}(x-n_{0}-1)+{\boldsymbol{\mu}}_{n_{0}}^{(p)}(x-n_{0}-1)={\bf
Z}_{n_{0}+1}(x).$

Sub case (b) (2). If $x\in[n_{0}+2-\tfrac{n-2}{2p},n_{0}+2)$, then, by (3.11),

$f_{R_{p,n+1}}(x)={\bf
Z}_{n_{0}+1}(x)+{\boldsymbol{\mu}}_{n_{0}-1}^{(p)}(x-n_{0}-1).$

Case (c). Let $i=n_{0}+2$ then by (3.10)

$f_{R_{p,n+1}}(x)={\bf
Y}_{n_{0}+2}(x)+{\boldsymbol{\mu}}_{n_{0}+1}^{(p)}(x-n_{0}-2).$

If in addition $x\in[n_{0}+2+\tfrac{n-2}{2p},n_{0}+3)$ then the last assertion
of Lemma 3.7 implies ${\boldsymbol{\mu}}_{n_{0}+1}^{(p)}(x-n_{0}-2)=0$.

Case (d). Let $i\in\\{n_{0}+3,\cdots,n-1\\}$. Then by Proposition 3.10

$f_{R_{p,n+1}}(x)={\boldsymbol{\nu}}_{i}^{(p)}(x-i)+{\boldsymbol{\mu}}_{i-1}^{(p)}(x-i)={\bf
Y}_{i}(x)\quad\mbox{for}\quad x\in[i,i+1).$

Case (e). Let $i\geq n$. Then by Lemma 3.5 (6),

$f_{R_{p,n+1}}(x)={\boldsymbol{\nu}}_{i}^{(p)}(x-i)+{\boldsymbol{\mu}}_{i-1}^{(p)}(x-i)=0.$

This completes the proof of the theorem when $n$ is an even number.

Assume $n\geq 3$ is odd. Let $x\in[i,i+1)$ Then by Proposition 3.10 (1) and
(6)

$\begin{array}[]{lcl}f_{R_{p,n+1}}(x)&=&{\bf Z}_{i}(x),\quad\mbox{if}\quad
i\leq n_{0}\\\ &=&{\bf Y}_{i}(x),\quad\mbox{if}\quad i\geq
n_{0}+2.\end{array}$

If $i=n_{0}+1$ then

$f_{R_{p,n+1}}(x)={\boldsymbol{\nu}}^{(p)}_{n_{0}+1}(x-n_{0}-1)+{\boldsymbol{\mu}}^{(p)}_{n_{0}}(x-n_{0}-1),\quad\mbox{if}\quad
x-n_{0}-1\in[0,1).$

Now applying Proposition 3.10 (2), (4) and (3) we get

$\begin{array}[]{lcl}f_{R_{p,n+1}}(x)&=&{\bf
Z}_{n_{0}+1}(x),\quad\mbox{if}\quad
x-n_{0}-1\in[0,\tfrac{1}{2}-\tfrac{n-2}{2p})\\\ &=&{\bf
Y}_{n_{0}+1}(x)+{\boldsymbol{\mu}}^{(p)}_{n_{0}}(x-n_{0}-1),\quad\mbox{if}\quad
x-n_{0}-1\in[\tfrac{1}{2}-\tfrac{n-2}{2p},\quad\tfrac{1}{2}+\tfrac{n-2}{2p})\\\
&=&{\bf Y}_{n_{0}+1}(x),\quad\mbox{if}\quad
x-n_{0}-1\in[\tfrac{1}{2}+\tfrac{n-2}{2p},\quad 1).\end{array}$

This completes the proof of the proposition.$\Box$

######  Theorem 4.1.

The function $f^{\infty}_{R_{n+1}}:[0,\infty)\longrightarrow[0,\infty)$ given
by

$f^{\infty}_{R_{n+1}}(x):=\lim_{p\to\infty}f_{R_{p,n+1}}(x)$

is partially symmetric continuous function, that is

$f^{\infty}_{R_{n+1}}(x)=f^{\infty}_{R_{n+1}}(n-x),\quad\mbox{for}\quad 0\leq
x\leq(n-2)/2$

and is described as follows:

  1. (1)

If $n\geq 4$ is even then

$f^{\infty}_{R_{n+1}}(x)=\begin{cases}{\bf Z}_{i}(x)&\quad\mbox{if}\quad i\leq
x<i+1\quad\mbox{and}\quad 0\leq i\leq n_{0}+1\\\ \\\ {\bf
Y}_{i}(x)&\quad\mbox{if}\quad i\leq x<i+1\quad\mbox{and}\quad n_{0}+2\leq
i<n\end{cases}$

and $f^{\infty}_{R_{n+1}}(x)=0$ otherwise.

  2. (2)

If $n\geq 3$ is an odd number then

$f^{\infty}_{R_{n+1}}(x)=\begin{cases}{\bf Z}_{i}(x)&\quad\mbox{if}\quad i\leq
x<i+1\quad\mbox{and}\quad 0\leq i\leq n_{0}\\\ \\\ {\bf
Z}_{n_{0}+1}(x)&\quad\mbox{if}\quad(n_{0}+1)\leq x<(n_{0}+\frac{3}{2})\\\ \\\
{\bf Y}_{n_{0}+1}(x)&\quad\mbox{if}\quad(n_{0}+\frac{3}{2})\leq x<(n_{0}+2)\\\
\\\ {\bf Y}_{i}(x)&\quad\mbox{if}\quad i\leq x<i+1\quad\mbox{and}\quad
n_{0}+2\leq i<n\end{cases}$

and $f^{\infty}_{R_{n+1}}(x)=0$ otherwise.

###### Proof.

The description of the function
$f^{\infty}_{R_{n+1}}:[0,\infty)\longrightarrow[0,\infty)$ follows from
Proposition 1.1. To prove the symmetry, we consider the ${\bf
Z}_{j}:[0,\infty)\longrightarrow[0,\infty)$ and ${\bf
Y}_{j}:[0,\infty)\longrightarrow[0\,\infty)$ and

Claim. ${\bf Z}_{j}(x)={\bf Y}_{n-1-j}(n-x)$, if $j\leq x<j+1$, where $j\leq
n_{0}-1$.

Proof of the claim: By induction on $j\geq 0$, first we prove the assertion
that

$\lim_{q\to\infty}Z_{-j}(a,q)/q^{n}=\lim_{q\to\infty}Y_{-(n-1-j,q)}(q-a,q)/q^{n}\quad\mbox{for}\quad
0\leq a<q.$

If $j=0$ then

$\lim_{q\to\infty}Z_{0}(a,q)/q^{n}=\lim_{q\to\infty}L_{a}/q^{n}=\lim_{q\to\infty}L_{a+n}/q^{n}=\lim_{q\to\infty}Y_{-(n-1)}(q-a,q)/q^{n}.$

Assume that the assertion holds for all $j$ where $0\leq j<i\leq n_{0}-1$. Now

$\lim_{q\to\infty}\frac{Z_{-i}(a,q)}{q^{n}}=\lim_{q\to\infty}{L_{a}}/{q^{n}}-{\left[L_{1}Z_{-i+1}(a,q)+\cdots+L_{i}Z_{0}(a,q)\right]}/q^{n}\\\
\
=\lim_{q\to\infty}{L_{a+n}}/{q^{n}}-\left[L_{1}Y_{-(n-i)}(q-a,q)+\cdots+L_{i}Y_{-(n-1)}(q-a,q)\right]/q^{n}\\\
\ =\lim_{q\to\infty}{Y_{-(n-1-i)}(q-a,q)}/{q^{n}},$

where the second equality follows from the induction hypothesis.

Now to prove the claim, it is enough to prove for $x=m/q$, where
$m\in{\mathbb{Z}}_{\geq 0}$. If $j\leq x<j+1$ then $m=a+jq$, where $0\leq
a<q$. Now

${\bf
Z}_{j}(x)=\lim_{q\to\infty}Z_{-j}(m-jq,q)/q^{n}=\lim_{q\to\infty}Y_{-(n-1-j)}((j+1)q-m),q)/q^{n}\\\
\ =\lim_{q\to\infty}Y_{-(n-1-j)}((n-m)q-(nq-q-jq),q)/q^{n}={\bf
Y}_{n-1-j}(n-x).$

This proves the claim.

If $n$ is even then $n_{0}=(n/2)-1$. Let $0\leq x<(n-2)/2=n_{0}$ then $i\leq
x<(i+1)$ for some $0\leq i<n_{0}$. Now

$f^{\infty}_{R_{n+1}}(x)={\bf Z}_{i}(x)={\bf
Y}_{n-1-i}(n-x)=f^{\infty}_{R_{n+1}}(n-x).$

where the second equality follows as $n-(i+1)<n-x\leq n-i$.

If $n$ is odd then $n_{0}=(n-1)/2$. Let $0\leq x<(n-2)/2=n_{0}-(1/2)$.

If $i\leq x<(i+1)$, where $i<n_{0}$ then

$f^{\infty}_{R_{n+1}}(x)={\bf Z}_{i}(x)={\bf
Y}_{n-1-i}(n-x)=f^{\infty}_{R_{n+1}}(n-x).$

If $(n_{0}-1)\leq x<n_{0}-1/2$ then again

$f^{\infty}_{R_{n+1}}(x)={\bf Z}_{n_{0}-1}(x)={\bf
Y}_{n_{0}+1}(n-x)=f^{\infty}_{R_{n+1}}(n-x).$

∎

######  Remark 4.2.

The same argument as above proves that $f_{R_{p,n+1}}$ is partially symmetric
and the symmetry is given by

$f_{R_{p,n+1}}(x)=f_{R_{p,n+1}}(n-x)\quad\mbox{for}\quad 0\leq
x\leq\tfrac{n-2}{2}\left(1-\tfrac{1}{p}\right).$

Proof of Corollary 1.2. If $n$ is even then $n_{0}=\frac{n-2}{2}$ and if $n$
is odd then $n_{0}=\frac{n-1}{2}$. Therefore we can express the intervals as
follows

$n\quad\mbox{is
even}\implies\mbox{$\left[n_{0}+2-\frac{n-2}{2p},~{}~{}n_{0}+2+\frac{n-2}{2p}\right)=\left[\frac{n+2}{2}-\frac{n-2}{2p},~{}~{}\frac{n+2}{2}+\frac{n-2}{2p}\right)$}$
$n\quad\mbox{is
odd}\implies\mbox{$\left[n_{0}+\frac{3}{2}-\frac{n-2}{2p},~{}~{}n_{0}+\frac{3}{2}+\frac{n-2}{2p}\right)=\left[\frac{n+2}{2}-\frac{n-2}{2p},~{}~{}\frac{n+2}{2}+\frac{n-2}{2p}\right)$}.$

Now the corollary follows from Proposition 1.1 and Theorem 4.1. $\Box$

######  Theorem 4.3.

Let $p\neq 2$ and let $p>n-2$. Then

$1+m_{n+1}+\tfrac{2n-4}{p}\geq e_{HK}(R_{p,n+1},{\bf m})\geq 1+m_{n+1}.$

In fact,

  1. (1)

if $n\geq 3$ odd and $p\geq 3n-4$ or

  2. (2)

if $n\geq 4$ even and $p\geq(3n-4)/2$

then $e_{HK}(R_{p,n+1},{\bf m})>1+m_{n+1}$.

###### Proof.

We prove the theorem for the case when $n$ is odd, as the proof for the case
when $n$ is even is very similar.

By Proposition 1.1 and Theorem 4.1

$0\leq\int_{0}^{\infty}f_{R_{p,n+1}}(x)dx-\int_{0}^{\infty}f_{R^{\infty}_{n+1}}(x)dx=\int_{\tfrac{1}{2}-\tfrac{n-2}{2p}}^{\tfrac{1}{2}+\tfrac{n-2}{2p}}{\boldsymbol{\mu}}_{n_{0}}^{(p)}(x))dx\leq\frac{2n-4}{p},$

where the last inequality follows from Lemma 3.8. On the other hand by Theorem
1.1 of [T1] we have

$e_{HK}({R_{p,n+1}},{\bf m})=\int_{0}^{\infty}f_{{R_{p,n+1}},{\bf m}}(x)dx$

and by the result of Gessel-Monsky [GM],
$\lim_{p\to\infty}e_{HK}({R_{p,n+1}},{\bf m})=1+m_{n+1}$. Hence by dominated
convergence theorem

$\int_{0}^{\infty}f_{R^{\infty}_{n+1},{\bf
m}}(x)dx=\lim_{p\to\infty}\int_{0}^{\infty}f_{{R_{p,n+1}},{\bf
m}}(x)dx=1+m_{n+1}.$

The strict inequality assertion follows from Remark 8.5. ∎

######  Corollary 4.4.

Let $p>2$ be a prime number such that $p>n-2$ then the $F$-threshold of the
ring $R_{p,n+1}$ is $c^{\bf m}({\bf m})=n$.

###### Proof.

By Theorem E of [TW1], the $F$-threshold $c^{\bf m}({\bf
m})=\mbox{max}~{}\\{x\mid f_{R_{p,n+1}}(x)\neq 0\\}$.

Now, by Proposition 1.1, $f_{R_{p,n+1}}(x)=0$, for $x\geq n$ and for $n-1\leq
x<n$,

$f_{R_{p,n+1}}(x)={\bf Y}_{n-1}(x)=\lim_{q\to\infty}\frac{Y_{-n+1}(\lfloor
xq\rfloor-(n-1)q,q)}{q^{n}}=\frac{2(n-x)^{n}}{n!}\neq 0,$

where the last equality follows as $Y_{-n+1}(a,p^{s})=L_{q-a-n}$. This proves
the corollary. ∎

##  5\. Generating polynomials for $f_{R_{p,n+1}}$

In this and the next section we setup the ground to give second formulation of
the HK density function $f_{R_{p,n+1}}$.

Throughout the section we denote $n_{0}=\lceil\tfrac{n}{2}\rceil-1$ and,
unless stated otherwise, assume that the characteristic $p\geq 3n-4$ if $n$ is
odd and $p\geq(3n-4)/2$ if $n$ is even.

By Proposition 3.10 we know that the every function
${\boldsymbol{\nu}}_{i}^{(p)}+{\boldsymbol{\mu}}_{i-1}^{(p)}:[0,1)\longrightarrow[0,\infty)$
is a polynomial outside the difficult range which is

  1. (1)

$[\frac{1}{2}-\frac{n-2}{2p},\quad\frac{1}{2}+\frac{n-2}{2p})$, if $n\geq 3$
is an odd integer and

  2. (2)

$[0,\quad\frac{n-2}{2p})\cup[1-\frac{n-2}{2p},~{}~{}~{}1)$, if $n\geq 4$ is an
even integer.

In Proposition 5.3 we consider the same property for
${\boldsymbol{\nu}}_{i}^{(p)}$ and ${\boldsymbol{\mu}}_{i-1}^{(p)}$ instead of
${\boldsymbol{\nu}}_{i}^{(p)}+{\boldsymbol{\mu}}_{i-1}^{(p)}$.

The formulation of $f_{R_{p,n+1}}$ in Proposition 1.1 suggest that we only
need to analyze the rank function ${\boldsymbol{\mu}}_{n_{0}}^{(p)}$ when $n$
is odd, and the rank functions ${\boldsymbol{\mu}}_{n_{0}-1}^{(p)}$ and
${\boldsymbol{\mu}}_{n_{0}+1}^{(p)}$ when $n$ is even. But the statements look
less technical if we consider all the rank functions together in a tuple.

###  5.1. Polynomials for the rank functions ${\boldsymbol{\nu}}^{(p)}_{i}$
and ${\boldsymbol{\mu}}^{(p)}_{j}$

######  Notations 5.1.

For the sake of uniformity in the indexing, henceforth we would write the
decomposition of $F_{*}^{s}({\mathcal{O}}(a))$ as follows: For $q=p^{s}$

$F_{*}^{s}({\mathcal{O}}(a))=\bigoplus_{i=0}^{n-1}{\mathcal{O}}(-i)^{l_{i}(a,q)}\oplus{\mathcal{S}}(-n_{0}+1)^{l_{n}(a,q)}\oplus{\mathcal{S}}(-n_{0})^{l_{n+1}(a,q)}\oplus{\mathcal{S}}(-n_{0}-1)^{l_{n+2}(a,q)}.$

We call the tuple $(l_{0}(a,q),\ldots,l_{n+2}(a,q))$ the rank tuple of
$F_{*}^{s}({\mathcal{O}}(a))$ and now onwards we relabel the rank functions
${\boldsymbol{\nu}}_{0}^{(p)}$, $\ldots$, ${\boldsymbol{\nu}}_{n-1}^{(p)}$,
${\boldsymbol{\mu}}_{n_{0}-1}^{(p)}$, ${\boldsymbol{\mu}}_{n_{0}}^{(p)}$,
${\boldsymbol{\mu}}_{n_{0}+1}^{(p)}$ as the functions
${\boldsymbol{r}}_{0}^{(p)}$, $\ldots,{\boldsymbol{r}}_{n-1}^{(p)}$,
${\boldsymbol{r}}_{n}^{(p)}$,${\boldsymbol{r}}_{n+1}^{(p)}$,
${\boldsymbol{r}}_{n+2}^{(p)}$ respectively.

Following set $\\{{\bf l}_{i}(x)\\}_{i}\cup\\{{\bf
r}_{i}(x)\\}_{i}\subset{\mathbb{Q}}[x]$ of polynomials is going to be the
first generating set of polynomials for the function $f_{R_{p,n+1}}$ when $n$
is odd. When $n$ is even the set will be $\\{{\bf
m}_{i}(x)\\}_{i}\subset{\mathbb{Q}}[x]$. The sets
$\\{F_{i}(t)\\}_{i},\\{G_{i}(t)\\}_{i},\\{{\mathchoice{\widetilde{F}}{\widetilde{F}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{F}$}}}$}}{\tilde{F}}}_{i}(t)\\}_{i}\subset{\mathbb{Q}}[t]$
come from integrating these polynomials which will be used for the formulation
of $e_{HK}(R_{p,n+1})$, but will list them here.

######  Definition 5.2.

Let ${\bf Z}_{i}(x)$ and ${\bf Y}_{i}(x)$ be as in Remark 3.4. We recall that
$n_{0}=\tfrac{n-1}{2}$ if $n$ is odd, and $n_{0}=\tfrac{n-2}{2}$ if $n$ is
even.

  1. (1)

Let $n\geq 3$ be an odd integer. Then we define

${\bf l}_{i}(x)=\begin{cases}{\bf Z}_{i}(x+i)&\mbox{if}\quad 0\leq i\leq
n_{0}\\\ {\bf Y}_{i}(x+i)&\mbox{if}\quad n_{0}+1\leq i<n\\\ 0&\mbox{if}\quad
i=n\\\ \frac{1}{2\lambda_{0}}\left[{\bf Z}_{n_{0}+1}(x+n_{0}+1)-{\bf
Y}_{n_{0}+1}(x+n_{0}+1)\right]&\mbox{if}\quad i=n+1\end{cases}.$

and

${\bf r}_{i}(x)=\begin{cases}{\bf Z}_{i}(x+i)&\mbox{if}\quad 0\leq i<n_{0}\\\
{\bf Z}_{n_{0}}(x+n_{0})-\left[{\bf Y}_{n_{0}+1}(x+n_{0}+1)-{\bf
Z}_{n_{0}+1}(x+n_{0}+1)\right]&\mbox{if}\quad i=n_{0}\\\ {\bf
Y}_{i}(x+i)&\mbox{if}\quad n_{0}+1\leq i<n\\\ \frac{1}{2\lambda_{0}}\left[{\bf
Y}_{n_{0}+1}(x+n_{0}+1)-{\bf Z}_{n_{0}+1}(x+n_{0}+1)\right]&\mbox{if}\quad
i=n\\\ 0&\mbox{if}\quad i=n+1.\end{cases}$

Moreover we consider another two sets of polynomials

(5.1)
$\\{F_{0}(t),F_{1}(t),\ldots,F_{n+1}(t)\\}\cup\\{G_{0}(t),G_{1}(t),\ldots,G_{n+1}(t)\\}\subset{\mathbb{Q}}[t]$

given by

$\int_{0}^{\tfrac{1}{2}-\tfrac{n-2}{2}t}{\bf
l}_{i}(x)dx=F_{i}(t),\quad\mbox{and}\quad\int_{\tfrac{1}{2}+\tfrac{n-2}{2}t}^{1}{\bf
r}_{i}(x)dx=G_{i}(t).$

  2. (2)

Let $n\geq 4$ be an even integer, we define polynomials in ${\mathbb{Q}}[x]$
as follows.

${\bf m}_{i}(x)=\begin{cases}{\bf Z}_{i}(x+i)&\mbox{if}\quad 0\leq i\leq
n_{0}\\\ {\bf Y}_{i}(x+i)&\mbox{if}\quad n_{0}+1\leq i<n\\\ 0&\mbox{if}\quad
i=n\\\ \frac{1}{2\lambda_{0}}\left[{\bf Z}_{n_{0}+1}(x+n_{0}+1)-{\bf
Y}_{n_{0}+1}(x+n_{0}+1)\right]&\mbox{if}\quad i=n+1\\\ 0&\mbox{if}\quad
i=n+2.\end{cases}$

Also another set of polynomials

(5.2) $\\{{\mathchoice{\widetilde{F}}{\widetilde{F}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{F}$}}}$}}{\tilde{F}}}_{0}(t),{\mathchoice{\widetilde{F}}{\widetilde{F}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{F}$}}}$}}{\tilde{F}}}_{1}(t),\ldots,{\mathchoice{\widetilde{F}}{\widetilde{F}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{F}$}}}$}}{\tilde{F}}}_{n+2}(t)\\}\subset{\mathbb{Q}}[t]$

given by

$\int_{\tfrac{n-2}{2}t}^{\tfrac{1}{2}-\tfrac{n-2}{2}t}{{\bf
m}}_{i}(x)dx={\mathchoice{\widetilde{F}}{\widetilde{F}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{F}$}}}$}}{\tilde{F}}}_{i}(t).$

Note that when $n$ is odd then ${\bf r}^{(p)}_{n+2}\equiv 0$ as
$\mu_{-n_{0}-1}^{(p)}(a)=0$ for $0\leq a<q$.

The proof of the following proposition is just the corollary of Lemma 3.7 and
Lemma 3.8.

######  Proposition 5.3.

Let $p\neq 2$ such that $p>n-2$.

  1. (1)

Let $n\geq 3$ be an odd integer then each rank function ${\bf
r}_{i}^{(p)}:[0,1]\longrightarrow[0,2)$ is a polynomial function on the
interval
$[0,\tfrac{1}{2}-\tfrac{n-2}{2p})\cup[\tfrac{1}{2}+\tfrac{n-2}{2p},1)$ and is
given by

${\boldsymbol{r}}_{i}^{(p)}(x)=\begin{cases}{\bf l}_{i}(x),\quad\mbox{ for
all}\quad x\in[0,\tfrac{1}{2}-\tfrac{n-2}{2p})\\\ {\bf r}_{i}(x),\quad\mbox{
for all}\quad x\in[\tfrac{1}{2}+\tfrac{n-2}{2p},~{}~{}~{}~{}1).\end{cases}$

Moreover

$\int_{0}^{\tfrac{1}{2}-\tfrac{n-2}{2}t}{\boldsymbol{r}}^{(p)}_{i}(x)dx=F_{i}(t)\quad\mbox{and}\quad\int_{\tfrac{1}{2}+\tfrac{n-2}{2}t}^{1}{{\boldsymbol{r}}}^{(p)}_{i}(x)dx=G_{i}(t).$

In particular all $F_{i}(1/{p})$ and $G_{i}(1/{p})$ are nonnegative for
$p>n-2$.

  2. (2)

Let $n\geq 4$ be even integer then each rank function
${\boldsymbol{r}}_{i}^{(p)}:[0,1)\longrightarrow[0,2)$ is a polynomial
function on the interval $[\tfrac{n-2}{2p},\quad 1-\tfrac{n-2}{2p})$ and is
given by

${{\boldsymbol{r}}}_{i}^{(p)}(x)={{\bf m}}_{i}(x),\quad\mbox{ for all}\quad
x\in[\tfrac{n-2}{2p},\quad 1-\tfrac{n-2}{2p}).$

Moreover

$\int_{\tfrac{n-2}{2}t}^{1-\tfrac{n-2}{2}t}{{\boldsymbol{r}}}^{(p)}_{i}(x)dx={\mathchoice{\widetilde{F}}{\widetilde{F}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{F}$}}}$}}{\tilde{F}}}_{i}(t)\geq 0.$

###  5.2. Polynomial matrices associated to $\nu^{1}_{-i}(a)$,
$\mu^{1}_{-i}(a)$, ${\mathchoice{\widetilde{\nu}}{\widetilde{\nu}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{\nu}$}}}$}}{\tilde{\nu}}}^{1}_{-i}(a)$ and
${\mathchoice{\widetilde{\mu}}{\widetilde{\mu}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{\mu}$}}}$}}{\tilde{\mu}}}^{1}_{-i}(a)$

Next we will consider a set of $n-2$ matrices with entries in
${\mathbb{Q}}[t]$. This set along with the set given in Definition 5.2 will be
the generating set of polynomials for the function $f_{R_{p,n+1}}$.

We recall two sets of matrices as given in Notations 10.5 from the Appendix,

  1. (1)

$\\{{{\mathbb{B}}}_{0}^{(p)},{{\mathbb{B}}}_{1}^{(p)},\ldots,{{\mathbb{B}}}_{n-3}^{(p)}\\}\subset
M_{n+2}({\mathbb{Z}}_{\geq 0})$ when $n\geq 3$ is odd and

  2. (2)

$\\{{{\mathbb{C}}}_{0}^{(p)},{{\mathbb{C}}}_{1}^{(p)},\ldots,{{\mathbb{C}}}_{n-3}^{(p)}\\}\subset
M_{n+2}({\mathbb{Z}}_{\geq 0})$, when $n\geq 4$ is even

We write

${{\mathbb{B}}}_{n_{i}}^{(p)}={[b_{k_{1},k_{2}}^{(n_{i})}(p)]}_{0\leq
k_{1},k_{2}\leq n+1},\quad
C_{n_{i}}^{(p)}=\bigl{[}c_{k_{1},k_{2}}^{(n_{i})}(p)\bigr{]}_{0\leq
k_{1},k_{2}\leq n+2}.$

######  Definition 5.4.

Lemma 10.6 and Lemma 10.8 imply that each $b_{k_{1},k_{2}}^{(n_{i})}(p)$ and
$c_{k_{1},k_{2}}^{(n_{i})}(p)$ has a polynomial expression, that is, there
exist polynomials

$\\{H_{k_{1},k_{2}}^{(n_{i})}(t)\in{\mathbb{Q}}[t]\mid 0\leq n_{i}\leq
n-3\quad\mbox{and}\quad 0\leq k_{1},k_{2}\leq n+1\\}$

and

$\\{{\mathchoice{\widetilde{H}}{\widetilde{H}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{H}$}}}$}}{\tilde{H}}}_{k_{1},k_{2}}^{(n_{i})}(t)\in{\mathbb{Q}}[t]\mid
0\leq n_{i}\leq n-3\quad\mbox{and}\quad 0\leq k_{1},k_{2}\leq n+2\\}$

such that

${b^{(n_{i})}_{k_{1},k_{2}}(p)}/{p^{n}}=H_{k_{1},k_{2}}^{(n_{i})}(t)|_{t=1/p},\quad\mbox{for
all}\quad p\geq 3n-4$

and

${c^{(n_{i})}_{k_{1},k_{2}}(p)}/{p^{n}}={\mathchoice{\widetilde{H}}{\widetilde{H}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{H}$}}}$}}{\tilde{H}}}_{k_{1},k_{2}}^{(n_{i})}(t)\bigr{\rvert}_{t=1/p},\quad\mbox{for
all}\quad p\geq(3n-4)/2.$

We denote

  1. (1)

if $n\geq 3$ is odd, then

${{\mathbb{H}}}^{(n_{i})}(t)=\left[H_{k_{1},k_{2}}^{(n_{i})}(t)\right]_{0\leq
k_{1},k_{2}\leq n+1}$

  2. (2)

$n\geq 4$ is even, then

${\mathchoice{\widetilde{{\mathbb{H}}}}{\widetilde{{\mathbb{H}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathbb{H}}}$}}}$}}{\tilde{{\mathbb{H}}}}}^{(n_{i})}(t)=\left[{\mathchoice{\widetilde{H}}{\widetilde{H}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{H}$}}}$}}{\tilde{H}}}_{k_{1},k_{2}}^{(n_{i})}(t)\right]_{0\leq
k_{1},k_{2}\leq n+2}.$

##  6\. Decomposition of the difficult range

We showed in the previous section that the rank functions ${\bf
r}^{(p)}_{i}:[0,1)\longrightarrow[0,\infty)$ are piecewise polynomial
functions in the complement of the difficult range, which is (1) the intervals
$[0,\frac{1}{2}-\frac{n-2}{2p})$ and $[\frac{1}{2}+\frac{n-2}{2p},1)$, when
$n\geq 3$ is odd, and (2) the interval
$[\tfrac{n-2}{2p},~{}~{}~{}1-\tfrac{n-2}{2p}]$ when $n\geq 4$ is even. Here
with the help of matrices $\\{{{\mathbb{H}}}^{(n_{i})}(t)\\}_{i}$ and
$\\{{\mathchoice{\widetilde{{\mathbb{H}}}}{\widetilde{{\mathbb{H}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathbb{H}}}$}}}$}}{\tilde{{\mathbb{H}}}}}^{(n_{i})}(t)\\}_{i}$
as given in Definition 5.4 we will be able to describe these functions in the
difficult range too.

To do this we almost cover the difficult range by countably infinitely many
intervals which are indexed combinatorially such that each ${\bf r}_{i}^{(p)}$
and hence the HK density function, when restricted to one such interval, has
neat expression in terms of the finite set of polynomials as given in
Definition 5.2 and Definition 5.4.

Let ${\mathcal{M}}=\\{0,1.\ldots,n-3\\}$. For the sake of abbreviation we will
denote a tuple $(n_{1},\ldots,n_{l})\in{\mathcal{M}}^{l}$ by
${\underline{n}}$, provided there is no confusion. If $n\geq 3$ is odd then we
construct a set of semi open intervals

$\\{I^{(p)}_{{\underline{n}}},J^{(p)}_{{\underline{n}}}\mid{\underline{n}}=(n_{1},\ldots,n_{l})\in\bigcup_{l\geq
1}{\mathcal{M}}^{l}\\}$

such that

  1. (1)

the indexing set is ${\mathcal{M}}^{l}$ which is independent of $p$, although
each interval $I^{(p)}_{\underline{n}}$ and $J^{(p)}_{\underline{n}}$ will
depend on $p$.

  2. (2)

These semi open intervals $I^{(p)}_{\underline{n}},J^{(p)}_{\underline{n}}$
almost cover (i.e., the uncovered part is of measure $0$) the difficult range
$[\tfrac{1}{2}-\tfrac{(n-2)}{2p},~{}~{}\tfrac{1}{2}+\tfrac{(n-2)}{2p})$.

  3. (3)

These semi open intervals are disjoint from each other.

Similarly, for even $n\geq 4$, we construct such a set of disjoint semi open
intervals $\\{{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{\underline{n}}\mid{\underline{n}}\in{{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}}}_{0}\cup{{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}}}_{1}=\cup_{l\in{\mathbb{N}}}{\mathcal{M}}^{l}\\}$
which almost cover the difficult range, namely, the interval
$[0,\quad\tfrac{n-2}{2p})\cup[1-\tfrac{n-2}{2p},\quad 1)$. Now we start with a
formal set up

######  Definition 6.1.

Let $n\geq 3$ be an integer. Let
${{{\mathcal{A}}}}\subseteq\\{0,1,\ldots,p-1\\}$ be a set of $n-2$ elements
indexed by the set ${\mathcal{M}}=\\{0,1,\ldots,n-3\\}$, that is every
$j_{i}\in{{{\mathcal{A}}}}$ is indexed by unique $n_{i}\in{\mathcal{M}}$. Let
${\mathcal{B}}=\\{b_{0},b_{0}+1,\ldots,b_{0}+t_{0}-1\\}$
$\subset\\{0,1,\ldots,p-1\\}$ be a fixed set of consecutive $t_{0}$ integers
such that ${{\mathcal{A}}}\cap{\mathcal{B}}=\phi$. Then, for $l\geq 1$, we
define

$I^{{\mathcal{B}}(p)}_{(n_{1},\ldots,n_{l})}=\bigl{[}\sum_{i=1}^{l}\tfrac{j_{i}}{p^{i}}+\tfrac{b_{0}}{p^{l+1}},\quad\tfrac{b_{0}+t_{0}}{p^{l+1}}+\sum_{i=1}^{l}\tfrac{j_{i}}{p^{i}}\bigr{)}.$

Similarly if ${\mathcal{D}}=\\{d_{0},d_{0}+1,\ldots,d_{0}+t_{1}-1\\}$ is
another such set of consecutive $t_{1}$ integers in $\\{0,\ldots,p-1\\}$ such
that ${{\mathcal{A}}}\cap{\mathcal{D}}=\phi$ then we define

$I^{{\mathcal{D}}(p)}_{(n_{1},\ldots,n_{l})}=[\sum_{i=1}^{l}\tfrac{j_{i}}{p^{i}}+\tfrac{d_{0}}{p^{l+1}},\quad\tfrac{d_{0}+t_{1}}{p^{l+1}}+\sum_{i=1}^{l}\tfrac{j_{i}}{p^{i}}).$

######  Lemma 6.2.

Let $n\geq 3$ be an integer. If the sets ${{\mathcal{A}}}$, ${\mathcal{D}}$
and ${\mathcal{B}}$, as given in Definition 6.1, are mutually disjoint sets
then the set of semi open intervals

$\\{I^{{\mathcal{B}}(p)}_{(n_{i_{1}},\ldots,n_{i_{l}})},\quad
I^{{\mathcal{D}}(p)}_{(n_{k_{1}},\ldots,n_{k_{l_{1}}})}\mid(n_{i_{1}},\ldots,n_{i_{l}})\in{\mathcal{M}}^{l},~{}~{}~{}(n_{k_{1}},\ldots,n_{k_{l_{1}}})\in{\mathcal{M}}^{l_{1}}\quad
l,l_{1}\geq 1\\}$

is a set of mutually disjoint intervals.

###### Proof.

By construction, each of these semi open intervals is a subset of $[0,1)$.
Therefore, for any $x$ in one of such intervals and for any $q=p^{s}$, we have
the $p$-adic expansion $\lfloor xq\rfloor=a_{0}+a_{1}p+\cdots+a_{s-1}p^{s-1}$.
On the other hand we can check that if $a$ is a nonnegative integer with the
$p$-adic expansion $a=a_{0}+a_{1}p+\cdots+a_{s-1}p^{s-1}$ and $l_{0}$ is an
integer then

  1. (1)

$a/p^{s}<l_{0}/p\mbox{ $\Longleftrightarrow$ }a_{s-1}<l_{0}$ and

  2. (2)

$a/p^{s}\geq l_{0}/p\mbox{ $\Longleftrightarrow$ }a_{s-1}\geq l_{0}$.

Let us fix an element $x\in
I^{{\mathcal{B}}(p)}_{(n_{i_{1}},\ldots,n_{i_{l}})}$.

Then for $q\geq p^{l+1}$, the $p$-adic expansion of $\lfloor xq\rfloor$ is
given by

$\lfloor
xq\rfloor=a_{0}+a_{1}p+\cdots+a_{s-l-1}p^{s-l-1}+j_{i_{l}}p^{s-l}+\cdots+j_{i_{1}}p^{s-1},$

where $a_{s-l-1}\in{\mathcal{B}}$ as $b_{0}\leq a_{s-l-1}<b_{0}+t_{0}$.

Now suppose $x\in I^{{\mathcal{B}}(p)}_{(n_{k_{1}},\ldots,n_{k_{l_{1}}})}$.
Without loss of generality we assume $l_{1}\geq l$, then $j_{i_{t}}=j_{k_{t}}$
for $t\leq l$, If $l_{1}>l$ then $j_{k_{l+1}}=a_{s-l-1}\in{{\mathcal{A}}}$,
which contradicts the assumption that ${{\mathcal{A}}}\cap{\mathcal{B}}=\phi$.
Therefore $l_{1}=l$ and
$(n_{k_{1}},\ldots,n_{k_{l_{1}}})=(n_{i_{1}},\ldots,n_{i_{l_{1}}})$.

If $x\in I^{{\mathcal{D}}(p)}_{(n_{k_{1}},\ldots,n_{k_{l_{1}}})}$, where
without loss of generality we assume $l_{1}\geq l$ then $j_{i_{t}}=j_{k_{t}}$
for $t\leq l$. Now if $l_{1}=l$ then $d_{0}\leq a_{s-l-1}<d_{0}+t_{1}$, which
implies $a_{s-l-1}\in{\mathcal{D}}$, otherwise $a_{s-l-1}\in{{\mathcal{A}}}$.
Both outcome contradict the mutual disjointness property of the sets
${{\mathcal{A}}}$, ${\mathcal{B}}$ and ${\mathcal{D}}$. Hence the lemma. ∎

######  Definition 6.3.

Let $S_{1}\subset S_{2}$ be two Lebesgue measurable set in ${\mathbb{R}}$. We
say $S_{1}$ almost covers $S_{2}$ if the Lebesgue measure of the set
$S_{2}\setminus S_{1}$ is $0$ in ${\mathbb{R}}$.

We use Lemma 6.2 to construct a set of semi open intervals which almost cover
the difficult range. We first construct such sets when $n\geq 4$ is an even
integer.

###  6.1. Decomposition of the difficult range when $n$ is even

######  Definition 6.4.

Let $n\geq 4$ be an even integer. Let $p>n-2$ be a prime. Let

${{\mathcal{A}}}_{e}=\\{0,1,\ldots,\tfrac{n}{2}-2\\}\cup\\{p+1-\tfrac{n}{2},\ldots,p-1\\}$

and let ${\mathcal{B}}=\\{\tfrac{n}{2}-1,\ldots,p-\tfrac{n}{2}\\}$ be the set
of consecutive integers. Then ${\mathcal{A}}_{e}$ and ${\mathcal{B}}$ are
disjoint.

We indexed the elements of the set ${{\mathcal{A}}}_{e}$ by the set
${\mathcal{M}}=\\{0,1,\ldots,n-3\\}$ as follows: An element
$j_{i}\in{{\mathcal{A}}}_{e}$ is indexed by $n_{i}\in{\mathcal{M}}$ if

$j_{i}=\left\\{\begin{matrix}n_{i}&\mbox{if}\quad
n_{i}\leq\left(\tfrac{n}{2}-2\right)\\\ \
p-(n_{i}-\tfrac{n}{2}-2)&\mbox{if}\quad
n_{i}>\left(\tfrac{n}{2}-2\right).\end{matrix}\right.$

Henceforth in this subsection we write the set
$I^{{\mathcal{B}}(p)}_{(n_{1},\ldots,n_{l})}$ as the set
${\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{(n_{1},\ldots,n_{l})}$.
Following the Definition 6.1, for $(n_{1},\ldots,n_{l})\in{\mathcal{M}}^{l}$,
where $l\geq 1$, we define

${\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{(n_{1},\ldots,n_{l})}=[\sum_{i=1}^{l}\tfrac{j_{i}}{p^{i}}+\tfrac{n-2}{2p^{l+1}},\quad\sum_{i=1}^{l}\tfrac{j_{i}}{p^{i}}+\tfrac{1}{p^{l}}-\tfrac{n-2}{2p^{l+1}})$

Let

${\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}}_{0}=\\{(n_{1},\ldots,n_{l})\in{\mathcal{M}}^{l}\mid
n_{1}\leq\tfrac{n}{2}-2\\}_{l\in{\mathbb{N}}}\quad\mbox{and}\quad{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}}_{1}=\\{(n_{1},\ldots,n_{l})\in{\mathcal{M}}^{l}\mid
n_{1}\geq\tfrac{n}{2}-1\\}_{l\in{\mathbb{N}}}.$

For the sake of abbreviation, provided there is no confusion, we will denote
an element $(n_{1},\ldots,n_{l})\in{\mathcal{M}}^{l}$ by ${\underline{n}}$

It is easy to check that

$\bigcup_{{\underline{n}}\in{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}}_{0}}{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{\underline{n}}\subset[0,\quad\tfrac{n-2}{2p})\quad\quad\mbox{and}\quad\quad\bigcup_{{\underline{n}}\in{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}}_{1}}{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{\underline{n}}\subset[1-\tfrac{n-2}{2p},\quad
1).$

######  Lemma 6.5.

  1. (1)

The set
$\bigcup_{{\underline{n}}\in{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}}_{0}}{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{\underline{n}}$ almost
covers the set $[0\quad\tfrac{n-2}{2p})$.

  2. (2)

The set
$\bigcup_{{\underline{n}}\in{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}}_{1}}{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{\underline{n}}$ almost
covers the set $[1-\tfrac{n-2}{2p},1)$.

###### Proof.

We prove the first assertion as the proof for the second assertion is very
similar. Let

$A_{k}=[0\quad\tfrac{n-2}{2p})\setminus\bigcup_{\\{(n_{1},\ldots,n_{l})\in{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}}_{0}\mid
l\leq k\\}}{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{(n_{1},\ldots,n_{l})}.$

It is enough to prove that the Lebesgue measure of the set $(A_{k})$ is

$\mu(A_{k})=(n-2)^{k+1}/2p^{k+1},\quad\mbox{for all}\quad k\geq 1.$

We prove the statement by induction on $k$. For $k=1$

$A_{1}=[0\quad\tfrac{n-2}{2p})\setminus\bigcup_{0\leq
n_{i}<(n-2)/2}\left[\tfrac{n_{i}}{p}+\tfrac{n-2}{2p^{2}},\quad\tfrac{n_{i}+1}{p}-\tfrac{n-2}{2p^{2}}\right).$

Hence

$\mu(A_{1})=\tfrac{n-2}{2p}-\tfrac{n-2}{2}\left[\tfrac{1}{p}-\tfrac{n-2}{p^{2}}\right]=\tfrac{(n-2)^{2}}{2p^{2}}.$

We assume the statement for $k-1$, that is $\mu(A_{k-1})=(n-2)^{k}/2p^{k}$.
Now

$A_{k}=A_{k-1}\setminus\bigcup_{(n_{i_{1}},\ldots,n_{i_{k}})\in{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}}_{0}}{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{(n_{i_{1}},\ldots,n_{i_{k}})}.$

Since, by Lemma 6.2, these semi open intervals are mutually disjoint we have

$\mu(A_{k})=\mu(A_{k-1})-\sum_{(n_{i_{1}},\ldots,n_{i_{k}})\in{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}}_{0}}\ell({\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{(n_{i_{1}},\ldots,n_{i_{k}})}).$

But

$\sum_{(n_{i_{1}},\ldots,n_{i_{k}})\in{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}}_{0}}\ell({\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{(n_{i_{1}},\ldots,n_{i_{k}})})=\tfrac{n-2}{2}(n-2)^{k-1}\left(\tfrac{1}{p^{k}}-\tfrac{n-2}{p^{k+1}}\right)=\tfrac{(n-2)^{k}}{2p^{k}}-\tfrac{(n-2)^{k+1}}{2p^{k+1}},$

which implies $\mu(A_{k})=\tfrac{(n-2)^{k+1}}{2p^{k+1}}$. By induction this
proves the result for all $k\geq 1$. ∎

###  6.2. Decomposition of the difficult range when $n$ is odd

######  Definition 6.6.

Let $n\geq 3$ be an odd integer and $p>n-2$ a prime. Let

$m_{0}=\frac{p}{2}-\frac{n-2}{2}\quad\mbox{and}\quad{{\mathcal{A}}}_{o}=\\{m_{0},m_{0}+1,\ldots,m_{0}+n-3\\}$

and let ${\mathcal{B}}=\\{0,1,\ldots,m_{0}-1\\}$ and
${\mathcal{D}}=\\{m_{0}+n-2,\ldots,,p-1\\}$ be two sets of consecutive
integers. We indexed the set ${{\mathcal{A}}}_{o}$ by the set
${\mathcal{M}}=\\{0,1,\ldots,n-3\\}$ so that an element $j_{i}$ of
${{\mathcal{A}}}_{o}$ is indexed by $n_{i}$ if $j_{i}=n_{i}+m_{0}$. Here
${\mathcal{A}}_{o}$, ${\mathcal{B}}$ and ${\mathcal{D}}$ are mutually disjoint
sets.

Henceforth in this subsection we denote the interval
$I^{{\mathcal{B}}(p)}_{(n_{1},\ldots,n_{l})}$ by
$I^{(p)}_{(n_{1},\ldots,n_{l})}$ and the interval
$I^{{\mathcal{D}}(p)}_{(n_{1},\ldots,n_{l})}$ by
$J^{(p)}_{(n_{1},\ldots,n_{l})}$. Following Definition 6.1, for
$(n_{1},\ldots,n_{l})\in{\mathcal{M}}^{l}$, where $l\geq 1$, we define two
semi open intervals

$I^{(p)}_{(n_{1},\ldots,n_{l})}=\left[\mbox{$\sum_{i=1}^{l}\frac{j_{i}}{p^{i}},\quad\sum_{i=1}^{l}\frac{j_{i}}{p^{i}}+\frac{m_{0}}{p^{l+1}}$}\right)$

and

$J^{(p)}_{(n_{1},\ldots,n_{l})}=\left[\mbox{$\frac{m_{0}+n-2}{p^{l+1}}+\sum_{i=1}^{l}\frac{j_{i}}{p^{i}},\quad\sum_{i=1}^{l}\frac{j_{i}}{p^{i}}+\frac{p}{p^{l+1}}$}\right).$

For the sake of abbreviation, provided there is no confusion, we will denote
an element $(n_{1},\ldots,n_{l})\in{\mathcal{M}}^{l}$ by ${\underline{n}}$.

######  Lemma 6.7.

Let $n\geq 3$ be an odd integer and $p>n-2$ be an odd prime. Then the set

$\bigcup_{{\underline{n}}\in\cup_{l\geq
1}{\mathcal{M}}^{l}}I^{(p)}_{{\underline{n}}}\cup\bigcup_{{\underline{n}}\in\cup_{l\geq
1}{\mathcal{M}}^{l}}J^{(p)}_{{\underline{n}}}\quad\mbox{ almost cover the
interval}\quad[\tfrac{1}{2}-\tfrac{n-2}{2p},\tfrac{1}{2}+\tfrac{n-2}{2p}).$

If $n=3$ then this covering misses the point $2.5$ as shown in Section 9.

###### Proof.

It is obvious that

$\bigcup_{{\underline{n}}\in{\mathcal{M}}^{l}}I^{(p)}_{\underline{n}}\cup
J^{(p)}_{\underline{n}}\subseteq[\tfrac{1}{2}-\tfrac{n-2}{2p},\quad\tfrac{1}{2}+\tfrac{n-2}{2p}).$

Let us denote

$A_{k}=[\tfrac{1}{2}-\tfrac{n-2}{2p},\quad\tfrac{1}{2}+\tfrac{n-2}{2p})\setminus\bigcup_{\\{(n_{1},\ldots,n_{l})\in{{\mathcal{M}}^{l}}\mid
l\leq k\\}}I^{(p)}_{\underline{n}}\cup J^{(p)}_{\underline{n}}$

It is enough to prove that, for each $k\geq 1$, the Lebesgue measure of the
set $A_{k}=((n-2)/{p})^{k+1}$, which can be checked using the induction
argument for $k$ as done in Lemma 6.5. ∎

##  7\. polynomial expression for the rank functions ${\bf r}_{i}^{(p)}$ and
$f_{R_{p,n+1}}$

In the previous section, for $n$ even and for $n$ odd each, we have
constructed an almost covering of the respective difficult range by semi open
intervals such that the intervals are disjoint and combinatorially indexed.

By construction, the elements belonging to the same subinterval have a
$p$-adic expression of certain ‘type’, we use this in the following key
Proposition to show that the rank functions ${\boldsymbol{r}}_{i}^{(p)}$ when
restricted to such intervals, are polynomials.

######  Notations 7.1.

For $n\geq 3$ odd, let

  1. (1)

$\\{{\bf l}_{0}(x),\ldots,{\bf l}_{n+1}(x)\\}\subset{\mathbb{Q}}[x]$ and
$\\{{\bf r}_{0}(x),\ldots,{\bf r}_{n+1}(x)\\}\subset{\mathbb{Q}}[x]$ be as in
Definition 5.2.

  2. (2)

Let $\\{{{\mathbb{H}}}^{(n_{i})}(t)\mid 0\leq n_{i}\leq n-3\\}\subset
M_{n+2}({\mathbb{Q}}[t])$ be as in Definition 5.4.

  3. (3)

Let $I^{(p)}_{(n_{1},\ldots,n_{l})}$ and $J^{(p)}_{(n_{1},\ldots,n_{l})}$ be
the semi open intervals as in Definition 6.6.

  4. (4)

Let

$\phi^{(p)}:\bigcup_{\\{{\underline{n}}\in{\mathcal{M}}^{l}\mid l\geq
1\\}}I^{(p)}_{\underline{n}}\cup
J^{(p)}_{\underline{n}}\longrightarrow[0,\tfrac{1}{2}-\tfrac{n-2}{2p})\cup[\tfrac{1}{2}+\tfrac{n-2}{2p},~{}~{}~{}1),$

such that for each ${\underline{n}}=(n_{1},\ldots,n_{l})$ the restriction maps

$\phi^{(p)}:I^{(p)}_{\underline{n}}\longrightarrow[0,\tfrac{1}{2}-\tfrac{n-2}{2p})\quad\mbox{and}\quad\phi^{(p)}:J^{(p)}_{\underline{n}}\longrightarrow[\tfrac{1}{2}+\tfrac{n-2}{2p},~{}~{}~{}1),$

are the surjective maps given by $x\to p^{l}(x-\sum_{i}\tfrac{j_{i}}{p^{i}})$,
where $j_{i}=n_{i}+\tfrac{p}{2}-\tfrac{n-2}{2}$.

######  Proposition 7.2.

Let $n\geq 3$ is an odd integer, where $p\geq 3n-4$. Then, for
${\underline{n}}=(n_{1},\ldots,n_{l})$ the rank functions
${\boldsymbol{r}}^{(p)}_{i}:I^{(p)}_{\underline{n}}\longrightarrow{\mathbb{R}}$
are given by the formula

(7.1)
$\Bigl{[}{\boldsymbol{r}}^{(p)}_{0}(x),\ldots,{\boldsymbol{r}}^{(p)}_{n+1}(x)\Bigr{]}_{1\times
n+2}=\\\ \left[{\bf l}_{0}(\phi^{(p)}(x)),\ldots,{\bf
l}_{n+1}(\phi^{(p)}(x))\right]_{1\times
n+2}\cdot{{\mathbb{H}}}^{(n_{l})}(t)\cdots{{\mathbb{H}}}^{(n_{1})}(t)\Bigr{\rvert}_{t=1/p}.$

Similarly the rank functions
${\boldsymbol{r}}^{(p)}_{i}:J^{(p)}_{(n_{1},\ldots,n_{l})}\longrightarrow{\mathbb{R}}$
are given by the formula

(7.2)
$\left[{\boldsymbol{r}}^{(p)}_{0}(x),\ldots,{\boldsymbol{r}}^{(p)}_{n+1}(x)\right]_{1\times
n+2}=\\\ \left[{\bf r}_{0}(\phi^{(p)}(x)),\ldots,{\bf
r}_{n+1}(\phi^{(p)}(x))\right]_{1\times
n+2}\cdot{{\mathbb{H}}}^{(n_{l})}(t)\cdots{{\mathbb{H}}}^{(n_{1})}(t)\Bigr{\rvert}_{t=1/p}.$

In particular we get a formula for
${\boldsymbol{\mu}}^{(p)}_{n_{0}}(x)={\boldsymbol{r}}^{(p)}_{n+1}(x)$.

###### Proof.

We fix $x\in I^{(p)}_{(n_{1},\ldots,n_{l})}$, then
$j_{i}=n_{i}+\tfrac{p}{2}-\tfrac{n-2}{2}$ and

$I^{(p)}_{(n_{1},\ldots,n_{l})}=\Bigl{[}\sum_{i=1}^{l}\tfrac{j_{i}}{p^{i}},\quad\sum_{i=1}^{l}\tfrac{j_{i}}{p^{i}}+\tfrac{m_{0}}{p^{l+1}}\Bigr{)}.$

Therefore for any given $q\geq p^{l+1}$ the $p$-adic expansion of $\lfloor
xq\rfloor$ is given by

$\lfloor
xq\rfloor=a_{0}+\cdots+a_{s-l-1}p^{s-l-1}+j_{l}p^{s-l}+\cdots+j_{1}p^{s-1},$

where $y:=\phi^{(p)}(x)\in[0,\tfrac{1}{2}-\tfrac{n-2}{2p})$ lies in the
complement of the difficult range. Hence by Proposition 5.3

${\boldsymbol{r}}_{i}^{(p)}(y)={\bf l}_{i}(y),\quad\mbox{for all}\quad 0\leq
i\leq n+1.$

Let us denote

$A_{s-l}=a_{0}+\cdots+a_{s-l-1}p^{s-l-1}=\lfloor yp^{s-l}\rfloor.$

Then ${\mathcal{O}}(\lfloor xq\rfloor)={\mathcal{O}}(A_{s-l})\otimes
F^{*s-l}{\mathcal{O}}(j_{l}+\cdots+j_{1}p^{l-1})$. Therefore, by projection
formula

$F_{*}^{s}({\mathcal{O}}(\lfloor
xq\rfloor))=F_{*}^{l}\Bigl{(}F_{*}^{s-l}({\mathcal{O}}(A_{s-l}))\otimes{\mathcal{O}}(j_{l}+\cdots+j_{1}p^{l-1})\Bigr{)}.$

Let the rank tuple of $F_{*}^{{s-l}}({\mathcal{O}}(A_{s-l}))$ be denoted as

$(l_{0},l_{1},\ldots,l_{n+1})=\left(l_{0}^{s-l}(A_{s-l}),l_{1}^{s-l}(A_{s-l}),\ldots,l_{n+1}^{s-l}(A_{s-l})\right)$

where

$F_{*}^{s-l}({\mathcal{O}}(A_{s-l}))={\mathcal{O}}^{l_{0}}\oplus\cdots\oplus{\mathcal{O}}(-n+1)^{l_{n-1}}\oplus{\mathcal{S}}(-n_{0}+1)^{l_{n}}\oplus{\mathcal{S}}(-n_{0})^{l_{n+1}}.$

Then we can write

$F_{*}^{s-l}({\mathcal{O}}(A_{s-l}))\otimes{\mathcal{O}}(j_{l}+\cdots+j_{1}p^{l-1})\\\
=\bigl{(}{\mathcal{O}}(j_{l})^{l_{0}}\oplus\cdots\oplus{\mathcal{O}}(j_{l}-n+1)^{l_{n-1}}\oplus{\mathcal{S}}(j_{l}-n_{0}+1)^{l_{n}}\oplus{\mathcal{S}}(j_{l}-n_{0})^{l_{n+1}}\bigr{)}\otimes
F^{*}{\mathcal{O}}(j_{l-1}+\cdots+j_{1}p^{l-2}).$

Again, by projection formula

$F_{*}^{s}({\mathcal{O}}(\lfloor
xq\rfloor))=F_{*}^{l-1}\Bigl{(}F_{*}\bigl{(}F_{*}^{s-l}({\mathcal{O}}(A_{s-l}))\otimes{\mathcal{O}}(j_{l}+\cdots+j_{1}p^{l-1})\bigr{)}\Bigr{)}\\\
=F_{*}^{l-1}\Bigl{(}\bigl{(}F_{*}{\mathcal{O}}(j_{l})^{l_{0}}\oplus\cdots\oplus
F_{*}S(-n_{0}+j_{l})^{l_{n+1}}\bigr{)}\otimes{\mathcal{O}}(j_{l-1}+\cdots+j_{1}p^{l-2})\Bigr{)}.$

Note that $p\geq 3n-4$ implies that, for all $0\leq t<n$ we have $0\leq
j_{i}-t<p$. Hence by Lemma 3.2 (2) the decomposition of each
$F_{*}{\mathcal{O}}(j_{l}-t)$ will only have bundles of type ${\mathcal{O}}$,
${\mathcal{O}}(-1)$, $\ldots,{\mathcal{O}}(1-n)$ and ${\mathcal{S}}(-n_{0}+1)$
and ${\mathcal{S}}(-n_{0})$. Therefore we can write

$F_{*}{\mathcal{O}}(j_{l})^{l_{0}}\oplus\cdots\oplus
F_{*}{\mathcal{O}}(j_{l}-n+1)^{l_{n-1}}\oplus
F_{*}S(-n_{0}+1+j_{l})^{l_{n}}\oplus F_{*}S(-n_{0}+j_{l})^{l_{n+1}}\\\ \
={\mathcal{O}}^{b_{0}}\oplus\cdots\oplus{\mathcal{O}}(1-n)^{b_{n-1}}\oplus{\mathcal{S}}(-n_{0}+1)^{b_{n}}\oplus{\mathcal{S}}(-n_{0})^{b_{n+1}},$

where as a matrix

$[b_{0},\ldots,b_{n+1}]_{1\times n+2}=[l_{0},\ldots,l_{n+1}]_{1\times
n+2}\cdot{{\mathbb{B}}}^{(p)}_{n_{l}}$, and where we recall

${{\mathbb{B}}}^{(p)}_{n_{l}}=\left[\begin{matrix}\nu_{0}(j_{l}),\ldots,\nu_{-n+1}(j_{l}),\mu_{-n_{0}+1}(j_{l}),\mu_{-n_{0}}(j_{l})\\\
\vdots\\\
\nu_{0}(j_{l}-n+1),\ldots,\nu_{-n+1}(j_{l}-n+1),\mu_{-n_{0}+1}(j_{l}-n+1),\mu_{-n_{0}}(j_{l}-n+1)\\\
{\mathchoice{\widetilde{\nu}}{\widetilde{\nu}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{\nu}$}}}$}}{\tilde{\nu}}}_{0}(j_{l}-n_{0}+1),\ldots,{\mathchoice{\widetilde{\nu}}{\widetilde{\nu}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{\nu}$}}}$}}{\tilde{\nu}}}_{-n+1}(j_{l}-n_{0}+1),{\mathchoice{\widetilde{\mu}}{\widetilde{\mu}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{\mu}$}}}$}}{\tilde{\mu}}}_{-n_{0}+1}(j_{l}-n_{0}+1),{\mathchoice{\widetilde{\mu}}{\widetilde{\mu}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{\mu}$}}}$}}{\tilde{\mu}}}_{-n_{0}}(j_{l}-n_{0}+1)\\\
{\mathchoice{\widetilde{\nu}}{\widetilde{\nu}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{\nu}$}}}$}}{\tilde{\nu}}}_{0}(j_{l}-n_{0}),\ldots,{\mathchoice{\widetilde{\nu}}{\widetilde{\nu}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{\nu}$}}}$}}{\tilde{\nu}}}_{-n+1}(j_{l}-n_{0}),{\mathchoice{\widetilde{\mu}}{\widetilde{\mu}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{\mu}$}}}$}}{\tilde{\mu}}}_{-n_{0}+1}(j_{l}-n_{0}),{\mathchoice{\widetilde{\mu}}{\widetilde{\mu}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{\mu}$}}}$}}{\tilde{\mu}}}_{-n_{0}}(j_{l}-n_{0})\end{matrix}\right].$

Iterating this we can write the rank tuple for
$F_{*}^{s}({\mathcal{O}}(\lfloor xq\rfloor))$ as

(7.3) $\left[\nu_{0}^{s}(\lfloor xq\rfloor),\ldots,\nu_{n-1}^{s}(\lfloor
xq\rfloor),\mu_{-n_{0}+1}^{s}(\lfloor xq\rfloor),\mu_{-n_{0}}^{s}(\lfloor
xq\rfloor)\right]_{1\times n+2}\\\
=\left[l_{0},\ldots,l_{n-1},l_{n},l_{n+1}\right]_{1\times
n+2}\cdot{{\mathbb{B}}}^{(p)}_{n_{l}}\cdots{{\mathbb{B}}}^{(p)}_{n_{1}}.$

We note that each matrix ${{\mathbb{B}}}_{n_{i}}^{(p)}$ is independent of $s$,
where $q=p^{s}$. Now

$\lim_{s\to\infty}\frac{l_{i}}{p^{sn}}=\lim_{s\to\infty}\frac{l_{i}^{s-l}(A_{s-l})}{p^{ln}(p^{s-l})^{n}}=\lim_{s\to\infty}\frac{l_{i}^{s-l}(\lfloor
yp^{s-l}\rfloor)}{p^{ln}(p^{s-l})^{n}}=\frac{{\bf
r}_{i}^{(p)}(y)}{p^{ln}}=\frac{{\bf l}_{i}(y)}{p^{ln}},$

where the second last equality follows from Lemma 3.8.

Hence multiplying both the sides of (7.3) by $q^{n}=p^{sn}$ and taking the
limit as $s\to\infty$ we get

$\left[{\boldsymbol{\nu}}^{(p)}_{0}(x),\ldots,{\boldsymbol{\nu}}^{(p)}_{n-1}(x),{\boldsymbol{\mu}}_{n_{0}-1}(x),{\boldsymbol{\mu}}_{n_{0}}(x)\right]_{1\times
n+2}\\\ =\left[{\bf l}_{0}(y),{\bf l}_{1}(y),\ldots,{\bf
l}_{n+1}(y)\right]_{1\times
n+2}\cdot(\tfrac{1}{p^{n}}\cdot{{\mathbb{B}}}^{(p)}_{n_{l}})\cdots(\tfrac{1}{p^{n}}\cdot{{\mathbb{B}}}^{(p)}_{n_{1}}),$

This gives (7.1).

We can argue similarly for the function
${\boldsymbol{r}}_{i}^{(p)}:J^{(p)}_{\underline{n}}\longrightarrow{\mathbb{R}}$.
where now $y$ lies in the interval $[\tfrac{1}{2}+\tfrac{n-2}{2p},1)$ and
hence ${\bf l}_{i}$ will be replaced by ${\bf r}_{i}$. ∎

The proof of the following proposition is along the same lines as for
Proposition 7.2. We recall the following.

######  Notations 7.3.

For $n\geq 4$ an even integer,

  1. (1)

let $\\{{{\bf m}}_{0}(x),\ldots,{{\bf m}}_{n+2}(x)\in{\mathbb{Q}}[x]\\}$ as in
Definition 5.2.

  2. (2)

Let
$\\{{\mathchoice{\widetilde{{\mathbb{H}}}}{\widetilde{{\mathbb{H}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathbb{H}}}$}}}$}}{\tilde{{\mathbb{H}}}}}^{(n_{i})}(t)\in
M_{n+3}({\mathbb{Q}}[t])\mid 0\leq n_{i}\leq n-3\\}$ as in Definition 5.4.

  3. (3)

Let ${\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{(n_{1},\ldots,n_{l})}$
be the interval as in Definition 6.4.

  4. (4)

Let

$\psi^{(p)}:\bigcup_{\\{{\underline{n}}\in{\mathcal{M}}^{l}\mid l\geq
1\\}}{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{\underline{n}}\longrightarrow[\tfrac{n-2}{2p},~{}~{}~{}1-\tfrac{n-2}{2p}),$

such that for each ${\underline{n}}=(n_{1},\ldots,n_{l})$ the restriction map

$\psi^{(p)}:{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{\underline{n}}\longrightarrow[\tfrac{n-2}{2p},~{}~{}~{}1-\tfrac{n-2}{2p}),\quad\mbox{given
by}\quad y\to p^{l}(x-\sum_{i}\tfrac{j_{i}}{p^{i}})$

is surjective.

######  Proposition 7.4.

Let $n\geq 4$ be an even integer, where $p\geq(3n-4)/2$. Then the functions
${\boldsymbol{r}}^{(p)}_{i}:{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{(n_{1},\ldots,n_{l})}\longrightarrow{\mathbb{R}}$
are given by the formula

$\left[{\boldsymbol{r}}^{(p)}_{0}(x),\ldots,{\boldsymbol{r}}^{(p)}_{n+2}(x)\right]_{1\times
n+3}=\\\ \left[{{\bf m}}_{0}\bigl{(}\psi^{(p)}(x)\bigr{)},\ldots,{{\bf
m}}_{n+2}\bigl{(}\psi^{(p)}(x)\bigr{)}\right]_{1\times
n+3}\cdot{\mathchoice{\widetilde{{\mathbb{H}}}}{\widetilde{{\mathbb{H}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathbb{H}}}$}}}$}}{\tilde{{\mathbb{H}}}}}^{(n_{l})}(t)\cdots{\mathchoice{\widetilde{{\mathbb{H}}}}{\widetilde{{\mathbb{H}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathbb{H}}}$}}}$}}{\tilde{{\mathbb{H}}}}}^{(n_{1})}(t)\Bigr{\rvert}_{t=1/p}.$

In particular we get formulas for
${\boldsymbol{\mu}}^{(p)}_{n_{0}+1}(x)={\boldsymbol{r}}^{(p)}_{n+2}(x)$ and
for ${\boldsymbol{\mu}}^{(p)}_{n_{0}-1}(x)={\boldsymbol{r}}^{(p)}_{n}(x)$.

Now we are ready to define the HK density function
$f_{R_{p,n+1}}:[0,\infty)\longrightarrow{\mathbb{R}}$ almost everywhere, that
is outside a set of measure $0$ in ${\mathbb{R}}$.

######  Notations 7.5.

We recall that $\\{{\bf Z}_{i}(x)\\}$ and $\\{{\bf Y}_{i}(x)\\}$ are
polynomials as in Notations 3.3 and $n_{0}=\lceil\tfrac{n}{2}\rceil-1$.

  1. (1)

If $n$ is odd then, by Lemma 6.7 the set
$[\tfrac{1}{2}-\tfrac{(n-2)}{2p},~{}~{}\tfrac{1}{2}+\tfrac{(n-2)}{2p})$ is
almost covered by the set of disjoint intervals
$\\{I^{(p)}_{\underline{n}},J^{(p)}_{\underline{n}}\mid{\underline{n}}\in\cup_{l\geq
1}{\mathcal{M}}^{l}\\}$, which are as in Definition 6.6.

  2. (2)

If $n\geq 4$ is even then, by Lemma 6.5 the set
$\left[1-\tfrac{n-2}{2p},1\right)$ is almost covered by the set
$\\{{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{\underline{n}}\mid{\underline{n}}\in{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}_{0}}\\},$
and the set $[1-\tfrac{(n-2)}{2p},~{}~{}~{}1)$ is almost covered by the set of
intervals $\\{{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{\underline{n}}\mid{\underline{n}}\in{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}_{1}}\\}$,
as in Definition 6.4.

Following theorem shows that outside a set of measure $0$ the HK density
function $f_{R_{p},n+1}$ is a piecewise polynomial function though they will
be infinitely many pieces.

######  Theorem 7.6.

Let $n\geq 3$ be an odd integer and let $p\geq 3n-4$. Then the HK density
function is defined outside a set of measure $0$ as follows:

$f_{R_{p},n+1}(x)=\begin{cases}{\bf Z}_{i}(x)\quad\mbox{if}\quad i\leq
x<i+1\quad\mbox{and}\quad 0\leq i\leq n_{0}\\\ \\\ {\bf
Z}_{n_{0}+1}(x),\quad\mbox{if}\quad(n_{0}+1)\leq
x<(n_{0}+\frac{3}{2})-\frac{n-2}{2p}\\\ \\\ {\bf
Y}_{n_{0}+1}(x)+{{\boldsymbol{\mu}}^{(p)}_{n_{0}}}(x-n_{0}-1),\quad\mbox{if}\quad\tfrac{1}{2}-\tfrac{n-2}{2p}\leq
x-(n_{0}+1)<\tfrac{1}{2}+\tfrac{n-2}{2p}\\\ \\\ {\bf
Y}_{n_{0}+1}(x),\quad\mbox{if}\quad(n_{0}+1)+\frac{1}{2}+\frac{n-2}{2p}\leq
x<(n_{0}+2)\\\ \\\ {\bf Y}_{i}(x),\quad\mbox{if}\quad i\leq
x<i+1\quad\mbox{and}\quad n_{0}+2\leq i<n\end{cases}$

where

  1. (i)

for $x-n_{0}-1\in I^{(p)}_{(n_{1},\ldots,n_{l})}$

(7.4) ${\boldsymbol{\mu}}_{n_{0}}^{(p)}(x-n_{0}-1)=(n+1)^{th}~{}~{}\mbox{entry
of the matrix}\quad\\\ \ \Bigl{[}{\bf
l}_{0}\bigl{(}\phi^{(p)}(x-n_{0}-1)\bigr{)},\ldots,{\bf
l}_{n+1}\bigl{(}\phi^{(p)}(x-n_{0}-1)\bigr{)}\Bigr{]}_{1\times
n+2}\cdot{{\mathbb{H}}}^{(n_{l})}(t)\cdots{{\mathbb{H}}}^{(n_{1})}(t)\Bigr{\rvert}_{t=1/{p}}.$

  2. (ii)

For $x-n_{0}-1\in J^{(p)}_{(n_{1},\ldots,n_{l})}$

(7.5) ${\boldsymbol{\mu}}_{n_{0}}^{(p)}(x-n_{0}-1)=(n+1)^{th}~{}~{}\mbox{entry
of the matrix}\quad\\\ \ \left[{\bf
r}_{0}\bigl{(}\phi^{(p)}(x-n_{0}-1)\bigr{)},\ldots,{\bf
r}_{n+1}\bigl{(}\phi^{(p)}(x-n_{0}-1)\bigr{)}\right]_{1\times
n+2}\cdot{{\mathbb{H}}}^{(n_{l})}(t)\cdots{{\mathbb{H}}}^{(n_{1})}(t)\Bigr{\rvert}_{t=1/{p}}.$

######  Theorem 7.7.

If $n\geq 4$ is an even number and $p\geq(3n-4)/2$ then the HK density
function is defined outside measure $0$ (in ${\mathbb{R}}$) as follows:

$f_{R_{p,n+1}}(x)=\begin{cases}{\bf Z}_{i}(x),\quad\mbox{if}\quad i\leq
x<i+1\quad\mbox{and}\quad 0\leq i\leq n_{0}\\\ \\\ {\bf
Z}_{n_{0}+1}(x),\quad\mbox{if}\quad(n_{0}+1)\leq x<(n_{0}+2)-\frac{n-2}{2p}\\\
\\\ {\bf
Z}_{n_{0}+1}(x)+{{\boldsymbol{\mu}}}^{(p)}_{n_{0}-1}(x-n_{0}-1),\quad\mbox{if\quad$1-\frac{n-2}{2p}\leq
x-(n_{0}+1)<1$}\\\ \\\ {\bf
Y}_{n_{0}+2}(x)+{{\boldsymbol{\mu}}}^{(p)}_{n_{0}+1}(x-n_{0}-2),\quad\mbox{if\quad$0\leq
x-(n_{0}+2)<\small{\frac{n-2}{2p}}$}\\\ \\\ {\bf
Y}_{n_{0}+2}(x),\quad\mbox{if}\quad(n_{0}+2)+\frac{n-2}{2p}\leq x<(n_{0}+3)\\\
\\\ {\bf Y}_{i}(x),\quad\mbox{if}\quad i\leq x<i+1\quad\mbox{and}\quad
n_{0}+3\leq i<n,\end{cases}$

where

  1. (1)

for $x-n_{0}-1\in{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{(n_{1},\ldots,n_{l})}$,
where
$(n_{1},\ldots,n_{l})\in{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}_{1}}$

(7.6) ${\boldsymbol{\mu}}_{n_{0}-1}^{(p)}(x-n_{0}-1)=n^{th}~{}~{}\mbox{entry
of the matrix}\quad\\\ \ \left[{{\bf
m}}_{0}\bigl{(}\psi^{(p)}(x-n_{0}-1)\bigr{)},\ldots,{\bf
m}_{n+2}\bigl{(}\psi^{(p)}(x-n_{0}-1)\bigr{)}\right]_{1\times
n+3}\cdot{\mathchoice{\widetilde{{\mathbb{H}}}}{\widetilde{{\mathbb{H}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathbb{H}}}$}}}$}}{\tilde{{\mathbb{H}}}}}^{(n_{l})}(t)\cdots{\mathchoice{\widetilde{{\mathbb{H}}}}{\widetilde{{\mathbb{H}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathbb{H}}}$}}}$}}{\tilde{{\mathbb{H}}}}}^{(n_{1})}(t)\Bigr{\rvert}_{t=1/{p}}.$

  2. (2)

For $x-n_{0}-2\in{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{(n_{1},\ldots,n_{l})}$,
where
$(n_{1},\ldots,n_{l})\in{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}_{0}}$

(7.7)
${\boldsymbol{\mu}}_{n_{0}+1}^{(p)}(x-n_{0}-2)=(n+2)^{th}~{}~{}\mbox{entry of
the matrix}\quad\\\ \ \left[{{\bf
m}}_{0}\bigl{(}\psi^{(p)}(x-n_{0}-2)\bigr{)},\ldots,{{\bf
m}}_{n+2}\bigl{(}\psi^{(p)}(x-n_{0}-2)\bigr{)}\right]_{1\times
n+3}\cdot{\mathchoice{\widetilde{{\mathbb{H}}}}{\widetilde{{\mathbb{H}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathbb{H}}}$}}}$}}{\tilde{{\mathbb{H}}}}}^{(n_{l})}(t)\cdots{\mathchoice{\widetilde{{\mathbb{H}}}}{\widetilde{{\mathbb{H}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathbb{H}}}$}}}$}}{\tilde{{\mathbb{H}}}}}^{(n_{1})}(t)\Bigr{\rvert}_{t=1/{p}}.$

##  8\. Applications

In the previous section we constructed an infinite set of semi open intervals
such that they cover the domain of definition of the HK density function
$f_{R_{p,n+1}}$ except on a set of measure $0$.

Since $f_{R_{p,n+1}}$ is a continuous function, we can ignore the measure $0$
part as far as the integration is concerned. Moreover $f_{R_{p,n+1}}$
restricted to such semi open interval is a polynomial functions which are
independent of $p$. Though the length of each interval depends on $p$, the set
of indexing is independent of $p$. These properties of the formulation of
$f_{R_{p,n+1}}$ allow us to compare the HK multiplicity $e_{HK}(R_{p,n+1})$ as
$p$ varies. As a consequence we settle an old conjecture of Yoshida
(Conjecture (2) in the introduction) in the following form

Theorem (C) Let $n\geq 3$ be an integer then there exists $\epsilon>0$ such
that for $p\geq 1/\epsilon$, $e_{HK}(R_{p_{1},n+1})$ is a strictly decreasing
function of $p$.

Moreover, following Notations 8.3, the $\epsilon=\epsilon_{1}$ or
$\epsilon=\epsilon_{2}$ depending on the parity of $n$.

Here we state and prove a basic result from analysis in the form which suits
our purpose.

######  Lemma 8.1.

If $H(t)=t^{i}(b_{0}+b_{1}t+\cdots+b_{m}t^{m})\in{\mathbb{Q}}[t]$ is a
polynomial such that $H(1/{p})\geq 0$ for all $p\gg 0$ then either $H(t)$ is a
zero polynomial or $tH(t)$ is a strictly increasing function on the interval
$[0,\epsilon_{H})$, where we define

$\epsilon_{H}=\mbox{min}~{}\\{1,\frac{b_{0}}{2|b_{1}|+3|b_{2}|+\cdots+(m+1)|b_{m}|}\\}.$

In particular $tH(t)>0$ for $t\in(0,\epsilon_{H})$.

###### Proof.

Suppose $H(t)$ is not a zero polynomial. Further we can assume $m>0$.

Claim. The number $b_{0}$ is strictly positive.

Proof of the claim: Suppose $b_{0}<0$.

Since $H(t)$ is a nonzero polynomial it has only finitely many zeroes.
Therefore there exists $\epsilon_{1}>0$ such that $H(t)$ has no zero on the
interval $(0,\epsilon_{1})$. Now by choosing $p_{0}\geq 1/\epsilon_{1}$ we can
ensure that $H(t)$ has no zero in the set $\\{1/{p}\mid p\geq p_{0}\\}$. Since
$-b_{0}>0$, we can further choose $\epsilon_{1}$ such that for all
$t\in(0,~{}~{}\epsilon_{1})$ we have
$t\left(|b_{1}|+\cdots+|b_{m}|t^{m-1}\right)<-b_{0}$ which implies

$b_{0}+b_{1}t+\cdots+b_{m}t^{m}\leq b_{0}+|b_{1}|t+\cdots+|b_{m}|t^{m}<0.$

But then it contradicts the hypothesis that $H\bigl{(}1/p)\bigr{)}\geq 0$ for
all $p\gg 0$. This proves the claim.

Let $G(t)=t(b_{0}+\cdots+b_{m}t^{m})$, then

$\frac{d(G(t))}{dt}=b_{0}+2b_{1}t+\cdots+(m+1)b_{m}t^{m}>0,\quad\mbox{for}\quad
t\in[0,\epsilon_{H}),$

as

$-2b_{1}t-3b_{2}t^{2}-\cdots-(m+1)b_{m}t^{m}<t\bigl{(}2|b_{1}|+\cdots+(m+1)|b_{m}|\bigr{)}<b_{0}.$

Hence $G(t)$ and therefore $t(b_{0}+\cdots+b_{m}t^{m})(t^{i})=tH(t)$ is a
strictly increasing function on $[0,\epsilon_{H})$. In particular $tH(t)>0$
for $t\in(0,\epsilon_{H})$. ∎

In the rest of the section we use the notations as in Definitions 5.4 and 5.2
and Proposition 5.3.

The following lemma proves that the integral of ${\boldsymbol{r}}_{i}^{(p)}$,
restricted to the each semi open interval, is determined by a polynomial in
${\mathbb{Q}}[t]$ evaluated at $t=1/{p}$.

######  Lemma 8.2.

Let $n\geq 3$ be an odd integer and let $p\geq 3n-4$. Then for
$(n_{1},\ldots,n_{l})\in{\mathcal{M}}^{l}$, where $l\geq 1$

(8.1)
$\Bigl{[}\int_{I^{(p)}_{(n_{1},\ldots,n_{l})}}{\boldsymbol{r}}^{(p)}_{0}(x)dx,\ldots,\int_{I^{(p)}_{(n_{1},\ldots,n_{l})}}{\boldsymbol{r}}^{(p)}_{n+1}(x)\Bigr{]}_{1\times
n+2}=\\\ \left[F_{0}(t),\ldots,F_{n+1}(t)\right]_{1\times
n+2}\cdot(t\cdot{{\mathbb{H}}}^{(n_{l})}(t))\cdots(t\cdot{{\mathbb{H}}}^{(n_{1})}(t))\Bigr{\rvert}_{t={1}/{{p}}}$

and

(8.2)
$[\int_{J^{(p)}_{(n_{1},\ldots,n_{l})}}{\boldsymbol{r}}^{(p)}_{0}(x)dx,\ldots,\int_{J^{(p)}_{(n_{1},\ldots,n_{l})}}{\boldsymbol{r}}^{(p)}_{n+1}(x)]_{1\times
n+2}=\\\ \left[G_{0}(t),\ldots,G_{n+1}(t)\right]_{1\times
n+2}\cdot\bigl{(}t\cdot{{\mathbb{H}}}^{(n_{l})}(t)\bigr{)}\cdots\bigl{(}t\cdot{{\mathbb{H}}}^{(n_{1})}(t)\bigr{)}\Bigr{\rvert}_{t={1}/{{p}}}.$

Let $n\geq 4$ be an even integer and let $p\geq(3n-4)/2$ then

(8.3) $\Bigl{[}\int_{{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{(n_{1},\ldots,n_{l})}}{\boldsymbol{r}}^{(p)}_{0}(x)dx,\ldots,\int_{I^{(p)}_{(n_{1},\ldots,n_{l})}}{\boldsymbol{r}}^{(p)}_{n+2}(x)\Bigr{]}_{1\times
n+3}=\\\ \left[{\mathchoice{\widetilde{F}}{\widetilde{F}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{F}$}}}$}}{\tilde{F}}}_{0}(t),\ldots,{\mathchoice{\widetilde{F}}{\widetilde{F}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{F}$}}}$}}{\tilde{F}}}_{n+2}(t)\right]_{1\times
n+3}\cdot\bigl{(}t\cdot{\mathchoice{\widetilde{{\mathbb{H}}}}{\widetilde{{\mathbb{H}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathbb{H}}}$}}}$}}{\tilde{{\mathbb{H}}}}}^{(n_{l})}(t)\bigr{)}\cdots\bigl{(}t\cdot{\mathchoice{\widetilde{{\mathbb{H}}}}{\widetilde{{\mathbb{H}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathbb{H}}}$}}}$}}{\tilde{{\mathbb{H}}}}}^{(n_{1})}(t)\bigr{)}\Bigr{\rvert}_{t={1}/{{p}}}.$

###### Proof.

For $I^{(p)}_{(n_{1},\ldots,n_{l})}$ and $J^{(p)}_{(n_{1},\ldots,n_{l})}$ as
in Definition 6.6

$\int_{I^{(p)}_{(n_{1},\ldots,n_{l})}}{\bf
l}_{i_{0}}\Bigl{(}p^{l}(x-\sum_{i}\frac{j_{i}}{p^{i}})\Bigr{)}dx=\frac{1}{p^{l}}\int_{0}^{\tfrac{1}{2}-\tfrac{n-2}{2p}}{\bf
l}_{i_{0}}(x)dx=t^{l}F_{i_{0}}(t)\Bigr{\rvert}_{t=1/{p}}$

and

$\int_{J^{(p)}_{(n_{1},\ldots,n_{l})}}{\bf
r}_{i_{0}}\Bigl{(}p^{l}(x-\sum_{i}\frac{j_{i}}{p^{i}})\Bigr{)}dx=\frac{1}{p^{l}}\int_{\tfrac{1}{2}+\tfrac{n-2}{2p}}^{1}{\bf
m}_{i_{0}}(x)dx=t^{l}\cdot G_{i_{0}}(t)\Bigr{\rvert}_{t=1/{p}}.$

Therefore by (7.1) and (7.2) of Proposition 7.2 the assertions follow.

For even $n$ the assertions follow by Proposition 7.4 ∎

######  Notations 8.3.

For a polynomial $H(t){\mathbb{Q}}[t]$ we define $\epsilon_{H}$ as given in
Lemma 8.1. Following Definitions 5.2 and Definition 5.4, let

${\mathcal{S}}_{gp}=\\{H_{k_{1},k_{2}}^{(n_{i})}(t)\mid 0\leq n_{i}\leq
n-3,~{}~{}0\leq k_{1},k_{2}\leq n+1\\}\cup\\{F_{i}(t),G_{i}(t)\mid 0\leq i\leq
n+1\\}$

and let $\epsilon_{1}=\min\\{\epsilon_{H}\mid H\in{\mathcal{S}}_{gp}\\}$.

Similarly let

${\mathchoice{\widetilde{{\mathcal{S}}}}{\widetilde{{\mathcal{S}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{S}}}$}}}$}}{\tilde{{\mathcal{S}}}}}_{gp}=\\{{\mathchoice{\widetilde{H}}{\widetilde{H}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{H}$}}}$}}{\tilde{H}}}_{k_{1},k_{2}}^{(n_{i})}(t)\mid
0\leq n_{i}\leq n-3,~{}~{}0\leq k_{1},k_{2}\leq
n+1\\}\cup\\{{\mathchoice{\widetilde{F}}{\widetilde{F}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{F}$}}}$}}{\tilde{F}}}_{0}(t),\ldots,{\mathchoice{\widetilde{F}}{\widetilde{F}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{F}$}}}$}}{\tilde{F}}}_{n+2}(t)\\}$

and let $\epsilon_{2}=\min\\{\epsilon_{H}\mid
H\in{\mathchoice{\widetilde{{\mathcal{S}}}}{\widetilde{{\mathcal{S}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{S}}}$}}}$}}{\tilde{{\mathcal{S}}}}}_{gp}\\}$.

######  Lemma 8.4.

If $n\geq 3$ is odd number and $p\geq 1/\epsilon_{1}$. Then

  1. (o1)

for $(n_{1},\ldots,n_{l})\in\cup_{l\geq 1}{\mathcal{M}}^{l}$

$\int_{I^{(p)}_{(n_{1},\ldots,n_{l})}}{\boldsymbol{r}}^{(p)}_{i}(x)dx\quad\mbox{and}\quad\int_{J^{(p)}_{(n_{1},\ldots,n_{l})}}{\boldsymbol{r}}^{(p)}_{i}(x)dx\quad\mbox{
are decreasing functions of}\quad p.$

  2. (o2)

Moreover, for each $n_{1}\in\\{0,\ldots,n-3\\}$

$\int_{I^{(p)}_{(n_{1})}}{\boldsymbol{r}}^{(p)}_{n}(x)dx=\int_{I^{(p)}_{(n_{1})}}{\boldsymbol{\mu}}^{(p)}_{n_{0}-1}(x)dx\quad\mbox{
is a strictly decreasing function of}~{}~{}p.$

If $n\geq 4$ is an even integer and $p\geq 1/\epsilon_{2}$. Then

  1. (e1)

for
$(n_{1},\ldots,n_{l})\in{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}}_{0}\cup{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}}_{1}$

$\int_{{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{(n_{1},\ldots,n_{l})}}{\boldsymbol{r}}^{(p)}_{i}(x)dx~{}~{}\mbox{is
a decreasing function of}~{}~{}p.$

  2. (e2)

Moreover, for each $n_{1}\in\\{\tfrac{n}{2}-1,\ldots,n-3\\}$

$\int_{{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{p}_{(n_{1})}}{\boldsymbol{\mu}}^{(p)}_{n_{0}-1}(x)dx\quad\mbox{
is a strictly decreasing function of}~{}~{}p.$

###### Proof.

Assertion (o1) and Assertion (e1) follow from Lemma 8.1 and Lemma 8.2.

Assertion (o2). We recall that
${\boldsymbol{\mu}}_{n_{0}}^{(p)}(x)={\boldsymbol{r}}_{n+1}^{(p)}(x)$.
Therefore, by Proposition 7.2, for $p\geq 3n-4$, we have

(8.4)
$\int_{I^{(p)}_{(n_{1})}}{\boldsymbol{\mu}}^{(p)}_{n_{0}}(x)dx=\sum_{i=0}^{n+1}t\Bigl{[}{{\mathbb{H}}}^{(n_{1})}(t)\Bigr{]}_{i,n+1}F_{i}(t)\Bigr{\rvert}_{t=1/p}=\sum_{i=0}^{n+1}tH^{(n_{1})}_{i,n+1}(t)F_{i}(t)\Bigr{\rvert}_{t=1/p},$

where $n_{1}\in\\{0,\ldots,n-3\\}$,

By Lemma 8.1, it is sufficient to prove
$tH^{(n_{1})}_{n-1,n+1}(t)F_{n-1}(t)\neq 0$, which follows as we have

$H^{(n_{1})}_{n-1,n+1}({1}/p)={b_{n-1,n+1}^{(n_{1})}}/{p^{n}}={\mu_{-n_{0}}(j_{1}-n+1)}/{p^{n}}>0,$

where the last inequality holds as, by Lemma 10.3, the condition $0\leq
j_{1}-n+1\leq m_{0}-2$ gives $\mu_{-n_{0}}(j_{1}-n+1)/p^{n}>0$. Also

$F_{n-1}(\tfrac{1}{p})=\int_{0}^{\tfrac{1}{2}-\tfrac{n-2}{2p}}{\bf
l}_{n-1}(x)dx=\int_{0}^{\tfrac{1}{2}-\tfrac{n-2}{2p}}\frac{2}{n!}(1-x)^{n}dx>0$

as $p>n-2$.

Assertion (e2). Now consider $n_{1}\in\\{\tfrac{n}{2}-1,\ldots,n-3\\}$ then
following the notations of Proposition 5.3

${\mathchoice{\widetilde{F}}{\widetilde{F}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{F}$}}}$}}{\tilde{F}}}_{0}(1/p)=\int_{\tfrac{n-2}{2p}}^{1-\frac{n-2}{2p}}{\bf
Z}_{0}(x)dx=\frac{2}{(n+1)!}\left[\left(1-\tfrac{n-2}{2p}\right)^{n+1}-\left(\tfrac{n-2}{2p}\right)^{n+1}\right]>0.$

Also, by Lemma 10.4, the condition
$j_{1}=p+\tfrac{n}{2}-2-n_{1}={\mathchoice{\widetilde{m}}{\widetilde{m}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{m}$}}}$}}{\tilde{m}}_{0}}+n-3-n_{1}\geq{\mathchoice{\widetilde{m}}{\widetilde{m}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{m}$}}}$}}{\tilde{m}}_{0}}$ gives

${\mathchoice{\widetilde{H}}{\widetilde{H}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{H}$}}}$}}{\tilde{H}}}^{(n_{1})}_{0,n}(1/p)={c_{0n}^{(n_{1})}}/{p^{n}}=\mu_{-n_{0}+1}(j_{1})/p^{n}>0.$

Whereas, by Proposition 7.4, for $p\geq(3n-4)/2$

(8.5) $\int_{{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{(n_{1})}}{\boldsymbol{\mu}}^{(p)}_{n_{0}-1}(x)dx=\sum_{i=0}^{n+2}\left(t{\mathchoice{\widetilde{H}}{\widetilde{H}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{H}$}}}$}}{\tilde{H}}}_{i,n}^{(n_{1})}(t){\mathchoice{\widetilde{F}}{\widetilde{F}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{F}$}}}$}}{\tilde{F}}}_{i}(t)\right)\mid_{t=1/p}.$

Hence, by the same logic as above for $n$ equal to odd case, this integral is
a strictly increasing function of $t$ for $t\in[0,\epsilon_{2})$, where
$\epsilon_{2}$ is as in the above notations. ∎

######  Remark 8.5.

By Definition 5.4 the polynomials $F_{i}(t)\bigr{\rvert}_{1/p}$,
$H_{k_{1},k_{2}}^{(n_{i})}(t)\bigr{\rvert}_{1/p}$ are nonnegative for $p\geq
3n-4$ and ${\mathchoice{\widetilde{F}}{\widetilde{F}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{F}$}}}$}}{\tilde{F}}}_{i}(t)\bigr{\rvert}_{1/p}$,
${\mathchoice{\widetilde{H}}{\widetilde{H}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{H}$}}}$}}{\tilde{H}}}_{k_{1},k_{2}}^{(n_{i})}(t)\bigr{\rvert}_{1/p}$
are nonnegative for $p\geq(3n-4)/2$. Therefore (8.4) and (8.5) imply that

  1. (1)

if $n\geq 3$ odd and $p\geq 3n-4$ then
$\int{\boldsymbol{\mu}}_{n_{0}}^{(p)}(x)dx>0$.

  2. (2)

If $n\geq 4$ even and $p\geq(3n-4)/2$ then
$\int{\boldsymbol{\mu}}_{n_{0}-1}^{(p)}(x)dx>0$.

Therefore there exists interval $I_{1}\subset[0,1)$ such that the rank
function ${\boldsymbol{\mu}}_{n_{0}}^{(p)}$ are strictly positive on $I_{1}$
in case $n$ is odd. Similar assertion holds for $n$ even.

######  Theorem 8.6.

If $n\geq 3$ then there exists $\epsilon>0$ such that for $p\geq 1/\epsilon$,
the HK multiplicity $e_{HK}(R_{p,n+1})$ of the ring $R_{p,n+1}$ is a strictly
decreasing function of $p$.

Moreover, following Notations 8.3, one can take $\epsilon=\epsilon_{1}$ or
$\epsilon=\epsilon_{2}$ depending on the parity of $n$.

###### Proof.

Suppose $n\geq 3$ is an odd number. Then

(8.6)
$e_{HK}(R_{p,n+1})=\int_{0}^{\infty}f_{R_{p},n+1}(x)dx=\int_{0}^{\infty}f^{\infty}_{R_{n+1}}(x)dx+\int_{\frac{1}{2}-\frac{n-2}{2p}}^{\frac{1}{2}+\frac{(n-2)}{2p}}{\boldsymbol{\mu}}^{(p)}_{n_{0}}(x)dx,$

where the first equality follows from Theorem 1.1 of [T1] and the second from
Proposition 1.1. Also by Proposition 1.1, the function
${\boldsymbol{\mu}}^{(p)}_{n_{0}}$ is a real valued continuous function on the
interval $[0,1)$. Hence by Lemmas 6.2 and 6.7

(8.7)
$\int^{\frac{1}{2}+\frac{(n-2)}{2p}}_{\frac{1}{2}-\frac{n-2}{2p}}{\boldsymbol{\mu}}^{(p)}_{n_{0}}(x)dx=\sum_{{\underline{n}}\in\cup_{l\geq
1}{\mathcal{M}}^{l}}\int_{I^{(p)}_{\underline{n}}}{\boldsymbol{\mu}}^{(p)}_{n_{0}}(x)dx+\sum_{{\underline{n}}\in\cup_{l\geq
1}{\mathcal{M}}^{l}}\int_{J^{(p)}_{\underline{n}}}{\boldsymbol{\mu}}^{(p)}_{n_{0}}(x)dx.$

Similarly, when $n\geq 4$ is even then

(8.8)
$e_{HK}(R_{p,n+1})=\displaystyle{\int_{0}^{\infty}f^{\infty}_{R_{n+1}}(x)dx+\int_{\frac{1}{2}-\frac{(n-2)}{2p}}^{1}{\boldsymbol{\mu}}^{(p)}_{n_{0}-1}(x)dx+\int_{0}^{\frac{(n-2)}{2p}}{\boldsymbol{\mu}}^{(p)}_{n_{0}+1}(x)dx},\\\
=\displaystyle{\int_{0}^{\infty}f^{\infty}_{R_{n+1}}(x)dx+\sum_{{\underline{n}}\in{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}}_{1}}\int_{{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{\underline{n}}}{\boldsymbol{\mu}}^{(p)}_{n_{0}-1}(x)dx+\sum_{{\underline{n}}\in{\mathchoice{\widetilde{{\mathcal{M}}}}{\widetilde{{\mathcal{M}}}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{{\mathcal{M}}}$}}}$}}{\tilde{{\mathcal{M}}}}}_{0}}\int_{{\mathchoice{\widetilde{I}}{\widetilde{I}}{\lower
2.0pt\hbox{$\textstyle{\tilde{\raise
2.0pt\hbox{$\scriptstyle{I}$}}}$}}{\tilde{I}}}^{(p)}_{\underline{n}}}{\boldsymbol{\mu}}^{(p)}_{n_{0}+1}(x)dx},$

where the last equality follows by Lemmas 6.2 and 6.5. Hence theorem follows
by Lemma 8.4. ∎

In fact for $p\gg 0$, the HK multiplicity $e_{HK}(R_{p,n+1})$ can be expressed
as a quotient of two rational polynomials. For this first we recall the
Cofactor Formula for the Inverse of a Matrix.

Let $R$ be a commutative ring with $1$.

$\mbox{Let}\quad{{\mathbb{B}}}\in M_{n+2}(R)\quad\mbox{and
let}\quad{{\mathbb{B}}_{c}}={[(-1)^{i+j}\det{\mathbb{B}}_{ji}]}_{i,j}\in
M_{n+2}(R),$

where ${{\mathbb{B}}}_{ij}\in M_{n+1}(R)$ is obtained from ${\mathbb{B}}$ by
deleting $i^{th}$ row and $j^{th}$ column. Then
${{\mathbb{B}}}{{\mathbb{B}}_{c}}={{\mathbb{B}}_{c}}{{\mathbb{B}}}=(\det{{\mathbb{B}}}){{\mathbb{I}}}_{n+2}$.

In particular $\det{{\mathbb{B}}}$ is a unit in $R$ if and only if
${{\mathbb{B}}}$ is unit in $M_{n+2}(R)$ and in that case we have
${{\mathbb{B}}}^{-1}=\frac{1}{\det{{\mathbb{B}}}}\cdot{{\mathbb{B}}_{c}}\in
M_{n+2}(R)$.

######  Notations 8.7.

Here $F_{i}(t)$ and $G_{i}(t)$ are polynomials in ${\mathbb{Q}}[t]$ as in
Proposition 5.3, and ${{\mathbb{H}}}^{(n_{i})}(t)\in M_{n+2}({\mathbb{Q}}[t])$
as in Definition 5.4. We define

${{\mathbb{B}}}(t)=t\cdot{{\mathbb{H}}}^{(0)}(t)+t\cdot{{\mathbb{H}}}^{(1)}(t)+\cdots+t\cdot{{\mathbb{H}}}^{(n-3)}(t),$

Also for any matrix ${\mathbb{B}}$, we will denote
$\frac{1}{\det{{\mathbb{B}}}}\cdot{{\mathbb{B}}_{c}}$ by
$\frac{{\mathbb{B}}_{c}}{\det{{\mathbb{B}}}}$.

######  Theorem 8.8.

Let $n\geq 3$ be an odd integer then let $p\geq 5$ and $p>2^{\lfloor
n/2\rfloor}(n-2)$. Then

$e_{HK}(R_{p,n+1})=(1+m_{n+1})+(n+2)^{th}~{}~{}\mbox{entry of the matrix}\\\ \
\bigl{[}F_{0}(t)+G_{0}(t),\ldots,F_{n+1}(t)+G_{n+1}(t)\bigr{]}_{1\times
n+2}\cdot{{\mathbb{B}}}(t)\cdot\frac{[{\mathbb{I}}_{n+2}-{\mathbb{B}}(t)]_{c}}{\det({{\mathbb{I}}}_{n+2}-{{\mathbb{B}}}(t))}\Bigr{\rvert}_{t=1/p},$

where $[{\mathbb{I}}_{n+2}-{{\mathbb{B}}}(t)]_{c}$ is the cofactor matrix of
${{\mathbb{I}}}_{n+2}-{{\mathbb{B}}}(t)$.

###### Proof.

By (8.6) and (8.7)

$e_{HK}(R_{p,n+1})=\int_{0}^{\infty}f^{\infty}_{R_{n+1}}(x)dx+\sum_{{\underline{n}}\in\cup_{l\geq
1}{\mathcal{M}}^{l}}\int_{I^{(p)}_{\underline{n}}}{\boldsymbol{\mu}}^{(p)}_{n_{0}}(x)dx+\sum_{{\underline{n}}\in\cup_{l\geq
1}{\mathcal{M}}^{l}}\int_{J^{(p)}_{\underline{n}}}{\boldsymbol{\mu}}^{(p)}_{n_{0}}(x)dx.$

If $p\geq 3n-4$ then by Lemma 8.2 (8.1)

$\sum_{{\underline{n}}\in{\mathcal{M}}^{l}}\int_{I^{(p)}_{\underline{n}}}{\boldsymbol{\mu}}^{(p)}_{n_{0}}(x)dx=(n+2)^{th}~{}~{}\mbox{entry
of}~{}~{}\\\ \ \left[F_{0}(t),\ldots,F_{n+1}(t)\right]_{1\times
n+2}\cdot\left(t\cdot{{\mathbb{H}}}^{(0)}(t)+t\cdot{{\mathbb{H}}}^{(1)}(t)+\cdots+t\cdot{{\mathbb{H}}}^{(n-3)}(t)\right)^{l}\Bigr{\rvert}_{t=1/{p}}$

Therefore

$\sum_{{\underline{n}}\in\cup_{l\geq
1}{\mathcal{M}}^{l}}\int_{I^{(p)}_{\underline{n}}}{\boldsymbol{\mu}}^{(p)}_{n_{0}}(x)dx=(n+2)^{th}~{}~{}\mbox{entry
of}~{}~{}\left[F_{0}(t),\ldots,F_{n+1}(t)\right]_{1\times n+2}\cdot\sum_{l\geq
1}{{\mathbb{B}}}(t)^{l}\Bigr{\rvert}_{t=1/{p}}.$

Similarly, by Lemma 8.2 (8.2)

$\sum_{{\underline{n}}\in\cup_{l\geq
1}{\mathcal{M}}^{l}}\int_{J^{(p)}_{\underline{n}}}{\boldsymbol{\mu}}^{(p)}_{n_{0}}(x)dx=(n+2)^{th}~{}~{}\mbox{entry
of}~{}~{}\left[G_{0}(t),\ldots,G_{n+1}(t)\right]_{1\times n+2}\cdot\sum_{l\geq
1}{{\mathbb{B}}}(t)^{l}\Bigr{\rvert}_{t=1/{p}}.$

Hence to prove the theorem it only remains to prove the following

Claim. $\det({\mathbb{I}}_{n+2}-{\mathbb{B}}(1/p))\neq 0$ if $p>2^{\lfloor
n/2\rfloor}(n-2)$ and $p\geq 5$.

Proof of the claim: We fix a prime $p$ as in the claim, and consider the
$l^{1}$ norm on ${\mathbb{Q}}^{n+2}$, that is for
$v=(v_{0},\ldots,v_{n+1})\in{\mathbb{Q}}^{n+2}$ the $l^{1}$ norm of $v$ is
given by $\|v\|=|v_{0}|+\cdots|v_{n+1}|$.

Note the-t if the claim does not hold true then there exists a nonzero vector
$v\in{\mathbb{Q}}^{n+2}$ such that
$v\cdot({\mathbb{I}}_{n+2}-{\mathbb{B}}(1/p))=0$, which implies
$\|v\|=\|v\cdot{\mathbb{B}}(1/p)\|$.

Now we show that $l^{1}$ norm of the matrix ${\mathbb{B}}(1/p)$ is $<1$, that
is $\|v\cdot{\mathbb{B}}(1/p)\|<\|v\|$, for any nonzero vector $v$. We recall

${\mathbb{B}}(1/p)={1}/{p}\sum_{i=0}^{n-3}{\mathbb{H}}^{(n_{i})}(1/p)={1}/{p}\sum_{i=0}^{n-3}{{\mathbb{B}}^{(p)}_{n_{i}}}/{p^{n}}.$

This gives