Notes on the Distribution of Roots Modulo a Prime of a Polynomial V: Weyl’s Criterion

Abstract

Let $f\left(x\right)$ be a monic integral polynomial of degree n and p a prime number, for which $f\left(x\right)$ is fully decomposable modulo p. Let $r_1,\dots ,r_n$ be the roots of $f\left(x\right)modp$ with $0\le r_1\le \cdots \le r_n<p$. We have conjectured that the sequence of $(r_1,\dots ,r_n)/p$ is uniformly distributed in some sense. We provide a clear explanation of this and generalize the Weyl criterion.

1. Introduction and Conjectures

For a monic polynomial $f\left(x\right)\in \mathbb{Z}\left[x\right]$, we proposed several conjectures about the distribution of roots of $f\left(x\right)\equiv 0modp$ for a prime number p ([1,2,3,4,5]). Very roughly speaking, they predict that the distribution is uniform. In order to explain the aim of this paper, we review them to some extent. Let a polynomial

$$f\left(x\right)=x^n+a_{n-1}{x}^{n-1}+\cdots +a_0(a_0,\cdots ,a_{n-1}\in \mathbb{Z})$$

be of degree n with complex roots ${\alpha }_1,\dots ,{\alpha }_n$. We fix the numbering of roots once and for all and define vector spaces $L{R}_0,LR$ over the rational number field $\mathbb{Q}$ by

$$\begin{array}{cc}L{R}_0\hfill & :=\{(l_1,\dots ,l_n)\in {\mathbb{Q}}^n\mid \sum _{i=1}^n{l}_i{\alpha }_i\in \mathbb{Q}\},\hfill \\ LR\hfill & :=\{(l_1,\dots ,l_{n+1})\in {\mathbb{Q}}^{n+1}\mid \sum _{i=1}^n{l}_i{\alpha }_i=l_{n+1}\}.\hfill \end{array}$$

The non-zero vector $(1,\dots ,1,-a_{n-1})$ is always in $LR$. Hence, we see clearly that

$$1\le t:={dim}_{\mathbb{Q}}L{R}_0={dim}_{\mathbb{Q}}LR\le n.$$

We say that a polynomial $f\left(x\right)$ has a non-trivial linear relation among roots if $t>1$. We know that if there is only a trivial linear relation, then the polynomial $f\left(x\right)$ is irreducible, and that if $f\left(x\right)$ is irreducible and $degf\left(x\right)=n$ is prime or the Galois group of $f\left(x\right)$ is isomorphic to the symmetric group $S_n$, then there is only a trivial linear relation among roots (cf. [1]).

We take and fix a $\mathbb{Z}$-basis of $L{R}_0\cap {\mathbb{Z}}^n$

$${\mathit{m}}_j:=(m_{j,1},\dots ,m_{j,n})(j=1,\dots ,t),$$

and write

$$m_j:=\sum _{i=1}^n{m}_{j,i}{\alpha }_i(\in \mathbb{Z})(j=1,\dots ,t).$$

We may take ${\mathit{m}}_1=(1,\dots ,1),m_1=-a_{n-1}$.

We introduce the following two groups associated with them:

$$\begin{array}{cc}\hfill \widehat{\mathit{G}}:& =\{\sigma \in S_n\mid \sigma \left(LR\right)\subset LR\},\hfill \\ \hfill \mathit{G}:& =\{\sigma \in S_n\mid \sigma \left(L{R}_0\right)\subset L{R}_0\},\hfill \end{array}$$

where the action of a permutation $\sigma \in S_n$ for $\mathit{x}=(x_1,\dots ,x_n)\in {\mathbb{R}}^n,x\in \mathbb{R}$ is defined by the following.

$$\begin{array}{cc}\hfill \sigma \left(\mathit{x}\right):=(x_{{\sigma }^{-1}\left(1\right)},\dots ,x_{{\sigma }^{-1}\left(n\right)}),\sigma \left((\mathit{x},x)\right)& :=\left(\sigma \right(\mathit{x}),x).\hfill \end{array}$$

If $f\left(x\right)$ has only a trivial linear relation among roots, then $(1,\dots ,1,-a_{n-1})$ is a basis of $LR\cap {\mathbb{Z}}^{n+1}$, hence $\widehat{\mathit{G}}=\mathit{G}=S_n$.

Next, in this paper, it is assumed that the letter p denotes a prime number, and put

$${\mathrm{Spl}}_X\left(f\right):=\left\{p\le X\mid f\left(x\right)\mathrm{is}\mathrm{fully}\mathrm{splitting}\mathrm{modulo}p\right\}$$

for a positive number X and $\mathrm{Spl}\left(f\right):={\mathrm{Spl}}_{\infty }\left(f\right)$.

We require the following two conditions on the local roots $r_1,\dots ,r_n(\in \mathbb{Z})$ of $f\left(x\right)\equiv 0modp$ for a prime $p\in \mathrm{Spl}\left(f\right)$:

$$\begin{array}{cc}\hfill & f\left(x\right)\equiv \prod _{i=1}^n(x-r_i)modp,\hfill \\ \hfill & 0\le r_1\le r_2\le \cdots \le r_n<p.\hfill \end{array}$$

We are concerned with the distribution of the vectors $(r_1/p,\dots ,r_n/p)$ and $(r_1,\dots ,r_n)$ for $p\in \mathrm{Spl}\left(f\right)$.

To state conjectures, we need to introduce the following notation.

$$\begin{array}{c}{\mathrm{Spl}}_X(f,\sigma ):=\left\{p\in {\mathrm{Spl}}_X\left(f\right)\left|\sum _{i=1}^n{m}_{j,i}{r}_{\sigma \left(i\right)}\equiv m_{j}modp(1\le {}^{\forall }j\le t)\right.\right\},\\ {\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\}):=\left\{p\in {\mathrm{Spl}}_X(f,\sigma )\left|\sum _{i=1}^n{m}_{j,i}{r}_{\sigma \left(i\right)}=m_j+k_{j}p(1\le {}^{\forall }j\le t)\right.\right\}.\end{array}$$

We note that

$${\mathrm{Spl}}_X\left(f\right)={\cup }_{\sigma \in S_n}{\mathrm{Spl}}_X(f,\sigma ),{\mathrm{Spl}}_X(f,\sigma )={\cup }_{\left\{k_j\right\}\in {\mathbb{Z}}^t}{\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\}),$$

where the set ${\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})$ is non-empty only for a finite number of $\left\{k_j\right\}\in {\mathbb{Z}}^t$. Furthermore, for integers $L(>1),R_1,\dots ,R_n$ we put

$$\begin{array}{c}{\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\},L,\left\{R_i\right\}):=\{p\in {\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})\mid r_i\equiv R_{i}modL(1\le {}^{\forall }i\le n)\}\end{array}$$

and we define densities by

$$\begin{array}{c}\mathrm{Pr}(f,\sigma ):=\underset{X\to \infty }{lim}{\displaystyle \frac{\#{\mathrm{Spl}}_X(f,\sigma )}{\#{\mathrm{Spl}}_X\left(f\right)}},\\ \mathrm{Pr}(f,\sigma ,\left\{k_j\right\}):=\underset{X\to \infty }{lim}{\displaystyle \frac{\#{\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})}{\#{\mathrm{Spl}}_X(f,\sigma )}},\\ \mathrm{Pr}(f,\sigma ,\left\{k_j\right\},L,\left\{R_i\right\}):=\underset{X\to \infty }{lim}{\displaystyle \frac{\#{\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\},L,\left\{R_i\right\})}{\#{\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})}},\end{array}$$

where for the last two, the denominators $\#{\mathrm{Spl}}_X(f,\sigma ),\#{\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})$ of the right-hand sides are supposed to tend to infinity, and the existence of the above three limits is the basic conjecture.

If there is only a trivial linear relation among roots, then ${\mathrm{Spl}}_X(f,\sigma )={\mathrm{Spl}}_X\left(f\right)$, i.e., $\mathrm{Pr}(f,\sigma )=1$ and ${\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})$ is independent of the permutation $\sigma$.

Next, we introduce the following geometric objects:

$$\begin{array}{cc}\hfill \Delta :& =\left\{\mathit{x}=(x_1,\dots ,x_n)\in {\mathbb{R}}^n\mid 0\le x_1\le \cdots \le x_n\le 1\right\},\hfill \\ \hfill {\widehat{\mathfrak{D}}}_n:& =\left\{(x_1\dots ,x_n)\in \Delta \left|\sum _{i=1}^n{x}_i\in \mathbb{Z}\right.\right\},\hfill \\ \hfill \mathfrak{D}(f,\sigma ):& =\left\{(x_1\dots ,x_n)\in \Delta \left|\sum _{i=1}^n{m}_{j,i}{x}_{\sigma \left(i\right)}\in \mathbb{Z}(1\le {}^{\forall }j\le t)\right.\right\},\hfill \\ \hfill \mathfrak{D}(f,\sigma ,\left\{k_j\right\}):& =\left\{(x_1,\dots ,x_n)\in \Delta \left|\sum _{i=1}^n{m}_{j,i}{x}_{\sigma \left(i\right)}=k_j(1\le {}^{\forall }j\le t)\right.\right\}.\hfill \end{array}$$

We note that $\mathfrak{D}(f,\sigma ,\left\{k_j\right\})\subset \mathfrak{D}(f,\sigma )\subset {\widehat{\mathfrak{D}}}_n\subset \Delta$ and

$$\mathfrak{D}(f,\sigma )={\cup }_{\left\{k_j\right\}\in {\mathbb{Z}}^t}\mathfrak{D}(f,\sigma ,\left\{k_j\right\})(\mathrm{a}\mathrm{finite}\mathrm{union}),$$

and so $\mathfrak{D}(f,\sigma )$ is disconnected, in general.

If the set $\mathrm{Spl}(f,\sigma ,\left\{k_j\right\})$ is an infinite set, every accumulation point $\mathit{x}=(x_1,\dots ,x_n)$ of $(r_1/p,\dots ,r_n/p)\in {[0,1)}^n$ for $p\in \mathrm{Spl}(f,\sigma ,\left\{k_j\right\})$ is in $\mathfrak{D}(f,\sigma ,\left\{k_j\right\})$ by ${\sum }_i{m}_{j,i}{r}_{\sigma \left(i\right)}=m_j+k_{j}p$. If $m_j=0$ holds for every j $(1\le j\le t)$, then $(r_1/p,\dots ,r_n/p)$ is in $\mathfrak{D}(f,\sigma ,\left\{k_j\right\})$. Otherwise, it is not in $\mathfrak{D}(f,\sigma ,\left\{k_j\right\})$. This is troublesome. The dimension of ${\widehat{\mathfrak{D}}}_n$ is $n-1$ and that of $\mathfrak{D}(f,\sigma )$ and $\mathfrak{D}(f,\sigma ,\left\{k_j\right\})$ is less than or equal to $n-t$. In the following, the volume $\mathrm{vol}$ of sets $\mathfrak{D}(f,\sigma )$ and $\mathfrak{D}(f,\sigma ,\left\{k_j\right\})$ means that as a subset of an $(n-t)$-dimensional set $\left\{(x_1\dots ,x_n)\in {\mathbb{R}}^n\left|{\sum }_{i=1}^n{m}_{j,i}{x}_{\sigma \left(i\right)}\in \mathbb{Z}(1\le {}^{\forall }j\le t)\right.\right\}$, that is, for a subset S in $\mathit{v}+\mathbb{R}[{\mathit{v}}_1,\dots ,{\mathit{v}}_{n-t}](\subset {\mathbb{R}}^n)$ with orthonormal vectors ${\mathit{v}}_1,\dots ,{\mathit{v}}_{n-t}\in {\mathbb{R}}^n$, we identify $\mathit{v}+\sum y_i{\mathit{v}}_i\in S$ with a point $(y_1,\dots ,y_{n-t})\in {\mathbb{R}}^{n-t}$. So, $\mathrm{vol}\left(\mathfrak{D}\right(f,\sigma \left)\right)=0$ holds if $dim\mathfrak{D}(f,\sigma )<n-t$.

If there is only a trivial linear relation among roots, then $\mathfrak{D}(f,\sigma )$ is equal to ${\widehat{\mathfrak{D}}}_n$ for every permutation $\sigma$, and ${\widehat{\mathfrak{D}}}_n$ is equal to

$$\{(0,\dots ,0),(1,\dots ,1)\}\cup {\cup }_{k=1}^{n-1}\{\mathit{x}\in \Delta \mid \sum x_i=k\}.$$

The first conjecture is the following:

Conjecture 1. 

For a permutation $\sigma \in S_n$ with $\mathrm{Pr}(f,\sigma )>0$, the ratio

$$c={\displaystyle \frac{\mathrm{vol}\left(\mathfrak{D}\right(f,\sigma \left)\right)}{\mathrm{Pr}(f,\sigma )}}$$

is independent of $\sigma$. If $\mathit{G}=\widehat{\mathit{G}}$ holds, then two conditions $\mathrm{Pr}(f,\sigma )>0$ and $\mathrm{vol}\left(\mathfrak{D}\right(f,\sigma \left)\right)>0$ are equivalent.

We note that if $f\left(x\right)$ has no rational root and ${\alpha }_i-{\alpha }_j\notin \mathbb{Q}$ for any distinct $i,j$, then the condition $\#\mathrm{Spl}(f,\sigma ,\left\{k_j\right\})=\infty$ implies $\mathrm{vol}\left(\mathfrak{D}(f,\sigma ,\left\{k_j\right\})\right)>0$, in particular $\mathrm{vol}\left(\mathfrak{D}\right(f,\sigma \left)\right)>0$ if $\mathrm{Pr}(f,\sigma )>0$, and that under the following three conditions: (i) $f\left(x\right)$ is not a product of linear forms, (ii) ${\alpha }_i-{\alpha }_j\notin \mathbb{Q}$ if $i\ne j$, and (iii) $\mathrm{vol}\left(\mathfrak{D}\right(f,\sigma \left)\right)=c\mathrm{Pr}(f,\sigma )$ holds for every $\sigma \in S_n$, we have

$$c={\displaystyle \frac{\sqrt{det{\left(({\mathit{m}}_i,{\mathit{m}}_j)\right)}_{1\le i,j\le t}}}{\#\widehat{\mathit{G}}}}.$$

If $f\left(x\right)$ has only a trivial linear relation among roots, then $\mathrm{Pr}(f,\sigma )$ is equal to 1, and by property $\mathfrak{D}(f,\sigma )={\widehat{\mathfrak{D}}}_n$ for all $\sigma \in S_n$ Conjecture 1 is obviously true with $c={\displaystyle \frac{\sqrt{n}}{n!}}$ (cf. Proposition 2).

The second is as follows:

Conjecture 2. 

Let a set D be of the form $D=\{\mathit{x}\in {\mathbb{R}}^n\mid {\sum }_{j=1}^n{d}_{i,j}{x}_j\le d_i(i=1,\dots ,l)\}$ such that all $d_{i,j},d_i$ are rational and ${\sum }_{j=1}^n{d}_{i,\mu \left(j\right)}{\alpha }_j\notin \mathbb{Q}$ holds for all $1\le i\le l$ and $\mu \in S_n$. Then for a permutation $\sigma$ such that $\mathrm{Pr}(f,\sigma )>0$ and $\mathrm{vol}\left(\mathfrak{D}\right(f,\sigma \left)\right)>0$, we have

$$\begin{array}{cc}\hfill {\mathrm{Pr}}_D(f,\sigma ):& {\displaystyle =\underset{X\to \infty }{lim}\frac{\#\{p\in {\mathrm{Spl}}_X(f,\sigma )\mid (r_1/p,\dots ,r_n/p)\in D\}}{\#{\mathrm{Spl}}_X(f,\sigma )}}\hfill \\ \hfill & {\displaystyle =\frac{\mathrm{vol}(D\cap \mathfrak{D}(f,\sigma \left)\right)}{\mathrm{vol}\left(\mathfrak{D}\right(f,\sigma \left)\right)}.}\hfill \end{array}$$

(1)

Next, to state the conjecture on $\mathrm{Pr}(f,\sigma ,\left\{k_j\right\},L,\left\{R_i\right\})$ for integers $L(>1)$, $R_i$$(i=1,\dots ,n)$ as above, we introduce the following condition $\left({\mathrm{C}}_1\right)$ to $L,R_i$, which is a necessary condition for $\#{\mathrm{Spl}}_{\infty }(f,\sigma ,\left\{k_j\right\},L,\left\{R_i\right\})=\infty$:

$\left({\mathrm{C}}_1\right)$ : $(k_j,L)=({\sum }_i{m}_{j,i}{R}_{\sigma \left(i\right)}-m_j,L)(=d_j$, say) for ${}^{\forall }j=1,\dots ,t$ and there is an integer q which satisfies (i) q is independent of j, (ii) q is relatively prime to L, (iii) ${\sum }_i{m}_{j,i}{R}_{\sigma \left(i\right)}-m_j\equiv k_j·qmodL$ and $\left[\right[q\left]\right]=\left[\right[1\left]\right]$ on $\mathbb{Q}\left(f\right)\cap \mathbb{Q}\left({\zeta }_{L/d_j}\right)$ for every $j=1,\dots ,t$.

• Here, ${\zeta }_l$ denotes a primitive lth root of unity, and for a subfield F in $\mathbb{Q}\left({\zeta }_l\right)$ and an integer q relatively prime to l, $\left[\right[q\left]\right]$ denotes the automorphism of F induced by ${\zeta }_l\to {\zeta }_l^q$. The condition $\left[\right[q\left]\right]=\left[\right[1\left]\right]$ on $\mathbb{Q}\left(f\right)\cap \mathbb{Q}\left({\zeta }_{L/d_j}\right)$ is trivially satisfied if $\mathbb{Q}\left(f\right)\cap \mathbb{Q}\left({\zeta }_{L/d_j}\right)=\mathbb{Q}$. To see that condition $\left({\mathrm{C}}_1\right)$ is necessary, it is sufficient to take a prime $p\in \mathrm{Spl}(f,\sigma ,\left\{k_j\right\})$ with $p>L$ as q in $\left({\mathrm{C}}_1\right)$, noting $(k_j,L)=(k_{j}p,L)$.

By writing

$$\begin{array}{c}\mathfrak{R}(f,\sigma ,\left\{k_j\right\},L):=\{\left\{R_i\right\}\in {[0,L-1]}^n\mid \left({\mathrm{C}}_1\right)\mathrm{is}\mathrm{satisfied}\},\end{array}$$

the last conjecture is the following.

Conjecture 3. 

Under the assumption $\#{\mathrm{Spl}}_{\infty }(f,\sigma ,\left\{k_j\right\})=\infty$,

$$\begin{array}{cc}\hfill & \mathrm{Pr}(f,\sigma ,\left\{k_j\right\},L,\left\{R_i\right\})\hfill \\ \hfill =& \left\{\begin{array}{cc}{\displaystyle \frac{1}{\#\mathfrak{R}(f,\sigma ,\left\{k_j\right\},L)}}\hfill & if\left\{R_i\right\}\in \mathfrak{R}(f,\sigma ,\left\{k_j\right\},L),\hfill \\ 0\hfill & \mathit{otherwise}.\hfill \end{array}\right.\hfill \end{array}$$

When $f\left(x\right)=x+a$, Conjectures 1 and 2 have no meaning, but it is easy to see that Conjecture 3 is equivalent to Dirichlet’s prime number theorem on arithmetic progressions.

Conjectures 2 and 3 are a kind of equi-distribution.

The author has no idea how to prove Conjectures, much less when the second and the third are combined together.

In this paper, we try to approach Weyl’s criterion for Conjecture 2. As shown in the next section, Weyl’s criterion in the theory of the traditional equi-distribution of a sequence of points ${\left\{x_n\right\}}_{n=1}^{\infty }$$(0\le x_n<1)$ on the interval $[0,1)$ states that step functions can be approximated by trigonometric functions via continuous functions. The difference between the traditional case and ours is that the sequence of points $(r_1/p,\dots ,r_n/p)\in {\mathbb{R}}^n$ accumulates a convex hull in a lower dimensional plane $S\subset {\mathbb{R}}^n$, but no points are in S. We will show how to confine the points in S and how to calculate the trigonometric functions on a simplex-a component of the convex hull.

With respect to Conjecture 3, there is nothing to say.

Conjectures with numerical examples were proposed in [1,3]. In [2,4], we proved Theorem 2 below, which shows that Conjectures imply the classical conjecture for 1-dimensional uniformity of the sequnece of $r_i/p$, and in [5], we confirmed that Conjectures appear to be true for a complicated domain $D=\{(x_1,\dots ,x_n)\in {(0,1)}^n\mid \sum x_i\in \mathbb{Z},\left\{N{x}_i\right\}<\left\{M{x}_i\right\}\left({}^{\forall }i\right)\}$ by evaluating the volume, where $\left\{x\right\}$ means the decimal part of x.

2. Weyl’s Criterion

Let us review the original Weyl criterion for a uniformly distributed sequence modulo 1 (see [6]): the Weyl criterion states that a sequence $\left\{x_n\right\}$, $x_n\in [0,1)$, $n=1,2,\dots$ is uniformly distributed modulo 1 if and only if the following equation

$$\underset{N\to \infty }{lim}{\displaystyle \frac{1}{N}}\sum _{n=1}^{N}F\left(x_n\right)={\int }_0^{1}F\left(x\right)dx$$

(2)

is satisfied for $F\left(x\right)=exp\left(2\pi ihx\right)$$({}^{\forall }h\in \mathbb{Z})$. Usually, the right-hand side is replaced by “0 for $h\ne 0$”, omitting the trivial case $h=0$. The proof proceeds as follows:

• A sequence $\left\{x_n\right\}$, $x_n\in [0,1)$, $n=1,2,\dots$ is uniformly distributed modulo 1 if and only if the Equation (2) holds for the characteristic function of every interval in $[0,1)$, which is identified with the compact space $\mathbb{R}/\mathbb{Z}$.
• Approximating a continuous function $F\left(x\right)$$\left(F\right(0)=F(1\left)\right)$ on $[0,1]$ by step functions from above and below by the supremum norm, one proves that Equation (2) is true for step functions if and only if it is true for continuous functions.
• Approximating a continuous function $F\left(x\right)$$\left(F\right(0)=F(1\left)\right)$ on $[0,1]$ by trigonometric functions $exp\left(2\pi ihx\right)$$(h\in \mathbb{Z})$, one shows that (2) is true for continuous functions $F\left(x\right)$ if and only if Equation (2) is true for trigonometric functions $exp\left(2\pi ihx\right)$$({}^{\forall }h\in \mathbb{Z})$.

To apply the idea to a general case, the following Stone-Weierstrass theorem may be helpful.

Theorem 1. 

Let T be a compact set and let $C\left(T\right)$ be the algebra of all $\mathbb{R}$-valued continuous functions on T. Suppose that $\mathfrak{A}$ is a subalgebra of $C\left(T\right)$ such that $\left(\mathrm{i}\right)$$\mathfrak{A}$ contains every constant function, $\left(\mathrm{ii}\right)$ for distinct points $x,y\in T$, there is a function $g\in \mathfrak{A}$ such that $g\left(x\right)\ne g\left(y\right)$. Then a continuous function f on T is approximated by functions in $\mathfrak{A}$ by the supremum norm $\left|\right|f\left|\right|:={sup}_{x\in T}\left|f\left(x\right)\right|$.

The proof in [7] is simple and easy.

Our target space for Conjecture 2 is not $[0,1)$ but $\mathfrak{D}(f,\sigma )$ or its component $\mathfrak{D}(f,\sigma ,\left\{k_j\right\})$.

3. Geometry of $\mathfrak{D}\mathbf{(}\mathit{f}\mathbf{,}\mathit{\sigma }\mathbf{,}\mathbf{\left\{}{\mathit{k}}_{\mathit{j}}\mathbf{\right\}}\mathbf{)}$

In this section, we suppose that the polynomial $f\left(x\right)$ has no rational root and is not of the form $g\left(x\right)g(x-c)h\left(x\right)$$\left(g\right(x),h(x)\in \mathbb{Z}[x],c\in \mathbb{Z})$, and write simply

$$\mathfrak{D}:=\mathfrak{D}(f,\sigma ,\left\{k_j\right\})=\left\{(x_1,\dots ,x_n)\in \Delta \left|\sum _{i=1}^n{m}_{j,i}{x}_{\sigma \left(i\right)}=k_j(1\le {}^{\forall }j\le t)\right.\right\}.$$

First, we note that the matrix whose rows are ${\mathit{m}}_1,\dots ,{\mathit{m}}_t$ defined in Section 1 is primitive, that is, elementary divisors of the $(t,n)$-matrix with the ith row ${\mathit{m}}_i$ are 1 only, hence we can supplement the bases ${\mathit{m}}_1,\dots ,{\mathit{m}}_t$ of linear relations among roots by integral vectors ${\mathit{m}}_{t+1},\dots ,{\mathit{m}}_n$ such that the $(n,n)$-matrix $M=\left(m_{i,j}\right)$ with the ith row ${\mathit{m}}_i$ is in $S{L}_n\left(\mathbb{Z}\right)$, and put

$$\sigma \left(M\right):=\left(m_{i,{\sigma }^{-1}\left(j\right)}\right),$$

that is, its i-th row is $\sigma \left({\mathit{m}}_i\right)$. Then, writing

$$\mathcal{S}:=\{\mathit{x}\in {\mathbb{R}}^n\mid (\mathit{x},\sigma \left({\mathit{m}}_i\right))=k_i(1\le {}^{\forall }i\le t)\},$$

we have

$$\begin{array}{c}\hfill \mathfrak{D}=\Delta \cap \mathcal{S},\end{array}$$

and by the transformation $\mathit{x}\mapsto \mathit{y}:=\mathit{x}{}^t\left(\sigma \left(M\right)\right)$ we see that, noting $(\mathit{x},\sigma \left({\mathit{m}}_i\right))=(\mathit{x}{}^t\left(\sigma \left(M\right)\right),\sigma \left({\mathit{m}}_i\right)\sigma {\left(M\right)}^{-1})=y_i$

$$\begin{array}{cc}\hfill \mathcal{S}{}^t\left(\sigma \left(M\right)\right)& =\{(y_1,\dots ,y_n)\in {\mathbb{R}}^n\mid y_i=k_i(1\le {}^{\forall }i\le t)\}.\hfill \end{array}$$

The dimension of the set $\mathfrak{D}$ is at most $n-t$, which is the dimension of the supporting space $\mathcal{S}$. Noting that, via

$$(x_1,\dots ,x_n)=(u_1,\dots ,u_n)\left(\begin{array}{cccc}1& 1& \dots & 1\\ 0& 1& \dots & 1\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \dots & 1\end{array}\right),$$

$$\begin{array}{c}\hfill \Delta =\{\mathit{x}\mid 0\le x_1\le \dots \le x_n\le 1\}=\left\{\sum _{i=1}^n{u}_i{\mathit{w}}_i\mid {}^{\forall }u_i\ge 0,\sum _{i=1}^n{u}_i\le 1\right\},\end{array}$$

with ${\mathit{w}}_1:=(1,\dots ,1),{\mathit{w}}_2:=(0,1,\dots ,1),\dots ,{\mathit{w}}_n:=(0,\dots ,0,1)$, the set $\Delta$ is the convex hull with vertices ${\mathit{w}}_1,\dots ,{\mathit{w}}_n$ and the origin ${\mathit{w}}_{n+1}:=\mathit{O}$, that is

$$\Delta =\left\{\sum _{i=1}^{n+1}{u}_i{\mathit{w}}_i\left|{}^{\forall }u_i\ge 0,\sum _{i=1}^{n+1}{u}_i=1\right.\right\},$$

hence, we see that $\mathfrak{D}(=\mathfrak{D}(f,\sigma ,\left\{k_j\right\}))$ is also a convex hull, so a finite union of simplexes.

Denote the i-th row of the matrix ${}^t{\left(\sigma \left(M\right)\right)}^{-1}$ by ${\mathit{m}}_i^{\prime }$, that is,

$$\left(\begin{array}{c}{\mathit{m}}_1^{\prime }\\ ⋮\\ {\mathit{m}}_n^{\prime }\end{array}\right)={}^t\sigma {\left(M\right)}^{-1},\mathrm{i}.\mathrm{e}.,({\mathit{m}}_i^{\prime },\sigma \left({\mathit{m}}_j\right))={\delta }_{i,j},$$

(3)

and put

$${\mathit{m}}_{\sigma }:=\sum _{i=1}^t{m}_i{\mathit{m}}_i^{\prime }.$$

Lemma 1. 

Write $\mathit{r}:=(r_1,\dots ,r_n)$ for local roots $r_i$ for $p\in \mathrm{Spl}(f,\sigma ,\left\{k_j\right\})$; then we have

$$\begin{array}{c}\mathcal{S}{}^t\left(\sigma \left(M\right)\right)=\{(k_1,\dots ,k_t,y_{t+1},\dots ,y_n)\mid y_{t+1},\dots ,y_n\in \mathbb{R}\},\\ {\mathit{m}}_{\sigma }{}^t\left(\sigma \left(M\right)\right)=(m_1,\dots ,m_t,0,\dots ,0),\\ (\mathit{r}-{\mathit{m}}_{\sigma })/p\in \mathcal{S}.\end{array}$$

and

$$(\mathit{r}-{\mathit{m}}_{\sigma })/p\notin \Delta$$

holds only for a finite number of primes under the assumption at the beginning of this section to the polynomial f.

Proof. 

The first is already given above, the second identity follows from (3) and the definition of ${\mathit{m}}_{\sigma }$, and the third comes from $({\mathit{m}}_{\sigma },\sigma \left({\mathit{m}}_j\right))=m_j$ and $(\mathit{r},\sigma \left({\mathit{m}}_j\right))=m_j+k_{j}p$$(j=1,\dots ,t)$ by the definition of the vector of local roots $r_i$ for $p\in \mathrm{Spl}(f,\sigma ,\left\{k_j\right\})$. Write ${\mathit{m}}_{\sigma }=(c_1,\dots ,c_n)$. If $(\mathit{r}-{\mathit{m}}_{\sigma })/p\notin \Delta$ happens for infinitely many primes p, then one of conditions $0\le r_1<c_1$, $0\le r_{i+1}-r_i<c_{i+1}-c_i$ or $0<p-r_n<-c_n$ occurs for infinitely many primes, hence one of $r_1=c$, $r_{i+1}-r_i=c$ or $r_n-p=c$ for some constant c occurs for infinitely many primes. This means $f\left(c\right)\equiv 0modp$, $f\left(r_i\right)\equiv f(r_i+c)\equiv 0modp$, $f\left(c\right)\equiv 0modp$, respectively. The first or third condition easily implies that the polynomial $f\left(x\right)$ has a rational root c. The second implies that the polynomial f is of the form $g\left(x\right)g(x-c)h\left(x\right)$. Because, if the identity $r_{i+1}-r_i=c$ holds for infinitely many primes, then $f(r_i+c)\equiv f\left(r_i\right)\equiv 0modp$ holds for infinitely many primes, which implies $\left(f\right(x),f(x+c\left)\right)\ne 1$ (resp. $(f\left(x\right),f^{\prime }\left(x\right))\ne 1$) for $c\ne 0$ (resp. $c=0$), Hence, in the case of $c\ne 0$, there is a decomposition $f\left(x\right)=g\left(x\right)h\left(x\right),f(x+c)=g\left(x\right)k\left(x\right)$ for an irreducible polynomial $g\left(x\right)$, i.e., $f\left(x\right)$ is divisible by $g\left(x\right)g(x-c)$. If $c=0$, then $f\left(x\right)$ is of the form $g{\left(x\right)}^{2}h\left(x\right)$. □

Proposition 1. 

Let D be the set of the form referred to in Conjecture 2, that is,

$$D=\left\{\mathit{x}\in {\mathbb{R}}^n\left|\sum _{j=1}^n{d}_{i,j}{x}_j\le d_i(i=1,\dots ,l)\right.\right\}$$

with all $d_{i,j},d_i$ being rational and suppose that ${\sum }_{j=1}^n{d}_{i,j}{\alpha }_{\mu \left(j\right)}\notin \mathbb{Q}$ holds for all $1\le i\le l$ and all permutations μ. Write

$$\begin{array}{cc}\hfill S& :=\{p\in \mathrm{Spl}(f,\sigma ,\left\{k_j\right\})\mid \mathit{r}/p\in D\},\hfill \\ \hfill T& :=\{p\in \mathrm{Spl}(f,\sigma ,\left\{k_j\right\})\mid (\mathit{r}-{\mathit{m}}_{\sigma })/p\in D\},\hfill \end{array}$$

then two sets $S\setminus T$ and $T\setminus S$ are finite sets, in particular, there exists a constant ${\kappa }_D$ dependent on D such that

$$\begin{array}{cc}\hfill & \{p\in \mathrm{Spl}(f,\sigma ,\left\{k_j\right\})\mid \mathit{r}/p\in D,p>{\kappa }_D\}\hfill \\ \hfill =& \{p\in \mathrm{Spl}(f,\sigma ,\left\{k_j\right\})\mid (\mathit{r}-{\mathit{m}}_{\sigma })/p\in D,p>{\kappa }_D\},\hfill \end{array}$$

is true for the set D.

Proof. 

Write ${\mathit{m}}_{\sigma }=(c_1,\dots ,c_n)$, and let d be a positive integer such that all $d{d}_{i,j},d{d}_i$ are integers. Let us show the contradiction under the supposition $\#(S\setminus T)=\infty$, that is, there are infinitely many primes $p\in \mathrm{Spl}\left(f\right)$ such that $\mathit{r}/p\in D$ and $(\mathit{r}-{\mathit{m}}_{\sigma })/p\notin D$; then the supposition implies ${\sum }_j{d}_{i,j}{r}_j\le p{d}_i$ for all i and ${\sum }_j{d}_{k,j}(r_j-c_j)>p{d}_k$ for some k, hence ${\sum }_j{d}_{k,j}{c}_j<{\sum }_j{d}_{k,j}{r}_j-p{d}_k\le 0$, where ${\sum }_j{d}_{k,j}{c}_j$ is a constant independent of p and $d({\sum }_j{d}_{k,j}{r}_j-p{d}_k)$ is an integer. Therefore, there are infinitely many primes $p\in \mathrm{Spl}\left(f\right)$ such that $d({\sum }_j{d}_{k,j}{r}_j-p{d}_k)=\kappa$ for some integer $\kappa$ satisfying $d{\sum }_j{d}_{k,j}{c}_j<\kappa \le 0$. Taking a prime ideal $\mathfrak{p}$ of $\mathbb{Q}({\alpha }_1,\dots ,{\alpha }_n)$ over p, we see that there are infinitely many prime ideals $\mathfrak{p}$ and a permutation $\mu$ such that ${\alpha }_{\mu \left(j\right)}\equiv r_{j}mod\mathfrak{p}$, hence $d{\sum }_j{d}_{k,j}{\alpha }_{\mu \left(j\right)}\equiv \kappa mod\mathfrak{p}$, which implies $d{\sum }_j{d}_{k,j}{\alpha }_{\mu \left(j\right)}=\kappa$, i.e., ${\sum }_j{d}_{k,j}{\alpha }_{\mu \left(j\right)}=\kappa /d\in \mathbb{Q}$, that is a contradiction. The case of $\mathit{r}/p\notin D$ and $(\mathit{r}-{\mathit{m}}_{\sigma })/p\in D$ is similarly proved. □

If a polynomial $f\left(x\right)$ has no rational root, then the typical example of the set D in the proposition is

$$\{\mathit{x}\in {\mathbb{R}}^n\mid a_i\le x_i\le b_i\mathrm{for} \mathrm{all} i=1,\dots ,n\}$$

with all $a_i,b_i$ being rational, since vectors $(d_{i,1},\dots ,d_{i,n})$ in Proposition 1 are $(0,\dots ,0,\pm 1,0,\dots ,0)$, which is in $L{R}_0$ if and only if $f\left(x\right)$ has a rational root.

In case of $m_j\ne 0$ for some $1\le j\le t$, every vector $\mathit{r}/p$ for $p\in \mathrm{Spl}(f,\sigma ,\left\{k_j\right\})$ is not in $\mathcal{S}$, but every accumulation point is in $\mathfrak{D}\subset \mathcal{S}$. So, Conjecture 2 is not within the scope of a traditional uniform distribution, and we need a reduction step. The condition on the set D in Proposition 1 is a sufficient condition to the following.

$$\begin{array}{cc}\hfill & {\displaystyle \frac{\#\{p\in {\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})\mid \mathit{r}/p\in D\}}{\#{\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})}}\hfill \\ \hfill =& {\displaystyle \frac{\#\{p\in {\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})\mid (\mathit{r}-{\mathit{m}}_{\sigma })/p\in D\}}{\#{\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})}}+o\left(1\right),\hfill \end{array}$$

(4)

where the error term $o\left(1\right)$ depends on D by Proposition 1, and every vector $(\mathit{r}-{\mathit{m}}_{\sigma })/p$ and every accumulation point are in the same set $\mathcal{S}$ by Proposition 1.

Thus, for the set D in Proposition 1, Equation (1) in Conjecture 2 is equivalent to

$$\begin{array}{c}\hfill \underset{X\to \infty }{lim}{\displaystyle \frac{\#\{{p\in \mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})\mid (\mathit{r}-{\mathit{m}}_{\sigma })/p\in D\cap \mathfrak{D}\}}{\#{\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})}}={\displaystyle \frac{\mathrm{vol}(D\cap \mathfrak{D})}{\mathrm{vol}\left(\mathfrak{D}\right)}},\end{array}$$

where it should be emphasized that the vector $(\mathit{r}-{\mathit{m}}_{\sigma })/p$ is on $\mathfrak{D}=\Delta \cap \mathcal{S}$ except for a finite number of primes in $\mathrm{Spl}(f,\sigma ,\left\{k_j\right\})$ that depend not on D but only on the polynomial $f\left(x\right)$, according to Lemma 1. So, it is equivalent to

$$\begin{array}{cc}\hfill & \underset{X\to \infty }{lim}{\displaystyle \frac{1}{\#{\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})}}\sum _{p\in {\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})}F((\mathit{r}-{\mathit{m}}_{\sigma })/p)\hfill \\ \hfill =& {\displaystyle \frac{1}{\mathrm{vol}\left(\mathfrak{D}\right)}}{\int }_{\mathfrak{D}}F\left(\mathit{x}\right)d\mathit{x},\hfill \end{array}$$

(5)

for the characteristic function F of the set $D\cap \mathfrak{D}$, hence, for a continuous function F on $\mathcal{S}$. The reduction step (4) seems to be unavoidable.

Applying Conjecture 2 to $D_{i,a}=\{(x_1,\dots ,x_n)\in {[0,1)}^n\mid x_i\le a\}$, we have the following theorem (cf. [2,4]).

Theorem 2. 

If a polynomial $f\left(x\right)$ has no rational root, ${\alpha }_i-{\alpha }_j\notin \mathbb{Q}$ holds for any distinct $i,j$, and the conditions $\mathrm{Pr}(f,\sigma )>0$ and $\mathrm{vol}\left(\mathfrak{D}\right(f,\sigma \left)\right)>0$ are equivalent, then Conjectures 1 and 2 imply the equi-distribution of $r_i/p$ for local roots $r_i$ of the polynomial.

For a polynomial $f\left(x\right)$ with no non-trivial linear relations among roots, the above can be proved by evaluating the density

$$\underset{X\to \infty }{lim}{\displaystyle \frac{\#\{p\in {\mathrm{Spl}}_X\left(f\right)\mid r_i/p<a\}}{\#{\mathrm{Spl}}_X\left(f\right)}}$$

explicitly, and the graph of the difference in the above density and the approximate values at $X={10}^{10}m$$(1\le m\le 300)$ for $f\left(x\right)=(x^7-1)/(x-1)$ is also provided in [2]. Some numerical data of Conjectures are also provided in [1,3].

The equi-distribution of $r_i/p$ for an irreducible polynomial $f\left(x\right)$ in Theorem 2 has been proven to hold without the assumptions of Conjectures 1, 2, but only for $n=2$ ([8,9]). For $n\ge 3$, however, it remains a difficult problem.

The condition on D in Proposition 1 is a sufficient condition for (4). How can one weaken the condition?

4. Integration

Let us rewrite the integral (5) on the set $\mathfrak{D}$ in a straightforward way to see. To perform this, let us restate the definitions:

$$\begin{array}{cc}\hfill \Delta & =\{\mathit{x}=(x_1,\dots ,x_n)\in {\mathbb{R}}^n\mid 0\le x_1\le \dots \le x_n\le 1\},\hfill \\ \hfill \mathcal{S}& =\{\mathit{x}\in {\mathbb{R}}^n\mid (\mathit{x},\sigma \left({\mathit{m}}_j\right))=k_j(j=1,\dots ,t)\},\hfill \\ \hfill \mathfrak{D}& =\mathfrak{D}(f,\sigma ,\left\{k_j\right\})\hfill \\ \hfill & =\{(x_1,\dots ,x_n)\in \Delta \mid \sum _{i=1}^n{m}_{j,i}{x}_{\sigma \left(i\right)}=k_j(1\le {}^{\forall }j\le t)\}\hfill \\ \hfill & =\Delta \cap \mathcal{S}.\hfill \end{array}$$

Writing

$$\begin{array}{cc}\hfill {\mathcal{S}}_0:& =\{\mathit{x}\in {\mathbb{R}}^n\mid (\mathit{x},\sigma \left({\mathit{m}}_j\right))=0(1\le {}^{\forall }j\le t)\}\hfill \\ \hfill & =\mathbb{R}[{\mathit{m}}_{t+1}^{\prime },\dots ,{\mathit{m}}_n^{\prime }]\left(\mathrm{by}\right(3\left)\right),\hfill \end{array}$$

we have

$$\begin{array}{c}\mathcal{S}={\mathit{x}}_0+{\mathcal{S}}_0={\mathit{x}}_0+\mathbb{R}[{\mathit{m}}_{t+1}^{\prime },\dots ,{\mathit{m}}_n^{\prime }]\mathrm{for} {}^{\forall }\mathit{x}_0\notin {\mathcal{S}}_0,\\ \mathfrak{D}=\Delta \cap \{{\mathit{x}}_0+\mathbb{R}[{\mathit{m}}_{t+1}^{\prime },\dots ,{\mathit{m}}_n^{\prime }]\},\end{array}$$

(6)

and since $(\mathit{x},\sigma \left({\mathit{m}}_j\right))=(\mathit{x}{}^t\sigma \left(M\right),\sigma \left({\mathit{m}}_j\right)\sigma {\left(M\right)}^{-1})$ is the j-th entry of $\mathit{x}{}^t\sigma \left(M\right)$, we see

$$\begin{array}{cc}\hfill {\mathcal{S}}_0{}^t\sigma \left(M\right)& =\{\mathit{x}\in {\mathbb{R}}^n\mid x_1=\dots =x_t=0\},\hfill \\ \hfill \mathcal{S}{}^t\sigma \left(M\right)& =\{\mathit{x}\in {\mathbb{R}}^n\mid x_j=k_j(1\le {}^{\forall }j\le t)\}.\hfill \end{array}$$

To specify the vector ${\mathit{x}}_0$, we introduce the following matrix.

$$\begin{array}{cc}\hfill \tilde{M}:& =\sigma \left(M\right){}^t\sigma \left(M\right)=M{}^{t}M=\left(\begin{array}{cc}{M}_1^{\left(t\right)}& M_2^{(t,n-t)}\\ {}^{t}M_2& M_4\end{array}\right)\hfill \\ \hfill & =\left(\begin{array}{cc}{M}_1^{\left(t\right)}& 0^{(t,n-t)}\\ 0& M_4-M_1^{-1}\left[M_2\right]\end{array}\right)\left[\left(\begin{array}{cc}{1}_t& M_1^{-1}{M}_2\\ 0^{(n-t,t)}& 1_{n-t}\end{array}\right)\right],\hfill \end{array}$$

where $A\left[B\right]$ means ${}^{t}BAB$, we have

$$\begin{array}{cc}\hfill {\tilde{M}}^{-1}& =\left(\begin{array}{cc}{M}_1^{-1}& 0\\ 0& {(M_4-M_1^{-1}\left[M_2\right])}^{-1}\end{array}\right)\left[\left(\begin{array}{cc}{1}_t& 0\\ -{}^t\left(M_1^{-1}{M}_2\right)& 1_{n-t}\end{array}\right)\right]\hfill \\ \hfill & =\left(\begin{array}{cc}\ast & \ast \\ \ast & {(M_4-M_1^{-1}\left[M_2\right])}^{-1}\end{array}\right).\hfill \end{array}$$

Here, the matrices $M_1$, $M_4-M_1^{-1}\left[M_2\right]$ are regular, since the matrix $M{}^{t}M$ is positive definite.

Using the above matrices, let us specify the vector ${\mathit{x}}_0$ as follows:

$$\begin{array}{cc}\hfill {\mathit{x}}_0& :=(k_1,\dots ,k_t)\left(\begin{array}{c}{\mathit{m}}_1^{\prime }\\ ⋮\\ {\mathit{m}}_t^{\prime }\end{array}\right)+(k_1,\dots ,k_t)M_1^{-1}{M}_2\left(\begin{array}{c}{\mathit{m}}_{t+1}^{\prime }\\ ⋮\\ {\mathit{m}}_n^{\prime }\end{array}\right)\hfill \\ \hfill & =(k_1,\dots ,k_t)((1_t,0^{(t,n-t)})+(0^{(t,t)},M_1^{-1}{M}_2)){}^t\sigma {\left(M\right)}^{-1}\hfill \\ \hfill & =(k_1,\dots ,k_t)M_1^{-1}(M_1,M_2){}^t\sigma {\left(M\right)}^{-1}\hfill \\ \hfill & =(k_1,\dots ,k_t)M_1^{-1}(1_t,0^{(t,n-t)})\tilde{M}{}^t\sigma {\left(M\right)}^{-1}\hfill \\ \hfill & =(k_1,\dots ,k_t)(M_1^{-1},0^{(t,n-t)})\sigma \left(M\right).\hfill \end{array}$$

Thus, we have ${\mathit{x}}_0{}^t\sigma \left(M\right)=(k_1,\dots ,k_t)(1_t,M_1^{-1}{M}_2)\in \mathcal{S}{}^t\sigma \left(M\right)$, which implies ${\mathit{x}}_0\in \mathcal{S}$. Let us see $(x_0,{\mathcal{S}}_0)=0$, which is equivalent to

$${\mathit{x}}_0({}^t\mathit{m}_{t+1}^{\prime },\dots ,{}^t\mathit{m}_n^{\prime })=0^{(1,n-t)},$$

which follows from

$$\begin{array}{cc}\hfill & {\mathit{x}}_0({}^t\mathit{m}_{t+1}^{\prime },\dots ,{}^t\mathit{m}_n^{\prime })\hfill \\ \hfill =& (k_1,\dots ,k_t)(M_1^{-1},0^{(t,n-t)})\sigma \left(M\right)({}^t\mathit{m}_{t+1}^{\prime },\dots ,{}^t\mathit{m}_n^{\prime })\hfill \\ \hfill =& (k_1,\dots ,k_t)(M_1^{-1},0^{(t,n-t)})\left(\begin{array}{c}{0}^{(t,n-t)}\\ 1_{n-t}\end{array}\right)\hfill \\ \hfill =& 0^{(1,n-t)}.\hfill \end{array}$$

Next, let the vectors ${\mathit{f}}_i\in {\mathbb{R}}^n$$(i=t+1,\dots ,n)$ be an orthonormal basis ${\mathit{f}}_i$ of ${\mathcal{S}}_0$, hence, there is a matrix $C^{(n-t,n-t)}$ such that

$$\begin{array}{c}\left(\begin{array}{c}{\mathit{m}}_{t+1}^{\prime }\\ ⋮\\ {\mathit{m}}_n^{\prime }\end{array}\right)=C\left(\begin{array}{c}{\mathit{f}}_{t+1}\\ ⋮\\ {\mathit{f}}_n\end{array}\right),\end{array}$$

therefore,

$$C{}^{t}C={\left(({\mathit{m}}_i^{\prime },{\mathit{m}}_j^{\prime })\right)}_{t+1\le i,j\le n}={(M_4-M_1^{-1}\left[M_2\right])}^{-1},$$

since the above matrix is the submatrix of

$${\left(({\mathit{m}}_i^{\prime },{\mathit{m}}_j^{\prime })\right)}_{1\le i,j\le n}={}^t\sigma {\left(M\right)}^{-1}\sigma {\left(M\right)}^{-1}={\tilde{M}}^{-1}.$$

Hence, we have

$$|detC|=\sqrt{det{M}_1}.$$

Let the vertices of $\mathfrak{D}$ be $\left\{{\mathit{u}}_i\right\}$, hence, $\mathfrak{D}=\{\sum c_i{\mathit{u}}_i\mid c_i\ge 0,\sum c_i=1\}$, and in view of (6) we write

$$X:=\{(x_{t+1},\dots ,x_n)\in {\mathbb{R}}^{n-t}\mid {\mathit{x}}_0+\sum _{j=t+1}^n{x}_j{\mathit{m}}_j^{\prime }\in \mathfrak{D}\}$$

and define the bijective mapping $\psi$ from X to $\mathfrak{D}$ by

$$\psi :\mathit{x}=(x_{t+1},\dots ,x_n)\in X\mapsto \mathit{u}={\mathit{x}}_0+\sum _{j=t+1}^n{x}_j{\mathit{m}}_j^{\prime }\in \mathfrak{D}.$$

Then, we see that for $\mathit{x}\in X$

$$\begin{array}{c}\hfill \psi \left(\mathit{x}\right)=\mathit{u}\iff (\mathit{u}-{\mathit{x}}_0){}^t\sigma \left(M\right)=(0^{(1,t)},\mathit{x}),\end{array}$$

and the vector ${\mathit{x}}_i\in X$ satisfying $\psi \left({\mathit{x}}_i\right)={\mathit{u}}_i$ is the vertex of X, that is, $X=\{\sum c_i{\mathit{x}}_i\mid c_i\ge 0,\sum c_i=1\}$.

Thus, for a function F on $\mathfrak{D}=\Delta \cap \mathcal{S}$, we see

$$\begin{array}{cc}\hfill & {\int }_{\mathfrak{D}}F\left(\mathit{x}\right)d\mathit{x}\hfill \\ \hfill =& {\int }_{{\mathit{x}}_0+{\sum }_{j=t+1}^n{y}_j{\mathit{f}}_j\in \mathfrak{D}}F({\mathit{x}}_0+\sum _{j=t+1}^n{y}_j{\mathit{f}}_j)d{y}_{t+1}\dots d{y}_n\hfill \\ \hfill =& \int F({\mathit{x}}_0+(y_{t+1},\dots ,y_n)C^{-1}\left(\begin{array}{c}{\mathit{m}}_{t+1}^{\prime }\\ ⋮\\ {\mathit{m}}_n^{\prime }\end{array}\right))d{y}_{t+1}\dots d{y}_n\hfill \\ \hfill =& \sqrt{det\left(M_1\right)}{\int }_{(x_{t+1},\dots ,x_n)\in X}F({\mathit{x}}_0+\sum _{j=t+1}^n{x}_j{\mathit{m}}_j^{\prime })d{x}_{t+1}\dots d{x}_n,\hfill \end{array}$$

(7)

hence, Equation (5) becomes

$$\begin{array}{cc}\hfill & \underset{X\to \infty }{lim}{\displaystyle \frac{1}{\#{\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})}}\sum _{p\in {\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})}F((\mathit{r}-{\mathit{m}}_{\sigma })/p)\hfill \\ \hfill =& {\displaystyle \frac{\sqrt{det\left(M_1\right)}}{\mathrm{vol}\left(\mathfrak{D}\right)}}{\int }_{(x_{t+1},\dots ,x_n)\in X}F({\mathit{x}}_0+\sum _{j=t+1}^n{x}_j{\mathit{m}}_j^{\prime })d{x}_{t+1}\dots d{x}_n,\hfill \end{array}$$

(8)

where ${\mathit{m}}_j^{\prime }$ is the jth row of the matrix ${}^t\sigma {\left(M\right)}^{-1}$ and $M_1={\left(({\mathit{m}}_i,{\mathit{m}}_j)\right)}_{1\le i,j\le t}$ as before.

Writing $X:={\cup }_k{\mathcal{X}}_k$ for the simplexes with vertices ${\mathit{x}}_{k,1},\dots ,{\mathit{x}}_{k,n-t+1}$, we have the following:

$$\begin{array}{cc}\hfill & \underset{X\to \infty }{lim}{\displaystyle \frac{1}{\#{\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})}}\sum _{p\in {\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})}F((\mathit{r}-{\mathit{m}}_{\sigma })/p)\hfill \\ \hfill =& {\displaystyle \frac{\sqrt{det\left(M_1\right)}}{\mathrm{vol}\left(\mathfrak{D}\right)}}\sum _k{\int }_{(x_{t+1},\dots ,x_n)\in {\mathcal{X}}_k}F({\mathit{x}}_0+\sum _{j=t+1}^n{x}_j{\mathit{m}}_j^{\prime })d{x}_{t+1}\dots d{x}_n.\hfill \end{array}$$

(9)

Here, we may suppose that the vertices ${\mathit{x}}_{k,i}$ of ${\mathcal{X}}_k$ are those of $X.$

The aim is to evaluate the integral for a trigonometric function as F on simplexes ${\mathcal{X}}_i$. Let $m:=n-t$ and $\mathcal{X}\subset {\mathbb{R}}^m$ be a simplex with vertices ${\mathit{x}}_1,\dots ,{\mathit{x}}_{m+1}$ for which m vectors ${\mathit{x}}_1-{\mathit{x}}_{m+1}$, $\dots ,$${\mathit{x}}_m-{\mathit{x}}_{m+1}$ are linearly independent. Then, we have

$$\begin{array}{cc}\hfill \mathcal{X}:& =\left\{\sum _{i=1}^{m+1}{y}_i{\mathit{x}}_i\mid {}^{\forall }y_i\ge 0,\sum _{i=1}^{m+1}{y}_i=1\right\}\hfill \\ \hfill & =\left\{\sum _{i=1}^m{x}_i({\mathit{x}}_i-{\mathit{x}}_{m+1})+{\mathit{x}}_{m+1}\mid {}^{\forall }x_i\ge 0,\sum _{i=1}^m{x}_i\le 1\right\}.\hfill \end{array}$$

Writing $V:=\left(\begin{array}{c}{\mathit{x}}_1-{\mathit{x}}_{m+1}\\ ⋮\\ {\mathit{x}}_m-{\mathit{x}}_{m+1}\end{array}\right)\in {\mathrm{GL}}_m\left(\mathbb{R}\right)$, we see that, first of all,

$${\int }_{(x_{t+1},\dots ,x_n)\in \mathcal{X}}1d{x}_{t+1}\dots d{x}_n={\displaystyle \frac{|det(V\left)\right|}{m!}},$$

because, putting $(x_{t+1},\dots ,x_n)=(y_{t+1},\dots ,y_n)V+{\mathit{x}}_{m+1}$, we see that the integral is equal to

$$\begin{array}{cc}\hfill & |det\left(V\right)|{\int }_{y_i\ge 0,\sum y_i\le 1}d{y}_{t+1}\dots d{y}_n\hfill \end{array}$$

transforming from $y_i$ to $z_{t+i}:=y_{t+1}+\cdots +y_{t+i}$ $(0\le z_{t+1}\le \cdots \le z_n\le 1)$

$$\begin{array}{c}=|det\left(V\right)|\mathrm{vol}\left({\Delta }_m\right)({\Delta }_m=\{(z_{t+1},\dots ,z_n)\mid 0\le z_{t+1}\le \cdots \le z_n\le 1\})\hfill \\ ={\displaystyle \frac{|det(V\left)\right|}{m!}}\sum _{\sigma \in S_m}\mathrm{vol}\left(\sigma \left({\Delta }_m\right)\right)\hfill \\ ={\displaystyle \frac{|det(V\left)\right|}{m!}}.\hfill \end{array}$$

Hereafter, we will specify the function $F\left(x\right)$ using the exponential function $e\left(x\right):=exp\left(2\pi \sqrt{-1}x\right)$ as follows:

$$F\left(\mathit{x}\right):=e\left((\mathit{x},\mathit{c})\right)\mathrm{for} \mathit{c}\in {\mathbb{R}}^n,$$

and we will evaluate the integral on the simplex in the last section. Then, the integral

$${\int }_{(x_{t+1},\dots ,x_n)\in \mathcal{X}}F({\mathit{x}}_0+\sum _{j=t+1}^n{x}_j{\mathit{m}}_j^{\prime })d{x}_{t+1}\dots d{x}_n$$

is

$$\begin{array}{cc}\hfill & e\left(({\mathit{x}}_0,\mathit{c})\right){\int }_{(x_{t+1},\dots ,x_n)\in \mathcal{X}}e\left(\sum _{j=t+1}^n{x}_j({\mathit{m}}_j^{\prime },\mathit{c})\right)d{x}_{t+1}\dots d{x}_n\hfill \end{array}$$

writing $\mathcal{C}:=(({\mathit{m}}_{t+1}^{\prime },\mathit{c}),\dots ,({\mathit{m}}_n^{\prime },\mathit{c}))$, we see ${\sum }_{j=t+1}^n{x}_j({\mathit{m}}_j^{\prime },\mathit{c})=(\mathit{x},\mathcal{C})=(\mathit{y}V+{\mathit{x}}_{m+1},\mathcal{C})$,

$$\begin{array}{cc}\hfill =& e(({\mathit{x}}_0,\mathit{c})+({\mathit{x}}_{m+1},\mathcal{C}))|det\left(V\right)|{\int }_{\begin{array}{c}{y}_i\ge 0,\\ \sum y_i\le 1\end{array}}e\left(\mathit{y}V{}^t\mathcal{C}\right)d{y}_{t+1}\dots d{y}_n\hfill \end{array}$$

Therefore, Equation (9) is, furthermore,

$$\begin{array}{cc}\hfill & \underset{X\to \infty }{lim}{\displaystyle \frac{1}{\#{\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})}}\sum _{p\in {\mathrm{Spl}}_X(f,\sigma ,\left\{k_j\right\})}e((\mathit{r}-{\mathit{m}}_{\sigma })/p,\mathit{c}))\hfill \\ \hfill =& {\displaystyle \frac{\sqrt{det\left(M_1\right)}e\left(({\mathit{x}}_0,\mathit{c})\right)}{\mathrm{vol}\left(\mathfrak{D}\right)}}\sum _{k}e\left(({\mathit{x}}_{k,m+1},\mathcal{C})\right)|det\left(V_k\right)|\times \hfill \\ \hfill & {\int }_{\begin{array}{c}{x}_i\ge 0,\\ \sum x_i\le 1\end{array}}e\left((x_{t+1},\dots ,x_n)V_k{}^t\mathcal{C}\right))d{x}_{t+1}\dots d{x}_n,\hfill \end{array}$$

(10)

where the simplex ${\mathcal{X}}_k$ and the vertex ${\mathit{x}}_{k,i}$ are those given at (9) and

$$V_k:=\left(\begin{array}{c}{\mathit{x}}_{k,1}-{\mathit{x}}_{k,m+1}\\ ⋮\\ {\mathit{x}}_{k,m}-{\mathit{x}}_{k,m+1}\end{array}\right)\mathrm{and} \mathcal{C}=(({\mathit{m}}_{t+1}^{\prime },\mathit{c}),\dots ,({\mathit{m}}_n^{\prime },\mathit{c}))$$

for $\mathit{c}\in {\mathbb{R}}^n$.

If Equation (10) is true for every $c\in {\mathbb{Z}}^n$, then Equation (8) is true for every continuous function F on $\mathfrak{D}$ by the Stone–Weierstrass theorem, and so is Conjecture 2. By Lemma 2 below, we see that the above integral vanishes if $({\mathit{x}}_{k,1},\mathcal{C}),\dots ,({\mathit{x}}_{k,m+1},\mathcal{C})$ are mutually distinct and all differences $({\mathit{x}}_{k,i},\mathcal{C})-({\mathit{x}}_{k,j},\mathcal{C})$ are integers. Note that every vertex ${\mathit{x}}_{k,j}$ can be chosen in the set of vertices of X. Denote by $\mathfrak{C}$ the set of vectors $\mathit{c}\in {\mathbb{R}}^n$ such that for $\mathcal{C}=(({\mathit{m}}_{t+1}^{\prime },\mathit{c}),\dots ,({\mathit{m}}_n^{\prime },\mathit{c}))$ and the vertices ${\mathit{x}}_i$ of X, all inner products $({\mathit{x}}_i,\mathcal{C})$ are mutually distinct and $({\mathit{x}}_i,\mathcal{C})-({\mathit{x}}_j,\mathcal{C})$ is an integer for every $i,j$. If the restrictions of functions $e((x_{t+1},\dots ,x_n)V_k{}^t\mathcal{C}))$ ($\mathit{c}\in \mathfrak{C}$) on ${\mathcal{X}}_k$ and the constant functions approximate all continuous functions on $\mathfrak{D}$, then we may say that the Weyl criterion is generalized. Unfortunately, they are not an algebra in the Stone–Weierstrass Theorem 1 and even if they approximate all continuous functions, the proof of (10) is another problem.

5. Example of $\mathfrak{D}(\mathit{f},\mathit{\sigma },\left\{{\mathit{k}}_{\mathit{j}}\right\})$

In this section, we will study $\mathfrak{D}=\mathfrak{D}(f,\sigma ,\left\{k_j\right\})$ explicitly in the special case. Since $\Delta$ is a convex hull, the set $\mathfrak{D}=\Delta \cap \mathcal{S}$ is also a convex hull, hence a finite union of simplexes.

First of all, we note that a point $\mathit{x}=(x_1,\dots ,x_n)\in \mathfrak{D}$ is not a vertex if and only if there is a non-zero vector $\mathit{v}=(v_1,\dots ,v_n)$ such that $\mathit{x}+\eta \mathit{v}\in \mathfrak{D}$ holds for any $\eta$ in a short open interval containing 0, that is, the following two hold for $\eta$:

$$\begin{array}{c}0\le x_1+\eta v_1\le \cdots \le x_n+\eta v_n\le 1,\\ \sum _i{m}_{j,{\sigma }^{-1}\left(i\right)}(x_i+\eta v_i)=k_j(1\le {}^{\forall }j\le t).\end{array}$$

Hence a point $\mathit{x}\in \mathfrak{D}$ is not a vertex if and only if there is a non-zero vector $\mathit{v}$ satisfying the following.

$$\begin{array}{c}\hfill \left\{\begin{array}{ccc}\sum _i{m}_{j,{\sigma }^{-1}\left(i\right)}{v}_i=0\hfill & \mathrm{for}& {}^{\forall }j=1,\dots ,t,\hfill \\ v_i=0\hfill & \mathrm{if}& x_i=0,\hfill \\ v_i=v_{i+1}\hfill & \mathrm{if}& x_i=x_{i+1},\hfill \\ v_i=0\hfill & \mathrm{if}& x_i=1.\hfill \end{array}\right.\end{array}$$

(11)

Another approach is as follows: Since the set $\Delta$ is the convex hull of ${\mathit{w}}_1:=(1,\dots ,1)$, ${\mathit{w}}_2:=(0,1,\dots ,1),\dots ,{\mathit{w}}_n:=(0,\dots ,0,1)$, and ${\mathit{w}}_{n+1}=(0,\dots ,0)$, the vertices of $\mathfrak{D}=\Delta \cap S$ are the set of the intersection of the line segment $\overline{{\mathit{w}}_i{\mathit{w}}_j}$ connecting the points ${\mathit{w}}_i$ and ${\mathit{w}}_j$, and the $(n-t)$-dimensional plane S. Hence, for $1\le i<j\le n+1$, a vector $\mathit{v}=u{\mathit{w}}_i+(1-u){\mathit{w}}_j$$(0\le {}^{\exists }u\le 1)$ on $\overline{{\mathit{w}}_i{\mathit{w}}_j}$ is on $\mathcal{S}$ if and only if $\mathit{v}{}^t\sigma \left(M\right)=(k_1,\dots ,k_t,\ast ,\dots ,\ast )$. Note that

$$\begin{array}{c}\hfill u{\mathit{w}}_i+(1-u){\mathit{w}}_j=\left\{\begin{array}{cc}(\underset{i-1}{\underbrace{0,\dots ,0}},\underset{n-i+1}{\underbrace{u,\dots ,u}})\hfill & \mathrm{if}j=n+1,\hfill \\ (\underset{i-1}{\underbrace{0,\dots ,0}},\underset{j-i}{\underbrace{u,\dots ,u}},\underset{n-j+1}{\underbrace{1,\dots ,1}})\hfill & \mathrm{if}j\le n.\hfill \end{array}\right.\end{array}$$

The shape of $\mathfrak{D}$ is highly dependent on the polynomial $f\left(x\right)$. We will give two examples.

Example 1. 

Suppose that the polynomial $f\left(x\right)$ has only a trivial linear relation among roots, i.e., $t=1$, and write

$$\begin{array}{c}\hfill {\Delta }_k:=\mathfrak{D}(f,\sigma ,k)=\left\{\mathit{x}\in \Delta \left|\sum _{i=1}^n{x}_i=k\right.\right\}(0\le k\le n),\end{array}$$

then

$$\begin{array}{c}\hfill {\widehat{\mathfrak{D}}}_n={\cup }_{k=0}^n{\Delta }_k({\Delta }_0=\left\{(0,\dots ,0)\right\},{\Delta }_n=\left\{(1,\dots ,1)\right\}).\end{array}$$

The vertices of ${\Delta }_k$ are the following $k(n-k)+1$ vectors:

$$\begin{array}{c}\hfill P(p,q,r):=(\underset{p}{\underbrace{0,\dots ,0}},\underset{q}{\underbrace{ϵ,\dots ,ϵ}},\underset{r}{\underbrace{1,\dots ,1}})\end{array}$$

with non-negative integers $p,q,r$ satisfying $p+q+r=n$ and

$$\left\{\begin{array}{cc}p=n-k,q=0\hfill & ifr=k,\hfill \\ k-r+1\le q\le n-r,ϵ={\displaystyle \frac{k-r}{q}}\hfill & if0\le r\le k-1.\hfill \end{array}\right.$$

Because, if $\mathit{x}\in {\Delta }_k$ satisfies three strict inequalities $x_i<x_{i+1}=\cdots =x_{i+a}<x_{i+a+1}=\cdots =x_{i+a+b}<x_{i+a+b+1}$, then for any η sufficiently close to 0, the vector $\mathit{x}\left(\eta \right)$, replacing $x_{i+1}=\cdots =x_{i+a},x_{i+a+1}=\cdots =x_{i+a+b}$ by $x_{i+a}+\eta /a,x_{i+a+1}-\eta /b$, respectively, is still in ${\Delta }_k$, that is, a short segment containing $\mathit{x}$ is in ${\Delta }_k$, thus $\mathit{x}$ is not a vertex. Therefore, if $\mathit{x}=(x_1,\dots ,x_n)$ is a vertex, then we have $\#\{x_i\mid i=1,\dots ,n\}\le 3$. Suppose that a point $\mathit{x}=(\underset{p}{\underbrace{x_1,\dots ,x_1}},\underset{q}{\underbrace{x_2,\dots ,x_2}},\underset{r}{\underbrace{x_3,\dots ,x_3}})\in {\Delta }_k$ ($0\le x_1<x_2<x_3\le 1$ and $p,q,r>0$) is a vertex. If $x_1>0$, then the vector $\mathit{v}=(1/p,\dots ,1/p,-1/q,\dots ,-1/q,0,\dots ,0)$ satisfies condition (11), therefore, $\mathit{x}$ is not a vertex. Thus, we have $x_1=0$, and similarly $x_3=1$, hence $0<x_2=(k-r)/q<1$. These are the $(k-1)(n-k-1)$ cases of $1\le r\le k-1$, $k-r+1\le q<n-r$, $p=n-q-r>0$. Next, suppose $\mathit{x}=(\underset{s}{\underbrace{x_1,\dots ,x_1}},\underset{t}{\underbrace{x_2,\dots ,x_2}})\in {\Delta }_k$ is a vertex with $0\le x_1<x_2\le 1$ and $s,t>0$. In case of $0<x_1<x_2<1$, $\mathit{x}$ is not a vertex, since the vector $\mathit{v}=(1/s,\dots ,1/s,-1/t,\dots ,-1/t)$ satisfies (11). Thus, $\mathit{x}$ is of the form $(0,\dots ,0,x_2,\dots ,x_2)$ or $(x_1,\dots ,x_1,1,\dots ,1)$, hence $\mathit{x}=(0,\dots ,0,k/q,\dots ,k/q)$ with $0<k<q,p>0,r=0$, $(0,\dots ,0,1,\dots ,1)$ with $r=k,q=0$ or $\left(\right(k-r)/q,\dots ,(k-r)/q,1,\dots ,1)$ with $0<(k-r)/q<1,p=0,q,r>0$. Lastly, in the case of $\mathit{x}=(x,\dots ,x)$, $\mathit{x}=(k/n,\dots ,k/n)$ with $q=n,p=r=0$ is obvious.

The convex hull ${\Delta }_k$ is a simplex if and only if $k=1,n-1$, and vertices of ${\Delta }_1$ are

$$(\underset{n-q}{\underbrace{0,\dots ,0}},\underset{q}{\underbrace{{\displaystyle \frac{1}{q}},\dots ,{\displaystyle \frac{1}{q}}}})(1\le q\le n),$$

and those of ${\Delta }_{n-1}$ are

$$(\underset{q}{\underbrace{1-{\displaystyle \frac{1}{q}},\dots ,1-{\displaystyle \frac{1}{q}}}},\underset{n-q}{\underbrace{1,\dots ,1}})(1\le q\le n).$$

Assume that the permutation $\sigma$ is the identity and $\mathfrak{D}={\Delta }_k$. We can take ${\mathit{m}}_1=(1,\dots ,1)$, ${\mathit{m}}_2=(0,1,0,\dots ,0)$, …, ${\mathit{m}}_n=(0,\dots ,0,1)$, thus,

$$M=\left(\begin{array}{cccc}1& 1& \dots & 1\\ 0& & & \\ ⋮& & 1_{n-1}& \\ 0& & \end{array}\right),M^{-1}=\left(\begin{array}{cccc}1& -1& \dots & -1\\ 0& & & \\ ⋮& & 1_{n-1}& \\ 0& & \end{array}\right)$$

and

$$M{}^{t}M=\left(\begin{array}{cccc}n& 1& \dots & 1\\ 1& & & \\ ⋮& & 1_{n-1}& \\ 1& & \end{array}\right).$$

Then, we have

$$\begin{array}{c}{\mathit{x}}_0=k(n^{-1},0^{(1,n-1)})M={\displaystyle \frac{k}{n}}{\mathit{m}}_1,\\ X=\left\{(x_2,\dots ,x_n)\in {\mathbb{R}}^{n-1}\left|{\displaystyle \frac{k}{n}}{\mathit{m}}_1+(-\sum _{j=2}^n{x}_j,x_2,\dots ,x_n)\in {\Delta }_k\right.\right\},\end{array}$$

and for $\mathit{x}\in X$ and $\mathit{u}\in \mathfrak{D}$, we see

$$\psi \left(\mathit{x}\right)=\mathit{u}\iff (\mathit{u}-{\displaystyle \frac{k}{n}}{\mathit{m}}_1){}^{t}M=(0,\mathit{x}),$$

and so $\mathrm{vol}\left(\mathfrak{D}\right)=\sqrt{n}·\mathrm{vol}\left(X\right)$ by (7).

In case of $k=1$, the vertices of X are

$$(0,\dots ,0),(\underset{n-q-1}{\underbrace{-{\displaystyle \frac{1}{n}},\dots ,-{\displaystyle \frac{1}{n}}}},\underset{q}{\underbrace{{\displaystyle \frac{1}{q}}-{\displaystyle \frac{1}{n}},\dots ,{\displaystyle \frac{1}{q}}-{\displaystyle \frac{1}{n}}}})(1\le q\le n-1).$$

In case of $k=n-1$, the vertices of X are

$$\left({\displaystyle \frac{1}{n}},\dots ,{\displaystyle \frac{1}{n}}\right),(\underset{q-1}{\underbrace{{\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{q}},\dots ,{\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{q}}}},\underset{n-q}{\underbrace{{\displaystyle \frac{1}{n}},\dots ,{\displaystyle \frac{1}{n}}}})(2\le q\le n).$$

To describe the volume of ${\Delta }_k$, let us introduce the Eulerian numbers $A(n,k)$ for $1\le k\le n$. They are defined recursively by

$$A(1,1):=1,A(n,k):=(n-k+1)A(n-1,k-1)+kA(n-1,k),$$

more explicitly

$$A(n,k)=\sum _{j=0}^k{(-1)}^j\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{j}}\right){(k-j)}^n$$

and the volume of

$$\{\mathit{x}\in {[0,1)}^{n-1}\mid \lceil x_1+\dots +x_{n-1}\rceil =k\}$$

is given by $E_n\left(k\right):=A(n-1,k)/(n-1)!$, where $\lceil x\rceil$ is the integer satisfying $x\le \lceil x\rceil <x+1$ ([10]).

Proposition 2. 

We see that

$$\mathrm{vol}\left({\widehat{\mathfrak{D}}}_n\right)={\displaystyle \frac{\sqrt{n}}{n!}}$$

and for an integer k $(1\le k\le n-1)$,

$$\mathrm{vol}\left({\Delta }_k\right)/\mathrm{vol}\left({\widehat{\mathfrak{D}}}_n\right)=E_n\left(k\right).$$

Proof. 

We note that $\mathrm{vol}$ is the $(n-1)$-dimensional volume here. Let $\theta$ be the angle between two hyperplanes defined by $x_n=0$ and $\sum x_i=k$, respectively, that is, the angle between the vectors $(0,\dots ,0,1)$ and $(1,\dots ,1)$, hence $cos\theta =1/\sqrt{n}$. Since a permutation acts on ${\mathbb{R}}^n$ and on the set $\{\mathit{x}\in {\mathbb{R}}^n\mid \sum x_i=k\}$ as an orthogonal transformation and the dimension of the set defined by $\sum x_i\in \mathbb{Z}$ and $x_i=x_{i+1}$ is $n-2$, $\mathrm{vol}\left({\widehat{\mathfrak{D}}}_n\right)$ is equal to the volume of the set

$$\{\mathit{x}\mid 0<x_1<\cdots <x_n<1,\sum x_i\in \mathbb{Z}\},$$

hence, we see that

$$\begin{array}{c}\hfill \mathrm{vol}\left({\widehat{\mathfrak{D}}}_n\right)={\displaystyle \frac{1}{n!}}\mathrm{vol}\left(\{(x_1,\dots ,x_n)\in {[0,1)}^n\mid \sum x_i\in \mathbb{Z}\}\right)\end{array}$$

and the projection $(x_1,\dots ,x_n)\mapsto (x_1,\dots ,x_{n-1})$ from $\{(x_1,\dots ,x_n)\in {[0,1)}^n\mid \sum x_i\in \mathbb{Z}\}$ to ${[0,1)}^{n-1}$ is bijective by $x_1+\dots +x_{n-1}+x_n=\lceil x_1+\dots +x_{n-1}\rceil$, we have $n!\mathrm{vol}\left({\widehat{\mathfrak{D}}}_n\right)cos\theta =\mathrm{vol}\left({[0,1)}^{n-1}\right)=1.$ Now, we have

$$\begin{array}{cc}\hfill & \mathrm{vol}\left({\Delta }_k\right)/\mathrm{vol}\left({\widehat{\mathfrak{D}}}_n\right)\hfill \\ \hfill =& n!cos\theta ·\mathrm{vol}\left(\{\mathit{x}\in {\widehat{\mathfrak{D}}}_n\mid \sum x_i=k\}\right)\hfill \\ \hfill =& cos\theta ·\mathrm{vol}\left(\{\mathit{x}\in {[0,1)}^n\mid \sum x_i=k\}\right)\hfill \\ \hfill =& cos\theta ·\mathrm{vol}(\{\mathit{x}\in {[0,1)}^n\mid ⌈\sum _{i=1}^{n-1}{x}_i⌉=k\})\hfill \\ \hfill =& \mathrm{vol}(\{\mathit{x}\in {[0,1)}^{n-1}\mid ⌈\sum _{i=1}^{n-1}{x}_i⌉=k\})\hfill \\ \hfill =& E_n\left(k\right).\hfill \end{array}$$

□

Last, letting $D=\{(x_1,\dots ,x_n)\in \Delta \mid |{\sum }_{i=1}^n{m}_{j,i}{x}_{\sigma \left(i\right)}-k_j|\le {\displaystyle \frac{1}{3}}(1\le {}^{\forall }j\le t)\}$ for integers $k_j$, we see that Conjecture 2 means

$$\begin{array}{cc}\hfill & {\displaystyle \underset{X\to \infty }{lim}\frac{\#\{p\in {\mathrm{Spl}}_X(f,\sigma )\mid |{\sum }_{i=1}^n{m}_{j,i}{r}_{\sigma \left(i\right)}/p-k_j|\le 1/3(1\le {}^{\forall }j\le t)\}}{\#{\mathrm{Spl}}_X(f,\sigma )}}\hfill \\ \hfill =& {\displaystyle \frac{\mathrm{vol}\left(\mathfrak{D}(f,\sigma ,\left\{k_j\right\})\right)}{\mathrm{vol}\left(\mathfrak{D}\right(f,\sigma \left)\right)}.}\hfill \end{array}$$

When the polynomial $f\left(x\right)$ has only a trivial linear relation among roots, Proposition 2 gives the above ratio.

Example 2. 

Suppose that an irreducible polynomial $f\left(x\right)$ is of degree 4 and has a non-trivial linear relations among roots, then it is of the form $g\left(h\right(x\left)\right)$ for quadratic polynomials $g\left(x\right),h\left(x\right)=x^2+ax$ ([1]). For roots ${\beta }_1,{\beta }_2$ of $g\left(x\right)$, take the complex numbers ${\alpha }_1,\dots ,{\alpha }_4$ such that $h\left({\alpha }_1\right)=h\left({\alpha }_4\right)={\beta }_1$, $h\left({\alpha }_2\right)=h\left({\alpha }_3\right)={\beta }_2$. Then, ${\alpha }_1,\dots ,{\alpha }_4$ are the roots of $f\left(x\right)$ which satisfy relations ${\alpha }_1+{\alpha }_4={\alpha }_2+{\alpha }_3=-a$. They are the basis of $L{R}_0\cap {\mathbb{Z}}^4$, i.e., $t=2$, ${\mathit{m}}_1=(1,0,0,1),{\mathit{m}}_2=(0,1,1,0)$. Hence, we may take ${\mathit{m}}_3:=(0,0,1,0),{\mathit{m}}_4:=(0,0,0,1)$; then we have

$$\begin{array}{cc}\hfill & \mathfrak{D}:=\mathfrak{D}(f,id,k_1,k_2)\hfill \\ \hfill =& \{(x_1,\dots ,x_4)\mid 0\le x_1\le \cdots \le x_4\le 1,x_1+x_4=k_1,x_2+x_3=k_2\}.\hfill \end{array}$$

It is easy to see that $dim\mathfrak{D}=2\iff k_1=k_2=1$, and then the vertices are ${\mathit{v}}_1=(0,0,1,1)$, ${\mathit{v}}_2=\left(0,{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}},1\right)$, ${\mathit{v}}_3=\left({\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}}\right)$.

Assume $\sigma =\mathrm{id}$ and $k_1=k_2=1$; then we find

$$\mathfrak{D}=\left\{(1-x_4,1-x_3,x_3,x_4)\in {\mathbb{R}}^4\left|{\displaystyle \frac{1}{2}}\le x_3\le x_4\le 1\right.\right\},$$

$$M=\left(\begin{array}{cc}{1}_2& H\\ 0& 1_2\end{array}\right),M^{-1}=\left(\begin{array}{cc}{1}_2& -H\\ 0& 1_2\end{array}\right) for H=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$$

and

$$M{}^{t}M=\left(\begin{array}{cc}{21}_2& H\\ H& 1_2\end{array}\right).$$

Then we see that

$$\begin{array}{c}\hfill {\mathit{x}}_0=(1,1)(2^{-1}{1}_2,0^{(2,2)})M=2^{-1}(1,1,1,1)\end{array}$$

and

$$\begin{array}{cc}\hfill X& =\{(x_3,x_4)\in {\mathbb{R}}^2\mid 2^{-1}(1,1,1,1)+(-x_4,-x_3,x_3,x_4)\in \mathfrak{D}\}\hfill \\ \hfill & =\left\{(x_3,x_4)\in {\mathbb{R}}^2\mid 0\le x_3\le x_4\le 2^{-1}\right\}.\hfill \end{array}$$

The vertices ${\mathit{x}}_i$ corresponding to ${\mathit{v}}_i$ are ${\mathit{x}}_1=\left(2^{-1},2^{-1}\right)$, ${\mathit{x}}_2=\left(0,2^{-1}\right)$, ${\mathit{x}}_3=(0,0)$.

Here, let us discuss the topic mentioned at the end of the third section.

We have

$$\begin{array}{c}{\mathit{m}}_3^{\prime }=(0,-1,1,0),{\mathit{m}}_4^{\prime }=(-1,0,0,1),\end{array}$$

and

$$\mathcal{C}=(({\mathit{m}}_3^{\prime },\mathit{c}),({\mathit{m}}_4^{\prime },\mathit{c}))=(c_3-c_2,c_4-c_1)\mathrm{for} \mathit{c}=(c_1,c_2,c_3,c_4),$$

hence,

$$({\mathit{x}}_1,\mathcal{C})=2^{-1}(c_3+c_4-c_1-c_2),({\mathit{x}}_2,\mathcal{C})=2^{-1}(c_4-c_1),({\mathit{x}}_3,\mathcal{C})=(0,0).$$

Thus, the condition that $({\mathit{x}}_i,\mathcal{C})$ are mutually distinct and $({\mathit{x}}_i,\mathcal{C})-({\mathit{x}}_j,\mathcal{C})$ is an integer for every $i,j$ is $c_1-c_4\ne 0$, $c_2-c_3\ne 0,$$c_1+c_2-(c_3+c_4)\ne 0$, $c_1-c_4\in 2\mathbb{Z}$ and $c_2-c_3\in 2\mathbb{Z}$. And we see by $V=\left(\begin{array}{c}{\mathit{x}}_1\\ {\mathit{x}}_2\end{array}\right)$

$$\begin{array}{cc}\hfill e\left((x_3,x_4)V{}^t\mathcal{C}\right)& =e(2^{-1}(c_3+c_4-c_1-c_2)x_3+2^{-1}(c_4-c_1)x_4)\hfill \\ \hfill & =e((n+m)x_3+n{x}_4),\hfill \end{array}$$

where $2n:=c_4-c_1,2m:=c_3-c_2$ with $n,m\in \mathbb{Z},$ and $n,m,n+m\ne 0$. Do these functions under the above restrictions on $c_i$ with constant functions approximate continuous functions on $X=\{(x_3,x_4)\mid 0\le x_3\le x_4\le 2^{-1}\}$ ?

6. Calculation of the Integral

Finally, we will evaluate the integral in (10).

Lemma 2. 

Let m be a positive integer. For mutually distinct real numbers ${\alpha }_1,\dots ,{\alpha }_{m+1}$, write $A_i:={\alpha }_i-{\alpha }_{m+1}$$(i=1,\dots ,m)$, we have

$$\begin{array}{cc}\hfill & {\int }_{x_i\ge 0,\sum x_i\le 1}e\left(\sum _{i=1}^m{A}_i{x}_i\right)d{x}_1\dots d{x}_m\hfill \\ \hfill =& {\displaystyle \frac{1}{{\left(2\pi i\right)}^m}}\left\{\sum _{i=1}^m{\displaystyle \frac{e\left(A_i\right)}{{\displaystyle A_i\prod _{\begin{array}{c}1\le j\le m,\\ j\ne i\end{array}}(A_i-A_j)}}}+{\displaystyle \frac{1}{{\displaystyle \prod _{1\le j\le m}(-A_j)}}}\right\},\hfill \\ \hfill =& {\displaystyle \frac{1}{{\left(2\pi i\right)}^m}}\sum _{i=1}^{m+1}{\displaystyle \frac{e({\alpha }_i-{\alpha }_{m+1})}{{\displaystyle \prod _{\begin{array}{c}1\le j\le m+1,\\ j\ne i\end{array}}({\alpha }_i-{\alpha }_j)}}}\hfill \\ \hfill =& {\displaystyle \frac{{(-1)}^{m}e(-{\alpha }_{m+1})}{{\left(2\pi i\right)}^m}}{\displaystyle \frac{\left|\begin{array}{cccc}e\left({\alpha }_1\right)& e\left({\alpha }_2\right)& \dots & e\left({\alpha }_{m+1}\right)\\ 1& 1& \dots & 1\\ {\alpha }_1& {\alpha }_2& \dots & {\alpha }_{m+1}\\ ⋮& ⋮& ⋮& ⋮\\ {\alpha }_1^{m-1}& {\alpha }_2^{m-1}& \dots & {\alpha }_{m+1}^{m-1}\end{array}\right|}{\left|\begin{array}{cccc}1& 1& \dots & 1\\ {\alpha }_1& {\alpha }_2& \dots & {\alpha }_{m+1}\\ ⋮& ⋮& ⋮& ⋮\\ {\alpha }_1^m& {\alpha }_2^m& \dots & {\alpha }_{m+1}^m\end{array}\right|}}\hfill \end{array}$$

and the integral vanishes if all $A_i$ are integers. Moreover, we have

$$\sum _{i=1}^m{\displaystyle \frac{1}{{\displaystyle A_i\prod _{\begin{array}{c}1\le j\le m,\\ j\ne i\end{array}}(A_i-A_j)}}}+{\displaystyle \frac{1}{{\displaystyle \prod _{1\le j\le m}(-A_j)}}}=0,$$

i.e.,

$$\sum _{i=1}^{m+1}{\displaystyle \frac{1}{{\displaystyle \prod _{\begin{array}{c}1\le j\le m+1,\\ j\ne i\end{array}}({\alpha }_i-{\alpha }_j)}}}=0,$$

and for a polynomial $g\left(x\right)=c_{m-1}{x}^{m-1}+c_{m-2}{x}^{m-2}+\dots$

$$\sum _{i=1}^m{\displaystyle \frac{g\left(A_i\right)}{{\displaystyle \prod _{\begin{array}{c}1\le j\le m,\\ j\ne i\end{array}}(A_i-A_j)}}}=c_{m-1}.$$

(12)

Proof. 

Denote the integral by $S(A_1,\dots ,A_m)$. We use induction on m. It is easy to see that $S\left(A_1\right)={\displaystyle \frac{e\left(A_1\right)-1}{2\pi i{A}_1}}$ and $S(A_1,\dots ,A_m)$ is equal to

$$\begin{array}{cc}\hfill & \underset{\begin{array}{c}{x}_1,\dots ,x_{m-1}\ge 0,\\ {\sum }_{i=1}^{m-1}{x}_i\le 1\end{array}}{\int }\left(\underset{0\le x_m\le 1-x_1-\dots -x_{m-1}}{\int }e(\sum A_i{x}_i)d{x}_m\right)d{x}_1\dots d{x}_{m-1}\hfill \\ \hfill =& \underset{\begin{array}{c}{x}_1,\dots ,x_{m-1}\ge 0,\\ {\sum }_{i=1}^{m-1}{x}_i\le 1\end{array}}{\int }{\displaystyle \frac{e\left({\sum }_{i=1}^{m-1}{A}_i{x}_i\right)}{2\pi i{A}_m}}\left(e(A_m(1-\sum _{i=1}^{m-1}{x}_i))-1\right)d{x}_1\dots d{x}_{m-1}\hfill \\ \hfill =& {\displaystyle \frac{e\left(A_m\right)}{2\pi i{A}_m}}S(A_1-A_m,\dots ,A_{m-1}-A_m)-{\displaystyle \frac{S(A_1,\dots ,A_{m-1})}{2\pi i{A}_m}}.\hfill \end{array}$$

The assumption of induction easily completes the proof of the first equality. With respect to the transform to ${\alpha }_i$, we use the Vandermonde determinant:

$$det\left(\begin{array}{cccc}1& 1& \dots & 1\\ {\alpha }_1& {\alpha }_2& \dots & {\alpha }_m\\ ⋮& ⋮& ⋮& ⋮\\ {\alpha }_1^{m-1}& {\alpha }_2^{m-1}& \dots & {\alpha }_m^{m-1}\end{array}\right)={(-1)}^{m(m-1)/2}\prod _{i<j}({\alpha }_i-{\alpha }_j).$$

For the second identities, replace $A_i$ by $ϵA_i$, then we have

$$\begin{array}{cc}\hfill & {\int }_{x_i\ge 0,\sum x_i\le 1}e(\sum _{i=1}^mϵA_i{x}_i)d{x}_1\dots d{x}_m\hfill \\ \hfill =& {\displaystyle \frac{1}{{\left(2\pi iϵ\right)}^m}}\left\{\sum _{i=1}^m{\displaystyle \frac{e\left(ϵA_i\right)}{{\displaystyle A_i\prod _{\begin{array}{c}j=1,\\ j\ne i\end{array}}^m(A_i-A_j)}}}+{\displaystyle \frac{1}{{\prod }_{j=1}^m(-A_j)}}\right\}\hfill \\ \hfill =& {\displaystyle \frac{1}{{\left(2\pi iϵ\right)}^m}}\left\{\sum _{i=1}^m{\displaystyle \frac{1}{{\displaystyle A_i\prod _{\begin{array}{c}j=1,\\ j\ne i\end{array}}^m(A_i-A_j)}}}+{\displaystyle \frac{1}{{\prod }_{j=1}^m(-A_j)}}\right.\hfill \\ \hfill & +\left.\sum _{k=1}^{\infty }{\displaystyle \frac{{\left(2\pi iϵ\right)}^k}{k!}}\sum _{i=1}^m{\displaystyle \frac{A_i^{k-1}}{{\displaystyle \prod _{\begin{array}{c}j=1,\\ j\ne i\end{array}}^m(A_i-A_j)}}}\right\}\hfill \\ \hfill & \to {\int }_{x_i\ge 0,\sum x_i\le 1}1d{x}_1\dots d{x}_m={\displaystyle \frac{1}{m!}}(ϵ\to 0).\hfill \end{array}$$

Therefore, the coefficient of ${ϵ}^{k-m}$ vanishes if $k-m<0$ and is ${\displaystyle \frac{1}{m!}}$ if $k-m=0$.

□

Theorem 3. 

Let ${\alpha }_1,\dots ,{\alpha }_{m+1}\in \mathbb{R}$ and let $a_1,\dots ,a_u$ be their distinct numbers, i.e., $\{{\alpha }_i\mid 1\le i\le m+1\}=\{a_i\mid 1\le i\le u\}$ and $T_J:=\{j\mid {\alpha }_j=a_J(1\le j\le m+1)\}$. Then, we have

$$\begin{array}{cc}\hfill & {\int }_{x_i\ge 0,\sum x_i\le 1}e(\sum _{i=1}^m({\alpha }_i-{\alpha }_{m+1})x_i)d{x}_1\dots d{x}_m\hfill \\ \hfill =& {\displaystyle \frac{1}{{\left(2\pi i\right)}^m}}\sum _{J=1}^u{\displaystyle \frac{e(a_J-{\alpha }_{m+1})}{\prod _{l\notin T_J}(a_J-{\alpha }_l)}}\left\{\sum _{\begin{array}{c}{n}_0,n_d\ge 0(d\notin T_J),\\ n_0+{\displaystyle \sum _{d\notin T_J}}{n}_d=\#T_J-1\end{array}}{\displaystyle \frac{{\left(2\pi i\right)}^{n_0}}{{\displaystyle n_0!\prod _{d\notin T_J}{({\alpha }_d-a_J)}^{n_d}}}}\right\},\hfill \end{array}$$

where the indexes $l,d$ run over $\{1,\dots ,m+1\}\setminus T_J$.

Proof. 

Let ${\delta }_i$ be mutually distinct numbers and put ${\beta }_i:={\alpha }_i+ϵ{\delta }_i$; then we see that ${\beta }_i-{\beta }_j={\alpha }_i-{\alpha }_j+({\delta }_i-{\delta }_j)ϵ$ is not 0 for a non-zero number $ϵ$ is sufficiently close to 0 if $i\ne j$. By Lemma 2, we have

$$\begin{array}{cc}\hfill & {\int }_{x_i\ge 0,\sum x_i\le 1}e(\sum _{i=1}^m({\alpha }_i-{\alpha }_{m+1}))d{x}_1\dots d{x}_m\hfill \\ \hfill =& \underset{ϵ\to 0}{lim}{\int }_{x_i\ge 0,\sum x_i\le 1}e(\sum _{i=1}^m({\beta }_i-{\beta }_{m+1}))d{x}_1\dots d{x}_m\hfill \\ \hfill & ={\displaystyle \frac{{(-1)}^m}{{\left(2\pi i\right)}^m}}\underset{ϵ\to 0}{lim}{\displaystyle \frac{N\left(ϵ\right)}{D\left(ϵ\right)}},\hfill \end{array}$$

where

$$N\left(ϵ\right)=\left|\begin{array}{cccc}e({\beta }_1-{\beta }_{m+1})& e({\beta }_2-{\beta }_{m+1})& \dots & e({\beta }_{m+1}-{\beta }_{m+1})\\ 1& 1& \dots & 1\\ {\beta }_1& {\beta }_2& \dots & {\beta }_{m+1}\\ ⋮& ⋮& ⋮& ⋮\\ {\beta }_1^{m-1}& {\beta }_2^{m-1}& \dots & {\beta }_{m+1}^{m-1}\end{array}\right|$$

and

$$D\left(ϵ\right)=\left|\begin{array}{cccc}1& 1& \dots & 1\\ {\beta }_1& {\beta }_2& \dots & {\beta }_{m+1}\\ ⋮& ⋮& ⋮& ⋮\\ {\beta }_1^m& {\beta }_2^m& \dots & {\beta }_{m+1}^m\end{array}\right|.$$

From now on, the notations ${{\displaystyle \dot{\sum }}}_{k,l}$ and ${{\displaystyle \dot{\prod }}}_{k,l}$ will mean the sum and the product on $k,l$ under the condition $1\le k<l\le m+1$. First, we see that

$$\begin{array}{cc}\hfill D\left(ϵ\right)& ={(-1)}^{(m+1)m/2}\underset{k,l}{\dot{\prod }}({\beta }_k-{\beta }_l)\hfill \\ \hfill & ={(-1)}^{(m+1)m/2}\underset{k,l,{\alpha }_k\ne {\alpha }_l}{\dot{\prod }}\left\{({\alpha }_k-{\alpha }_l)(1+{\displaystyle \frac{{\delta }_k-{\delta }_l}{{\alpha }_k-{\alpha }_l}}ϵ)\right\}\underset{k,l,{\alpha }_k={\alpha }_l}{\dot{\prod }}\left\{({\delta }_k-{\delta }_l)ϵ\right\}\hfill \\ \hfill & =\{{(-1)}^{(m+1)m/2}\underset{k,l,{\alpha }_k\ne {\alpha }_l}{\dot{\prod }}({\alpha }_k-{\alpha }_l)·\underset{k,l,{\alpha }_k={\alpha }_l}{\dot{\prod }}({\delta }_k-{\delta }_l)\}{ϵ}^{\kappa }+{ϵ}^{\kappa +1}O\left(1\right),\hfill \end{array}$$

where $\kappa :=\#\{(k,l)\mid 1\le k<l\le m+1,{\alpha }_k={\alpha }_l\}$.

Next, we see that

$$\begin{array}{cc}\hfill N\left(ϵ\right)& =\sum _{j=1}^{m+1}{(-1)}^{j-1}e({\beta }_j-{\beta }_{m+1})·{(-1)}^{m(m-1)/2}\underset{k,l\ne j}{\dot{\prod }}({\beta }_k-{\beta }_l)\hfill \\ \hfill =& {(-1)}^{m(m-1)/2}\sum _{j=1}^{m+1}{(-1)}^{j-1}e({\alpha }_j-{\alpha }_{m+1})\times \hfill \\ \hfill & \underset{k,l\ne j,{\alpha }_k\ne {\alpha }_l}{\dot{\prod }}({\alpha }_k-{\alpha }_l)·\underset{k,l\ne j,{\alpha }_k={\alpha }_l}{\dot{\prod }}({\delta }_k-{\delta }_l)\times \hfill \\ \hfill & e\left(({\delta }_j-{\delta }_{m+1})ϵ\right)·\underset{k,l\ne j,{\alpha }_k\ne {\alpha }_l}{\dot{\prod }}(1+{\displaystyle \frac{{\delta }_k-{\delta }_l}{{\alpha }_k-{\alpha }_l}}ϵ)·{ϵ}^{{\kappa }_j}\hfill \end{array}$$

where ${\kappa }_j:=\#\{(k,l)\mid 1\le k<l\le m+1,k,l\ne j,{\alpha }_k={\alpha }_l\}$. Here, we note that ${\kappa }_j$ depends only on $T_J$ containing j by the equation $\kappa -{\kappa }_j=\#\{(k,l)\mid 1\le k<l\le m+1,{\alpha }_k={\alpha }_l,k\mathrm{or}l=j\}=\#T_J-1$. Moreover, by

$$\begin{array}{cc}\hfill & \{(k,l)\mid 1\le k<l\le m+1,{\alpha }_k\ne {\alpha }_l\}\hfill \\ \hfill =& \{(k,l)\mid 1\le k<l\le m+1,{\alpha }_k\ne {\alpha }_l,k,l\ne j\}\hfill \\ \hfill & \cup \{(j,l)\mid 1\le j<l\le m+1,{\alpha }_l\ne {\alpha }_j\}\hfill \\ \hfill & \cup \{(k,j)\mid 1\le k<j\le m+1,{\alpha }_k\ne {\alpha }_j\},\hfill \\ \hfill & \{(k,l)\mid 1\le k<l\le m+1,{\alpha }_k={\alpha }_l\}\hfill \\ \hfill =& \{(k,l)\mid 1\le k<l\le m+1,{\alpha }_k={\alpha }_l,k,l\ne j\}\hfill \\ \hfill & \cup \{(j,l)\mid 1\le j<l\le m+1,{\alpha }_l={\alpha }_j\}\hfill \\ \hfill & \cup \{(k,j)\mid 1\le k<j\le m+1,{\alpha }_k={\alpha }_j\},\hfill \end{array}$$

we see

$$\begin{array}{cc}\hfill & {\displaystyle {\dot{\prod }}_{\begin{array}{c}k,l,\\ {\alpha }_k\ne {\alpha }_l\end{array}}({\alpha }_k-{\alpha }_l)}\hfill \\ \hfill =& {(-1)}^{{\eta }_j}{\dot{\prod }}_{\begin{array}{c}k,l,\\ {\alpha }_k\ne {\alpha }_l,k,l\ne j\end{array}}({\alpha }_k-{\alpha }_l)\prod _{1\le l\le m+1,{\alpha }_l\ne {\alpha }_j}({\alpha }_j-{\alpha }_l),\hfill \end{array}$$

where ${\eta }_j=\#\{k\mid 1\le k<j,{\alpha }_k\ne {\alpha }_j\}$, and

$$\begin{array}{cc}\hfill & {\dot{\prod }}_{\begin{array}{c}k,l,\\ {\alpha }_k={\alpha }_l\end{array}}({\delta }_k-{\delta }_l)\hfill \\ \hfill =& {(-1)}^{{\gamma }_j}{\dot{\prod }}_{\begin{array}{c}k,l,\\ {\alpha }_k={\alpha }_l,k,l\ne j\end{array}}({\delta }_k-{\delta }_l)\prod _{1\le l\le m+1,l\ne j,{\alpha }_l={\alpha }_j}({\delta }_j-{\delta }_l),\hfill \end{array}$$

where ${\gamma }_j=\#\{k\mid 1\le k<j,{\alpha }_k={\alpha }_j\}$, therefore, we have

$$\begin{array}{cc}\hfill & {(-1)}^{j-1}\underset{k,l\ne j,{\alpha }_k\ne {\alpha }_l}{\dot{\prod }}({\alpha }_k-{\alpha }_l)·\underset{k,l\ne j,{\alpha }_k={\alpha }_l}{\dot{\prod }}({\delta }_k-{\delta }_l)\hfill \\ \hfill =& {\displaystyle \frac{{\displaystyle \underset{k,l,{\alpha }_k\ne {\alpha }_l}{\dot{\prod }}}({\alpha }_k-{\alpha }_l)·{\displaystyle \underset{k,l,{\alpha }_k={\alpha }_l}{\dot{\prod }}}({\delta }_k-{\delta }_l)}{{\displaystyle \prod _{l,{\alpha }_l\ne {\alpha }_j}}({\alpha }_j-{\alpha }_l)·{\displaystyle \prod _{l\ne j,{\alpha }_l={\alpha }_j}}({\delta }_j-{\delta }_l)}}.\hfill \end{array}$$

Thus, we have

$$\begin{array}{cc}\hfill & N\left(ϵ\right)\hfill \\ \hfill =& {(-1)}^{m(m-1)/2}\underset{k,l,{\alpha }_k\ne {\alpha }_l}{\dot{\prod }}({\alpha }_k-{\alpha }_l)·\underset{k,l,{\alpha }_k={\alpha }_l}{\dot{\prod }}({\delta }_k-{\delta }_l)\times \hfill \\ \hfill & \sum _{J=1}^{u}e(a_J-{\alpha }_{m+1})\sum _{j\in T_J}{\displaystyle \frac{1}{{\displaystyle \prod _{l,{\alpha }_l\ne {\alpha }_j}}({\alpha }_j-{\alpha }_l)·{\displaystyle \prod _{l\ne j,{\alpha }_l={\alpha }_j}}({\delta }_j-{\delta }_l)}}\times \hfill \\ \hfill & e\left(({\delta }_j-{\delta }_{m+1})ϵ\right)\underset{k,l\ne j,{\alpha }_k\ne {\alpha }_l}{\dot{\prod }}(1+{\displaystyle \frac{{\delta }_k-{\delta }_l}{{\alpha }_k-{\alpha }_l}}ϵ)·{ϵ}^{{\kappa }_j}\hfill \\ \hfill =& {(-1)}^{m(m-1)/2}\underset{k,l,{\alpha }_k\ne {\alpha }_l}{\dot{\prod }}({\alpha }_k-{\alpha }_l)·\underset{k,l,{\alpha }_k={\alpha }_l}{\dot{\prod }}({\delta }_k-{\delta }_l)\times \hfill \\ \hfill & \sum _{J=1}^u{\displaystyle \frac{e(a_J-{\alpha }_{m+1})}{\left\{{\displaystyle \prod _{l\notin T_J}}(a_J-{\alpha }_l)\right\}}}\sum _{j\in T_J}{\displaystyle \frac{e\left(({\delta }_j-{\delta }_{m+1})ϵ\right)}{\left\{{\displaystyle \prod _{l\in T_J,l\ne j}}({\delta }_j-{\delta }_l)\right\}}}\left\{\underset{k,l\ne j,{\alpha }_k\ne {\alpha }_l}{\dot{\prod }}(1+{\displaystyle \frac{{\delta }_k-{\delta }_l}{{\alpha }_k-{\alpha }_l}}ϵ)\right\}{ϵ}^{{\kappa }_j}.\hfill \end{array}$$

Here, by the equality

$$\begin{array}{cc}\hfill & \{(k,l)\mid k,l\ne j,{\alpha }_k\ne {\alpha }_l\}\hfill \\ \hfill =& \{(k,l)\mid k,l\notin T_J,{\alpha }_k\ne {\alpha }_l\}\hfill \\ \hfill & \cup \{(k,l)\mid k\notin T_J,l\in T_J,l\ne j\}\hfill \\ \hfill & \cup \{(k,l)\mid k\in T_J,l\notin T_J,k\ne j\},\hfill \end{array}$$

the product ${\displaystyle \underset{k,l\ne j,{\alpha }_k\ne {\alpha }_l}{\dot{\prod }}}(1+{\displaystyle \frac{{\delta }_k-{\delta }_l}{{\alpha }_k-{\alpha }_l}}ϵ)$ is equal to

$$\begin{array}{cc}\hfill & \underset{k,l\notin T_J,{\alpha }_k\ne {\alpha }_l}{\dot{\prod }}(1+{\displaystyle \frac{{\delta }_k-{\delta }_l}{{\alpha }_k-{\alpha }_l}}ϵ)\underset{k\notin T_J,l\in T_J,l\ne j}{\dot{\prod }}(1+{\displaystyle \frac{{\delta }_k-{\delta }_l}{{\alpha }_k-{\alpha }_l}}ϵ)\times \hfill \\ \hfill & \underset{k\in T_J,l\notin T_J,k\ne j}{\dot{\prod }}(1+{\displaystyle \frac{{\delta }_k-{\delta }_l}{{\alpha }_k-{\alpha }_l}}ϵ)\hfill \\ \hfill =& \underset{k,l\notin T_J,{\alpha }_k\ne {\alpha }_l}{\dot{\prod }}(1+{\displaystyle \frac{{\delta }_k-{\delta }_l}{{\alpha }_k-{\alpha }_l}}ϵ)\prod _{\begin{array}{c}1\le c,d\le m+1,\\ c\in T_J,c\ne j,d\notin T_J\end{array}}(1+{\displaystyle \frac{{\delta }_c-{\delta }_d}{{\alpha }_c-{\alpha }_d}}ϵ)\hfill \\ \hfill =& \underset{k,l\notin T_J,{\alpha }_k\ne {\alpha }_l}{\dot{\prod }}(1+{\displaystyle \frac{{\delta }_k-{\delta }_l}{{\alpha }_k-{\alpha }_l}}ϵ){\displaystyle \frac{{\displaystyle \prod _{\begin{array}{c}1\le c,d\le m+1,\\ c\in T_J,d\notin T_J\end{array}}}(1+\frac{{\delta }_c-{\delta }_d}{{\alpha }_c-{\alpha }_d}ϵ)}{{\displaystyle \prod _{d\notin T_J}}(1+\frac{{\delta }_j-{\delta }_d}{{\alpha }_j-{\alpha }_d}ϵ)}}\hfill \end{array}$$

hence,

$$\begin{array}{cc}\hfill & N\left(ϵ\right)\hfill \\ \hfill =& {(-1)}^{m(m-1)/2}\underset{k,l,{\alpha }_k\ne {\alpha }_l}{\dot{\prod }}({\alpha }_k-{\alpha }_l)·\underset{k,l,{\alpha }_k={\alpha }_l}{\dot{\prod }}({\delta }_k-{\delta }_l)\times \hfill \\ \hfill & \sum _{J=1}^u{\displaystyle \frac{e(a_J-{\alpha }_{m+1})}{{\displaystyle \prod _{l\notin T_J}}(a_J-{\alpha }_l)}}\underset{k,l\notin T_J,{\alpha }_k\ne {\alpha }_l}{\dot{\prod }}(1+{\displaystyle \frac{{\delta }_k-{\delta }_l}{{\alpha }_k-{\alpha }_l}}ϵ)\prod _{\begin{array}{c}1\le c,d\le m+1,\\ c\in T_J,d\notin T_J\end{array}}(1+{\displaystyle \frac{{\delta }_c-{\delta }_d}{{\alpha }_c-{\alpha }_d}}ϵ)\times \hfill \\ \hfill & {ϵ}^{\kappa -\#T_J+1}\sum _{j\in T_J}\{{\displaystyle \frac{e\left(({\delta }_j-{\delta }_{m+1})ϵ\right)}{{\displaystyle \prod _{l\in T_J,l\ne j}}({\delta }_j-{\delta }_l)}}{\displaystyle \frac{1}{{\displaystyle \prod _{\begin{array}{c}1\le d\le m+1,\\ d\notin T_J\end{array}}}(1+\frac{{\delta }_j-{\delta }_d}{a_J-{\alpha }_d}ϵ)}}\}.\hfill \end{array}$$

The partial sum on $j\in T_J$ is

$$\begin{array}{cc}\hfill & \sum _{j\in T_J}{\displaystyle \frac{e\left(({\delta }_j-{\delta }_{m+1})ϵ\right)}{{\displaystyle \prod _{l\in T_J,l\ne j}}({\delta }_j-{\delta }_l)}}{\displaystyle \frac{1}{{\displaystyle \prod _{\begin{array}{c}1\le d\le m+1,\\ d\notin T_J\end{array}}}(1+\frac{{\delta }_j-{\delta }_d}{a_J-{\alpha }_d}ϵ)}}\hfill \\ \hfill =& \sum _{j\in T_J}{\displaystyle \frac{e\left(({\delta }_j-{\delta }_{m+1})ϵ\right)}{{\displaystyle \prod _{l\in T_J,l\ne j}}({\delta }_j-{\delta }_l)}}\prod _{\begin{array}{c}1\le d\le m+1,\\ d\notin T_J\end{array}}\left\{\sum _{q\ge 0}{(-{\displaystyle \frac{{\delta }_j-{\delta }_d}{a_J-{\alpha }_d}}ϵ)}^q\right\}\hfill \\ \hfill =& \sum _{\begin{array}{c}{n}_0,n_d\ge 0,\\ (1\le d\le m+1,d\notin T_J)\end{array}}\left\{\sum _{j\in T_J}{\displaystyle \frac{{\left(2\pi i({\delta }_j-{\delta }_{m+1})\right)}^{n_0}{\displaystyle \prod _{d\notin T_J}}{\left(\frac{{\delta }_j-{\delta }_d}{{\alpha }_d-a_J}\right)}^{n_d}}{n_0!{\displaystyle \prod _{l\in T_J,l\ne j}}({\delta }_j-{\delta }_l)}}\right\}{ϵ}^{n_0+{\sum }_{d\notin T_J}{n}_d},\hfill \end{array}$$

applying (12) to the sum on $j\in T_J$ with ${\delta }_j-{\delta }_{m+1}$ for $A_i$ there, which is equal to

$$\begin{array}{cc}\hfill & \sum _{\begin{array}{c}{n}_0,n_d\ge 0,d\notin T_J,\\ n_0+{\sum }_{d\notin T_J}{n}_d=\#T_J-1\end{array}}{\displaystyle \frac{{\left(2\pi i\right)}^{n_0}}{n_0!{\displaystyle \prod _{d\notin T_J}}{({\alpha }_d-a_J)}^{n_d}}}{ϵ}^{\#T_J-1}+{ϵ}^{\#T_J}O\left(1\right).\hfill \end{array}$$

Thus, we have

$$\begin{array}{cc}\hfill N\left(ϵ\right)=& {(-1)}^{m(m-1)/2}\underset{k,l,{\alpha }_k\ne {\alpha }_l}{\dot{\prod }}({\alpha }_k-{\alpha }_l)·\underset{k,l,{\alpha }_k={\alpha }_l}{\dot{\prod }}({\delta }_k-{\delta }_l)\times \hfill \\ \hfill & \sum _{J=1}^u{\displaystyle \frac{e(a_J-{\alpha }_{m+1})}{{\displaystyle \prod _{l\notin T_J}}(a_J-{\alpha }_l)}}\left\{\sum _{\begin{array}{c}{n}_0,n_d\ge 0,d\notin T_J,\\ n_0+{\sum }_{d\notin T_J}{n}_d=\#T_J-1\end{array}}{\displaystyle \frac{{\left(2\pi i\right)}^{n_0}}{n_0!{\displaystyle \prod _{d\notin T_J}}{({\alpha }_d-a_J)}^{n_d}}}\right\}{ϵ}^{\kappa }+\hfill \\ \hfill & {ϵ}^{\kappa +1}O\left(1\right),\hfill \end{array}$$

which completes the proof. □