1 Introduction
Let
$E,F$
be two non-empty compact subsets of
$\mathbb{R}^{d}$
. We say that
$E$
and
$F$
are Lipschitz equivalent and denote it by
$E\sim F$
if there is a bi-Lipschitz map
$\unicode[STIX]{x1D719}$
from
$E$
onto
$F$
, i.e.,
$\unicode[STIX]{x1D719}$
is a bijection and there is a constant
$C>0$
such that
$$\begin{eqnarray}C^{-1}|x-y|\leqslant |\unicode[STIX]{x1D719}(x)-\unicode[STIX]{x1D719}(y)|\leqslant C|x-y|\quad \text{for all }x,y\in E.\end{eqnarray}$$
Lipschitz equivalence is an interesting topic in geometric measure theory and fractal geometry. It is well-known that the Hausdorff dimension is a Lipschitz invariant. Since the late 1980s, there have been a lot of studies devoted to the topic (see [Reference Cooper and Pignataro1, Reference Falconer and Marsh4, Reference Falconer and Marsh5, Reference Lau and Luo7, Reference Llorente and Mattila9, Reference Rao, Ruan and Wang11–Reference Xi and Xiong18]). A pioneering work on Cantor sets was done by Falconer and Marsh [Reference Falconer and Marsh5], where they gave some elementary conditions on the contraction ratios to determine the Lipschitz equivalence between two dust-like self-similar sets (also called Cantor sets). Recently, Rao et al [Reference Rao, Ruan and Wang11] extended their result and developed several elegant algebraic criteria to characterize the Lipschitz equivalence of Cantor sets.
Let
$\{f_{i}\}_{i=1}^{m}$
with
$f_{i}(x)=\unicode[STIX]{x1D6FC}_{i}R_{i}(x+d_{i})$
be an iterated function system (IFS) on
$\mathbb{R}^{d}$
, where
$0<\unicode[STIX]{x1D6FC}_{i}<1$
are contraction ratios,
$R_{i}$
are orthogonal matrices, and
$d_{i}\in \mathbb{R}^{d}$
. Then there exists a unique non-empty compact subset
$E$
[Reference Falconer3] such that
$$\begin{eqnarray}E=\mathop{\bigcup }_{i=1}^{m}f_{i}(E).\end{eqnarray}$$
We call such
$E$
a self-similar set, and call
$E$
dust-like if it further satisfies
$f_{i}(E)\cap f_{j}(E)=\emptyset$
for
$i\not =j$
. Given
$\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{m}\in (0,1)$
with
$\sum _{i=1}^{m}\unicode[STIX]{x1D6FC}_{i}^{d}<1$
, we call
$\boldsymbol{\unicode[STIX]{x1D6FC}}=(\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{m})$
a contraction vector, and denote by
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})$
the collection of all dust-like self-similar sets satisfying (1.1). Clearly, all sets in
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})$
have the same Hausdorff dimension
$s$
[Reference Falconer3] which is the unique solution of
$$\begin{eqnarray}\mathop{\sum }_{i=1}^{m}\unicode[STIX]{x1D6FC}_{i}^{s}=1.\end{eqnarray}$$
Moreover, any two sets in
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})$
are always Lipschitz equivalent. We say that
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})$
and
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
are Lipschitz equivalent, and often denote it by
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
if
$E\sim F$
for some (and thus, for all)
$E\in {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}}),F\in {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
.
Let
$E\in {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})$
and
$\unicode[STIX]{x1D6F4}_{m}^{\ast }=\bigcup _{k=0}^{\infty }\unicode[STIX]{x1D6F4}_{m}^{k}$
be the symbolic space representing the IFS as in (1.1), where
$\unicode[STIX]{x1D6F4}_{m}=\{1,\ldots ,m\}$
,
$\unicode[STIX]{x1D6F4}_{m}^{k}=\{(i_{1},\ldots ,i_{k}):\;i_{j}\in \unicode[STIX]{x1D6F4}_{m}\text{ for all }1\leqslant j\leqslant k\}$
and
$\unicode[STIX]{x1D6F4}_{m}^{0}=\emptyset$
. Given
$\mathbf{i}=(i_{1},\ldots ,i_{n})\in \unicode[STIX]{x1D6F4}_{m}^{\ast }$
, we denote
$f_{\mathbf{i}}=f_{i_{1}}\circ f_{i_{2}}\circ \cdots \circ f_{i_{n}}$
and
$\unicode[STIX]{x1D6FC}_{\mathbf{i}}=\unicode[STIX]{x1D6FC}_{i_{1}}\cdots \unicode[STIX]{x1D6FC}_{i_{n}}$
. A finite subset
$\unicode[STIX]{x1D6EC}$
of
$\unicode[STIX]{x1D6F4}_{m}^{\ast }$
is called a partition of
$\unicode[STIX]{x1D6F4}_{m}^{\ast }$
with respect to
$\boldsymbol{\unicode[STIX]{x1D6FC}}$
if it satisfies
$E=\bigcup _{\mathbf{i}\in \unicode[STIX]{x1D6EC}}f_{\mathbf{i}}(E)$
, where the union is disjoint.
Let
$\boldsymbol{\unicode[STIX]{x1D6FC}}=(\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{m}),~\boldsymbol{\unicode[STIX]{x1D6FD}}=(\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{n})$
be two contraction vectors. We say
$\boldsymbol{\unicode[STIX]{x1D6FC}}$
is derived from
$\boldsymbol{\unicode[STIX]{x1D6FD}}$
if there exists a partition
$\{\mathbf{j}_{1},\ldots ,\mathbf{j}_{m}\}$
of
$\unicode[STIX]{x1D6F4}_{n}^{\ast }$
with respect to
$\boldsymbol{\unicode[STIX]{x1D6FD}}$
such that
$\boldsymbol{\unicode[STIX]{x1D6FC}}=(\unicode[STIX]{x1D6FD}_{\mathbf{j}_{1}},\ldots ,\unicode[STIX]{x1D6FD}_{\mathbf{j}_{m}})$
. Then,
$\boldsymbol{\unicode[STIX]{x1D6FC}}$
and
$\boldsymbol{\unicode[STIX]{x1D6FD}}$
are called equivalent, denoted by
$\boldsymbol{\unicode[STIX]{x1D6FC}}\sim \boldsymbol{\unicode[STIX]{x1D6FD}}$
, if there exists a sequence
$$\begin{eqnarray}\boldsymbol{\unicode[STIX]{x1D6FC}}=\boldsymbol{\unicode[STIX]{x1D6FC}}_{1},\boldsymbol{\unicode[STIX]{x1D6FC}}_{2},\ldots ,\boldsymbol{\unicode[STIX]{x1D6FC}}_{N}=\boldsymbol{\unicode[STIX]{x1D6FD}}\end{eqnarray}$$
such that
$\boldsymbol{\unicode[STIX]{x1D6FC}}_{j+1}$
is derived from
$\boldsymbol{\unicode[STIX]{x1D6FC}}_{j}$
or vice versa for
$1\leqslant j\leqslant N$
. Trivially, if
$\boldsymbol{\unicode[STIX]{x1D6FC}}\sim \boldsymbol{\unicode[STIX]{x1D6FD}}$
then
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
. A natural question is the following problem.
Problem 1.1 [Reference Rao, Ruan and Wang12, Problem 1.6].
Find non-trivial sufficient conditions and necessary conditions on
$\boldsymbol{\unicode[STIX]{x1D6FC}}$
and
$\boldsymbol{\unicode[STIX]{x1D6FD}}$
such that
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
. In particular, is it true that
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
if and only if
$\boldsymbol{\unicode[STIX]{x1D6FC}}\sim \boldsymbol{\unicode[STIX]{x1D6FD}}$
?
Write
$\mathbb{Q}(\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{m})$
for the sub-field of
$(\mathbb{R},+,\times )$
formed by the rational functions of
$\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{m}$
, and sgp
$(\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{m})$
for the sub-semigroup of
$(\mathbb{R}^{+},\times )$
generated by
$\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{m}$
. Falconer and Marsh [Reference Falconer and Marsh5] provided some algebraic conditions for the above problem.
Theorem 1.2 [Reference Falconer and Marsh5].
Let
$\boldsymbol{\unicode[STIX]{x1D6FC}}=(\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{m}),\boldsymbol{\unicode[STIX]{x1D6FD}}=(\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{n})$
be two contraction vectors. If
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
with common Hausdorff dimension
$s$
, then the following holds:
(1)
$\mathbb{Q}(\unicode[STIX]{x1D6FC}_{1}^{s},\ldots ,\unicode[STIX]{x1D6FC}_{m}^{s})=\mathbb{Q}(\unicode[STIX]{x1D6FD}_{1}^{s},\ldots ,\unicode[STIX]{x1D6FD}_{n}^{s})$
;(2) there exist
$p,q\in \mathbb{Z}^{+}$
such that $$\begin{eqnarray}\displaystyle \text{sgp}(\unicode[STIX]{x1D6FC}_{1}^{p},\ldots ,\unicode[STIX]{x1D6FC}_{m}^{p}) & \subset \text{sgp}(\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{n}), & \displaystyle \nonumber\\ \displaystyle \text{sgp}(\unicode[STIX]{x1D6FD}_{1}^{q},\ldots ,\unicode[STIX]{x1D6FD}_{n}^{q}) & \subset \text{sgp}(\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{m}). & \displaystyle \nonumber\end{eqnarray}$$
Let
$\langle \boldsymbol{\unicode[STIX]{x1D6FC}}\rangle$
denote the free abelian group of
$(\mathbb{R}^{+},\times )$
generated by
$\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{m}$
. In [Reference Rao, Ruan and Wang11], Rao et al defined a concept of rank for
$\boldsymbol{\unicode[STIX]{x1D6FC}}$
, say
$\text{rank}\langle \boldsymbol{\unicode[STIX]{x1D6FC}}\rangle$
, to be the cardinality of the basis for the free abelian group
$\langle \boldsymbol{\unicode[STIX]{x1D6FC}}\rangle$
. They partially improved Falconer and Marsh’s theorem and obtained the necessary and sufficient condition on
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
in particular cases.
Theorem 1.3 [Reference Rao, Ruan and Wang11].
Let
$\boldsymbol{\unicode[STIX]{x1D6FC}}=(\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{m}),\boldsymbol{\unicode[STIX]{x1D6FD}}=(\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{m})$
be two contraction vectors. If
$\text{rank}\langle \boldsymbol{\unicode[STIX]{x1D6FC}}\rangle =m$
, then
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
if and only if
$\boldsymbol{\unicode[STIX]{x1D6FC}}$
is a permutation of
$\boldsymbol{\unicode[STIX]{x1D6FD}}$
.
If
$m=2$
and assume that
$\unicode[STIX]{x1D6FC}_{1}\leqslant \unicode[STIX]{x1D6FC}_{2}$
,
$\unicode[STIX]{x1D6FD}_{1}\leqslant \unicode[STIX]{x1D6FD}_{2}$
,
$\unicode[STIX]{x1D6FC}_{1}\leqslant \unicode[STIX]{x1D6FD}_{1}$
, then
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
if and only if
$\boldsymbol{\unicode[STIX]{x1D6FC}}=\boldsymbol{\unicode[STIX]{x1D6FD}}$
, or there exists a real number
$0<\unicode[STIX]{x1D706}<1$
, such that
$(\unicode[STIX]{x1D6FC}_{1},\unicode[STIX]{x1D6FC}_{2})=(\unicode[STIX]{x1D706}^{5},\unicode[STIX]{x1D706})$
and
$(\unicode[STIX]{x1D6FD}_{1},\unicode[STIX]{x1D6FD}_{2})=(\unicode[STIX]{x1D706}^{3},\unicode[STIX]{x1D706}^{2})$
.
Moreover, if one of the two contraction vectors is homogeneous, then Rao et al gave a complete characterization on the Lipschitz equivalence as below.
Theorem 1.4 [Reference Rao, Ruan and Wang11].
Let
$\boldsymbol{\unicode[STIX]{x1D6FC}}=(\unicode[STIX]{x1D6FC},\ldots ,\unicode[STIX]{x1D6FC})\in \mathbb{R}^{m},\boldsymbol{\unicode[STIX]{x1D6FD}}=(\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{n})\in \mathbb{R}^{n}$
be two contraction vectors. If
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})$
and
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
have the same dimension
$s$
. Then
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
if and only if there exist
$q,p_{1},\ldots ,p_{n}\in \mathbb{Z}^{+}$
such that
$m^{1/q}\in \mathbb{Z}^{+}$
and
$\unicode[STIX]{x1D6FD}_{j}=\unicode[STIX]{x1D6FC}^{p_{j}/q}$
for all
$j=1,\ldots ,n$
.
In this paper, by investigating the irreducibility of trinomials and quadrinomials, we provide an effective criterion for the Lipschitz equivalence of certain Cantor sets.
Theorem 1.5. Let
$\boldsymbol{\unicode[STIX]{x1D6FC}}=(\unicode[STIX]{x1D706}^{a},\unicode[STIX]{x1D706}^{b},\unicode[STIX]{x1D706}^{c})$
and
$\boldsymbol{\unicode[STIX]{x1D6FD}}=(\unicode[STIX]{x1D706}^{d},\unicode[STIX]{x1D706}^{e})$
be two contraction vectors with
$0<\unicode[STIX]{x1D706}<1,~a,b,c,d,e\in \mathbb{Z}^{+}$
. If
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
, then the polynomial
$f(x)=x^{a}+x^{b}+x^{c}-1$
is reducible.
In some special case, we can improve the theorem to get a necessary and sufficient condition.
Theorem 1.6. Let
$\boldsymbol{\unicode[STIX]{x1D6FC}}=(\unicode[STIX]{x1D706}^{a},\unicode[STIX]{x1D706}^{b},\unicode[STIX]{x1D706}^{c})$
,
$\boldsymbol{\unicode[STIX]{x1D6FD}}=(\unicode[STIX]{x1D706}^{d},\unicode[STIX]{x1D706}^{e})$
be two contraction vectors where
$0<\unicode[STIX]{x1D706}<1$
,
$a=b+c$
,
$\gcd (a,b,c)=1$
. Then
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
if and only if
$a=4$
with
$\{b,c\}=\{1,3\}$
and
$\{d,e\}=\{1,2\}$
; or
$a=8$
with
$\{b,c\}=\{1,7\}$
and
$\{d,e\}=\{1,5\}$
,
$\{2,3\}$
.
Finally, inspired by Theorem 1.4, we give an affirmative answer to the later part of Problem 1.1 when
$\boldsymbol{\unicode[STIX]{x1D6FC}}$
(or
$\boldsymbol{\unicode[STIX]{x1D6FD}}$
) is homogeneous.
Theorem 1.7. Let
$\boldsymbol{\unicode[STIX]{x1D6FC}}=(\unicode[STIX]{x1D6FC},\ldots ,\unicode[STIX]{x1D6FC})\in \mathbb{R}^{m},~\boldsymbol{\unicode[STIX]{x1D6FD}}=(\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{n})\in \mathbb{R}^{n}$
be two contraction vectors. Then
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
if and only if
$\boldsymbol{\unicode[STIX]{x1D6FC}}\sim \boldsymbol{\unicode[STIX]{x1D6FD}}$
.
The paper is organized as follows: in §2, we prove Theorems 1.5 and 1.6 by applying the irreducibility of integer polynomials, and also provide several easy criteria to judge the non-Lipschitz equivalence of Cantor sets. A short proof of Theorem 1.7 will be included in §3.
2 Irreducibility of polynomials
The irreducibility of polynomials is a classical subject and there are lots of related works in the literature (please refer to [Reference Dubickas2, Reference Finch and Jones6, Reference Ljunggren8, Reference Mills10]). In this section, we recall some results on the irreducibility of certain trinomials and quadrinomials. Then we establish the relationship between the irreducibility of polynomials and the Lipschitz equivalence of Cantor sets.
Proposition 2.1 [Reference Ljunggren8].
Let
$d\geqslant 2e>0$
. Write
$d=d_{1}\ell ,e=e_{1}\ell$
, where
$\ell =\gcd (d,e)$
. Then the polynomial
$g(x)=x^{d}+\unicode[STIX]{x1D716}x^{e}+\unicode[STIX]{x1D6FF},~\unicode[STIX]{x1D716},\unicode[STIX]{x1D6FF}\in \{1,-1\}$
is irreducible unless
$d_{1}+e_{1}=0~(\text{mod}~3)$
and one of the following three conditions holds:
$d_{1},e_{1}$
are both odd and
$\unicode[STIX]{x1D716}=1$
;
$d_{1}$
is even and
$\unicode[STIX]{x1D6FF}=1$
;
$e_{1}$
is even and
$\unicode[STIX]{x1D716}=\unicode[STIX]{x1D6FF}$
.
In any of these exceptional cases,
$g(x)$
is the product of the polynomial
$x^{2\ell }+\unicode[STIX]{x1D716}^{e_{1}}\unicode[STIX]{x1D6FF}^{d_{1}}x^{\ell }+1$
and a second irreducible polynomial.
Proposition 2.2 [Reference Mills10].
Suppose that
$f(x)$
is a polynomial over the rationals of the form
$$\begin{eqnarray}f(x)=x^{a}+\unicode[STIX]{x1D716}_{1}x^{b}+\unicode[STIX]{x1D716}_{2}x^{c}+\unicode[STIX]{x1D716}_{3},\end{eqnarray}$$
where
$a>b>c>0$
and
$\unicode[STIX]{x1D716}_{1},\unicode[STIX]{x1D716}_{2},\unicode[STIX]{x1D716}_{3}\in \{1,-1\}$
. Let
$f(x)=A(x)B(x)$
where every root of
$A(x)$
and no root of
$B(x)$
is a root of unity. Then
$A(x)$
is the greatest common divisor of
$f(x)$
and
$f^{\ast }(x)$
, where
$f^{\ast }(x)$
denote the reciprocal polynomial
$f^{\ast }(x)=x^{a}f(x^{-1})$
. The second factor
$B(x)$
is irreducible except when
$f(x)$
is one of the following four forms:
$$\begin{eqnarray}\displaystyle x^{8r}+x^{7r}+x^{r}-1 & = & \displaystyle (x^{2r}+1)(x^{3r}+x^{2r}-1)(x^{3r}-x^{r}+1),\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle x^{8r}-x^{7r}-x^{r}-1 & = & \displaystyle (x^{2r}+1)(x^{3r}-x^{2r}+1)(x^{3r}-x^{r}-1),\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle x^{8r}+x^{4r}+x^{2r}-1 & = & \displaystyle (x^{2r}+1)(x^{3r}+x^{2r}-1)(x^{3r}-x^{2r}+1),\end{eqnarray}$$
$$\begin{eqnarray}\displaystyle x^{8r}+x^{6r}+x^{4r}-1 & = & \displaystyle (x^{2r}+1)(x^{3r}-x^{r}-1)(x^{3r}-x^{r}+1).\end{eqnarray}$$
In the four cases above, the factors of degree
$3r$
are irreducible.
Proposition 2.3 [Reference Ljunggren8].
If
$a=a_{1}t$
,
$b=b_{1}t$
,
$c=c_{1}t$
, and
$\gcd (a_{1},b_{1},c_{1})=1$
,
$\gcd (a_{1},b_{1}-c_{1})=t_{1}$
,
$\gcd (b_{1},a_{1}-c_{1})=t_{2}$
,
$\gcd (c_{1},a_{1}-b_{1})=t_{3}$
;
$\unicode[STIX]{x1D716}_{1},\unicode[STIX]{x1D716}_{2},\unicode[STIX]{x1D716}_{3}\in \{1,-1\}$
then all possible roots of unity of
$f(x)=x^{a}+\unicode[STIX]{x1D716}_{1}x^{b}+\unicode[STIX]{x1D716}_{2}x^{c}+\unicode[STIX]{x1D716}_{3}$
are simple zeros, which are to be found among the zeros of
$$\begin{eqnarray}\displaystyle x^{tt_{1}}=\pm 1,\qquad x^{tt_{2}}=\pm 1,\qquad x^{tt_{3}}=\pm 1. & & \displaystyle\end{eqnarray}$$
Proof of Theorem 1.5.
If
$d=e$
, then from Theorem 1.4, we have
$d|a,~d|b,~d|c.$
Hence, we need only to consider the case that
$\boldsymbol{\unicode[STIX]{x1D6FC}}=(\unicode[STIX]{x1D706}^{a},\unicode[STIX]{x1D706}^{b},\unicode[STIX]{x1D706}^{c})$
and
$\boldsymbol{\unicode[STIX]{x1D6FD}}=(\unicode[STIX]{x1D706},\unicode[STIX]{x1D706})$
by letting
$d=e=1$
. If
$D(\boldsymbol{\unicode[STIX]{x1D6FC}})\sim D(\boldsymbol{\unicode[STIX]{x1D6FD}})$
, it follows from (1.2) that the common Hausdorff dimension
$s$
satisfies both
$$\begin{eqnarray}2\unicode[STIX]{x1D706}^{s}=1\quad \text{and}\quad \unicode[STIX]{x1D706}^{sa}+\unicode[STIX]{x1D706}^{sb}+\unicode[STIX]{x1D706}^{sc}=1.\end{eqnarray}$$
That is,
$2^{-a}+2^{-b}+2^{-c}=1$
, and thus
$(a,b,c)=(1,2,2)$
or its permutations. This implies that
$f(x)$
is the polynomial
$2x^{2}+x-1$
, which is reducible.
In what follows, without loss of generality, we assume that
$d>e$
. Let
$g(x)=x^{d}+x^{e}-1$
. Then
$g(x)$
and
$f(x)$
have the same root
$\unicode[STIX]{x1D706}^{s}$
. Obviously, if
$g(x)$
is irreducible, then
$g(x)|f(x)$
so that
$f(x)$
is reducible. If
$g(x)$
is reducible, we may consider the following two situations: when
$d\geqslant 2e$
,
$g(x)$
satisfies the first exceptional case of Proposition 2.1, then
$g(x)=(x^{2\ell }-x^{\ell }+1)h_{1}(x)$
, where
$\ell =\gcd (d,e)$
and
$h_{1}(x)$
is irreducible; when
$d<2e$
, we consider the reciprocal polynomial
$-g^{\ast }(x)=-x^{d}g(x^{-1})=x^{d}-x^{d-e}-1$
which is reducible because so is
$g$
. Moreover it satisfies
$d\geqslant 2(d-e)$
, so by Proposition 2.1 it has the form
$-x^{d}g(x^{-1})=(x^{2\ell }+\unicode[STIX]{x1D716}x^{\ell }+1)h_{2}(x)$
, where
$\unicode[STIX]{x1D716}\in \{-1,1\}$
and
$h_{2}(x)$
is irreducible. Therefore,
$g(x)=(1+\unicode[STIX]{x1D716}x^{\ell }+x^{2\ell })(-x^{d-2\ell }h_{2}(x^{-1}))$
and the second factor is irreducible. In both cases we have
$$\begin{eqnarray}g(x)=(x^{2\ell }+\unicode[STIX]{x1D716}x^{\ell }+1)h(x),\end{eqnarray}$$
where
$h(x)$
is irreducible. Since all zeros of
$x^{2\ell }\pm x^{\ell }+1$
are non-real,
$\unicode[STIX]{x1D706}^{s}$
is not a root of the first factor. Hence, it must be a root of
$h(x)$
. It follows that
$h(x)|f(x)$
.
If
$f(x)$
were irreducible, then
$f(x)=h(x)$
and
$$\begin{eqnarray}\displaystyle x^{d}+x^{e}-1=(x^{2\ell }+\unicode[STIX]{x1D716}x^{\ell }+1)(x^{a}+x^{b}+x^{c}-1). & & \displaystyle \nonumber\end{eqnarray}$$
It is easy to check that for both
$\unicode[STIX]{x1D716}=1$
and
$\unicode[STIX]{x1D716}=-1$
, the above two sides are always not equal when we set
$x=1$
, which is a contradiction. Therefore,
$f(x)$
must be reducible.◻
The following is a sufficient condition for the irreducibility of quadrinomials.
Lemma 2.4 [Reference Dubickas2, Reference Ljunggren8].
Let
$a,b,c$
be three distinct positive integers. If they are all odd, then the polynomial
$f(x)=x^{a}+x^{b}+x^{c}\pm 1$
is irreducible over
$\mathbb{Q}$
.
Suppose
$\text{gcd}(a,b,c)=2^{k}m$
, where
$m$
is odd. Define
$$\begin{eqnarray}a^{\prime }=a/2^{k},\qquad b^{\prime }=b/2^{k},\qquad c^{\prime }=c/2^{k}\end{eqnarray}$$
and
$$\begin{eqnarray}\bar{a}=\text{gcd}(a^{\prime },b^{\prime }-c^{\prime }),\qquad \bar{b}=\text{gcd}(b^{\prime },a^{\prime }-c^{\prime }),\qquad \bar{c}=\text{gcd}(c^{\prime },a^{\prime }-b^{\prime }).\end{eqnarray}$$
We have a simple criterion of the irreducibility of quadrinomials.
Lemma 2.5 [Reference Finch and Jones6].
If
$f(x)=x^{a}+x^{b}+x^{c}-1$
, then
$f(x)$
is irreducible over
$\mathbb{Q}$
if and only if
$a^{\prime }\neq 0~(\text{mod}~2\bar{a}),~b^{\prime }\neq 0~(\text{mod}~2\bar{b}),~c^{\prime }\neq 0~(\text{mod}~2\bar{c})$
, where
$a^{\prime },b^{\prime },c^{\prime }$
are as above.
Following the above notation and combining Theorem 1.5, Lemmas 2.4 and 2.5, we obtain an easy way to verify that two Cantor sets are non-Lipschitz equivalent.
Theorem 2.6. Let
$\boldsymbol{\unicode[STIX]{x1D6FC}}=(\unicode[STIX]{x1D706}^{a},\unicode[STIX]{x1D706}^{b},\unicode[STIX]{x1D706}^{c})$
and
$\boldsymbol{\unicode[STIX]{x1D6FD}}=(\unicode[STIX]{x1D706}^{d},\unicode[STIX]{x1D706}^{e})$
be two contraction vectors with
$0<\unicode[STIX]{x1D706}<1,~a,b,c,d,e\in \mathbb{Z}^{+}$
. Then
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})\nsim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
if any one of the following conditions holds:
(1)
$a,b,c$
are distinct odd integers;(2)
$a^{\prime }\neq 0~(\text{mod}~2\bar{a}),~b^{\prime }\neq 0~(\text{mod}~2\bar{b}),~c^{\prime }\neq 0~(\text{mod}~2\bar{c})$
, where
$a^{\prime },b^{\prime },c^{\prime }$
and
$\bar{a}$
,
$\bar{b}$
,
$\bar{c}$
are as above.
Let
$a>b>c>0$
be integers;
$\unicode[STIX]{x1D6FD},\unicode[STIX]{x1D6FE},\unicode[STIX]{x1D6FF}\in \{-1,1\}$
; and let
$f(x)=x^{a}+\unicode[STIX]{x1D6FD}x^{b}+\unicode[STIX]{x1D6FE}x^{c}+\unicode[STIX]{x1D6FF}$
be a quadrinomial. It is shown in [Reference Ljunggren8] that
$f(x)$
is reducible over
$\mathbb{Q}$
if and only if
$f(\unicode[STIX]{x1D702})=0$
for some root of unity
$\unicode[STIX]{x1D702}$
. By using this, finally we can prove our second main result. In the proof, we need the following iterating operation, which is very natural according to the self-similarity of the IFS.
Given a contraction vector
$\boldsymbol{\unicode[STIX]{x1D6FC}}=(\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{m})$
. For any
$i\in \{1,\ldots ,m\}$
, we call
$$\begin{eqnarray}\boldsymbol{\unicode[STIX]{x1D6FC}}^{i}=(\unicode[STIX]{x1D6FC}_{1},\unicode[STIX]{x1D6FC}_{2},\ldots ,\unicode[STIX]{x1D6FC}_{i-1},\unicode[STIX]{x1D6FC}_{i}\unicode[STIX]{x1D6FC}_{1},\unicode[STIX]{x1D6FC}_{i}\unicode[STIX]{x1D6FC}_{2},\ldots ,\unicode[STIX]{x1D6FC}_{i}\unicode[STIX]{x1D6FC}_{m},\unicode[STIX]{x1D6FC}_{i+1},\unicode[STIX]{x1D6FC}_{i+2},\ldots ,\unicode[STIX]{x1D6FC}_{m})\end{eqnarray}$$
the contraction vector obtained by iterating the
$\unicode[STIX]{x1D6FC}_{i}$
term in
$\boldsymbol{\unicode[STIX]{x1D6FC}}$
. Similarly, we call
$$\begin{eqnarray}\displaystyle \boldsymbol{\unicode[STIX]{x1D6FC}}^{i,i} & = & \displaystyle (\unicode[STIX]{x1D6FC}_{1},\unicode[STIX]{x1D6FC}_{2},\ldots ,\unicode[STIX]{x1D6FC}_{i-1},\unicode[STIX]{x1D6FC}_{i}\unicode[STIX]{x1D6FC}_{1}^{2},\unicode[STIX]{x1D6FC}_{i}\unicode[STIX]{x1D6FC}_{1}\unicode[STIX]{x1D6FC}_{2},\ldots ,\unicode[STIX]{x1D6FC}_{i}\unicode[STIX]{x1D6FC}_{m-1}\unicode[STIX]{x1D6FC}_{m},\unicode[STIX]{x1D6FC}_{i}\unicode[STIX]{x1D6FC}_{m}^{2},\nonumber\\ \displaystyle & & \displaystyle \quad \unicode[STIX]{x1D6FC}_{i+1},\unicode[STIX]{x1D6FC}_{i+2},\ldots ,\unicode[STIX]{x1D6FC}_{m} )\nonumber\end{eqnarray}$$
the contraction vector obtained by iterating the
$\unicode[STIX]{x1D6FC}_{i}$
term twice in
$\boldsymbol{\unicode[STIX]{x1D6FC}}$
. Clearly, both
$\boldsymbol{\unicode[STIX]{x1D6FC}}^{i}$
and
$\boldsymbol{\unicode[STIX]{x1D6FC}}^{i,i}$
are derived from
$\boldsymbol{\unicode[STIX]{x1D6FC}}$
so that
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}}^{i})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})$
and
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}}^{i,i})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})$
.
Proof of Theorem 1.6.
First we prove the sufficient part. Iterating the
$\unicode[STIX]{x1D706}^{2}$
term in
$(\unicode[STIX]{x1D706}^{2},\unicode[STIX]{x1D706})$
, we obtain the contraction vector
$(\unicode[STIX]{x1D706}^{4},\unicode[STIX]{x1D706}^{3},\unicode[STIX]{x1D706})$
. Thus,
${\mathcal{D}}(\unicode[STIX]{x1D706}^{4},\unicode[STIX]{x1D706}^{3},\unicode[STIX]{x1D706})\sim {\mathcal{D}}(\unicode[STIX]{x1D706}^{2},\unicode[STIX]{x1D706})$
. Iterating the
$\unicode[STIX]{x1D706}$
term in
$(\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{7},\unicode[STIX]{x1D706})$
, we obtain
${\mathcal{D}}(\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{7},\unicode[STIX]{x1D706})\sim {\mathcal{D}}(\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{7},\unicode[STIX]{x1D706}^{9},\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{2})$
. Iterating the
$\unicode[STIX]{x1D706}^{3}$
term in
$(\unicode[STIX]{x1D706}^{3},\unicode[STIX]{x1D706}^{2})$
twice yields
$$\begin{eqnarray}{\mathcal{D}}(\unicode[STIX]{x1D706}^{3},\unicode[STIX]{x1D706}^{2})\sim {\mathcal{D}}(\unicode[STIX]{x1D706}^{6},\unicode[STIX]{x1D706}^{5},\unicode[STIX]{x1D706}^{2})\sim {\mathcal{D}}(\unicode[STIX]{x1D706}^{9},\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{7},\unicode[STIX]{x1D706}^{2})\end{eqnarray}$$
so that
${\mathcal{D}}(\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{7},\unicode[STIX]{x1D706})\sim {\mathcal{D}}(\unicode[STIX]{x1D706}^{3},\unicode[STIX]{x1D706}^{2})$
. By Theorem 1.3, we know that
${\mathcal{D}}(\unicode[STIX]{x1D706}^{5},\unicode[STIX]{x1D706})\sim {\mathcal{D}}(\unicode[STIX]{x1D706}^{3},\unicode[STIX]{x1D706}^{2})$
. Thus,
${\mathcal{D}}(\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{7},\unicode[STIX]{x1D706})\sim {\mathcal{D}}(\unicode[STIX]{x1D706}^{5},\unicode[STIX]{x1D706})$
.
Now we prove the necessary part. From
$a=b+c,~\gcd (a,b,c)=1$
, we know
$b\neq c$
. Without loss of generality we may assume
$a>b>c$
. If
$d=e$
, by the proof of Theorem 1.5, then
$(a,b,c)$
should be
$(1,2,2)$
or its permutations, contradicting the assumption of
$a=b+c$
. Hence, we may assume
$d>e$
.
Define
$f(x)=x^{a}+x^{b}+x^{c}-1$
and
$g(x)=x^{d}+x^{e}-1$
. If
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
, then
$g(x)$
and
$f(x)$
have the same root
$\unicode[STIX]{x1D706}^{s}$
where
$s$
is the common Hausdorff dimension. Moreover, by Theorem 1.5,
$f(x)$
is reducible. Hence,
$f(x)=A(x)B(x)$
as in Proposition 2.2 where every root of
$A(x)$
and no root of
$B(x)$
is a root of unity,
$B(x)$
is irreducible except when
$f(x)$
is one of the four forms (2.1)–(2.4).
Case 1. Suppose
$B(x)$
is reducible. From
$\gcd (a,b,c)=1$
and Proposition 2.2, we have
$$\begin{eqnarray}f(x)=x^{8}+x^{7}+x-1=(x^{2}+1)(x^{3}+x^{2}-1)(x^{3}-x+1).\end{eqnarray}$$
If
$g(x)$
is irreducible, then
$g(x)=x^{3}+x^{2}-1$
, so we have
$(a,b,c,d,e)=(8,7,1,3,2)$
. If
$g(x)$
is reducible, then by (2.6),
$g(x)=(x^{2\ell }+\unicode[STIX]{x1D716}x^{\ell }+1)h(x)$
, where
$\unicode[STIX]{x1D716}\in \{-1,1\}$
,
$\ell =\gcd (d,e)$
and
$h(x)$
is irreducible. Since all zeros of
$x^{2\ell }\pm x^{\ell }+1$
are the roots of unity and
$\unicode[STIX]{x1D706}^{s}\in (0,1)$
, we only have
$h(\unicode[STIX]{x1D706}^{s})=0$
, and
$h(x)|f(x)$
. Moreover,
$x^{2}+1$
and
$x^{3}-x+1$
have no roots in
$(0,1)$
, it follows that
$h(x)|x^{3}+x^{2}-1$
. Hence,
$h(x)=x^{3}+x^{2}-1$
by the irreducibility of
$x^{3}+x^{2}-1$
over the rationals. Therefore,
$$\begin{eqnarray}\displaystyle x^{d}+x^{e}-1=(x^{2\ell }+\unicode[STIX]{x1D716}x^{\ell }+1)(x^{3}+x^{2}-1). & & \displaystyle \nonumber\end{eqnarray}$$
By letting
$x=1$
, it can be easily seen that
$\unicode[STIX]{x1D716}=-1$
. This yields
$d=3+2\ell$
and
$$\begin{eqnarray}x^{e}=x^{2+2\ell }-x^{2\ell }-x^{3+\ell }-x^{2+\ell }+x^{\ell }+x^{3}+x^{2}.\end{eqnarray}$$
It follows that
$\ell =1$
,
$e=1$
,
$d=5$
, then we have
$(a,b,c,d,e)=(8,7,1,5,1)$
.
Case 2. Suppose
$B(x)$
is irreducible. Let
$t_{1}=\gcd (a,b-c)$
,
$t_{2}=b$
and
$t_{3}=c$
. From
$a=b+c$
,
$\gcd (a,b,c)=1$
and Proposition 2.3, we know that all possible roots of unity of
$f(x)$
are simple zeros, which are to be found among the zeros of
$$\begin{eqnarray}\displaystyle x^{t_{1}}=\pm 1;\qquad x^{b}=\pm 1;\qquad x^{c}=\pm 1. & & \displaystyle \nonumber\end{eqnarray}$$
Let
$\unicode[STIX]{x1D702}$
be a root of unity of
$f(x)$
. Since
$a=b+c$
, we have
$f(x)=(x^{b}+1)(x^{c}+1)-2$
. If
$\unicode[STIX]{x1D702}$
is the zero of
$x^{b}=1$
, from
$f(\unicode[STIX]{x1D702})=(\unicode[STIX]{x1D702}^{b}+1)(\unicode[STIX]{x1D702}^{c}+1)-2=0$
, it follows that
$\unicode[STIX]{x1D702}^{c}=0$
, that is impossible; if
$\unicode[STIX]{x1D702}$
is the zero of
$x^{b}=-1$
, then
$f(\unicode[STIX]{x1D702})=-2\neq 0$
, which is a contradiction. Similarly,
$\unicode[STIX]{x1D702}$
cannot be the zero of
$x^{c}=\pm 1$
. So all the roots of unity of
$f(x)$
can only be found in the zeros of
$x^{t_{1}}=\pm 1$
.
Let
$p(x)=(x^{t_{1}}-1)(x^{t_{1}}+1)$
. Then
$A(x)|p(x)$
as all possible roots of unity of
$A(x)$
(or
$f(x)$
) are simple zeros.
By the assumptions that
$a=b+c$
and
$\gcd (a,b,c)=1$
, we have
$\gcd (b,c)=1$
, and
$t_{1}=\gcd (a,b-c)=\gcd (b+c,b-c)$
. If
$b,c$
are different from odevity, then
$t_{1}=1$
. If
$b,c$
are both odd, then
$t_{1}=2$
. If
$t_{1}=1$
, then
$p(x)=(x-1)(x+1)$
. If
$t_{1}=2$
, then
$p(x)=(x^{2}-1)(x^{2}+1)=(x-1)(x+1)(x^{2}+1)$
. Due to
$A(x)|p(x)$
and the fact
$f(1)\neq 0$
(hence,
$A(1)\neq 0$
), all possible forms of
$A(x)$
could be
$$\begin{eqnarray}x+1;\qquad x^{2}+1;\quad (x+1)(x^{2}+1).\end{eqnarray}$$
Case 2.1. If
$A(x)=x+1$
, then
$f(x)=A(x)B(x)=(x+1)B(x)$
. Assume that
$g(x)$
is irreducible. Since
$g(x)$
and
$f(x)$
have the same zero
$\unicode[STIX]{x1D706}^{s}$
, we have
$g(x)=B(x)$
so that
$$\begin{eqnarray}\displaystyle x^{a}+x^{b}+x^{c}-1 & = & \displaystyle (x+1)(x^{d}+x^{e}-1)\nonumber\\ \displaystyle & = & \displaystyle x^{1+d}+x^{1+e}-x+x^{d}+x^{e}-1.\nonumber\end{eqnarray}$$
By comparing the powers of
$x$
on both sides, if
$e<c$
, then
$e=1$
,
$a=d+1$
and
$b+c=1+e+d=d+2$
, contradicting the assumption of
$a=b+c$
; if
$e=c$
, then
$x$
on the right side cannot be canceled by any other term on both sides; if
$e>c$
, then
$x^{c}$
cannot be canceled by any other term on both sides. All are impossible. Assume that
$g(x)$
is reducible. By (2.6),
$g(x)=(x^{2\ell }+\unicode[STIX]{x1D716}x^{\ell }+1)h(x)$
, where
$\unicode[STIX]{x1D716}\in \{-1,1\}$
,
$\ell =\gcd (d,e)$
and
$h(x)$
is irreducible. Notice that
$h(\unicode[STIX]{x1D706}^{s})=B(\unicode[STIX]{x1D706}^{s})=0$
. Thus,
$h(x)=B(x)$
so that
$$\begin{eqnarray}(x^{a}+x^{b}+x^{c}-1)(x^{2\ell }+\unicode[STIX]{x1D716}x^{\ell }+1)=(x+1)(x^{d}+x^{e}-1).\end{eqnarray}$$
If
$\unicode[STIX]{x1D716}=1$
, letting
$x=1$
, then we get
$6=2$
which is a contradiction. If
$\unicode[STIX]{x1D716}=-1$
, we denote
$C_{l}(x)$
and
$C_{r}(x)$
to be the coefficients of
$x$
on the left-hand side and right-hand side of (2.7), respectively. In the case where
$e\geqslant 2$
, we have
$C_{l}(x)\geqslant 0$
and
$C_{r}(x)=-1$
. In the case where
$e=1$
, we have
$\ell =1$
so that
$C_{l}(x)>0$
and
$C_{r}(x)=0$
. Therefore, all are impossible.
Case 2.2. If
$A(x)=x^{2}+1$
, then
$f(x)=(x^{2}+1)B(x)$
. If
$g(x)$
is irreducible, then
$g(x)=B(x)$
so that
$$\begin{eqnarray}\displaystyle x^{a}+x^{b}+x^{c}-1=x^{2+d}+x^{2+e}-x^{2}+x^{d}+x^{e}-1. & & \displaystyle\end{eqnarray}$$
If
$e<c$
, then
$e=2,a=d+2$
and
$b+c=2+e+d=d+4$
, contradicting the assumption of
$a=b+c$
. If
$e>c$
, then
$x^{c}$
on the left-hand side of (2.8) cannot be canceled by any other term on both sides of (2.8). That is impossible. If
$e=c$
, then (2.8) can be reduced to
$$\begin{eqnarray}x^{a}+x^{b}=x^{2+d}+x^{2+e}-x^{2}+x^{d}.\end{eqnarray}$$
Hence,
$d=2,a=d+2=4$
and
$b=2+c$
. From
$a=b+c$
, it follows that
$c=1$
. Thus,
$(a,b,c,d,e)=(4,3,1,2,1)$
. Assume that
$g(x)$
is reducible. Similarly as in Case 2.1, there exists
$\unicode[STIX]{x1D716}\in \{-1,1\}$
such that
$$\begin{eqnarray}(x^{a}+x^{b}+x^{c}-1)(x^{2\ell }+\unicode[STIX]{x1D716}x^{\ell }+1)=(x^{2}+1)(x^{d}+x^{e}-1),\end{eqnarray}$$
where
$\ell =\gcd (d,e)$
. If
$\unicode[STIX]{x1D716}=1$
, letting
$x=1$
, then we get
$6=2$
, which is a contradiction. If
$\unicode[STIX]{x1D716}=-1$
, we consider the following two possibilities: when
$\ell \geqslant 2$
, we can use a similar method as in Case 2.1 to obtain a contradiction by comparing the coefficients of
$x^{2}$
in (2.9); when
$\ell =1$
, factoring the two sides of (2.9) and eliminating the same terms, we have
$$\begin{eqnarray}x^{a+2}+x^{b+2}+x^{c+2}+x^{a}+x^{b}+x^{c}+x-x^{a+1}-x^{b+1}-x^{c+1}=x^{d+2}+x^{e+2}+x^{d}+x^{e}.\end{eqnarray}$$
By comparing the powers of
$x$
, we obtain
$a+2=d+2,a+1=b+2$
and
$e=1$
. Hence,
$a=d=b+1$
. From the assumption that
$a=b+c$
, it follows that
$c=1=e$
. That is impossible as
$\unicode[STIX]{x1D706}^{sa}+\unicode[STIX]{x1D706}^{sb}+\unicode[STIX]{x1D706}^{sc}=1=\unicode[STIX]{x1D706}^{sd}+\unicode[STIX]{x1D706}^{se}$
.
Case 2.3. If
$A(x)=(x+1)(x^{2}+1)$
, then
$f(x)=(x+1)(x^{2}+1)B(x)$
. If
$g(x)$
is irreducible, we have
$g(x)=B(x)$
so that
$$\begin{eqnarray}\displaystyle x^{a}+x^{b}+x^{c}-1=(x+1)(x^{2}+1)(x^{d}+x^{e}-1). & & \displaystyle \nonumber\end{eqnarray}$$
Let
$x=1$
, then the left side is
$2$
while the right side is
$4$
. That is impossible. Assume that
$g(x)$
is reducible. Similarly as in Case 2.1, there exists
$\unicode[STIX]{x1D716}\in \{-1,1\}$
such that
$$\begin{eqnarray}(x^{a}+x^{b}+x^{c}-1)(x^{2\ell }+\unicode[STIX]{x1D716}x^{\ell }+1)=(x+1)(x^{2}+1)(x^{d}+x^{e}-1),\end{eqnarray}$$
where
$\ell =\gcd (d,e)$
. Let
$x=1$
, then
$2(2+\unicode[STIX]{x1D716})=4$
, implying
$\unicode[STIX]{x1D716}=0$
, which contradicts that
$\unicode[STIX]{x1D716}\in \{-1,1\}$
.◻
We remark that Theorem 1.6 might be true if we remove the assumption of
$\gcd (a,b,c)=1$
by using Theorem 2.6, but the proof will become tedious. As for the exceptional form (2.3) not included in the theorem, we have the following complement.
Proposition 2.7. Let
$0<\unicode[STIX]{x1D706}<1$
. Then
${\mathcal{D}}(\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{4},\unicode[STIX]{x1D706}^{2})\sim {\mathcal{D}}(\unicode[STIX]{x1D706}^{3},\unicode[STIX]{x1D706}^{2})$
.
Proof. Let
$f_{1}(x)=\unicode[STIX]{x1D706}^{8}x,f_{2}(x)=\unicode[STIX]{x1D706}^{4}x+\unicode[STIX]{x1D706}^{3}-\unicode[STIX]{x1D706}^{4}$
and
$f_{3}(x)=\unicode[STIX]{x1D706}^{2}x+1-\unicode[STIX]{x1D706}^{2}$
be an IFS on
$\mathbb{R}$
and let
$E$
be the associated self-similar set; let
$g_{1}(x)=\unicode[STIX]{x1D706}^{3}x,g_{2}(x)=\unicode[STIX]{x1D706}^{2}x+1-\unicode[STIX]{x1D706}^{2}$
be another IFS on
$\mathbb{R}$
and let
$F$
be the associated self-similar set. Obviously, both
$E$
and
$F$
are dust-like and
$E\in {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}}),F\in {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
. Hence, in order to show
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
, it suffices to show
$E\sim F$
. Indeed, we can find that
$E$
and
$F$
have the same graph-directed structure in the following way: let
$E_{1}=E;E_{2}=\unicode[STIX]{x1D706}^{5}E\cup (\unicode[STIX]{x1D706}E+1-\unicode[STIX]{x1D706})$
, then
$$\begin{eqnarray}\displaystyle E_{1} & = & \displaystyle (\unicode[STIX]{x1D706}^{2}E_{1}+1-\unicode[STIX]{x1D706}^{2})\cup (\unicode[STIX]{x1D706}^{3}E_{2});\nonumber\\ \displaystyle E_{2} & = & \displaystyle (\unicode[STIX]{x1D706}^{5}E_{1})\cup (\unicode[STIX]{x1D706}^{3}E_{1}+1-\unicode[STIX]{x1D706}^{3})\cup (\unicode[STIX]{x1D706}^{4}E_{2}+1-\unicode[STIX]{x1D706}).\nonumber\end{eqnarray}$$
Similarly for
$F$
, we let
$F_{1}=F_{2}=F$
, then
$$\begin{eqnarray}\displaystyle F_{1} & = & \displaystyle (\unicode[STIX]{x1D706}^{2}F_{1}+1-\unicode[STIX]{x1D706}^{2})\cup (\unicode[STIX]{x1D706}^{3}F_{2});\nonumber\\ \displaystyle F_{2} & = & \displaystyle (\unicode[STIX]{x1D706}^{3}F_{1})\cup (\unicode[STIX]{x1D706}^{5}F_{1}+1-\unicode[STIX]{x1D706}^{2})\cup (\unicode[STIX]{x1D706}^{4}F_{2}+1-\unicode[STIX]{x1D706}^{4}).\nonumber\end{eqnarray}$$
Both
$\{E_{1},E_{2}\}$
and
$\{F_{1},F_{2}\}$
are dust-like graph-directed sets and satisfy the conditions of [Reference Rao, Ruan and Xi13, Theorem 2.1], and thus,
$E_{i}\sim F_{i}$
for
$i=1,2$
. Therefore,
$E\sim F$
.
We also give an alternative proof. Iterating the terms in
$(\unicode[STIX]{x1D706}^{5},\unicode[STIX]{x1D706})$
as follows:
$$\begin{eqnarray}\displaystyle (\unicode[STIX]{x1D706}^{5},\unicode[STIX]{x1D706}) & \rightarrow & \displaystyle (\unicode[STIX]{x1D706}^{10},\unicode[STIX]{x1D706}^{6},\unicode[STIX]{x1D706}^{6},\unicode[STIX]{x1D706}^{2})\rightarrow (\unicode[STIX]{x1D706}^{10},\unicode[STIX]{x1D706}^{6},\unicode[STIX]{x1D706}^{6},\unicode[STIX]{x1D706}^{7},\unicode[STIX]{x1D706}^{3})\nonumber\\ \displaystyle & \rightarrow & \displaystyle (\unicode[STIX]{x1D706}^{10},\unicode[STIX]{x1D706}^{6},\unicode[STIX]{x1D706}^{6},\unicode[STIX]{x1D706}^{12},\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{4}).\nonumber\end{eqnarray}$$
Then
$$\begin{eqnarray}{\mathcal{D}}(\unicode[STIX]{x1D706}^{5},\unicode[STIX]{x1D706})\sim {\mathcal{D}}(\unicode[STIX]{x1D706}^{10},\unicode[STIX]{x1D706}^{6},\unicode[STIX]{x1D706}^{6},\unicode[STIX]{x1D706}^{12},\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{4}).\end{eqnarray}$$
Iterating
$\unicode[STIX]{x1D706}^{4}$
and
$\unicode[STIX]{x1D706}^{2}$
in
$(\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{4},\unicode[STIX]{x1D706}^{2})$
yields
$(\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{12},\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{6},\unicode[STIX]{x1D706}^{10},\unicode[STIX]{x1D706}^{6},\unicode[STIX]{x1D706}^{4})$
. Hence,
$$\begin{eqnarray}{\mathcal{D}}(\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{4},\unicode[STIX]{x1D706}^{2})\sim {\mathcal{D}}(\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{12},\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{6},\unicode[STIX]{x1D706}^{10},\unicode[STIX]{x1D706}^{6},\unicode[STIX]{x1D706}^{4}).\end{eqnarray}$$
Combining this with Theorem 1.3, we obtain that
${\mathcal{D}}(\unicode[STIX]{x1D706}^{8},\unicode[STIX]{x1D706}^{4},\unicode[STIX]{x1D706}^{2})\sim {\mathcal{D}}(\unicode[STIX]{x1D706}^{5},\unicode[STIX]{x1D706})\sim {\mathcal{D}}(\unicode[STIX]{x1D706}^{3},\unicode[STIX]{x1D706}^{2})$
.◻
3 Cantor sets with homogeneous contraction vectors
Lemma 3.1. Let
$\boldsymbol{\unicode[STIX]{x1D6FC}}=(\unicode[STIX]{x1D6FC},\ldots ,\unicode[STIX]{x1D6FC})\in \mathbb{R}^{m}$
be a contraction vector, and let
$s$
be the Hausdorff dimension of
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})$
. If there is a finite sequence of positive integers
$\{a_{i}\}_{i=1}^{n}$
satisfying
$\sum _{i=1}^{n}\unicode[STIX]{x1D6FC}^{sa_{i}}=1$
, then there exists a partition
$\unicode[STIX]{x1D6EC}=\{\mathbf{i}_{1},\ldots ,\mathbf{i}_{n}\}$
of
$\unicode[STIX]{x1D6F4}_{m}^{\ast }$
with respect to
$\boldsymbol{\unicode[STIX]{x1D6FC}}$
such that
$\unicode[STIX]{x1D6FC}_{\mathbf{i}_{i}}=\unicode[STIX]{x1D6FC}^{a_{i}}$
,
$1\leqslant i\leqslant n$
.
Proof. Let
$\ell =\max _{1\leqslant i\leqslant n}a_{i}$
. We shall show the lemma by induction. If
$\ell =1$
, then
$n=m$
and
$\unicode[STIX]{x1D6EC}=\unicode[STIX]{x1D6F4}_{m}$
is a partition. If
$\ell =k$
, the lemma is true, for
$\ell =k+1$
, write
$$\begin{eqnarray}\unicode[STIX]{x1D6EC}_{1}=\{i:a_{i}=k+1\}\quad \text{and}\quad \unicode[STIX]{x1D6EC}_{2}=\{i:a_{i}\leqslant k\}.\end{eqnarray}$$
Then
$$\begin{eqnarray}1=\mathop{\sum }_{i=1}^{n}\unicode[STIX]{x1D6FC}^{sa_{i}}=\mathop{\sum }_{i=1}^{n}m^{-a_{i}}=\mathop{\sum }_{i\in \unicode[STIX]{x1D6EC}_{1}}m^{-a_{i}}+\mathop{\sum }_{i\in \unicode[STIX]{x1D6EC}_{2}}m^{-a_{i}}=\#\unicode[STIX]{x1D6EC}_{1}m^{-(k+1)}+\mathop{\sum }_{i\in \unicode[STIX]{x1D6EC}_{2}}m^{-a_{i}}\end{eqnarray}$$
which implies
$$\begin{eqnarray}\#\unicode[STIX]{x1D6EC}_{1}=m\biggl(m^{k}-\mathop{\sum }_{i\in \unicode[STIX]{x1D6EC}_{2}}m^{k-a_{i}}\biggr).\end{eqnarray}$$
Let
$r=(m^{k}-\sum _{i\in \unicode[STIX]{x1D6EC}_{2}}m^{k-a_{i}})$
, and hence,
$$\begin{eqnarray}1=rm^{-k}+\mathop{\sum }_{i\in \unicode[STIX]{x1D6EC}_{2}}m^{-a_{i}}.\end{eqnarray}$$
By the assumption, there exists a partition
$\unicode[STIX]{x1D6EC}=\{\mathbf{j}_{i}:1\leqslant i\leqslant r\}\cup \{\mathbf{i}_{i}:i\in \unicode[STIX]{x1D6EC}_{2}\}$
such that
$\unicode[STIX]{x1D6FC}_{\mathbf{j}_{i}}=\unicode[STIX]{x1D6FC}^{k},i=1,\ldots ,r$
and
$\unicode[STIX]{x1D6FC}_{\mathbf{i}_{i}}=\unicode[STIX]{x1D6FC}^{a_{i}},~i\in \unicode[STIX]{x1D6EC}_{2}$
. Take
$$\begin{eqnarray}\unicode[STIX]{x1D6EC}^{\prime }=\biggl(\mathop{\bigcup }_{i=1}^{r}\mathbf{j}_{i}\unicode[STIX]{x1D6F4}_{m}\biggr)\cup \{\mathbf{i}_{i}:i\in \unicode[STIX]{x1D6EC}_{2}\}.\end{eqnarray}$$
That is a partition as desired. Therefore, the result follows. ◻
The next lemma is a special case of [Reference Rao, Ruan and Xi13].
Lemma 3.2. Let
$\boldsymbol{\unicode[STIX]{x1D6FC}}=(\unicode[STIX]{x1D6FC}_{1},\ldots ,\unicode[STIX]{x1D6FC}_{m}),\boldsymbol{\unicode[STIX]{x1D6FD}}=(\unicode[STIX]{x1D6FD}_{1},\ldots ,\unicode[STIX]{x1D6FD}_{n})$
be two contraction vectors. If there exist two partitions
$\unicode[STIX]{x1D6EC}_{1}=\{\mathbf{i}_{k}\}_{1\leqslant k\leqslant N}$
of
$\unicode[STIX]{x1D6F4}_{m}^{\ast }$
with respect to
$\boldsymbol{\unicode[STIX]{x1D6FC}}$
, and
$\unicode[STIX]{x1D6EC}_{2}=\{\mathbf{j}_{k}\}_{1\leqslant k\leqslant N}$
of
$\unicode[STIX]{x1D6F4}_{n}^{\ast }$
with respect to
$\boldsymbol{\unicode[STIX]{x1D6FD}}$
, such that
$(\unicode[STIX]{x1D6FC}_{\mathbf{i}_{1}},\ldots ,\unicode[STIX]{x1D6FC}_{\mathbf{i}_{N}})$
is a permutation of
$(\unicode[STIX]{x1D6FD}_{\mathbf{j}_{1}},\ldots ,\unicode[STIX]{x1D6FD}_{\mathbf{j}_{N}})$
. Then
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
.
Proof of Theorem 1.7.
The sufficient part is obvious by Lemma 3.2. We only need to prove the necessary part. By Theorem 1.2, there exists
$q_{0}\in \mathbb{Z}^{+}$
such that
$$\begin{eqnarray}\text{sgp}(\unicode[STIX]{x1D6FD}_{1}^{q_{0}},\ldots ,\unicode[STIX]{x1D6FD}_{n}^{q_{0}})\subset \text{sgp}(\unicode[STIX]{x1D6FC},\ldots ,\unicode[STIX]{x1D6FC})=\{\unicode[STIX]{x1D6FC}^{p}:p\in \mathbb{Z}^{+}\}.\end{eqnarray}$$
Thus, there exists
$p_{j}\in \mathbb{Z}^{+}$
such that
$\unicode[STIX]{x1D6FD}_{j}=\unicode[STIX]{x1D6FC}^{p_{j}/q_{0}}$
for all
$j.$
Let
$d=\text{gcd}(q_{0},p_{1},\ldots ,p_{n})$
and
$q=q_{0}/d$
,
$p_{j}^{\prime }=p_{j}/d$
for all
$j=1,\ldots ,n$
. Then
$\text{gcd}(q,p_{1}^{\prime },\ldots ,p_{n}^{\prime })=1$
. Denote by
$s$
the common Hausdorff dimension of
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})$
and
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
. Since
$$\begin{eqnarray}\mathbb{Q}(\unicode[STIX]{x1D6FD}_{1}^{s},\ldots ,\unicode[STIX]{x1D6FD}_{n}^{s})=\mathbb{Q}(\unicode[STIX]{x1D6FC}^{s},\ldots ,\unicode[STIX]{x1D6FC}^{s})=\mathbb{Q}\biggl(\frac{1}{m}\biggr)=\mathbb{Q},\end{eqnarray}$$
we have
$\unicode[STIX]{x1D6FD}_{j}^{s}\in \mathbb{Q}$
, i.e.,
$\unicode[STIX]{x1D6FC}^{p_{j}s/q_{0}}=m^{-p_{j}^{\prime }/q}\in \mathbb{Q}$
for all
$j$
. Combining this with
$\text{gcd}(q,p_{1}^{\prime },\ldots ,p_{n}^{\prime })=1$
, we know that
$m^{1/q}\in \mathbb{Q}$
so that
$m^{1/q}\in \mathbb{Z}^{+}$
.
Let
$k=m^{1/q}$
and
$\unicode[STIX]{x1D706}=\unicode[STIX]{x1D6FC}^{1/q}$
. Then
$k\unicode[STIX]{x1D706}^{s}=1$
. Define another contraction vector
$\boldsymbol{\unicode[STIX]{x1D706}}=(\unicode[STIX]{x1D706},\ldots ,\unicode[STIX]{x1D706})\in \mathbb{R}^{k}$
. Then
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D706}})$
has Hausdorff dimension
$s$
. Notice that
$\unicode[STIX]{x1D6EC}=\{1,\ldots ,k\}^{q}$
is a partition of
$\unicode[STIX]{x1D6F4}_{k}^{\ast }$
with respect to
$\boldsymbol{\unicode[STIX]{x1D706}}$
with
$\#\unicode[STIX]{x1D6EC}=k^{q}=m$
and
$\unicode[STIX]{x1D706}_{\mathbf{i}}=\unicode[STIX]{x1D706}^{q}=\unicode[STIX]{x1D6FC}$
for
$\mathbf{i}\in \unicode[STIX]{x1D6EC}$
. Thus, by Lemmas 3.1 and 3.2, we have
$\boldsymbol{\unicode[STIX]{x1D6FC}}\sim \boldsymbol{\unicode[STIX]{x1D706}}$
. Similarly, from
$\unicode[STIX]{x1D6FD}_{j}=\unicode[STIX]{x1D6FC}^{p_{j}/q}=\unicode[STIX]{x1D706}^{p_{j}}$
and
$\sum _{j=1}^{n}\unicode[STIX]{x1D706}^{sp_{j}}=\sum _{j=1}^{n}\unicode[STIX]{x1D6FD}_{j}^{s}=1$
, we have
$\boldsymbol{\unicode[STIX]{x1D706}}\sim \boldsymbol{\unicode[STIX]{x1D6FD}}$
as well. Therefore,
$\boldsymbol{\unicode[STIX]{x1D6FC}}\sim \boldsymbol{\unicode[STIX]{x1D6FD}}$
.◻
The theorem easily yields the following useful result, which can also be induced by Theorem 1.2.
Corollary 3.3. Let
$\boldsymbol{\unicode[STIX]{x1D6FC}}=(\unicode[STIX]{x1D6FC},\ldots ,\unicode[STIX]{x1D6FC})\in \mathbb{R}^{m},~\boldsymbol{\unicode[STIX]{x1D6FD}}=(\unicode[STIX]{x1D6FD},\ldots ,\unicode[STIX]{x1D6FD})\in \mathbb{R}^{n}$
, and
$m\unicode[STIX]{x1D6FC}^{s}=n\unicode[STIX]{x1D6FD}^{s}=1$
. Then
${\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FC}})\sim {\mathcal{D}}(\boldsymbol{\unicode[STIX]{x1D6FD}})$
if and only if
$\log m/\text{log}\,n\in \mathbb{Q}$
.
Acknowledgements
The first author gratefully acknowledges the support of K. C. Wong Education Foundation and DAAD. The authors are also grateful to the referee for valuable suggestions.
