The aim of this article is to show that the multifractal Hausdorff and packing measures are mutually singular, which in particular provides an answer to Olsen's questions.
Citation: Zied Douzi, Bilel Selmi. On the mutual singularity of multifractal measures[J]. Electronic Research Archive, 2020, 28(1): 423-432. doi: 10.3934/era.2020024
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The aim of this article is to show that the multifractal Hausdorff and packing measures are mutually singular, which in particular provides an answer to Olsen's questions.
Cluster algebras were invented by Fomin and Zelevinsky in a series of papers [9,2,10,11]. A cluster algebra is a
We first recall the definition of cluster automorphisms, which were introduced by Assem, Schiffler and Shamchenko in [1].
Definition 1.1 ([1]). Let
Cluster automorphisms and their related groups were studied by many authors, and one can refer to [6,7,8,14,13,4,5,16] for details.
The following very insightful conjecture on cluster automorphisms is by Chang and Schiffler, which suggests that we can weaken the conditions in Definition 1.1. In particular, it suggests that the second condition in Definition 1.1 can be obtained from the first one and the assumption that
Conjecture 1. [5,Conjecture 1] Let
The following is our main result, which affirms the Conjecture 1.
Theorem 3.6 Let
In this section, we recall basic concepts and important properties of cluster algebras. In this paper, we focus on cluster algebras without coefficients (that is, with trivial coefficients). For a positive integer
Recall that
Fix an ambient field
x′k=n∏i=1x[bik]+i+n∏i=1x[−bik]+ixk |
and
b′ij={−bij,if i=k or j=k;bij+sgn(bik)[bikbkj]+,otherwise. |
where
It can be seen that
Let
Lemma 2.1 ([2]). Let
(1)
(2)
(3)
Definition 2.2 ([9,11]).
xt=(x1;t,x2;t,…,xn;t),Bt=(btij). |
The cluster algebra
Theorem 2.3 (Laurent phenomenon and positivity [11,15,12]). Let
Z≥0[x±11;t0,x±12;t0,…,x±1n;t0]. |
In this section, we will give our main result, which affirms the Conjecture 1.
Lemma 3.1. Let
Proof.. Since
B′=aE(B′)F=a2E2(B′)F2=⋯=asEs(B′)Fs, |
where
Assume by contradiction that
A square matrix
Lemma 3.2. Let
Proof. If there exists
Let
BD=B′AD=(B′D)A. |
By the definition of mutation, we know that
Lemma 3.3. Let
Proof. By the definition of mutation, we know that
Lemma 3.4. Let
Proof. After permutating the rows and columns of
B=diag(B1,B2,⋯,Bs), |
where
Without loss of generality, we assume that
Let
xkx′k=n∏i=1x[bik]+i+n∏i=1x[−bik]+i,andzkz′k=n∏i=1z[b′ik]+i+n∏i=1z[−b′ik]+i. |
Thus
f(x′k)=f(n∏i=1x[bik]+i+n∏i=1x[−bik]+ixk)=n∏i=1z[bik]+i+n∏i=1z[−bik]+izk=n∏i=1z[bik]+i+n∏i=1z[−bik]+in∏i=1z[b′ik]+i+n∏i=1z[−b′ik]+iz′k. |
Note that the above expression is the expansion of
f(x′k)∈f(A)=A⊂Z[z±11,…,(z′k)±1,…,z±1n], |
we can get
n∏i=1z[bik]+i+n∏i=1z[−bik]+in∏i=1z[b′ik]+i+n∏i=1z[−b′ik]+i∈Z[z±11,⋯,z±1k−1,z±1k+1,⋯,z±1n]. |
Since both
n∏i=1z[bik]+i+n∏i=1z[−bik]+in∏i=1z[b′ik]+i+n∏i=1z[−b′ik]+i∈Z[z1,⋯,zk−1,zk+1,⋯,zn]. |
So for each
B=diag(B1,B2,⋯,Bs), |
where
(b1k,b2k,…,bnk)T=ak(b′1k,b′2k,…,b′nk)T=±(b′1k,b′2k,…,b′nk)T. |
Hence,
n∏i=1z[bik]+i+n∏i=1z[−bik]+in∏i=1z[b′ik]+i+n∏i=1z[−b′ik]+i=1. |
Thus we get
f(x′k)=n∏i=1z[bik]+i+n∏i=1z[−bik]+in∏i=1z[b′ik]+i+n∏i=1z[−b′ik]+iz′k=z′k. |
So
Lemma 3.5. Let
(i)
(ii)
Proof. (ⅰ) Since
Since
Hence,
(ⅱ) follows from (ⅰ) and the definition of cluster automorphisms.
Theorem 3.6. Let
Proof. "Only if part": It follows from the definition of cluster automorphism.
"If part": It follows from Lemma 3.4 and Lemma 3.5.
This project is supported by the National Natural Science Foundation of China (No.11671350 and No.11571173) and the Zhejiang Provincial Natural Science Foundation of China (No.LY19A010023).
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