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On the mutual singularity of multifractal measures

  • Received: 01 February 2020 Revised: 01 February 2020
  • Primary: 28A20; Secondary: 28A80

  • 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 Z-subalgebra of an ambient field F=Q(u1,,un) generated by certain combinatorially defined generators (i.e., cluster variables), which are grouped into overlapping clusters. Many relations between cluster algebras and other branches of mathematics have been discovered, for example, Poisson geometry, discrete dynamical systems, higher Teichmüller spaces, representation theory of quivers and finite-dimensional algebras.

    We first recall the definition of cluster automorphisms, which were introduced by Assem, Schiffler and Shamchenko in [1].

    Definition 1.1 ([1]). Let A=A(x,B) be a cluster algebra, and f:AA be an automorphism of Z-algebras. f is called a cluster automorphism of A if there exists another seed (z,B) of A such that

    (1)f(x)=z;

    (2) f(μx(x))=μf(x)(z) for any xx.

    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 f is a Z-algebra homomorphism.

    Conjecture 1. [5,Conjecture 1] Let A be a cluster algebra, and f:AA be a Z-algebra homomorphism. Then f is cluster automorphism if and only if there exist two clusters x and z such that f(x)=z.

    The following is our main result, which affirms the Conjecture 1.

    Theorem 3.6 Let A be a cluster algebra, and f:AA be a Z-algebra homomorphism. Then f is a cluster automorphism if and only if there exist two clusters x and z such that f(x)=z.

    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 n, we will always denote by [1,n] the set {1,2,,n}.

    Recall that B is said to be skew-symmetrizable if there exists an positive diagonal integer matrix D such that BD is skew-symmetric.

    Fix an ambient field F=Q(u1,u2,,un). A labeled seed is a pair (x,B), where x is an n-tuple of free generators of F, and B is an n×n skew-symmetrizable integer matrix. For k[1,n], we can define another pair (x,B)=μk(x,B), where

    (1) x=(x1,,xn) is given by

    xk=ni=1x[bik]+i+ni=1x[bik]+ixk

    and xi=xi for ik;

    (2) B=μk(B)=(bij)n×n is given by

    bij={bij,if i=k or j=k;bij+sgn(bik)[bikbkj]+,otherwise.

    where [x]+=max{x,0}. The new pair (x,B)=μk(x,B) is called the mutation of (x,B) at k. We also denote B=μk(B).

    It can be seen that (x,B) is also a labeled seed and μk is an involution.

    Let (x,B) be a labeled seed. x is called a labeled cluster, elements in x are called cluster variables, and B is called an exchange matrix. The unlabeled seeds are obtained by identifying labeled seeds that differ from each other by simultaneous permutations of the components in x, and of the rows and columns of B. We will refer to unlabeled seeds and unlabeled clusters simply as seeds and clusters respectively, when there is no risk of confusion.

    Lemma 2.1 ([2]). Let B be an n×n skew-symmetrizable matrix. Then μk(B)=(Jk+Ek)B(Jk+Fk), where

    (1) Jk denotes the diagonal n×n matrix whose diagonal entries are all 1s, except for 1 in the k-th position;

    (2) Ek is the n×n matrix whose only nonzero entries are eik=[bik]+;

    (3) Fk is the n×n matrix whose only nonzero entries are fkj=[bkj]+.

    Definition 2.2 ([9,11]). (1) Two labeled seeds (x,B) and (x,B) are said to be mutation equivalent if (x,B) can be obtained from (x,B) by a sequence of mutations;

    (2) Let Tn be an n-regular tree and valencies emitting from each vertex are labelled by 1,2,,n. A cluster pattern is an n-regular tree Tn such that for each vertex tTn, there is a labeled seed Σt=(xt,Bt) and for each edge labelled by k, two labeled seeds in the endpoints are obtained from each other by seed mutation at k. We always write

    xt=(x1;t,x2;t,,xn;t),Bt=(btij).

    The cluster algebra A=A(xt0,Bt0) associated with the initial seed (xt0,Bt0) is a Z-subalgebra of F generated by cluster variables appeared in Tn(xt0,Bt0), where Tn(xt0,Bt0) is the cluster pattern with (xt0,Bt0) lying in the vertex t0Tn.

    Theorem 2.3 (Laurent phenomenon and positivity [11,15,12]). Let A=A(xt0,Bt0) be a cluster algebra. Then each cluster variable xi,t is contained in

    Z0[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 0B be a skew-symmetrizable integer matrix. If B is obtained from B by a sequence of mutations and B=aB for some aZ, then a=±1 and B=±B.

    Proof.. Since B is obtained from B by a sequence of mutations, there exist integer matrices E and F such that B=EBF, by Lemma 2.1. If B=aB, then we get B=aE(B)F. Also, we can have

    B=aE(B)F=a2E2(B)F2==asEs(B)Fs,

    where s0. By B0, we know a0. Thus 1asB=Es(B)Fs holds for any s0.

    Assume by contradiction that a±1, then when s is large enough, 1asB will not be an integer matrix. But Es(B)Fs is always an integer matrix. This is a contradiction. So we must have a=±1 and thus B=±B.

    A square matrix A is decomposable if there exists a permutation matrix P such that PAPT is a block-diagonal matrix, and indecomposable otherwise.

    Lemma 3.2. Let 0B be an indecomposable skew-symmetrizable matrix. If B is obtained from B by a sequence of mutations and B=BA for some integer diagonal matrix A=diag(a1,,an), then A=±In and B=±B.

    Proof. If there exists i0 such that ai0=0, then the i0-th column vector of B is zero, by B=BA. This contradicts that B is indecomposable and B0. So each ai0 is nonzero for i0=1,,n.

    Let D=diag(d1,,dn) be a skew-symmetrizer of B. By B=BA and AD=DA, we know that

    BD=BAD=(BD)A.

    By the definition of mutation, we know that D is also a skew-symmetrizer of μk(B), k=1,,n. Since B is obtained from B by a sequence of mutations, we get that D is a skew-symmetrizer of B. Namely, we have that both BD and BD=(BD)A are skew-symmetric. Since 0B is indecomposable, we must have a1==an. So A=aIn for some aZ, and B=aB. Then by Lemma 3.1, we can get A=±In and B=±B.

    Lemma 3.3. Let B=diag(B1,,Bs), where each Bi is a nonzero indecomposable skew-symmetrizable matrix of size ni×ni. If B is obtained from B by a sequence of mutations and B=BA for some integer diagonal matrix A=diag(a1,,an), then aj=±1 for j=1,,n.

    Proof. By the definition of mutation, we know that B has the form of B=diag(B1,,Bs), where each Bi is obtained from Bi by a sequence of mutations. We can write A as a block-diagonal matrix A=diag(A1,,As), where Ai is a ni×ni integer diagonal matrix. By B=BA, we know that Bi=BiAi. Then by Lemma 3.2, we have Ai=±Ini and Bi=±Bi for i=1,,s. In particular, we get aj=±1 for j=1,,n.

    Lemma 3.4. Let A=A(x,B) be a cluster algebra, and f:AA be a Z-homomorphism of A. If there exists another seed (z,B) of A such that such that f(x)=z, then f(μx(x))=μf(x)(z) for any xx.

    Proof. After permutating the rows and columns of B, it can be written as a block-diagonal matrix as follows.

    B=diag(B1,B2,,Bs),

    where B1 is an n1×n1 zero matrix and Bj is nonzero indecomposable skew-symmetrizable matrix of size nj×nj for j=2,,s.

    Without loss of generality, we assume that f(xi)=zi for 1in.

    Let xk and zk be the new obtained variables in μk(x,B) and μk(z,B). So we have

    xkxk=ni=1x[bik]+i+ni=1x[bik]+i,andzkzk=ni=1z[bik]+i+ni=1z[bik]+i.

    Thus

    f(xk)=f(ni=1x[bik]+i+ni=1x[bik]+ixk)=ni=1z[bik]+i+ni=1z[bik]+izk=ni=1z[bik]+i+ni=1z[bik]+ini=1z[bik]+i+ni=1z[bik]+izk.

    Note that the above expression is the expansion of f(xk) with respect to the cluster μk(z). By

    f(xk)f(A)=AZ[z±11,,(zk)±1,,z±1n],

    we can get

    ni=1z[bik]+i+ni=1z[bik]+ini=1z[bik]+i+ni=1z[bik]+iZ[z±11,,z±1k1,z±1k+1,,z±1n].

    Since both ni=1z[bik]+i+ni=1z[bik]+i and ni=1z[bik]+i+ni=1z[bik]+i is not divided by any zi, we actually have

    ni=1z[bik]+i+ni=1z[bik]+ini=1z[bik]+i+ni=1z[bik]+iZ[z1,,zk1,zk+1,,zn].

    So for each k, there exists an integer akZ such that (b1k,b2k,,bnk)T=ak(b1k,b2k,,bnk)T. Namely, we have B=BA, where A=diag(a1,,an). Note that B has the form of

    B=diag(B1,B2,,Bs),

    where B1 is an n1×n1 zero matrix and Bj is a nonzero indecomposable skew- symmetrizable matrix of size nj×nj for j=2,,s. Applying Lemma 3.3 to the skew-symmetrizable matrix diag(B2,,Bs), we can get aj=±1 for n1+1,,n. Since the first n1 column vectors of both B and B are zero vectors, we can just take a1==an1=1. So for each k, we have ak=±1 and

    (b1k,b2k,,bnk)T=ak(b1k,b2k,,bnk)T=±(b1k,b2k,,bnk)T.

    Hence,

    ni=1z[bik]+i+ni=1z[bik]+ini=1z[bik]+i+ni=1z[bik]+i=1.

    Thus we get

    f(xk)=ni=1z[bik]+i+ni=1z[bik]+ini=1z[bik]+i+ni=1z[bik]+izk=zk.

    So f(μx(x))=μf(x)(z) for any xx.

    Lemma 3.5. Let A=A(x,B)F be a cluster algebra, and f be an automorphism of the ambient field F. If there exists another seed (z,B) of A such that f(x)=z and f(μx(x))=μf(x)(z) for any xx. Then

    (i) f is an automorphism of A;

    (ii) f is a cluster automorphism of A.

    Proof. (ⅰ) Since f is an automorphism of the ambient field F, we know that f is injective.

    Since f commutes with mutations, we know that f restricts to a surjection on X, where X is the set of cluster variables of A. Because A is generated by X, we get that f restricts to an epimorphism of A.

    Hence, f is an automorphism of A.

    (ⅱ) follows from (ⅰ) and the definition of cluster automorphisms.

    Theorem 3.6. Let A be a cluster algebra, and f:AA be a Z-algebra homomorphism. Then f is a cluster automorphism if and only if there exist two clusters x and z such that f(x)=z.

    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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