On a characterization of shifts of Haar distributions on

compact open subgroups of a compact Abelian group

G.M. Feldman

Abstract. Let $X$ be a compact Abelian group. In the article we obtain a
characterization of shifts of Haar distributions on compact open subgroups of
the group $X$ by the symmetry of the conditional distribution of one linear
form of independent random variables taking values in $X$ given another.
Coefficients of the linear forms are topological automorphisms of the group
$X$. This result can be viewed as an analogue for compact Abelian groups of
the well-known Heyde theorem, where the Gaussian distribution on the real line
is characterized by the symmetry of the conditional distribution of one linear
form of independent random variables given another.

Keywords. compact Abelian group, conditional distribution, Haar distribution

Mathematics Subject Classification. 43A35, 60B15, 62E10

##  1  Introduction

Let $X$ be a compact Abelian group. The purpose of this article is to obtain a
characterization of shifts of Haar distributions on compact open subgroups of
the group $X$ by the symmetry of the conditional distribution of one linear
form of independent random variables taking values in $X$ given another.
Coefficients of considered linear forms are topological automorphisms of the
group $X$. This result can be viewed as an analogue for compact Abelian groups
of the well-known Heyde theorem ([16], see also [17, Theorem 13.4.1]), where
Gaussian distributions on the real line are characterized by the symmetry of
the conditional distribution of one linear form of real independent random
variables given another. On other analogues of the Heyde theorem for different
classes of locally compact Abelian groups see e.g. [2, 3, 4, 10, 6, 8, 7, 9,
11, 19, 20]. We note that shifts of Haar distributions on compact open
subgroups of a compact Abelian group $X$ have not previously appeared in
characterization theorems on groups.

In the article we use standard results of abstract harmonic analysis, in
particular the duality theory for locally compact Abelian groups (see [15]).
Let us agree on the following notation. Let $X$ be a second countable locally
compact Abelian group. We consider only such groups without mentioning it
specifically. Denote by $Y$ the character group of the group $X$, and by
$(x,y)$ the value of a character $y\in Y$ at an element $x\in X$. If $K$ is a
closed subgroup of $X$, denote by $A(Y,K)=\\{y\in Y:(x,y)=1$ for all  $x\in
K\\}$ its annihilator. Denote by ${\rm Aut}(X)$ the group of all topological
automorphisms of the group $X$. Let $\alpha\in{\rm Aut}(X)$. The adjoint
automorphism $\tilde{\alpha}\in{\rm Aut}(Y)$ is defined by the formula
$(\alpha x,y)=(x,\tilde{\alpha}y)$ for all $x\in X$, $y\in Y$. Denote by $I$
the identity automorphism. Denote by $c_{X}$ the connected component of zero
of the group $X$, and by $b_{Y}$ the subgroup consisting of all compact
elements of the group $Y$. Then $b_{Y}=A(Y,c_{X})$. Let
$\\{K_{\iota}:\iota\in{\cal I}\\}$ be a not empty set of compact Abelian
groups. Denote by $\mathop{\mbox{\rm\bf P}}\limits_{\iota\in{\cal
I}}K_{\iota}$ the direct product of the groups $K_{\iota}$. Unless otherwise
stated, we will consider the group $\mathop{\mbox{\rm\bf
P}}\limits_{\iota\in{\cal I}}K_{\iota}$ in the product topology. Let
$\\{L_{\iota}:\iota\in{\cal I}\\}$ be a not empty set of discrete Abelian
groups. Denote by $\mathop{\mbox{\rm\bf P}^{*}}\limits_{\iota\in{\cal
I}}L_{\iota}$ the weak direct product of the groups $L_{\iota}$ considering in
the discrete topology. A $p$-group is defined to be a group whose elements
have order equal to a nonnegative power of $p$. Denote by
$\mathbb{T}=\\{z\in\mathbb{C}:|z|=1\\}$ the circle group (the one dimensional
torus), by $\mathbb{Z}$ the group of integers, by
$\mathbb{Z}(n)=\\{0,1,\dots,n-1\\}$ the group of residue classes modulo $n$,
and by $\mathbb{Q}$ the group of rational numbers considering in the discrete
topology. Denote by $\cal P$ the set of prime numbers, and by $p_{n}$ the
$n$th prime number. Let $p\in\cal P$, and let $\Delta_{p}$ be the group of
$p$-adic integers. Denote by ${\mathbb{Z}}(p^{\infty})$ the set of rational
numbers of the form ${\\{{k/p^{n}}:k=0,1,\dots,p^{n}-1,\ n=0,1,\dots\\}}$. If
we define the operation in ${\mathbb{Z}}(p^{\infty})$ as addition modulo 1,
then ${\mathbb{Z}}(p^{\infty})$ is transformed into an Abelian group which we
consider in the discrete topology. Obviously, this group is isomorphic to the
multiplicative group of $p^{n}$th roots of unity considering in the discrete
topology, where $n$ goes through the nonnegative integers. The group
${\mathbb{Z}}(p^{\infty})$ is topologically isomorphic the character group of
the group $\Delta_{p}$ ([15, (25.2)]).

Denote by ${\rm M}^{1}(X)$ the convolution semigroup of probability
distributions on the group $X$. Let $\mu\in{\rm M}^{1}(X)$. Denote by

$\hat{\mu}(y)=\int_{X}(x,y)d\mu(x)$

the characteristic function (Fourier transform) of the distribution $\mu$, and
by $\sigma(\mu)$ the support of $\mu$. Define the distribution
$\bar{\mu}\in{\rm M}^{1}(X)$ by the formula $\bar{\mu}(B)=\mu(-B)$ for any
Borel subset $B$ of the group $X$. Then
$\hat{\bar{\mu}}(y)=\overline{\hat{\mu}(y)}$.

######  Definition 1.1 ([21, Chapter IV, §6]).

A distribution $\gamma\in{\rm M}^{1}(X)$ is called Gaussian if its
characteristic function is represented in the form

$\hat{\gamma}(y)=(x,y)\exp\\{-\varphi(y)\\},$

where $x\in X$ and $\varphi(y)$ is a continuous nonnegative function on the
group $Y$ satisfying the equation

$\varphi(u+v)+\varphi(u-v)=2[\varphi(u)+\varphi(v)],\ \ u,v\in Y.$

Denote by $E_{x}$ the degenerate distribution concentrated at an element $x\in
X$. Denote by $m_{X}$ a Haar measure on the group $X$. If $X$ is a compact
group, we assume that $m_{X}$ is a distribution. If $K$ is a compact subgroup
of the group $X$, the characteristic function of the distribution $m_{K}$ is
of the form

$\hat{m}_{K}(y)=\begin{cases}1,&\text{\ if\ }\ y\in A(Y,K),\\\ 0,&\text{\ if\
}\ y\not\in A(Y,K).\end{cases}$ (1)

##  2  Main theorem

Let $\xi_{1}$ and $\xi_{2}$ be independent random variables with values in a
locally compact Abelian group $X$ and distributions $\mu_{1}$ and $\mu_{2}$.
Let $\alpha_{j},\beta_{j}\in{\rm Aut}(X)$. Assume that the conditional
distribution of the linear form $L_{2}=\beta_{1}\xi_{1}+\beta_{2}\xi_{2}$
given $L_{1}=\alpha_{1}\xi_{1}+\alpha_{2}\xi_{2}$ is symmetric. Obviously, if
we are interested in the description of $\mu_{j}$, then we can suppose without
loss of generality that $L_{1}=\xi_{1}+\xi_{2}$ and
$L_{2}=\xi_{1}+\alpha\xi_{2}$, where $\alpha\in{\rm Aut}(X)$.

Let $X$ be a nondiscrete locally compact Abelian group and let $\mu$ be a
distribution on $X$. Then $\mu$ is represented as a sum

$\mu=\mu_{ac}+\mu_{s}+\mu_{d},$

where $\mu_{ac}$ is an absolutely continuous measure with respect to $m_{X}$,
$\mu_{s}$ is a singular measure with respect to $m_{X}$, and $\mu_{d}$ is a
discrete measure. The measure $\mu_{ac}$ is called the absolutely continuous
component of the distribution $\mu$ with respect to $m_{X}$.

The main result of this article is the following theorem, which can be
regarded as an analogue for compact Abelian groups of the well-known Heyde
theorem.

######  Theorem 2.1.

Let $X$ be a nondiscrete compact Abelian group, let $\alpha$ be a topological
automorphism of $X$, and let $\xi_{1}$ and $\xi_{2}$ be independent random
variables with values in $X$ and distributions $\mu_{1}$ and $\mu_{2}$.
Suppose that the following conditions are satisfied:

  1. (i)

the factor-group $X/c_{X}$ is topologically isomorphic to a group of the form

$\mathop{\mbox{\rm\bf P}}\limits_{p\in{\cal P},\ p>2}{X}_{p},$ (2)

where $X_{p}=\Delta_{p}^{n_{p}}\times F_{p}$, $n_{p}$ is a nonnegative integer
and $F_{p}$ is a finite Abelian $p$-group;

  2. (ii)

$I\pm\alpha\in{\rm Aut}(X)$;

  3. (iii)

at least one of the distributions $\mu_{j}$ has a nonzero absolutely
continuous component with respect to $m_{X}$.

If the conditional distribution of the linear form
$L_{2}=\xi_{1}+\alpha\xi_{2}$ given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric, then
$\mu_{j}=m_{G}*E_{x_{j}}$, where $G$ is a compact open subgroup of $X$,
$x_{j}\in X$, $j=1,2$. Moreover, $\alpha(G)=G$.

To prove Theorem 2.1, we need the following lemmas.

######  Lemma 2.2 ([5, Lemma 16.1]).

Let $X$ be a locally compact Abelian group and let $\alpha$ be a topological
automorphism of $X$. Let $\xi_{1}$ and $\xi_{2}$ be independent random
variables with values in the group $X$ and distributions $\mu_{1}$ and
$\mu_{2}$. The conditional distribution of the linear form
$L_{2}=\xi_{1}+\alpha\xi_{2}$ given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric if
and only if the characteristic functions $\hat{\mu}_{j}(y)$ satisfy the
equation

$\hat{\mu}_{1}(u+v)\hat{\mu}_{2}(u+\tilde{\alpha}v)=\hat{\mu}_{1}(u-v)\hat{\mu}_{2}(u-\tilde{\alpha}v),\quad
u,v\in Y.$ (3)

It is convenient for us to formulate the following well-known statements in
the form of a lemma.

######  Lemma 2.3.

Let $X$ be a locally compact Abelian group and let $K$ be a closed subgroup of
$X$. Let $\alpha$ be a topological automorphism of the group $X$. Then the
following statements are true:

  1. (i)

$\alpha(K)=K$ if and only if $\tilde{\alpha}(A(Y,K))=A(Y,K)$.

  2. (ii)

If $\alpha(K)=K$, then $\alpha$ induces a topological automorphism
$\hat{\alpha}$ on the factor-group $X/K$ by the formula
$\hat{\alpha}[x]=[\alpha x]$, $[x]\in X/K$. The adjoint automorphism to
$\hat{\alpha}$ is the restriction of $\tilde{\alpha}$ to $A(Y,K)$.

  3. (iii)

Let $\alpha(K)=K$ and let $\mu\in{\rm M}^{1}(X)$. The restriction of the
characteristic function $\hat{\mu}(y)$ to the subgroup $A(Y,K)$ is the
characteristic function of a distribution on the factor-group $X/K$. It is the
image of $\mu$ under the natural homomorphism $X\rightarrow X/K$.

The following lemma results from a special case of Corollary 17.2 and Remark
17.5 in [5].

######  Lemma 2.4.

Let $X$ a finite Abelian group containing no elements of order $2$. Let
$\alpha$ be an automorphism of the group $X$ satisfying the condition
$I+\alpha\in{\rm Aut}(X)$. Assume that $\mu_{1}$ and $\mu_{2}$ are
distributions on $X$ such that the characteristic functions $\hat{\mu}_{j}(y)$
are nonnegative and satisfy equation $(\ref{1})$. Then
$\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)$ for all $y\in Y$, and $\hat{\mu}_{j}(y)$
take only the values $0$ and $1$. Moreover, $\tilde{\alpha}(E)=E$, where
$E=\\{y\in Y:\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)=1\\}$.

######  Lemma 2.5.

Let $X$ be a compact totally disconnected Abelian group topologically
isomorphic to a group of the form $(\ref{2})$. Let $\alpha$ be a topological
automorphism of the group $X$ satisfying the condition $I+\alpha\in{\rm
Aut}(X)$. Assume that $\mu_{1}$ and $\mu_{2}$ are distributions on $X$ such
that the characteristic functions $\hat{\mu}_{j}(y)$ are nonnegative and
satisfy equation $(\ref{1})$. Then $\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)$ for all
$y\in Y$, and $\hat{\mu}_{j}(y)$ take only the values $0$ and $1$. Moreover,
$\tilde{\alpha}(E)=E$, where $E=\\{y\in
Y:\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)=1\\}$.

Proof. Since the group $X$ is compact and topologically isomorphic to a group
of the form $(\ref{2})$, the group $Y$ is discrete and topologically
isomorphic to a group of the form

$\mathop{\mbox{\rm\bf P}^{*}}\limits_{p\in{\cal P},\ p>2}{Y}_{p},$ (4)

where $Y_{p}=(\mathbb{Z}(p^{\infty}))^{n_{p}}\times H_{p}$, and $H_{p}$ is the
character group of the group $F_{p}$. Put

$Y_{m,n}=\\{y\in Y:(p_{2}p_{3}\cdots p_{n})^{m}y=0,\
m=1,2,\dots,n=2,3,\dots,\\}.$

Then $Y_{n,m}$ is a finite subgroup of the group $Y$. Moreover, it follows
from (4) that

$Y=\bigcup\limits_{m\geq 1,\ n\geq 2}Y_{m,n}.$

It is obvious that the each of the subgroups $Y_{m,n}$ is carried into itself
by every automorphism of the group $Y$. We have

$\tilde{\alpha}(Y_{m,n})=Y_{m,n}.$ (5)

It follows from $I+\alpha\in{\rm Aut}(X)$ that $I+\tilde{\alpha}\in{\rm
Aut}(Y)$. Hence,

$(I+\tilde{\alpha})(Y_{m,n})=Y_{m,n}.$ (6)

Since $Y_{m,n}$ is a finite and hence a compact group, its annihilator
$A(X,Y_{m,n})$ is an open subgroup of $X$. This implies that the factor-group
$X/A(X,Y_{m,n})$ is a discrete, and hence a finite group. The group $Y_{m,n}$
is isomorphic to the character group of the factor-group $X/A(X,Y_{m,n})$.
Obviously, the group $X/A(X,Y_{m,n})$ contains no elements of order 2. By
Lemma 2.2, the symmetry of the conditional distribution of the linear form
$L_{2}$ given $L_{1}$ implies that the characteristic functions
$\hat{\mu}_{j}(y)$ satisfy equation (3). In view of (5), we can consider the
restriction of equation (3) to the subgroup $Y_{m,n}$. Taking into account (5)
and (6), it follows from Lemma 2.3 that we can apply Lemma 2.4 to the finite
Abelian group $X/A(X,Y_{m,n})$ and the induced automorphism
$\hat{\alpha}\in{\rm Aut}(X/A(X,Y_{m,n}))$. By Lemma 2.4,
$\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)$ for all $y\in Y_{m,n}$, and the
characteristic functions $\hat{\mu}_{j}(y)$ take only values 0 and 1 on
$Y_{m,n}$. It follows from this that $\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)$ for
all $y\in Y$, and the functions $\hat{\mu}_{j}(y)$ take only values 0 and 1 on
$Y$. Put $E=\\{y\in Y:\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)=1\\}$ and
$E_{m,n}=E\cap Y_{m,n}$. By Lemma 2.4, we have
$\tilde{\alpha}(E_{m,n})=E_{m,n}$. This implies that $\tilde{\alpha}(E)=E$.
$\Box$

The following statement is well known. For its proof see, for example, ([5,
Proposition 2.13]).

######  Lemma 2.6.

Let $X$ be a locally compact Abelian group and let $\mu$ be a distribution on
$X$. Put $E=\\{y\in Y:\hat{\mu}(y)=1\\}$. Then $E$ is a closed subgroup of the
group $Y$. Put $G=A(X,E)$. Then $\sigma(\mu)\subset G$, and if we consider
$\mu$ as a distribution on the group $G$, then its characteristic function is
equal to $1$ only at zero of the character group of the group $G$.

Proof of Theorem 2.1. By Lemma 2.2, the symmetry of the conditional
distribution of the linear form $L_{2}$ given $L_{1}$ implies that the
characteristic functions $\hat{\mu}_{j}(y)$ satisfy equation (3). Put
$\nu_{j}=\mu_{j}*\bar{\mu}_{j}$. Then
$\hat{\nu}_{j}(y)=|\hat{\mu}_{j}(y)|^{2}\geq 0$ and
$\hat{\nu}_{j}(-y)=\hat{\nu}_{j}(y)$, $y\in Y$. The characteristic functions
$\hat{\nu}_{j}(y)$ also satisfy equation (3). In view of (1), if we prove that
$\nu_{1}=\nu_{2}=m_{K}$, where $K$ is a compact open subgroup of the group
$X$, then the required representation for the distributions $\mu_{j}$ follows
from this. For this reason, without loss of generality, we can prove the
theorem under the assumption that

$\hat{\mu}_{j}(y)\geq 0,\quad\hat{\mu}_{j}(-y)=\hat{\mu}_{j}(y),\quad j=1,2.$
(7)

Since $X$ is a compact group, $Y$ is a discrete group. Therefore, the subgroup
$b_{Y}$ consists of elements of finite order of the group $Y$. It is obvious
that the subgroup $b_{Y}$ is carried into itself by every automorphism of the
group $Y$. We have

$\tilde{\alpha}(b_{Y})=b_{Y}.$ (8)

It follows from $I+\alpha\in{\rm Aut}(X)$ that $I+\tilde{\alpha}\in{\rm
Aut}(Y)$. Hence,

$(I+\tilde{\alpha})(b_{Y})=b_{Y}.$ (9)

Since $c_{X}=A(X,b_{Y})$, the group $b_{Y}$ is topologically isomorphic to the
character group of the factor-group $X/c_{X}$. In view of (8), we can consider
the restriction of equation (3) to the subgroup $b_{Y}$. Taking into account
(8) and (9), it follows from Lemma 2.3, that we can apply Lemma 2.5 to the
compact totally disconnected Abelian group $X/c_{X}$ and the induced
topological automorphism $\hat{\alpha}\in{\rm Aut}(X/c_{X})$. Lemma 2.5
implies that the restrictions of the characteristic functions
$\hat{\mu}_{j}(y)$ to the subgroup $b_{Y}$ take only the values $0$ and $1$.
Moreover,

$\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)\ \text{\ for all\ }\ y\in b_{Y}.$ (10)

Denote by $G_{j}$ the minimal closed subgroup of the group $X$ containing the
support $\sigma(\mu_{j})$, $j=1,2$. By the conditions of the theorem at least
one of the distributions $\mu_{j}$ has a non-zero absolutely continuous
component with respect to $m_{X}$. Assume for definiteness that it is
$\mu_{1}$. Hence, $m_{X}(\sigma(\mu_{1}))>0$. This implies that the difference
set $D_{1}=\\{x=x_{1}-x_{2}:x_{i}\in\sigma(\mu_{1})\\}$ contains a
neighbourhood of zero of the group $X$. It follows from $D_{1}\subset G_{1}$
that $G_{1}$ is an open subgroup. Hence, its annihilator $H_{1}=A(Y,G_{1})$ is
a compact subgroup of the group $Y$. Since $Y$ is a discrete group, $H_{1}$ is
a finite group, and hence, $H_{1}\subset b_{Y}$.

Put $E_{j}=\\{y\in Y:\hat{\mu}_{j}(y)=1\\}$, $j=1,2$. Then by Lemma 2.6, the
sets $E_{j}$ are closed subgroups of the group $Y$, and
$\sigma(\mu_{j})\subset A(X,E_{j})$. Therefore $G_{j}\subset A(X,E_{j})$, and
hence $E_{1}\subset H_{1}\subset b_{Y}$. It follows from this that
$\hat{\mu}_{1}(y)<1$ for all $y\notin b_{Y}$. Since $E_{1}\subset b_{Y}$, we
have $E_{1}=\\{y\in b_{Y}:\hat{\mu}_{1}(y)=1\\}$. Hence, in view of (10), we
have $E_{1}=\\{y\in b_{Y}:\hat{\mu}_{2}(y)=1\\}\subset E_{2}$. Taking this
into account, put

$E=\\{y\in b_{Y}:\mu_{1}(y)=\hat{\mu}_{2}(y)=1\\},\quad G=A(X,E).$

Thus, the independent random variables $\xi_{1}$ and $\xi_{2}$ take values in
the group $G$. If we consider $\xi_{1}$ and $\xi_{2}$ as independent random
variables with values in $G$, then by Lemma 2.6, the characteristic function
$\hat{\mu}_{1}(y)$ takes the value 1 only at zero of the character group of
the group $G$. Therefore $G$ is the the minimal closed subgroup of the group
$X$ containing the supports of the distributions $\mu_{j}$. Applying Lemma 2.5
to the compact totally disconnected Abelian group $X/c_{X}$, we get that
$\tilde{\alpha}(E)=E$. By Lemma 2.3, it follows from this that $\alpha(G)=G$.
Hence, the restriction of $\alpha$ to $G$ is a topological automorphism of the
group $G$. We note that $E$ is a finite group, and for this reason $E$ is a
compact group. Inasmuch the subgroup $G$ is the annihilator of $E$, this
implies that $G$ is an open group of the group $X$.

Let us verify that conditions (i)–(iii) of the theorem are valid if we
consider the subgroup $G$ instead of the group $X$.

Note that $c_{X}\subset G$. Hence $c_{G}=c_{X}$, and the factor-group
$G/c_{G}=G/c_{X}$ is topologically isomorphic to a closed subgroup of the
factor-group $X/c_{X}$. But each closed subgroup of a group which is
topologically isomorphic to a group of the form $(\ref{2})$ is also
topologically isomorphic to a group of the form $(\ref{2})$. Hence, the group
$G$ satisfies condition (i).

By the condition of the theorem $I\pm\alpha\in{\rm Aut}(X)$. This implies that
$I\pm\tilde{\alpha}\in{\rm Aut}(Y)$. Since $E$ is a finite subgroup of the
group $Y$ and $\tilde{\alpha}\in{\rm Aut}(E)$, we have
$I\pm\tilde{\alpha}\in{\rm Aut}(E)$. Hence, $(I\pm\tilde{\alpha})(E)=E$. By
Lemma 2.3, it follows from this that $(I\pm\alpha)(G)=G$. It means that the
restrictions of the topological automorphisms $I\pm\alpha$ to $G$ are
topological automorphisms of the group $G$. Hence, the restriction of the
topological automorphism $\alpha$ to $G$ satisfies condition (ii).

The subgroup $G$ is open in $X$. This implies that if the distribution
$\mu_{1}$ has a non-zero absolutely continuous component with respect to
$m_{X}$, then $\mu_{1}$ has a non-zero absolutely continuous component with
respect to $m_{G}$. Hence, if we consider $\mu_{j}$ as distributions on the
group $G$, the distributions $\mu_{j}$ satisfy condition (iii).

Thus, we proved that conditions (i)–(iii) of the theorem are satisfied if we
consider the subgroup $G$ instead of the group $X$. Hence, we can prove the
theorem for the group $G$. In other words, we can assume that the
characteristic function $\hat{\mu}_{1}(y)$ satisfies the condition

$\\{y\in Y:\hat{\mu}_{1}(y)=1\\}=\\{0\\}.$ (11)

Let us prove that in this case $\mu_{1}=\mu_{2}=m_{X}$. Thus, the theorem will
be proved.

Substitute $u=\tilde{\alpha}y$, $v=-y$ in equation (3). In view of
$\hat{\mu}_{j}(-y)=\hat{\mu}_{j}(y)$, $j=1,2$, we obtain

$\hat{\mu}_{1}((I-\tilde{\alpha})y)=\hat{\mu}_{1}((I+\tilde{\alpha})y)\hat{\mu}_{2}(2\tilde{\alpha}y),\quad
y\in Y.$ (12)

Similarly, substituting $u=y$, $v=-y$ in equation (3), we find

$\hat{\mu}_{2}((I-\tilde{\alpha})y)=\hat{\mu}_{1}(2y)\hat{\mu}_{2}((I+\tilde{\alpha})y),\quad
y\in Y.$ (13)

In view of $I-\tilde{\alpha}\in{\rm Aut}(Y)$, it follows from (12) and (13)
that

$\hat{\mu}_{1}(y)=\hat{\mu}_{1}((I+\tilde{\alpha})(I-\tilde{\alpha})^{-1}y)\hat{\mu}_{2}(2\tilde{\alpha}(I-a)^{-1}y),\quad
y\in Y.$ (14)
$\hat{\mu}_{2}(y)=\hat{\mu}_{1}(2(I-\tilde{\alpha})^{-1}y)\hat{\mu}_{2}((I+\tilde{\alpha})(I-\tilde{\alpha})^{-1}y),\quad
y\in Y.$ (15)

Put $b=(I+\tilde{\alpha})(I-\tilde{\alpha})^{-1}$,
$c=2\tilde{\alpha}(I-\tilde{\alpha})^{-1}$, $d=2(I-\tilde{\alpha})^{-1}$. Let
us check that the group $Y$ contains no elements of order 2. On the one hand,
condition (i) implies that the character group of the factor-group $X/c_{X}$
is topologically isomorphic to a group of the form (4). Since $p>2$, it
contains no elements of order 2. On the other hand, note that the character
group of the factor-group $X/c_{X}$ is topologically isomorphic to the
annihilator $A(Y,c_{X})=b_{Y}$. Since the subgroup $b_{Y}$ consists of all
elements of finite order of the group $Y$, the group $Y$ contains no elements
of order 2. It follows from condition (ii) that $I\pm\tilde{\alpha}\in{\rm
Aut}(Y)$. This implies that $b$, $c$ and $d$ are monomorphisms. It is obvious
that the monomorphisms $b$, $c$ and $d$ commutate.

It follows from (14) and (15) that the representations

$\hat{\mu}_{1}(y)=\prod_{i=1}^{2^{n-1}}\hat{\mu}_{1}(z_{1}(n,i)y)\hat{\mu}_{2}(w_{1}(n,i)y),\quad
y\in Y,$ (16)

and

$\hat{\mu}_{2}(y)=\prod_{i=1}^{2^{n-1}}\hat{\mu}_{1}(z_{2}(n,i)y)\hat{\mu}_{2}(w_{2}(n,i)y),\quad
y\in Y,$ (17)

where

$z_{j}(n,i)=b^{k_{j}({n,i})}c^{l_{j}({n,i})}d^{m_{j}({n,i})},\quad
w_{j}(n,i)=b^{p_{j}({n,i})}c^{q_{j}({n,i})}d^{r_{j}({n,i})},$

and $k_{j}({n,i})$, $l_{j}({n,i})$, $m_{j}({n,i})$, $p_{j}({n,i})$,
$q_{j}({n,i})$, $r_{j}({n,i})$, $j=1,2,$ are nonnegative integers, holds for
any natural $n$.

Since $Y$ is a discrete group and the distribution $\mu_{1}$ has a non-zero
absolutely continuous component with respect to $m_{X}$, it follows from (7)
and (11) that

$\sup_{y\in Y,\ y\neq 0}\hat{\mu}_{1}(y)=C<1.$ (18)

Take $y\in Y$, $y\neq 0$. Since $b$, $c$ and $d$ are monomorphisms, for any
natural $n$ and $1\leq i\leq 2^{n-1}$ the endomorphisms $z_{1}(n,i)$ and
$z_{2}(n,i)$ are also monomorphisms. Hence, $z_{j}(n,i)y\neq 0$, $j=1,2$, and
therefore in view of (18), we have $\hat{\mu}_{1}(z_{j}(n,i)y)<C$,
$i=1,2,\dots,2^{n-1}$, $j=1,2$. This implies from (16) and (17) that

$\hat{\mu}_{1}(y)\leq C^{2^{n-1}},\quad\hat{\mu}_{2}(y)\leq C^{2^{n-1}},\quad
y\neq 0.$

Since $n$ is arbitrary and $C<1$, we obtain that
$\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)=0$ for all $y\neq 0$.

As a result, for the characteristic functions $\hat{\mu}_{j}(y)$ we receive
the representation

$\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)=\begin{cases}1,&\text{\ if\ }\ y=0,\\\
0,&\text{\ if\ }\ y\neq 0.\end{cases}$

It follows from (1) that $\mu_{1}=\mu_{2}=m_{X}$. $\Box$

We note that compact totally disconnected Abelian groups of the form (2) first
arose in connection with the study of the Skitovich–Darmois theorem on locally
compact Abelian groups ([12], see also [13]). It turns out that the
following statement holds. Let $X$ be a compact totally disconnected Abelian
group and let $\alpha$ be a topological automorphisms of the group $X$. Let
$\xi_{1}$ and $\xi_{2}$ be independent random variables with values in $X$ and
distributions $\mu_{1}$ and $\mu_{2}$. The independence of the linear forms
$L_{1}=\xi_{1}+\xi_{2}$ and $L_{2}=\ \xi_{1}+\alpha_{2}\xi_{2}$ implies that
$\mu_{j}$ are shifts of the Haar distribution on a compact subgroup of the
group $X$ if and only if $X$ is topologically isomorphic to a group of the
form (2).

Let us formulate two corollaries of Theorem 2.1. Suppose that $X$ is a compact
connected Abelian group. Since in this case the only compact open subgroup of
the group $X$ is the group $X$ itself, condition (i) of Theorem 2.1 holds
true. We get from Theorem 2.1 the following characterization of the Haar
distribution on the group $X$.

######  Corollary 2.7.

Let $X$ be a compact connected Abelian group, and let $\alpha$ be a
topological automorphism of the group $X$ satisfying the condition
$I\pm\alpha\in{\rm Aut}(X)$. Let $\xi_{1}$ and $\xi_{2}$ be independent random
variables with values in the group $X$ and distributions $\mu_{1}$ and
$\mu_{2}$. Assume that at least one of the distributions $\mu_{j}$ has a non-
zero absolutely continuous component with respect to $m_{X}$. If the
conditional distribution of the linear form $L_{2}=\xi_{1}+\alpha\xi_{2}$
given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric, then $\mu_{1}=\mu_{2}=m_{X}$.

We note that J. H. Stapleton proved in [22] that the Haar distribution on a
connected compact Abelian group $X$ is characterized by the independence of
$n$ linear forms of $n$ independent random variables with values in the group
$X$. In so doing, coefficients of the linear forms are integers. This result
was strengthened by I. P. Mazur in [18].

Consider the finite-dimensional torus $\mathbb{T}^{n}$. Taking into account
that $\mathbb{T}^{n}$ is a compact connected Abelian group, Theorem 2.1
implies the following statement.

######  Corollary 2.8.

Let $\alpha$ be a topological automorphism of the group $\mathbb{T}^{n}$
satisfying the condition $I\pm\alpha\in{\rm Aut}(\mathbb{T}^{n})$. Let
$\xi_{1}$ and $\xi_{2}$ be independent random variables with values in the
group $\mathbb{T}^{n}$ and having Gaussian distributions $\gamma_{1}$ and
$\gamma_{2}$. If the conditional distribution of the linear form
$L_{2}=\xi_{1}+\alpha\xi_{2}$ given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric, then
the distributions $\gamma_{j}$ are singular with respect to
$m_{\mathbb{T}^{n}}$.

Put $X^{(2)}=\\{2x:x\in X\\}$, $Y_{(2)}=\\{y\in Y:2y=0\\}$ and prove the
following statement (compare with Corollary 2.7).

######  Proposition 2.9.

Let $X$ be a compact Abelian group and let $\alpha$ be a topological
automorphism of the group $X$. Let $\xi_{1}$ and $\xi_{2}$ be independent
identically distributed random variables with values in the group $X$ and
distribution $m_{X}$. The conditional distribution of the linear form
$L_{2}=\xi_{1}+\alpha\xi_{2}$ given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric if
and only if

$X^{(2)}\subset(I-\alpha)(X).$ (19)

Proof. Assume that the conditional distribution of the linear form
$L_{2}=\xi_{1}+\alpha\xi_{2}$ given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric. By
Lemma 2.2, the characteristic function $\hat{m}_{X}(y)$ satisfies equation (3)
which takes the form

$\hat{m}_{X}(u+v)\hat{m}_{X}(u+\tilde{\alpha}v)=\hat{m}_{X}(u-v)\hat{m}_{X}(u-\tilde{\alpha}v),\
\ u,v\in Y.$ (20)

Taking into account (1), the left-hand side of (20) is equal to 1 if and only
if $u+v=0$ and $u+\tilde{\alpha}v=0$. In other words, if and only if $u=-v$
and $v\in{\rm Ker}(I-\tilde{\alpha})$. The right-hand side of (20) is equal to
1 if and only if $u-v=0$ and $u-\tilde{\alpha}v=0$. In other words, if and
only if $u=v$ and $v\in{\rm Ker}(I-\tilde{\alpha})$. Obviously, the left-hand
and right-hand sides of (20) are equal to 1 simultaneously if and only if

${\rm Ker}(I-\tilde{\alpha})\subset Y_{(2)}.$ (21)

In view of $A(X,{\rm Ker}(I-\tilde{\alpha}))=(I-\alpha)(X)$ and
$A(X,Y_{(2)})=X^{(2)}$, (21) implies (19).

Assume that (19) is true. Then (21) is fulfilled, and, as noted above, the
left-hand and right-hand sides of (20) are equal to 1 simultaneously. Thus,
the characteristic function $\hat{m}_{X}(y)$ satisfies equation (20). Hence,
by Lemma 2.2, the conditional distribution of the linear form
$L_{2}=\xi_{1}+\alpha\xi_{2}$ given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric.

Note that (19) is satisfied if $I-\alpha\in{\rm Aut}(X)$. It should also be
noted that in Proposition 2.9 there are no restrictions either on the compact
Abelian group $X$ or on the topological automorphism $\alpha$. $\hfill\Box$

##  3 Remarks on Theorem 2.1

######  Remark 3.1.

Let us check that Theorem 2.1 can not be strengthened by narrowing the class
of distributions which are characterized by the symmetry of the conditional
distribution of one linear form of independent random variables given another.
Namely, the following statement is true.

######  Proposition 3.2.

Let $X$ be a compact Abelian group, and let $K$ be a compact open subgroup of
$X$. Let $\alpha$ be a topological automorphism of the group $X$ satisfying
the condition $I\pm\alpha\in{\rm Aut}(X)$, and such that $\alpha(K)=K$. Let
$\xi_{1}$ and $\xi_{2}$ be independent identically distributed random
variables with values in the group $X$ and distribution $m_{X}$. Then the
conditional distribution of the linear form $L_{2}=\xi_{1}+\alpha\xi_{2}$
given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric.

For the proof, we need the following lemma ([5, Proposition 17.6]).

######  Lemma 3.3.

Let $X$ be a locally compact Abelian group and let $K$ be a compact subgroup
of $X$. Let $\alpha$ be a topological automorphism of the group $X$ satisfying
the condition $I\pm\alpha\in{\rm Aut}(X)$. Let $\xi_{1}$ and $\xi_{2}$ be
independent identically distributed random variables with values in the group
$X$ and distribution $m_{K}$. The conditional distribution of the linear form
$L_{2}=\xi_{1}+\alpha\xi_{2}$ given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric if
and only if

$K\subset(I+\alpha)^{-1}(I-\alpha)(K).$ (22)

Proof of Proposition 3.2. Since $X$ is a compact group, $Y$ is a discrete
group. Note that $(I+\alpha)(K)=K$. Indeed, consider the annihilator
$A(Y,K)=L$. Since $K$ is an open subgroup, $L$ is a compact subgroup. Taking
into account that $Y$ is a discrete group, $L$ is a finite subgroup. It
follows from $I+\alpha\in{\rm Aut}(X)$ that $I+\tilde{\alpha}\in{\rm Aut}(Y)$.
By the condition $\alpha(K)=K$. By Lemma 2.3, this implies that
$\tilde{\alpha}(L)=L$, and hence, $(I+\tilde{\alpha})(L)\subset L$. Since
$I+\tilde{\alpha}\in{\rm Aut}(Y)$ and $L$ is a finite subgroup, we have
$(I+\tilde{\alpha})(L)=L$, and hence, by Lemma 2.3, $(I+\alpha)(K)=K$. Arguing
similarly we obtain that $(I-\alpha)(K)=K$. Thus, $(I\pm\alpha)(K)=K$. This
implies that $K=(I+\alpha)^{-1}(I-\alpha)(K).$ Hence (22) is fulfilled. Then
by Lemma 3.3, the conditional distribution of the linear form
$L_{2}=\xi_{1}+\alpha\xi_{2}$ given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric.

Note that since $K$ is an open subgroup, the distribution $m_{K}$ is
absolutely continuous with respect to $m_{X}$.

Using Lemma 2.2, it is easy to check that if $\xi_{j}$ are independent
identically distributed random variables with values in the group $X$ and
distributions $\mu_{j}=m_{K}*E_{x_{j}}$, $j=1,2$, where $x_{j}$ are elements
of the group $X$ such that $2(x_{1}+\alpha x_{2})\in K$, then conditional
distribution of the linear form $L_{2}=\xi_{1}+\alpha\xi_{2}$ given
$L_{1}=\xi_{1}+\xi_{2}$ is symmetric. $\Box$

######  Remark 3.4.

Show that Theorem 2.1 fails if we omit one of conditions (i)–(iii).

1\. Let $X=(\mathbb{Z}(2))^{\aleph_{0}}$. Then $X$ is a nondiscrete compact
totally disconnected Abelian group. It is obvious that condition (i) is not
satisfied. Denote by $x=(x_{1},x_{2},\dots,x_{n}\dots)$, where
$x_{i}\in\mathbb{Z}(2)$, elements of the group $X$. Consider the mapping
$\alpha:X\rightarrow X$, defined by the formula

$\alpha(x_{1},x_{2},\dots,x_{n},\dots)=(x_{2},x_{1}+x_{2},x_{4},x_{3}+x_{4},\dots,x_{2n},x_{2n-1}+x_{2n},\dots).$

It is obvious that $\alpha\in{\rm Aut}(X)$. Let $\mu_{1}$ and $\mu_{2}$ be
arbitrary distributions on the group $X$. Since $y=-y$ for all $y\in Y$, the
characteristic functions $\hat{\mu}_{j}(y)$ satisfy equation (3). Let
$\xi_{1}$ and $\xi_{2}$ be independent random variables with values in the
group $X$ and distributions $\mu_{1}$ and $\mu_{2}$. By Lemma 2.2, the
conditional distribution of the linear form $L_{2}=\xi_{1}+\alpha\xi_{2}$
given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric. It is easy to see that
$I\pm\alpha\in{\rm Aut}(X)$. Thus, condition (ii) is satisfied. Obviously, we
can suppose that each of the distributions $\mu_{j}$ has a non-zero absolutely
continuous component with respect to $m_{X}$. Hence, condition (iii) is also
satisfied. This example shows that Theorem 2.1 fails if we omit condition (i).

2\. Let $X=\mathbb{T}^{2}$. Denote by $x=(z,w)$, where $z,w\in\mathbb{T}$,
elements of the group $X$. The group $Y$ is isomorphic to the group
$\mathbb{Z}^{2}$. Denote by $y=(m,n)$, where $m,n\in\mathbb{Z}$, elements of
the group $Y$. It follows from Definition 1.1 that the characteristic function
of a Gaussian distribution $\gamma$ on the two-dimensional torus
$\mathbb{T}^{2}$ is of the form

$\hat{\gamma}(y)=(x,y)\exp\\{-\langle{}{}Ay,y\rangle\\},\quad y\in Y,$

where $x\in X$ and $A=(a_{ij})_{i,j=1}^{2}$ is a symmetric positive
semidefinite matrix, $\langle.,.\rangle$ is the scalar product in
$\mathbb{R}^{2}$. Each topological automorphism $\alpha$ of the group
$\mathbb{T}^{2}$ is defined by an integer-valued matrix
$\left(\begin{matrix}a&b\\\ c&d\end{matrix}\right)$ such that $|ad-bc|=1$, and
$\alpha$ acts on $\mathbb{T}^{2}$ as following

$\alpha(z,w)=(z^{a}w^{c},z^{b}w^{d}),\quad(z,w)\in\mathbb{T}^{2}.$

The adjoint automorphism $\tilde{\alpha}\in{\rm Aut}(\mathbb{Z}^{2})$ is of
the form

$\tilde{\alpha}(m,n)=(am+bn,cm+dn),\quad(m,n)\in\mathbb{Z}^{2}.$

We will identify the automorphisms $\alpha$ and $\tilde{\alpha}$ with the
matrix $\left(\begin{matrix}a&b\\\ c&d\end{matrix}\right).$

Let $\alpha=\left(\begin{matrix}-1&1\\\ 1&-2\end{matrix}\right)\in{\rm
Aut}(X)$. Put $A_{1}=\left(\begin{matrix}1&-1\\\ -1&2\end{matrix}\right)$,
$A_{2}=\left(\begin{matrix}1&0\\\ 0&1\end{matrix}\right)$. Then $A_{j}$ are
symmetric positive semidefinite matrices. Consider Gaussian distributions
$\mu_{j}$ on the group $X$ with the characteristic functions

$\hat{\mu}_{1}(y)=\exp\\{-\langle{}{}A_{1}y,y\rangle\\},\quad\hat{\mu}_{2}(y)=\exp\\{-\langle{}{}A_{2}y,y\rangle\\},\quad
y\in Y.$

Direct verification shows that the characteristic functions $\hat{\mu}_{j}(y)$
satisfy equation (3). Let $\xi_{1}$ and $\xi_{2}$ be independent random
variables with values in the group $X$ and distributions $\mu_{1}$ and
$\mu_{2}$. By Lemma 2.2, the conditional distribution of the linear form
$L_{2}=\xi_{1}+\alpha\xi_{2}$ given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric.
Since $X$ is a connected Abelian group, condition (i) is satisfied. Since
$\det A_{j}>0$, $j=1,2$, both of the distributions $\mu_{j}$ are absolutely
continuous with respect to $m_{X}$, i.e. condition (iii) is also satisfied.
Note that $I+\alpha\in{\rm Aut}(X)$, but $I-\alpha\notin{\rm Aut}(X)$, i.e.
condition (ii) is not satisfied. This example shows that Theorem 2.1 fails if
we omit condition (ii).

3\. Let $\text{\boldmath$a$}=(a_{0},a_{1},\dots,a_{n},\dots)$ be an arbitrary
sequence of integers such that all $a_{j}>1$. Let
$\Delta_{\text{\boldmath$a$}}$ be the group of $a$-adic integers ([15,
(10.2)]). Consider the group $\mathbb{R}\times\Delta_{{\text{\boldmath$a$}}}$.
Denote by $B$ the subgroup of the group
$\mathbb{R}\times\Delta_{{\text{\boldmath$a$}}}$ of the form
$B=\\{(n,n\mathbf{u}):n\in\mathbb{Z}\\}$, where
$\mathbf{u}=(1,0,\dots,0,\dots)\in\Delta_{\text{\boldmath$a$}}$. The factor-
group
$\Sigma_{{\text{\boldmath$a$}}}=(\mathbb{R}\times\Delta_{{\text{\boldmath$a$}}})/B$
is called the $a$-adic solenoid. The group $\Sigma_{{\text{\boldmath$a$}}}$ is
compact connected and has dimension 1 ([15, (10.12), (10.13), (24.28)]). The
character group of the group $\Sigma_{{\text{\boldmath$a$}}}$ is topologically
isomorphic to the discrete group of the form

$H_{\text{\boldmath$a$}}=\left\\{{m\over a_{0}a_{1}\cdots a_{n}}:\
n=0,1,\dots;\ m\in{\mathbb{Z}}\right\\}.$

It follows from Definition 1.1 that the characteristic function of a Gaussian
distribution $\gamma$ on an $a$-adic solenoid $\Sigma_{\text{\boldmath$a$}}$
is of the form

$\hat{\gamma}(y)=(x,y)\exp\\{-\sigma y^{2}\\},\quad y\in
H_{\text{\boldmath$a$}},$

where $x\in\Sigma_{\text{\boldmath$a$}}$, $\sigma\geq 0$.

Let ${\text{\boldmath$a$}}=(2,3,\dots,n,\dots)$. Consider the corresponding
$a$-adic solenoid $X=\Sigma_{\text{\boldmath$a$}}$. Then the group $Y$ is
topologically isomorphic to the group of rational numbers $\mathbb{Q}$
considering in the discrete topology. Every topological automorphism of the
group $X$ is a multiplication by a non-zero rational number.

Since $X$ is a connected Abelian group, condition (i) is satisfied. Let
$\alpha$ be a negative rational number such that $\alpha\neq-1$. Then
multiplication by $\alpha$ is a topological automorphism of the group $X$ such
that $I\pm\alpha\in{\rm Aut}(X)$. Hence, condition (ii) is also satisfied.

Consider Gaussian distributions $\mu_{j}$ on the group $X$ with the
characteristic functions

$\hat{\mu}_{1}(y)=\exp\\{-\sigma_{1}y^{2}\\},\quad\hat{\mu}_{2}(y)=\exp\\{-\sigma_{2}y^{2}\\},\quad
y\in\mathbb{Q},$

where $\sigma_{1}+\alpha\sigma_{2}=0$. Then the characteristic functions
$\hat{\mu}_{j}(y)$ satisfy equation (3). Let $\xi_{1}$ and $\xi_{2}$ be
independent random variables with values in the group $X$ and distributions
$\mu_{1}$ and $\mu_{2}$. By Lemma 2.2, the conditional distribution of the
linear form $L_{2}=\xi_{1}+\alpha\xi_{2}$ given $L_{1}=\xi_{1}+\xi_{2}$ is
symmetric. The group $X$ is not locally connected. Then, as proven in [1], see
also [5, Proposition 3.14], each Gaussian distribution on the group $X$ is
singular with respect to $m_{X}$, i.e. condition (iii) is not satisfied. This
example shows that Theorem 2.1 fails if we omit condition (iii).

######  Remark 3.5.

Let us make the following remark about groups satisfying condition (i) of
Theorem 2.1. Obviously, these are compact connected Abelian groups, compact
totally disconnected Abelian groups that are topologically isomorphic to a
group of the form (2), and also their direct products. These groups do not
exhaust all groups satisfying condition (i) of Theorem 2.1. There exist also
other compact Abelian groups such that condition (i) of Theorem 2.1 is
satisfied for $X$ whereas the group $X$ is not topologically isomorphic to the
direct product of a totally disconnected and a connected Abelian group. To
construct an example consider the group

$L=\mathop{\mbox{\rm\bf P}^{*}}\limits_{p\in{\cal P},\ p>3}\mathbb{Z}(p).$
(23)

Since $L$ is a reduced unbounded group, there exist a mixed Abelian group such
that its torsion part is isomorphic to $L$, and which is not isomorphic to the
direct product of its torsion part and a torsion-free Abelian group. Then
there is a mixed Abelian group $Y$ such that its torsion part $M$ is
isomorphic to $L$, the factor group $Y/M$ is isomorphic to the group of
rational numbers $\mathbb{Q}$, and which is not isomorphic to the direct
product of $L$ and $\mathbb{Q}$ ([14, §54]). Obviously, the group $Y$ is
countable. Consider the group $Y$ in the discrete topology. Denote by $X$ the
character group of the group $Y$. Then $X$ is a compact Abelian group. The
character group of the factor-group $X/c_{X}$ is topologically isomorphic to
the group $b_{Y}$, which is isomorphic to $L$. For this reason the factor-
group $X/c_{X}$ is topologically isomorphic to the group

$\mathop{\mbox{\rm\bf P}}\limits_{p\in{\cal P},\ p>3}\mathbb{Z}(p),$

i.e. a group of the form (2). So, the group $X$ satisfies condition (i) of
Theorem 2.1. Since the group $Y$ is not isomorphic to the direct product of
its torsion part and a torsion free group, the group $X$ is not topologically
isomorphic to the direct product of a totally disconnected and a connected
Abelian group.

Denote by $\alpha$ multiplication on 3 in the group $X$. Since $p>3$ in (23),
it is easy to see that $\alpha\in{\rm Aut}(X)$ and $I\pm\alpha\in{\rm
Aut}(X)$. So, $\alpha$ satisfies condition (ii) of Theorem 2.1.

##  4 Generalisation of Theorem 2.1

The following statement is a generalization of Theorem 2.1 for a wide class of
locally compact Abelian groups.

######  Theorem 4.1.

Let $X=\mathbb{R}^{n}\times K\times D$, where $K$ is a nondiscrete compact
Abelian group such that the factor-group $K/c_{K}$ is topologically isomorphic
to a group of the form $(\ref{2})$, and $D$ is a discrete Abelian group
containing no elements of order $2$. Let $\alpha$ be a topological
automorphism of the group $X$ satisfying the condition $I\pm\alpha\in{\rm
Aut}(X)$. Let $\xi_{1}$ and $\xi_{2}$ be independent random variables with
values in the group $X$ and distributions $\mu_{1}$ and $\mu_{2}$ such that at
least one of the distributions $\mu_{j}$ has a non-zero absolutely continuous
component with respect to $m_{X}$.

If the conditional distribution of the linear form
$L_{2}=\xi_{1}+\alpha\xi_{2}$ given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric, then
$\mu_{j}=\gamma_{j}*m_{G}*E_{x_{j}}$, where $\gamma_{j}$ is a Gaussian
distribution in $\mathbb{R}^{n}$, $G$ is a compact open subgroup of the group
$K\times D$, $x_{j}\in K\times D$, $j=1,2$. Moreover
$\alpha(\mathbb{R}^{n}\times G)=\mathbb{R}^{n}\times G$.

To prove Theorem 4.1, we need the following lemmas.

######  Lemma 4.2.

Let $X=\mathbb{R}^{n}\times G$, where $G$ is a locally compact Abelian group.
Let $\nu=\mu*\bar{\mu}$, where $\mu\in{\rm M}^{1}(X)$. If $\nu=\gamma*m_{K}$,
where $\gamma$ is a symmetric Gaussian distribution in $\mathbb{R}^{n}$, and
$K$ is a compact subgroup of the group $G$, then $\mu=\lambda*m_{K}*E_{g}$,
where $\lambda$ a Gaussian distribution in $\mathbb{R}^{n}$, $g\in G$.

Proof. The group $Y$ is topologically isomorphic to the group
$\mathbb{R}^{n}\times H$, where $H$ is the character group of the group $G$.
Denote by $y=(s,h)$, where $s\in\mathbb{R}^{n}$, $h\in H$, elements of the
group $Y$. We can assume without loss of generality that
$\\{s\in\mathbb{R}^{n}:\hat{\gamma}(s)=1\\}=\\{0\\}$. Then, obviously, the
support of the distribution $\nu$ coincides with the subgroup
$F=\mathbb{R}^{n}\times K$ of the group $X$. It is easy to see that we can
substitute the distribution $\mu$ by its shift $\mu^{\prime}$ in such a way
that $\sigma(\mu^{\prime})\subset F$, and
$\nu=\mu^{\prime}*\bar{\mu}^{\prime}$. If we consider $\nu$, as a distribution
on $F$, then its characteristic function is equal to $1$ only at zero of the
character group of the group $F$. Therefore, we can prove the lemma assuming
that $X=\mathbb{R}^{n}\times K$ and $\\{(s,h)\in
Y:\hat{\nu}(s,h)=1\\}=\\{0\\}$. Then the characteristic function
$\hat{\nu}(s,h)$ is of the form

$\hat{\nu}(s,h)=\begin{cases}\hat{\gamma}(s),&\text{\ if\ }\
s\in\mathbb{R}^{n},\ h=0,\\\ 0,&\text{\ if\ }\ s\in\mathbb{R}^{n},\ h\neq
0.\end{cases}$

The statement of the lemma now follows from Cramér’s theorem on the
decomposition of a Gaussian distribution in $\mathbb{R}^{n}$: if $\gamma$ is a
Gaussian distribution in $\mathbb{R}^{n}$ and $\gamma=\gamma_{1}*\gamma_{2}$,
where $\gamma_{j}\in{\rm M}^{1}(\mathbb{R}^{n})$, $j=1,2$, then $\gamma_{j}$
are Gaussian distributions in $\mathbb{R}^{n}$. $\hfill\Box$

######  Lemma 4.3 ([4, Theorem 1]).

Let $X$ be a discrete Abelian group containing no elements of order $2$. Let
$\alpha$ be an automorphism of the group $X$ satisfying the condition
$I\pm\alpha\in{\rm Aut}(X)$. Assume that $\mu_{1}$ and $\mu_{2}$ are
distributions on $X$ such that the characteristic functions $\hat{\mu}_{j}(y)$
are nonnegative and satisfy equation $(\ref{1})$. Then
$\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)$ for all $y\in Y$ and $\hat{\mu}_{j}(y)$
take only the values $0$ and $1$. Moreover,
$\tilde{\alpha}(E)=(I\pm\tilde{\alpha})(E)=E$, where $E=\\{y\in
Y:\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)=1\\}$.

Taking into account Lemma 2.2, the following statement results from Theorem 1
in [8].

######  Lemma 4.4.

Let $\alpha$ be a topological automorphism of $\mathbb{R}^{n}$ satisfying the
condition $I+\alpha\in{\rm Aut}(\mathbb{R}^{n})$. Assume that $\gamma_{1}$ and
$\gamma_{2}$ are distributions on $\mathbb{R}^{n}$ such that the
characteristic functions $\hat{\gamma}_{j}(s)$ are nonnegative and satisfy
equation $(\ref{1})$. Then $\gamma_{j}$ are symmetric Gaussian distributions
in $\mathbb{R}^{n}$.

Proof of Theorem 4.1. The group $Y$ is topologically isomorphic to the group
$\mathbb{R}^{n}\times L\times C$, where $L$ is the character group of the
group $K$, and $C$ is the character group of the group $D$. The group $L$ is
discrete and the group $C$ is compact. Passing from the distributions
$\mu_{j}$ to the distributions $\nu_{j}=\mu_{j}*\bar{\mu}_{j}$ and applying
Lemma 4.2, we see that it suffices to prove the theorem, assuming that
conditions (7) hold.

Reduce the proof of the theorem to the case when $D=\\{0\\}$. Since the
character group of the factor-group $K/c_{K}$ is topologically isomorphic to
the torsion part $b_{L}$ of the group $L$, this implies that $b_{L}$ is
isomorphic to a group of the form (4). Put

$(b_{L})_{m,n}=\\{l\in b_{L}:(p_{2}p_{3}\cdots p_{n})^{m}l=0,\
m=1,2,\dots,n=2,3,\dots,\\}.$

Then $(b_{L})_{n,m}$ is a finite subgroup of the group $b_{L}$. Moreover, it
follows from (4) that

$b_{L}=\bigcup\limits_{m\geq 1,\ n\geq 2}(b_{L})_{m,n}.$

It is obvious that the each of the subgroups $(b_{L})_{m,n}$ is carried into
itself by every automorphism of the group $b_{L}$.

Denote by $(l,c)$, where $l\in b_{L}$, $c\in C$, elements of the group
$b_{L}\times C$, and consider the continuous homomorphism $\pi:b_{L}\times
C\rightarrow b_{L}$ defined by the formula $\pi(l,c)=l$, $l\in b_{L}$, $c\in
C$. Note that $\tilde{\alpha}(C)\subset b_{L}\times C$. Consider the subgroup
$\pi(\tilde{\alpha}(C))$. Since $C$ is a compact subgroup and $b_{L}$ is a
discrete subgroup, $\pi(\tilde{\alpha}(C))$ is a finite subgroup. This implies
that there is a finite subgroup $M=(b_{L})_{m,n}$ for some $m$ and $n$ such
that $\pi(\tilde{\alpha}(C))\subset M$. It is easy to see that
$\tilde{\alpha}(M)\subset M\times C$ and $\tilde{\alpha}(C)\subset M\times C$.
We have

$\tilde{\alpha}(M\times C)\subset\tilde{\alpha}(M)+\tilde{\alpha}(C)\subset
M\times C+M\times C=M\times C.$ (24)

If we suppose that the subgroup $M$ also contains the subgroup
$\pi(\tilde{\alpha}^{-1}(C))$, then, in view of $\tilde{\alpha}^{-1}(M)\subset
M\times C$ and $\tilde{\alpha}^{-1}(C)\subset M\times C$, we get

$\tilde{\alpha}^{-1}(M\times
C)\subset\tilde{\alpha}^{-1}(M)+\tilde{\alpha}^{-1}(C)\subset M\times
C+M\times C=M\times C.$ (25)

It follows from (24) and (25) that

$\tilde{\alpha}(M\times C)=M\times C.$ (26)

If we suppose that the subgroup $M$ also contains the subgroups
$\pi((I\pm\tilde{\alpha})(C))$ and $\pi((I\pm\tilde{\alpha})^{-1}(C))$, then
arguing similarly, we obtain that

$(I\pm\tilde{\alpha})(M\times C)=M\times C.$ (27)

So, we see that the restriction of the topological automorphism
$\tilde{\alpha}$ to the subgroup $M\times C$ is a topological automorphism of
the group $M\times C$ satisfying condition (27).

Observe that $A(X,M\times C)=\mathbb{R}^{n}\times A(K,M)$, and the group
$M\times C$ is topologically isomorphic to the character group of the factor-
group $X/A(X,M\times C)$. Since $X/A(X,M\times C)=(K/A(K,M))\times D$, we have
that the factor-group $X/A(X,M\times C)$ is discrete and contains no elements
of order 2. By Lemma 2.2, the symmetry of the conditional distribution of the
linear form $L_{2}$ given $L_{1}$ implies that the characteristic functions
$\hat{\mu}_{j}(y)$ satisfy equation (3). In view of (26), we can consider the
restriction of equation (3) to the subgroup $M\times C$. Taking into account
(26) and (27), it follows from Lemma 2.3 that we can apply Lemma 4.3 to the
discrete Abelian group $X/A(X,M\times C)$ and the induced automorphism
$\hat{\alpha}\in{\rm Aut}(X/A(X,M\times C))$. We obtain that
$\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)$ for all $y\in M\times C$, and
$\hat{\mu}_{j}(y)$ take only the values $0$ and $1$ on $M\times C$. Put

$E=\\{y\in M\times C:\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)=1\\},\quad N=A(K\times
D,E).$

By Lemma 4.3, $\tilde{\alpha}(E)=(I\pm\tilde{\alpha})(E)=E$. Since $E$ is a
compact subgroup of the group $M\times C$, it follows that $N$ is an open
subgroup of $K\times D$. Note that $\\{y\in M\times
C:\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)=0\\}$ is a closed set. Hence, $E$ is an
open subgroup of $M\times C$. Inasmuch the subgroup $N$ is the annihilator of
$E$, this implies that $N$ is a compact subgroup of $K\times D$. So, $N$ is a
compact open subgroup of the group $K\times D$. In view of $c_{N}=c_{K}$, it
is easy to see that the factor-group $N/c_{N}$ is topologically isomorphic to
a group of the form $(\ref{2})$. Hence, condition (i) of the theorem is
satisfied for the group $\mathbb{R}^{n}\times N$.

By Lemma 2.6, $\sigma(\mu_{j})\subset A(X,E)=\mathbb{R}^{n}\times N$, $j=1,2$.
By Lemma 2.3, it follows from $\tilde{\alpha}(E)=E$ and
$(I\pm\tilde{\alpha})(E)=E$ that $\alpha(\mathbb{R}^{n}\times
N)=\mathbb{R}^{n}\times N$ and

$(I\pm\alpha)(\mathbb{R}^{n}\times N)=\mathbb{R}^{n}\times N.$ (28)

It means that the restriction of $\alpha$ to the subgroup
$\mathbb{R}^{n}\times N$ is a topological automorphism of the group
$\mathbb{R}^{n}\times N$, and in view of (28), this restriction satisfies
condition (ii) of the theorem.

The subgroup $\mathbb{R}^{n}\times N$ is open in $X$. This implies that if the
distribution $\mu_{j}$ has a non-zero absolutely continuous component with
respect to $m_{X}$, then $\mu_{j}$ has a non-zero absolutely continuous
component with respect to $m_{\mathbb{R}^{n}\times N}$. Hence, the
distributions $\mu_{j}$ satisfy condition (iii) of the theorem, if we consider
$\mu_{j}$ as distributions on the group $\mathbb{R}^{n}\times N$. The above
reasoning shows that we can prove the theorem, assuming that $D=\\{0\\}$.

We will follow now the scheme of the proof of Theorem 2.1. Assume for
definiteness that the distributions $\mu_{1}$ has a non-zero absolutely
continuous component with respect to $m_{X}$. The reduction of the proof of
the theorem to the case when condition (11) holds, is carried out in the same
way as in Theorem 2.1. In so doing, we note that $c_{X}=\mathbb{R}^{n}\times
c_{K}$, and for this reason the factor-group $X/c_{X}$ is topologically
isomorphic to a group of the form (2), and the group $b_{Y}$ consists of all
elements of finite order of the group $Y$.

Thus, we can prove the theorem assuming that $X=\mathbb{R}^{n}\times K$,
conditions (7) and (11) are fulfilled, and we will prove that
$\mu_{j}=\gamma_{j}*m_{K}$, $j=1,2$, where $\gamma_{j}$ are symmetric Gaussian
distributions in $\mathbb{R}^{n}$.

Denote by $y=(s,l)$, where $s\in\mathbb{R}^{n}$, $l\in L$, elements of the
group $Y$. By Lemma 2.2, the symmetry of the conditional distribution of the
linear form $L_{2}$ given $L_{1}$ implies that the characteristic functions
$\hat{\mu}_{j}(s,l)$ satisfy equation (3). Since $\mathbb{R}^{n}$ is the
connected component of zero of the group $Y$, we have
$\tilde{\alpha}(\mathbb{R}^{n})=\mathbb{R}^{n}$, i.e. the restriction of
$\tilde{\alpha}$ to $\mathbb{R}^{n}$ is a topological automorphism of
$\mathbb{R}^{n}$. Denote by $\tilde{\alpha}_{\mathbb{R}^{n}}$ this
restriction. Then the restriction of equation (3) to $\mathbb{R}^{n}$ is of
the form

$\hat{\mu}_{1}(s_{1}+s_{2},0)\hat{\mu}_{2}(s_{1}+\tilde{\alpha}_{\mathbb{R}^{n}}s_{2},0)=\hat{\mu}_{1}(s_{1}-s_{2},0)\hat{\mu}_{2}(s_{1}-\tilde{\alpha}_{\mathbb{R}^{n}}s_{2},0),\quad
s_{1},s_{2}\in\mathbb{R}^{n}.$ (29)

It follows from $I+\tilde{\alpha}\in{\rm Aut}(Y)$ that
$I+\tilde{\alpha}_{\mathbb{R}^{n}}\in{\rm Aut}(\mathbb{R}^{n})$. By Lemma 4.4,
we obtain from (29) that

$\hat{\mu}_{j}(s,0)=\hat{\gamma}_{j}(s),\quad s\in\mathbb{R}^{n},\quad j=1,2,$
(30)

where $\gamma_{j}$ are symmetric Gaussian distributions in $\mathbb{R}^{n}$.

Next, arguing as in the proof of Theorem 2.1, we get (16) and (17). Note that
generally speaking, (18) is not fulfilled. But the fact that the distributions
$\mu_{1}$ has a non-zero absolutely continuous component with respect to
$m_{X}$, (7) and (11) imply that

$\sup_{(s,l)\in Y,\ l\neq 0}\hat{\mu}_{1}(s,l)=C<1.$ (31)

Take $y=(s,l)\in Y$ such that $l\neq 0$. It follows from (16), (17) and (31)
that

$\hat{\mu}_{1}(s,l)\leq C^{2^{n-1}},\quad\hat{\mu}_{2}(s,l)\leq
C^{2^{n-1}},\quad s\in\mathbb{R}^{n},\ l\neq 0.$

Hence, $\hat{\mu}_{1}(s,l)=\hat{\mu}_{2}(s,l)=0$ for all $s\in\mathbb{R}^{n}$,
$l\neq 0$.

In view of (30), as a result we obtain for the characteristic functions
$\hat{\mu}_{j}(s,l)$ the following representation

$\hat{\mu}_{1}(s,l)=\begin{cases}\hat{\gamma}_{1}(s),&\text{\ if\ }\
s\in\mathbb{R}^{n},\ l=0,\\\ 0,&\text{\ if\ }\ s\in\mathbb{R}^{n},\ l\neq
0,\end{cases}\quad\quad\hat{\mu}_{2}(s,l)=\begin{cases}\hat{\gamma}_{2}(s),&\text{\
if\ }\ s\in\mathbb{R}^{n},\ l=0,\\\ 0,&\text{\ if\ }\ s\in\mathbb{R}^{n},\
l\neq 0.\end{cases}$

Taking into account (1), this implies that $\mu_{j}={\gamma_{j}*m_{K}}$,
$j=1,2$. $\Box$

######  Remark 4.5.

Theorem 4.1 is valid not only for locally compact Abelian groups
$X=\mathbb{R}^{n}\times K\times D$, where $K$ is a nondiscrete compact Abelian
group such that the factor-group $K/c_{K}$ is topologically isomorphic to a
group of the form $(\ref{2})$, and $D$ is a discrete Abelian group containing
no elements of order $2$. Following Proposition 3.4 in [10], we will give an
example of a nondiscrete locally compact totally disconnected Abelian group
$X$ such that $X$ is not topologically to a topological direct product of a
nondiscrete compact and a discrete Abelian group, whereas Theorem 4.1 holds
for $X$.

Put

$T=\mathop{\mbox{\rm\bf P}}\limits_{n=3}^{\infty}{\mathbb{Z}}(p_{n}^{2}).$

We consider $T$ only as algebraical group. Put
$K_{n}=\\{0,p_{n},2p_{n},\dots,(p_{n}-1)p_{n}\\}\subset{\mathbb{Z}}(p^{2}_{n})$.
It is obvious that $K_{n}$ is a subgroup, and $K_{n}$ is isomorphic to
${\mathbb{Z}}(p_{n})$. Denote by $Y$ a subgroup of $T$ consisting of all
elements $y=(y_{3},\dots,y_{n},\dots)$, where
$y_{n}\in{\mathbb{Z}}(p^{2}_{n})$, such that $y_{n}\notin K_{n}$ only for a
finite number indexes $n$. Put

$F=\mathop{\mbox{\rm\bf P}}\limits_{n=3}^{\infty}K_{n}.$

Then $F$ is a subgroup of $Y$. Consider the group $F$ in the product topology.
Then $F$ is a compact group. Define a topology on $Y$ in such a way that $F$
is an open subgroup of $Y$. Then $Y$ is a second countable locally compact
Abelian group. It is obvious that $Y$ is a nondiscrete locally compact totally
disconnected Abelian group. Denote by $X$ the character group of the group
$Y$. Then $X$ is also a nondiscrete locally compact totally disconnected
Abelian group. The standard reasoning show that the group $X$ is not
topologically isomorphic to a topological direct product of a nondiscrete
compact and a discrete Abelian group.

Let $\alpha$ be a topological automorphism of the group $X$ satisfying the
condition $I\pm\alpha\in{\rm Aut}(X)$. Let $\xi_{1}$ and $\xi_{2}$ be
independent random variables with values in the group $X$ and distributions
$\mu_{1}$ and $\mu_{2}$ such that at least one of the distributions $\mu_{j}$
has a non-zero absolutely continuous component with respect to $m_{X}$. We
will prove that if the conditional distribution of the linear form
$L_{2}=\xi_{1}+\alpha\xi_{2}$ given $L_{1}=\xi_{1}+\xi_{2}$ is symmetric, then
$\mu_{j}=m_{K}*E_{x_{j}}$, where $K$ is a compact open subgroup of $X$,
$x_{j}\in X$, $j=1,2$. Moreover $\alpha(K)=K$.

By Lemma 2.2, the symmetry of the conditional distribution of the linear form
$L_{2}$ given $L_{1}$ implies that the characteristic functions
$\hat{\mu}_{j}(y)$ satisfy equation (3). Passing from the distributions
$\mu_{j}$ to the distributions $\nu_{j}=\mu_{j}*\bar{\mu}_{j}$ and reasoning
as in the proof of Theorem 2.1, we can assume that conditions (7) hold. It
follows from $I\pm\alpha\in{\rm Aut}(X)$ that $I\pm\tilde{\alpha}\in{\rm
Aut}(Y)$. It is easy to see that the subgroup $F$ is carried into itself by
every topological automorphism of the group $Y$. For this reason

$\tilde{\alpha}(F)=F$ (32)

and

$(I\pm\tilde{\alpha})(F)=F.$ (33)

Since $F$ is a compact subgroup, its annihilator $A(X,F)$ is an open subgroup
of the group $X$, and $X/A(X,F)$ is a discrete group. The group $F$ is
topologically isomorphic to the character group of the factor-group
$X/A(X,F)$. In view of (32), we can consider the restriction of equation (3)
to the subgroup $F$. It is obvious that the group $X/A(X,F)$ contains no
elements of order 2. Taking into account (32) and (33), it follows from Lemma
2.3 that we can apply Lemma 4.3 to the discrete Abelian group $X/A(X,F)$ and
the induced automorphism $\hat{\alpha}\in{\rm Aut}(X/A(X,F))$. By Lemma 4.3,
$\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)$ for all $y\in F$, and $\hat{\mu}_{j}(y)$
take only the values $0$ and $1$ on $F$. Put

$E=\\{y\in F:\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)=1\\},\quad G=A(X,E).$

Note that $\\{y\in F:\hat{\mu}_{1}(y)=\hat{\mu}_{2}(y)=0\\}$ is a closed set.
Hence, $E$ is an open subgroup of $F$. Since $E$ is a compact open subgroup,
it results that $G$ is also a compact open subgroup. The character group of
the group $G$ is topologically isomorphic to the factor-group $Y/E$. Since $E$
is a compact open subgroup of $Y$, this implies that the factor-group $Y/E$ is
discrete and topologically isomorphic to a group of the form (4). Hence the
group $G$ is topologically isomorphic to a group of the form (2).

By Lemma 4.3, $\tilde{\alpha}(E)=(I\pm\tilde{\alpha})(E)=E$. Hence, by Lemma
2.3, $\alpha(G)=(I\pm\alpha)(G)=G$. It means, in particular, that the
restriction of the topological automorphism $\alpha$ to $G$ is a topological
automorphism of the group $G$. Denote by $\alpha_{G}$ this restriction. It
follows from $(I\pm\alpha)(G)=G$ that $I\pm\alpha_{G}\in{\rm Aut}(G)$. By
Lemma 2.6, $\sigma(\mu_{j})\subset G$, $j=1,2$. Since $G$ is an open subgroup
of $X$, it results that at least one of the distributions $\mu_{j}$ has a non-
zero absolutely continuous component with respect to $m_{G}$. It follows from
the above that if we consider $\xi_{j}$ as independent random variables with
values in the group $G$, then conditions (ii) and (iii) of Theorem 2.1 are
satisfied. By Theorem 2.1, $\mu_{1}=\mu_{2}=m_{K}$, where $K$ is a compact
open subgroup of the group $G$, and $\alpha(K)=K$. Obviously, $K$ is a compact
open subgroup of the group $X$.

Acknowledgements

This article was written during my stay at the Department of Mathematics
University of Toronto as a Visiting Professor. I am very grateful to Ilia
Binder for his invitation and support.

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Department of Mathematics University of Toronto
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