Abstract Generated abstract
This paper studies invariant indecomposable measures for continuous one-parameter dynamical systems on compact phase spaces and relates them to indecomposable measures for individual time maps. It proves that, for any nonzero time, an indecomposable invariant measure of the flow can be represented as the average over one period of an indecomposable invariant measure for the corresponding homeomorphism, and describes the ambiguity in this representation. The paper further characterizes when such a measure becomes decomposable for a time map in terms of eigenfrequencies of the flow of the form 2k pi divided by the chosen time. A final result connects measurable eigenfunctions of a single time map with eigenfunctions of the full flow, except in the case where the flow has one of these resonant eigenfrequencies.
Full Text
M. I. GRABAR
INDECOMPOSABLE MEASURES IN DYNAMICAL SYSTEMS
(Presented by Academician A. N. Kolmogorov on 17 VIII 1962)
Let us consider a dynamical system with compact phase space, i.e., a continuous one-parameter group \(\{S_t\}\) of homeomorphisms of a compact set \(R\) onto itself. As is known, a normalized measure invariant for the system \(\{S_t\}\) is called indecomposable if, for every measurable set invariant with respect to the system, this measure is equal either to \(0\) or to \(1\). The indecomposability of a measure with respect to a single homeomorphism \(S_t\) is defined analogously.
The following theorem establishes a connection between measures invariant and indecomposable for the whole system \(\{S_t\}\) and for its individual homeomorphisms. All measures are assumed to be normalized.
Theorem 1. \(1^\circ\). For any fixed \(t_0 \ne 0\) and any measure \(m\), invariant and indecomposable for \(\{S_t\}\), there exists a measure \(\mu\), invariant and indecomposable for \(S_{t_0}\), such that
\[ m(A)=\frac{1}{t_0}\int_0^{t_0}\mu(S_tA)\,dt \tag{1} \]
for any Borel set \(A \subset R\).
\(2^\circ\). The totality of all measures \(\mu\) for which the representation (1) holds for a given measure \(m\) is given by the formula
\[ \mu_t(A)=\mu(S_tA), \tag{2} \]
where \(\mu\) is any one of such measures.
\(3^\circ\). Every measure \(\mu\), invariant and indecomposable for \(S_{t_0}\), determines by formula (1) some measure \(m\), invariant and indecomposable for the system \(\{S_t\}\).
Proof. We first prove assertion \(3^\circ\). The fact that formula (1) defines some measure in the space \(R\) is obvious. Further:
\[ m(S_\tau A)=\frac{1}{t_0}\int_0^{t_0}\mu(S_tS_\tau A)\,dt =\frac{1}{t_0}\int_0^{t_0}\mu(S_{t+\tau}A)\,dt= \]
\[ =\frac{1}{t_0}\int_\tau^{t_0+\tau}\mu(S_tA)\,dt =\frac{1}{t_0}\int_0^{t_0}\mu(S_tA)\,dt =m(A), \]
since the function \(\mu(S_tA)\) has period \(t_0\) by virtue of the invariance of the measure \(\mu\) with respect to \(S_{t_0}\). This proves the invariance of the measure \(m\) with respect to the system \(\{S_t\}\). If now \(A\) is a set invariant for \(\{S_t\}\), then \(S_tA=A\) for every \(t\), and consequently,
\[ m(A)=\frac{1}{t_0}\int_0^{t_0}\mu(A)\,dt=\mu(A)= \left\{ \begin{array}{l} 0\\ 1 \end{array} \right\}, \]
for \(\mu\) is indecomposable.
We pass to assertion \(1^\circ\). Let \(m\) be any indecomposable measure in the system \(\{S_t\}\), and let \(E\) be its ergodic set \((^1)\). The set \(E\), being invariant for the system, is also invariant for \(S_{t_0}\). The same is true for the measure \(m\). Therefore the measure \(m\) is the limit of some sequence of linear combinations of measures indecomposable for \(S_{t_0}\). Consequently, there is a measure \(\mu\), indecomposable for \(S_{t_0}\), such that \(\mu(E)=1\). Then the measure \(m'\), defined from the measure \(\mu\) by formula (1), will coincide with the measure \(m\) on the set \(E\), since, obviously, \(m'(E)=1\), and an ergodic set determines only one indecomposable measure. \(1^\circ\) is proved.
Finally, suppose that
\[ m(A)=\frac{1}{t_0}\int_0^{t_0}\mu_1(S_tA)\,dt =\frac{1}{t_0}\int_0^{t_0}\mu_2(S_tA)\,dt, \tag{3} \]
where \(\mu_1,\mu_2\) are measures indecomposable for \(S_{t_0}\). Let \(\mathscr E_1\) and \(\mathscr E_2\) be the ergodic sets corresponding to these measures. Put \(E_1=\bigcup_t S_t(\mathscr E_1)\), \(E_2=\bigcup_t S_t(\mathscr E_2)\). Since \(\mu_1(\mathscr E_1)=1\) and \(\mu_2(\mathscr E_2)=1\), it follows, by (3), that \(m(E_1)=m(E_2)=1\), and consequently \(E_1\cap E_2\ne 0\). If \(x\in E_1\cap E_2\), then \(x=S_{t_1}x_1=S_{t_2}x_2\), where \(x_1\in\mathscr E_1\) and \(x_2\in\mathscr E_2\). Hence \(x_2=S_{t_1-t_2}x_1\), and this means that \(\mathscr E_2=S_t(\mathscr E_1)\) and, consequently, \(\mu_1(A)=\mu_2(S_tA)\), where \(t=t_1-t_2\). \(2^\circ\) is proved.
As is known \((^2)\), if a measure \(m\), indecomposable and invariant with respect to the system \(\{S_t\}\), is decomposable with respect to some \(S_{t_0}\), where \(t_0\ne0\), then the system \(\{S_t\}\) has at least one nonzero eigenfrequency. The following theorem refines this assertion.
Theorem 2. Let \(t_0\ne0\) be fixed, and let \(m\) be any measure invariant and indecomposable for the system \(\{S_t\}\). If the measure \(m\) is decomposable for \(S_{t_0}\), then there is an integer \(k\ne0\) such that \(\lambda=2k\pi/t_0\) is an eigenfrequency of the system \(\{S_t\}\). If, on the other hand, the measure \(m\) is indecomposable for \(S_{t_0}\), then for \(k\ne0\) there are no numbers of the form \(2k\pi/t_0\) among the eigenfrequencies of the system \(\{S_t\}\).
Proof. If the measure \(m\) is decomposable for \(S_{t_0}\), then any measure \(\mu\), invariant and indecomposable for \(S_{t_0}\) and connected with \(m\) by formula (1), will be distinct from \(m\). The latter means that the measure \(\mu\) cannot be invariant for all \(S_t\). Denote by \(I(\mu)\) the set of all numbers \(t\) such that \(\mu\) is invariant with respect to \(S_t\). \(I(\mu)\) is a closed subgroup of the group of all real numbers.
The group property is obvious. Let us prove closedness. If \(t_n\in I(\mu)\) and \(t_n\to t\), then for any continuous function \(f(x)\)
\[ \int_R f(x)\,d\mu = \int_R f(S_{t_n}x)\,d\mu \to \int_R f(S_tx)\,d\mu = \int_R f(x)\,d\mu . \]
Consequently, \(t\in I(\mu)\).
Since, as was already noted, \(I(\mu)\) cannot contain all real numbers, \(I(\mu)\) is a set of numbers of the form \(n\tau_0\), where \(\tau_0\ne0\) and \(n\) is an integer. In particular, there is an integer \(k\ne0\) such that \(t_0=k\tau_0\). This means that \(S_{t_0}=(S_{\tau_0})^k\), and consequently the measure \(\mu\) will be indecomposable also for \(S_{\tau_0}\). Let \(\mathscr E\) be the ergodic set of the homeomorphism \(S_{\tau_0}\) corresponding to this measure. Put \(E=\bigcup_t S_t(\mathscr E)\). Then, by (1), \(m(E)=1\), since \(\mu(\mathscr E)=1\). We now prove that if \(S_t(\mathscr E)\cap\mathscr E\ne0\), then \(t=n\tau_0\), where \(n\) is an integer. Let \(x_1\in S_t(\mathscr E)\cap\mathscr E\). Then \(x_1=S_tx_0\), where \(x_0\in\mathscr E\). Since the individual measures of the points \(x_0\) and \(x_1\) coincide with the measure \(\mu\), for any continuous-
continuous function \(f(x)\)
\[ \int_R f(S_t x)\,d\mu = \lim_{T\to\infty}\frac{1}{T}\int_0^T f(S_t S_\tau x_0)\,d\tau = \lim_{T\to\infty}\frac{1}{T_0}\int_0^T f(S_\tau S_t x_0)\,d\tau = \]
\[ = \lim_{T\to\infty}\frac{1}{T}\int_0^T f(S_\tau x_1)\,d\tau = \int_R f(x)\,d\mu . \]
This means that \(t\in I(\mu)\) and, consequently, \(t=n\tau_0\). Let now \(x\in E\). Then, if \(x=S_{t_1}x_1=S_{t_2}x_2\), where \(x_1,x_2\in\mathcal E\), then \(x_2=S_{t_1-t_2}x_1\) and, consequently, \(t_1-t_2=n\tau_0\). Therefore there exists a number \(\alpha(x)\), least in absolute value, such that \(x=S_{\alpha(x)}\tilde x\), where \(\tilde x\in\mathcal E\). It is easy to verify that
\[ \alpha(S_t x)=\alpha(x)+t+n\tau_0, \tag{4} \]
where \(n\) is an integer. Finally, put, for \(x\in E\),
\[ \varphi(x)=e^{\,i\frac{2\pi\alpha(x)}{\tau_0}} . \]
Then, by virtue of (4),
\[ \varphi(S_t x)=e^{i\lambda t}\varphi(x), \qquad \text{where } \lambda=\frac{2\pi}{\tau_0}=\frac{2k\pi}{t_0}. \]
Thus, \(\varphi(x)\) is an eigenfunction of the system \(\{S_t\}\) with eigenfrequency \(\lambda=2k\pi/t_0\).
Let now the measure \(m\) be indecomposable for \(S_{t_0}\). Suppose that there exists an eigenfunction \(\varphi(x)\) of the system \(\{S_t\}\) with eigenfrequency \(\lambda=2k\pi/t_0\), where \(k\ne 0\) is an integer. Put \(\tau_0=t_0/k\). Then \(S_{t_0}=(S_{\tau_0})^k\) and, consequently, \(m\) will also be indecomposable for \(S_{\tau_0}\). Since, obviously, \(\varphi(S_{\tau_0}x)=\varphi(x)\), it follows that \(\varphi(x)\) is constant almost everywhere (with respect to the measure \(m\)), which is possible only when \(k=0\). Theorem 2 is proved.
The following theorem is also adjacent to Theorem 2.
Theorem 3. If, for some \(t_0\ne 0\), the homeomorphism \(S_{t_0}\) has a nonconstant and measurable \((m)\) eigenfunction with eigenfrequency \(\alpha\), then either this function is also an eigenfunction for the system \(\{S_t\}\), with frequency \(\lambda=\alpha+2k\pi/t_0\), where \(k\) is an integer, or the system \(\{S_t\}\) will have an eigenfrequency \(\lambda=2k\pi/t_0\) with an integer \(k\ne 0\).
Proof. Let \(\varphi(x)\) be an eigenfunction of \(S_{t_0}\) with frequency \(\alpha\). Then almost everywhere with respect to the measure \(m\)
\[ \varphi(S_{t_0}x)=e^{i\alpha t_0}\varphi(x). \tag{5} \]
Assume first that the measure \(m\) is indecomposable for \(S_{t_0}\). Since it follows from (5) that \(|\varphi(S_{t_0}x)|=|\varphi(x)|\), i.e. \(|\varphi(x)|\) is a function invariant with respect to \(S_{t_0}\), almost everywhere \(|\varphi(x)|=\mathrm{const}\), and one may assume that \(|\varphi(x)|=1\). Further one may assume that (5) holds for all \(x\in R\). Then, for arbitrary \(t\),
\[ \varphi(S_t S_{t_0}x)=\varphi(S_{t_0}S_t x)=e^{i\alpha t_0}\varphi(S_t x). \tag{6} \]
From (5) and (6) it follows that the function \(\varphi(S_t x)/\varphi(x)\) is invariant with respect to \(S_{t_0}\), and therefore, for every \(t\), almost everywhere
\[ \varphi(S_t x)=z(t)\varphi(x), \tag{7} \]
where \(z(t)\) does not depend on \(x\). From (7) it is easy to obtain that for any \(t_1\) and \(t_2\)
\(z(t_1+t_2)=z(t_1)z(t_2)\). Let us note further that equality (7) in the space \(L_m^2(R)\) is equivalent to
\[ U_t\varphi=z(t)\varphi, \tag{8} \]
where \(U_t\) is the unitary operator corresponding to \(S_t\). It follows from (8) that \((U_t\varphi,\varphi)=z(t)(\varphi,\varphi)\), and, consequently, \(z(t)\) is a continuous function of \(t\). Finally, since \(|\varphi(x)|=1\), we have \(|z(t)|=1\), and therefore \(z(t)=e^{i\lambda t}\), where \(\lambda\) is a real number. Thus (7) takes the form
\[ \varphi(S_t x)=e^{i\lambda t}\varphi(x), \tag{9} \]
and comparison of (9) and (5) shows that \(\lambda=\alpha+2k\pi/t_0\). The first assertion of the theorem is proved. As for the second, it follows directly from Theorem 2 if the measure \(m\) is decomposable for \(S_{t_0}\).
Remark. If the system \(\{S_t\}\) is strictly ergodic, i.e., admits a unique invariant normalized measure, then Theorem 1 of Note \(^3\) follows in an obvious way from Theorem 2.
Received
17 VIII 1962
REFERENCES
\(^1\) V. V. Nemytskii, V. V. Stepanov, Qualitative Theory of Differential Equations, Moscow—Leningrad, 1949, pp. 529—531. \(^2\) V. A. Rokhlin, Uspekhi Mat. Nauk, 4, no. 2 (30) (1949). \(^3\) M. I. Grabar’, Dokl. Akad. Nauk SSSR, 95, no. 1 (1954).