# Isomophism between two irrational rotations

Define two dynamical systems $([0,1), \mathbb{B}_1, \mathbb{L}, T_1(x)=x+\alpha_1\mod 1)$ and $([0,1), \mathbb{B}_2, \mathbb{L}, T_2(x)=x+\alpha_2\mod 1)$, where $\alpha_1,\alpha_2$ are two irrational numbers, $\mathbb{L}$ is the Lebesgue measure and $\mathbb{B}_i$ are the sigma algebras. Now if these two systems are isomorphic, then what the relationship between $\alpha_1$ and $\alpha_2$? I do know that the irrational rotation is isomorphic to the Sturmian orbit dynamical systems.

• $\alpha_1=\pm\alpha_2$: look at eigenfunctions. – Anthony Quas Sep 22 '14 at 16:19
• Doug/Anthony, I suppose the criterion for Kakutani equivalence of the rotations is that $\alpha_1$ and $\alpha_2$ be in the same orbit under the action of SL(2,Z) (as fractional linear transformations)? – David Handelman Sep 22 '14 at 22:41
• @David No, unless you put restrictions on the inducing set, any two irrational rotations are Kakutani equivalent. If you confine yourself to inducing on intervals, then I believe something like being in the same $SL(2,\mathbb{Z})$ orbit is true. – Douglas Lind Sep 23 '14 at 15:00

The first major classification result in ergodic theory was found by Halmos and von Neumann in 1942 (Operator methods in classical mechanics II, Ann. of  Math. (2) 43 (1942), 332–350.). They showed that for ergodic rotations of compact abelian groups, spacial isomorphism (i.e. measure-theoretic isomorphism) is the same as spectral isomorphism (i.e. unitary equivalence of the associated operators). For irrational rotations $\alpha_1$ and $\alpha_2$ of the circle, this amounts saying that the subgroups generated by $\alpha_1$ and $\alpha_2$ are equal. From this it is easy to deduce that $\alpha_1 = \pm \alpha_2$. This is simply an expanded version of Anthony's initial answer.
• By the famous Discrete spectrum theorem two dynamical systems with discrete spectrum are measure theoretically isomorphic iff they have the same group of eigenvalues. Luckily, the irrational rotation has discrete spectrum. The point spectrum of $T_1$ is $\{\alpha_1^k:k\in \mathbb{Z}\}$, similarly, the point spectrum of $T_2$ is $\{\alpha_2^k:k\in \mathbb{Z}\}$. Hence if $T_1$ and $T_2$ are isomorphic if and only if these two group are the same. Hence if and only if $\alpha_1=\alpha_2$. But I cannot understand why $\alpha_2=-\alpha_1$. – Ben Ben Sep 23 '14 at 11:22
• It looks like you're confusing the additive circle with the multiplicative circle; let's use the additive version, since that is how your original question is phrased. The second to last sentence in your comment isn't correct: both $\alpha$ and $-\alpha$ generate the same subgroup. As Anthony points out, the map $t\mapsto -t$ conjugates rotation by $\alpha$ to rotation by $-\alpha$. – Douglas Lind Sep 23 '14 at 14:36
• I'm not sure this answers the question. I was assuming measure-theoretically isomorphic. Even if you want topologically conjugate, +$\alpha$ and $-\alpha$ are conjugate by flipping [ wiki might have screwed up on this one ]. – Anthony Quas Sep 22 '14 at 17:28
• @AnthonyQuas Don't $\pm \alpha$ give the same rotation number? As for the notion of isomorphism, I have no idea what the OP meant, but the "usual" dynamics definition was the first on my list of what he might have meant... – Igor Rivin Sep 22 '14 at 17:31
• No. $+\alpha$ gives the rotation number $\alpha$; and $-\alpha$ gives the rotation number $-\alpha$. Also the fact that the OP mentions measures and $\sigma$-algs and uses the word isomorphic suggests to me that we're talking about measure-theoretic isomorphism. – Anthony Quas Sep 22 '14 at 17:33