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First the eigenvalues of Hecke operators are real. Here is the needed Petersson inner product argument: unfold the finite coset sum defining $T_n$ into its correspondence domain, then change variables by each matrix. The determinant-normalized slash action preserves $y^{k}dxdy/y^2$ in the paired integrand, and transfers a representative to its inverse. Multiplying that inverse by the positive scalar $n$ gives its adjugate without changing the slash action. Adjugation reverses the integer determinant-$n$ correspondence, so the resulting sum is again $T_n$. Consequently $⟨T_{n}f,g⟩=⟨f,T_{n}g⟩$. Taking $g=f$ proves reality of $\lambda (n)$, since $⟨f,f⟩>0$.

Fix one prime $p$ and set $x=\lambda (p)/p^{(k-1)/2}$ and $b_e=\lambda (p^e)/p^{e(k-1)/2}$. The Hecke multiplication relations proved in (a) give $b_0=1$, $b_1=x$ and $b_{e+2}=x{b}_{e+1}-b_e$. The distinct-root assumption excludes $x=\pm 2$.

If $|x|<2$, write $x=2\cos \theta$, $0<\theta <\pi$. Solving the linear recurrence gives $b_e=\sin ((e+1)\theta )/\sin \theta$. If $\theta /(2\pi )$ is rational, infinitely many $e$ satisfy $(e+1)\theta \equiv \theta (\mathrm{mod}2\pi )$, hence $b_e=1$. If it is irrational, density of the corresponding irrational rotation gives infinitely many $e$ in the open arc where $\sin ((e+1)\theta )>\sin \theta /2$.

If $|x|>2$, let $\beta ,{\beta }^{-1}$ be the real roots of $X^2-xX+1$, choosing $|\beta |>1$. Then $b_e=({\beta }^{e+1}-{\beta }^{-e-1})/(\beta -{\beta }^{-1})$. Its even-index terms are positive and tend to infinity, whether $\beta$ is positive or negative. Thus the prime-power coefficients of a level-one eigenform are often large and positive, and in either case

Take the separate oscillator energiesThey depend on disjoint canonical coordinate pairs, soThus they are first integrals in involution. Their differentialsare nonzero precisely when $(x_i,y_i)\ne (0,0)$ and have disjoint supports. Hence they are functionally independent on the specified open set.

To understand all further global integrals of motion, use the actionsDirect differentiation gives $d{x}_i\wedge d{y}_i=d{\theta }_i\wedge d{R}_i$, so these are action-angle variables withWhen ${\alpha }_2/{\alpha }_1$ is irrational, every orbit is dense in its fixed-action two-dimensional flat torus. One elementary proof samples the orbit at times $2\pi j/{\alpha }_1$: the first angle returns, and the second undergoes an irrational rotation. These samples are dense in the second circle. Allowing a fixed additional time supplies any desired first angle, proving density in the whole flat torus.

A continuous, globally defined integral of motion must have the same value on an orbit and its closure. Therefore it is constant on every fixed-action flat torus. A $C^1$ such integral has the form $F=\varphi (R_1,R_2)$, with $\varphi$ $C^1$, by evaluating $F$ at a fixed choice of angles. HenceIt cannot be independent of $E_1,E_2$. This proves there are at most two globally independent differentiable integrals. If a third global integral were independent at a point on an excluded coordinate plane, independence would persist in a neighbourhood and hence at nearby points of the regular open set, again a contradiction.

The word global matters: on an angular chart, ${\alpha }_2{\theta }_1-{\alpha }_1{\theta }_2$ is a local integral of motion independent of the actions. It is not single-valued on the flat torus when the frequency ratio is irrational. Thus the assertion would be false for arbitrary local, rather than global, integrals.

Fix a real angle fraction $\alpha$, and repeatedly multiply by $g=e^{2\pi i\alpha }$ in the circle group. The generated cyclic subgroup is $H_{\alpha }=\{g^k:k\in \mathbb{Z}\}$. Equivalently, the angle fractions are the fractional parts of $k\alpha$, with one full turn identified with zero.

If $\alpha =p/q$ is a rational number in lowest terms, with $q>0$ and $p,q$ coprime, then $g^q=1$ and no smaller positive power is $1$. Hence $H_{\alpha }$ is a finite cyclic group of order $q$, consisting of equally spaced points. For $\alpha =2/7$, the successive angle fractions, starting at $k=0$, areThe subgroup is finite, so it is a closed subgroup of the circle group; it is also a discrete subgroup. It is not an open subgroup, since every ambient neighbourhood of any of its points contains an arc, rather than only these seven points.

If $\alpha$ is an irrational number, no nonzero integer $k$ satisfies $k\alpha \in \mathbb{Z}$, so the powers never repeat. The stronger statement is that $H_{\alpha }$ is a dense set in the circle group: every nonempty open arc contains a power $g^k$. Thus its closure is all of $\mathbb{T}$, while $H_{\alpha }$ itself is a countable proper subgroup and is not a closed subgroup or a discrete subgroup. See irrational rotation and everywhere interval frequency under an irrational rotation for the corresponding motion and its long-term frequencies.

For $\alpha =\sqrt{2}-1$, direct computation of the fractional parts of $k\alpha$ gives, to six decimal places,For example, $\{3\alpha \}=3\sqrt{2}-4\approx 0.242641$ and $\{5\alpha \}=5\sqrt{2}-7\approx 0.071068$. These calculations specify exactly which points the diagram displays.

This example shows why algebraic generation and closure must be distinguished: one irrational rotation generates a countable cyclic subgroup, yet its closure is the whole uncountable compact group.