Golden mean rotation.ipynb

Kernel: SageMath (stable)

Golden Mean Rotation

Defining the map

The following defines the golden mean as an algebraic real number:

phi = AA((1+sqrt(5))/2)

Rotation by $\phi$ mod $1$:

def R(x):
    xx = x+phi
    return xx - floor(xx)

The graph of $R$:

plot(R,(x,0,1))

The map $R$ modified to act intervals (pairs of points) as well as numbers.

def R(x):
    try:
        x = AA(x) # Attempt to convert to an algebraic real.
        xx = x+phi
        return xx - floor(xx)
    except TypeError:
        # Conversion failed. Assume we have an interval.
        a,b = x
        # We will normalize so that the right endpoint takes values in (0,1]
        Rb = R(b)
        if Rb==0:
            Rb=1
        return [R(a),Rb]

Demonstration of the new $R$:

R(0)

0.618033988749895?

R([0,1-phi])

[0.618033988749895?, 1]

Drawing pictures of the intervals

def draw_interval(I, y, **options):
    # Draw the interval subset of [0,1] on the horizontal line at height y.    
    a,b = I      # Get the endpoints
    plt = Graphics()
    if b<a:
        # The interval wraps around the circle.
        plt +=line2d([(a,y),(1,y)],**options)+line2d([(0,y),(b,y)],**options)
    else:
        plt += line2d([(a,y),(b,y)],**options)
    # Now draw the endpoints
    plt += point2d([(a,y),(b,y)], color="black", zorder=10, size=20)
    return plt

draw_interval([0,1],0,color="green", thickness="1", axes=False) + \
    draw_interval([1/4,1/2],0,color="blue", thickness="3") +\
    draw_interval([3/4,1/8],0,color="red", thickness="3")

A0=[0,1-1/phi]
A0

[0, 0.3819660112501051?]

B0=[1-1/phi,1]
B0

[0.3819660112501051?, 1]

In the following picture, we can see the action of $R$. Intervals are translated.

draw_interval( A0, 0, color="red", thickness="3", axes=False, aspect_ratio=1) +\
    draw_interval( B0, 0, color="blue", thickness="3") +\
    draw_interval( R(A0), 0.2, color="red", thickness="3") +\
    draw_interval( R(B0), 0.2, color="blue", thickness="3")

The renormalization

The map $\psi:[1-\phi^{-1},1] \to [0,1]$ is defined by $\psi(x)=\phi*(1-x)$.

def psi(x):
    try:
        x = AA(x) # Attempt to convert to an algebraic real.
        xx = phi*(1-x)
        if xx==1:
            xx = 0
        return xx
    except TypeError:
        # Conversion failed. Assume we have an interval.
        a,b = x
        # Note that the map reverses orientation.
        left_endpoint = psi(b)
        right_endpoint = psi(a)
        if right_endpoint == 0:
            right_endpoint = AA(1)
        return [left_endpoint, right_endpoint]

Checking:

psi(1)

0

psi(1-phi^(-1))

0

psi([1-phi^(-1),1])

[0, 1]

Compute $\psi^{-1}$.

def psi_inv(x):
    try:
        x = AA(x) # Attempt to convert to an algebraic real.
        xx = 1-(x/phi)
        if xx==1:
            xx = 0
        return xx
    except TypeError:
        # Conversion failed. Assume we have an interval.
        a,b = x
        # Note that the map reverses orientation.
        left_endpoint = psi_inv(b)
        right_endpoint = psi_inv(a)
        if right_endpoint == 0:
            right_endpoint = AA(1)
        return [left_endpoint, right_endpoint]

A1 = psi_inv(A0)
A1

[0.763932022500211?, 1]

B1 = psi_inv(B0)
B1

[0.3819660112501051?, 0.763932022500211?]

draw_interval( A0, 0, color="red", thickness="3", axes=False, aspect_ratio=1) +\
    draw_interval( B0, 0, color="blue", thickness="3") +\
    draw_interval( A1, 0.2, color="red", thickness="3") +\
    draw_interval( B1, 0.2, color="blue", thickness="3")

The following depicts the orbit of A1 and B1 up to the first return.

draw_interval( A0, 0, color="red", thickness="3", axes=False, aspect_ratio=1) +\
    draw_interval( B0, 0, color="blue", thickness="3") +\
    draw_interval( A1, 0.1, color="red", thickness="3") +\
    draw_interval( B1, 0.1, color="blue", thickness="3") +\
    draw_interval( A1, 0.2, color="red", thickness="3") +\
    draw_interval( B1, 0.2, color="blue", thickness="3") +\
    draw_interval( R(B1), 0.2, color="blue", thickness="3") +\
    draw_interval( R(A1), 0.3, color="red", thickness="3") +\
    draw_interval( R(R(B1)), 0.3, color="blue", thickness="3")

Check the renormalization. The following two lines show it works on A1 up to some considerations about the endpoint (which results in the stray dot in the figure above).

# A1 returns immediately to the interval
print( R(psi(A1))==psi(R(A1)) )
R(psi(A1))

True

[0.618033988749895?, 1]

# B1 returns in two steps.
print( R(psi(B1)) == psi(R(R(B1))) )
R(psi(B1))

True

[0, 0.618033988749895?]

Observe that $A_0 = R(B_1)$ and $B_0=A_1 \cup B_1$. Therefore if the measure is $R$-invariant we can deduce the measures of $A_0$ and $B_0$ from those of $A_1$ and $B_1$. The relationship is governed by the following matrix.

M = matrix([\
            [0, 1], \
            [1, 1]])
show(M)

For example if the measures of $A_1$ and $B_1$ were $a_1$ and $b_1$ respectively then the measures of $A_0$ and $B_0$ would be given by the following vector:

a1=var("a1")
b1=var("b1")
M*vector([a1,b1])

(b1, a1 + b1)

Repeating the renormalization.

A_list = [A0,A1]
B_list = [B0,B1]

Augment the list by repeating the renormalization $k$ times.

k=6
for i in xrange(k):
    A_list.append( psi_inv(A_list[-1]) ) # A_list[-1] returns the last element.
    B_list.append( psi_inv(B_list[-1]) )

plt = Graphics()
for i in xrange(len(A_list)):
    IA = A_list[i]
    IB = B_list[i]
    v = M^i*vector([1,1])
    # The first entry of v tells us how many images of IA to draw.
    for j in xrange(v[0]):
        # We will draw the first one thicker.
        if j==0:
            plt += draw_interval( IA, i/10, color="red", thickness="6", axes=False, aspect_ratio=1)
        else:
            plt += draw_interval( IA, i/10, color="red", thickness="3", axes=False, aspect_ratio=1)
        IA = R(IA)
    for j in xrange(v[1]):
        if j==0:
            plt += draw_interval( IB, i/10, color="blue", thickness="6", axes=False, aspect_ratio=1)
        else:
            plt += draw_interval( IB, i/10, color="blue", thickness="3", axes=False, aspect_ratio=1)
        IB = R(IB)
plt

Understanding the above picture plus the use of Caratheodory's Extension Theorem implies that the rotation $R$ is uniquely ergodic, i.e., there is only one $R$-invariant probability measure (namely Lebesgue measure on the circle).