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Topological Conjugacy for homeomorphisms of R\mathbb R.

Let f:IIf:I \to I and g:JJg: J \to J be continuous maps. We say they are topologically conjugate if there is a homeomorphism h:IJh:I \to J so that gh(x)=hf(x)g \circ h(x)=h \circ f(x) for all xIx \in I.

We will demonstrate the idea behind the following result:

Theorem. Suppose I=(a,b)I=(a,b) and J=(c,d)J=(c,d) are intervals in R\mathbb R. Suppose f:IIf:I \to I and g:JJg:J \to J are orientation-preserving homeomorphisms so that

  • f(x)>xf(x)>x for each xIx \in I, and
  • g(y)>yg(y)>y for each yJy \in J. Then ff and gg are topologically conjugate.

To demonstrate this, we will consider two such maps.

f(x) = sqrt(2*x-x^2) # Consider over the interval (0,pi)
plot(f, 0, 1, aspect_ratio=1) + plot(x,(x, 0, 1), color="red")
# Here we work out the inverse map
x=var("x")
y=var("y")
assume(y>0) # Used to help Sage find the solution we want below.
assume(y<1) 
show((f(x)==y).solve(x))
[x=y2+1+1,x=y2+1+1]\left[x = -\sqrt{-y^{2} + 1} + 1, x = \sqrt{-y^{2} + 1} + 1\right]

Note that the inverse must be the first one since the second takes values greater than one.

# Here we define the inverse map
finv(y) = -sqrt(-y^2 + 1) + 1

Lets plot f1f^{-1} with ff to be sure.

plot(finv, 0, 1, color="green", aspect_ratio=1) + plot(f,(x,0,1), color="blue")
g(x) = 1/2*(x*(3-x)) # Consider over the interval (0,pi)

plot(g, 0, 1, aspect_ratio=1) + plot(x,(x,0,1), color="red")
# Here we work out the inverse map
x=var("x")
y=var("y")
assume(y>0) # Used to help Sage find the solution we want below.
assume(y<1)
(g(x)==y).solve(x)
[x == -1/2*sqrt(-8*y + 9) + 3/2, x == 1/2*sqrt(-8*y + 9) + 3/2]
ginv(y) = -1/2*sqrt(-8*y + 9) + 3/2
plot(ginv, 0, 1, color="green", aspect_ratio=1) + plot(g,(x,0,1), color="blue")

Defining the topological conjugacy:

First we pick a points afa_f and aga_g in the domains of ff and gg:

a_f = 1/2
a_g = 1/2

We define bf=f(af)b_f = f(a_f) and bg=g(ag)b_g = g (a_g):

b_f = f(a_f)
print("b_f = %s"%b_f)
b_g = g(a_g)
print("b_g = %s"%b_g)
b_f = 1/2*sqrt(3) b_g = 5/8

The intervals [af,bf)[a_f, b_f) is a fundamental domains for ff. This means for each x(0,1)x \in (0,1), there is a unique nZn \in \mathbb Z so that fn(x)[af,bf)f^n(x) \in [a_f, b_f). Similarly, [ag,bg)[a_g, b_g) is a fundamental domain for gg.

We define a homeomorphism h0:[af,bf)[ag,bg).h_0:[a_f,b_f) \to [a_g, b_g).

h_0(x) = (b_g - a_g)/(b_f - a_f)*(x - a_f) + a_g
show(h_0(x))
assert h_0(a_f)==a_g # Prints errors if false.
assert h_0(b_f)==b_g 
2x18(31)+12\frac{2 \, x - 1}{8 \, {\left(\sqrt{3} - 1\right)}} + \frac{1}{2}

Note that this function is more complex, so we define it using a Python type function. This allows us to use any Python or Sage type expression we want, including if statements and loops.

def h(x):
    assert 0 < x < 1 # Cause an error if not in the domain of f.
    if a_f <= x < b_f:
        # Use h_0:
        return h_0(x)
    if x >= b_f:
        count = 0
        while x >= b_f:        # Apply f^-1 until we land in the fundamental
            x = finv(x)        # domain and count the number of times
            count = count + 1  # we apply f^-1.
        assert a_f <= x < b_f
        y = h_0(x)  # Move to the domain of g using h_0.
        for i in range(count): # Now apply g to y, the same number of times.
            y = g(y)           
        return y
    if x < a_f:
        count = 0
        while x < a_f:
            x = f(x)
            count = count + 1
        assert a_f <= x < b_f
        y = h_0(x)
        for i in range(count):
            y = ginv(y)
        return y
# Plot h.
# Note that h is not defined at zero or at one, so
# we have shrunk the interval we are plotting slightly.
# Calling plot(h, 0, 1) will give rise to errors.
plot(h, 0.001, 0.999)
# Check one value larger than b_f:
show( h(9/10) )
# Check the conjugacy equation:
assert( h(f(9/10)) == g(h(9/10)) )
13200(19531+100)(1953120)-\frac{1}{3200} \, {\left(\frac{\sqrt{19} - 5}{\sqrt{3} - 1} + 100\right)} {\left(\frac{\sqrt{19} - 5}{\sqrt{3} - 1} - 20\right)}
# Check one value larger than b_f:
show( h(1/4) )
# Check the conjugacy equation:
assert( h(f(1/4)) == g(h(1/4)) )
12722(31)+5+32-\frac{1}{2} \, \sqrt{-\frac{\sqrt{7} - 2}{2 \, {\left(\sqrt{3} - 1\right)}} + 5} + \frac{3}{2}

We can graphically check the conjugacy. For plot1 we will plot hfh \circ f and for plot2 we will plot ghg \circ h.

plot1 = plot(lambda x: h(f(x)), 0.001, 0.999)
show(plot1)
plot2 = plot(lambda x: g(h(x)), 0.001, 0.999, color="red")
show(plot2)
# Plot them on top of each other
plot1 + plot2
# Note that h is continuous but not differentiable:
def approximate_derivative_of_h(x, epsilon=0.0001):
    return (h(x+epsilon)-h(x)) / epsilon
plot(approximate_derivative_of_h,0.1,0.9)

By fiddling appropriately with our function h0(x)h_0(x) we could get hh to be smooth. The main issue is derivatives at afa_f. At other points, hh is defined to be h0h_0 or compositions of h0h_0 with powers of ff and gg. Note that for values slightly bigger than afa_f, hh is given by h0h_0. While for values slightly to the left of afa_f, hh is given by g1h0fg^{-1} \circ h_0 \circ f. Thus if we want the derivative to match at afa_f, we would have to have h0(af)=(g1)(hf(af))h0(f(af))f(af)=(g1)(bg)h0(bf)f(af)=f(af)g(ag)h0(bf).h_0'(a_f)= (g^{-1})'\big(h \circ f(a_f)\big) \cdot h_0'\big(f(a_f)\big) \cdot f'(a_f)= (g^{-1})'(b_g) h_0'(b_f) f'(a_f)= \frac{f'(a_f)}{g'(a_g)} h_0'(b_f). So we would have to choose h0h_0 to satisfy g(ag)h0(af)=f(af)h0(bf)g'(a_g) h_0'(a_f) = f'(a_f) h_0'(b_f).