Escher’s “Print Exhibition”

In 1956 the dutch artist M. C. Escher created the lithography “Print Exhibition”. This famous picture visualizes an interesting two-dimensional Riemannian manifold which attracts attention not only to the surprised spectator, but also to many mathematicians. The fascination of this picture springs from Escher’s art to select a scene for his lithography which is locally embedded naturally in this non-euclidean space, i. e. in every convex sector not including the central point in a realistic manner. His basic idea was to connect the “true” environment with its own image presented by a picture in an exhibition, which is this picture, too! Thus it is possible to walk inside Escher’s Print Exhibition which actually is only a print exhibited in itself, its own exhibition. This paradox is possible by the special manifold Escher chose. We like to deduce this exciting mapping

Φ: ℝ^{2} → ℝ^{2} with (x, y) ↦ (u, v)  (1)

mathematically precise. It appears quite smooth, such that we can suppose that all our considered functions may be at least differentiable, if not even two-times differentiable — except maybe in the center of the picture. At a first glance, there is some kind of rotational symmetry around the center of the picture. Therefore let us select this center pointer as the origin of the coordinate system and also introduce polar-coordinates (ρ, ϑ) in the image with

ρ = \sqrt{u^{2} + v^{2}}, ϑ = arctan \frac{v}{u}  (2)

and

r = \sqrt{x^{2} + y^{2}}, φ = arctan \frac{y}{x}  (3)

for the domain. Starting with ρ as an unknown function ρ(x, y), we notice that

ϑ = φ + μ(ρ) with μ: ℝ^{+} → ℝ^{+} strictly monotonious. (4)

This is a fundamental property of our transformation Φ. Further, we are interested only in a “nice” analytic solution Φ; hence we demand that Φ is holomorphic in the whole ℝ^{2}-plane except the point (0, 0). The immediate consequence for this function Φ, regarded as a mapping into polarcoordinates, is

\frac{1}{ρ} \frac{∂ρ}{∂x} = \frac{∂ϑ}{∂y}  and  − \frac{1}{ρ} \frac{∂ρ}{∂y} = \frac{∂ϑ}{∂x} ,  (5)

which is isomorphic to the “Riemann-Cauchy differential equation” for Φ ≡ (ρ, ϑ). Substituting now (3) and (4) into (5), we get with

\frac{∂φ}{∂x} = \frac{−\frac{y}{x^{2}}}{1 + (\frac{y}{x})^{2}} = \frac{−y}{x^{2} + y^{2}}  and  \frac{∂φ}{∂y} = \frac{\frac{1}{x}}{1 + (\frac{y}{x})^{2}} = \frac{x}{x^{2} + y^{2}}

the conditions

\frac{1}{ρ} \frac{∂ρ}{∂x} = \frac{x}{x^{2} + y^{2}} + \frac{dμ}{dρ} \frac{∂ρ}{∂y}  and  − \frac{1}{ρ} \frac{∂ρ}{∂y} = −\frac{y}{x^{2} + y^{2}} + \frac{dμ}{dρ} \frac{∂ρ}{∂x} (6)

which in turn mean

\frac{∂ρ}{∂x}(\frac{1}{ρ} \frac{∂ρ}{∂x} − \frac{x}{x^{2} + y^{2}} ) = \frac{∂ρ}{∂x} \frac{dμ}{dρ} \frac{∂ρ}{∂y} = \frac{∂ρ}{∂y}( − \frac{1}{ρ} \frac{∂ρ}{∂y} + \frac{y}{x^{2} + y^{2}} ) . (7)

Thus, eliminating the function μ from these equations and substituting

ρ = e^{f} ⇔ f = ln ρ

we get

\frac{∂f}{∂x}(\frac{∂f}{∂x} − \frac{x}{x^{2} + y^{2}} ) + \frac{∂f}{∂y} (\frac{∂f}{∂y} − \frac{y}{x^{2} + y^{2}} ) = 0 .  (8)

and this can be transformed with a “quadratic expansion” into

(\frac{∂f}{∂x} − \frac{\frac{1}{2}x}{x^{2} + y^{2}} )^{2} + (\frac{∂f}{∂y} − \frac{\frac{1}{2}y}{x^{2} + y^{2}} )^{2} = \frac{\frac{1}{4}x^{2} + \frac{1}{4}y^{2}}{(x^{2} + y^{2})^{2}} .  (9)

Finally we set

f = \frac{1}{2}g + \frac{1}{2} ln \sqrt{x^{2} + y^{2}}  (10)

and get the simple first order partial differential equation of degree two

(\frac{∂g}{∂x})^{2} + (\frac{∂g}{∂y} )^{2} = \frac{1}{x^{2} + y^{2}} .  (11)

Next, we “guess” two algebraic independent solutions g_{1} and g_{2} of (11), namely

g_{1} = ln \sqrt{x^{2} + y^{2}} = ln r and g_{2} = arctan \frac{y}{x} = φ .  (12)

The theory of partial differential equations tells us that the “complete integral” of (11) is

g(x, y) = αg_{1} + α′g_{2} + β with α, α′, β ∈ ℝ and α^{2} + α′^{2} = 1 . (13)

Here the constants α, α′ and β are already separated, thus the general solution of (11) is already (13) with α, α′ and β considered as constant functions. Because α^{2} + α′^{2} = 1, it is naturally to express these constants with trigonometric functions and thus set α = cos δ and α′ = −sin δ. Before we continue, let us switch again to polar-coordinates which will be handier for the following calculations. Now let us re-write the general solution (13) as

g(r, φ) = cos δ ln r − sin δ φ + β  with δ, β ∈ ℝ .  (14)

Next let us determine the singular solutions g_{s} which are yielded by elimination δ in equation (14) by the condition \frac{∂g}{∂δ} = 0. A short calculation results for these singular solutions

tan δ_{s} = −\frac{φ}{ln r} .  (15)

Substituting the δ_{s} in (15) for δ and simplifying yields

g_{s}(r, φ) = ±\sqrt{( ln r)^{2} + φ^{2}} + β  with β ∈ ℝ .  (16)

Now we take back the substitution (10). So we get

f(r, φ) = \frac{1}{2}( ln r + cos δ ln r − sin δ φ + β) ,  (17)

respectively

f_{s}(r, φ) = \frac{1}{2}( ln r ± \sqrt{( ln r)^{2} + φ^{2}} + β) ,  (18)

and finally resubstituting f = ln ρ, we get with B = e^{\frac{β}{2}}

ρ(r, φ) = Br^{\frac{1+cos δ}{2}} e^{−\frac{sin δ}{2} φ} ,  (19)

respectively

ρ_{s}(r, φ) = B\sqrt{re^{±\sqrt{(ln r)^{2}+φ^{2}}}} .  (20)

Next we use (6) to get a linear partial differential equation in μ:

\frac{1}{ρ} \frac{∂ρ}{∂x} − \frac{x}{x^{2} + y^{2}} = \frac{∂μ}{∂y}  and  − \frac{1}{ρ} \frac{∂ρ}{∂y} + \frac{y}{x^{2} + y^{2}} = \frac{∂μ}{∂x} .  (21)

Substitution the known function ρ(x, y) yields an exact differential equation for μ with the general solution

μ(x, y) = \frac{1}{2}( cos δ arctan \frac{y}{x} − sin δ ln \sqrt{x^{2} + y^{2}} − arctan \frac{y}{x}) + μ_{0} (22)

also writen as

μ(r, φ) = \frac{1}{2}((cos δ − 1)φ + sin δ ln r) + μ_{0}  (23)

which immediatle yields

ϑ(r, φ) = \frac{1}{2}((1 + cos δ)φ + sin δ ln r) + ϑ_{0} .  (24)

Putting the singular solution ρ_{s} given by (20) into (5) yields two partiel equations for ϑ_{s}:

\frac{1}{2}(\frac{x}{r^{2}} ±\frac{ln r\frac{x}{r^{2}} + φ\frac{−y}{r^{2}}}{\sqrt{(ln r)^{2} + φ^{2}}} ) = \frac{∂ϑ_{s}}{∂y} and −\frac{1}{2}(\frac{y}{r^{2}} ±\frac{ln r\frac{y}{r^{2}} + φ\frac{x}{r^{2}}}{\sqrt{(ln r)^{2} + φ^{2}}} ) = \frac{∂ϑ_{s}}{∂x} .

(25)

But the necessary condition \frac{∂^{2}ϑ_{s}}{∂x∂y} = \frac{∂^{2}ϑ_{s}}{∂y∂x} for the existence of ϑ_{s} leads to r\sqrt{(ln r)^{2} + φ^{2}} = 0. Thus, there exists no further holomorphic solution Φ_{s}.

Finally, we like to give μ explicitly as demanded in (4) as μ(ρ). From (19) and (23) we get after some calculating

μ(ρ) = μ_{0} + (ln ρ − \frac{β}{2} ) tan \frac{δ}{2}  with μ_{0}, β ∈ ℝ  (26)

which means according to (4) also

dϑ = dϕ + tan \frac{δ}{2} d ln ρ .  (27)

Here we see the intrinsic factor tan \frac{δ}{2} which we will interpretate below.

Joining (19) and (24) at last, we can write Φ as a complex transformation ℂ → ℂ depending on two parameters C and δ by

z ↦ C\sqrt{z^{1+cos δ+i sin δ}} = C\sqrt{z^{1+e^{iδ}}}  with δ ∈ ℝ, C ∈ ℂ . (28)

This is a bijective, holomorphic transformation of ℂ\{0} onto itself, if δ ≢ π (mod 2π). Moreover the Jacobi-determinant of it, respectively (1), is

| \begin{matrix}\frac{∂u}{∂x} & \frac{∂u}{∂y} \\ \frac{∂v}{∂x} & \frac{∂v}{∂y} \end{matrix} | = \frac{1 + cos δ}{2} (\frac{ρ}{r})^{2}

which is positively as long as δ ≢ π (mod 2π).

Next we recognize that the factor C reflects a fixed rotation and/or scaling of the mapping Φ and the angle δ reflects the magnitude of zooming if we circle the center. This becomes clear by considering the transformed coordinates (ρ, ϑ) and (ρ′, ϑ′) which correspond to (r, φ) and (r′, φ′) = (r, φ + 2π). For these we get the relations

ϑ′ − ϑ = 2π \frac{1 + cos δ}{2}  and  \frac{ρ′}{ρ} = e^{−π sin δ}  (29)

or

\frac{ln \frac{ρ′}{ρ}}{ϑ′ − ϑ} = \frac{− sin δ}{1 + cos δ} .  (30)

This means, if we set ϑ′ − ϑ = −2π for circling the center exactly onetime clockwise, then \frac{ρ′}{ρ} reflects the zooming factor. Hence the relation between the zooming factor t and the angle δ is given by

\frac{ln t}{2π} = tan \frac{δ}{2}  ∀ t ∈ ℝ^{+} ∀ δ ∈]−π, π[ .  (31)

The extrem case is δ ≡ π (mod 2π), which causes a collapse of the bijectivity of the mapping Φ , i.e. Φ(z) = C for all points z. In contrast, the most innocent case occurs for δ ≡ 0 (mod 2π), giving Φ(z) = Cz, a simple scaling and/or rotating of the mapping.

In Escher’s “Print Exhibition” we have a zooming by a factor of four for a quarter rotation, hence here t = 4^{4} = 256, resulting in δ = 2 arctan \frac{2 ln 4}{π}. The constant C in Escher’s drawing seems to be a positive reell number, because the picture scene appears upright and not rotated. :)

See: \underline{http://www.uni-bielefeld.de/~achim/lib/bitmaps/exhibition.gif}

Recorded in L^{A}TEX 1998–12–13, firstly documentated 1988–12–30.