That Can't Be Right

In the following post, we go through an example computation of the K-theory of the tiling space of the Fibonacci tiling of $\R$ . The main reference for these notes will be the book of Sadun [1]. In particular pages 27-29 and Theorem 2.1. We adopt the following conventions: The Fibonacci substitution rule with alphabet

$\{a, b\}$

is given by

$\sigma(a) = b \qquad \sigma(b) = ab.$

The nomenclature likely comes from the fact that the length of the word ${\sigma^n(a)}$ is determined by the n’th Fibonacci number.We let ${\sigma}$ act on (possibly infinite) words by acting on each letter

$\sigma(... abaab...) := ... \sigma(a) \sigma(b) \sigma(a) \sigma(a) \sigma(b) ...$

If we let ${a}$ and ${b}$ correspond to half open intervals, then up to translation a tiling of ${\R}$ corresponds to a bi-infinite word in ${a, b}$ . We may determine the tiling completely by adding a “point” to the bi-infinite word indicating where ${0}$ lies in the tiling. We have a natural action of ${\R}$ on the space of tilings given by translation which will briefly make its appearance at the end of these notes.

If ${T}$ is a tile, then ${\sigma^n(T)}$ is called the level-n supertile of ${T}$ . A tiling is called admissible if every finite patch of the tile is contained in some supertile ${\sigma^n(a)}$ for some integer ${n > 0}$ (note that ${\sigma^n(a)}$ spans the alphabet). The tiling space of ${\sigma}$ is then defined to be

$\Omega := \Omega_\sigma = \{\text{Tilings } T ~|~ \text{every patch of T is found in some supertile}~~ \sigma^n(a) , ~\text{for some } n\in \N\}.$

These tilings we refer to as Fibonacci tilings of ${\R}$ . Clearly if ${T}$ is an admissible tiling, then so is ${\sigma(T)}$ , so this set is closed under the Fibonacci substitution rule. Similarly ${\Omega}$ is clearly closed under the action of ${\R}$ by translation and we may thus form the corresponding action groupoid ${\Omega \rtimes \R}$ . Note that ${baab}$ and ${bbb}$ are not admissible tiles, hence neither are their ${ababab}$ being a supertile of ${bbb}$ .

A concrete realization of a Fibonacci tiling of ${\R}$ is given by letting ${a}$ be intervals of varying length. If we set $a$ to be the half open interval of lenth $1$ and ${b}$ the interval of length ${\tau = \frac{1 = \sqrt{5}}{2}}$ (the golden mean) then ${\sigma}$ can be taken to be the multiplication by ${\tau}$ map on ${\R}$ .

To compute the K-theory of the tiling space of an admissible tiling using the Anderson-Putnam complex below, we will need our substitution tiling to “force the border”. A substitution rule is said to force the border if there is an ${n \in \N}$ such that a supertile of level-n uniquely determine its adjacent letters. Looking at the words

$\sigma(a) = b, \quad \sigma(b) = ab, \quad \sigma(ab) = bab, \quad \sigma(bab) = abbab ...$

one can deduce that every supertile has a ${b}$ to its right, and alternates between ${a}$ and ${b}$ to the left. Thus if we pick ${n}$ such that ${\sigma^n(T)}$ starts with an ${a}$ , we know that it must be padded to the left by the letter ${b}$ as no admissible tiling has two consecutive ${a}$ ‘s. However we cannot determine the letter to the right of ${\sigma^n(T)}$ for a general tile ${T}$ as both ${ba}$ and ${bb}$ may occur.

So the Fibanacci tiling does not “force the border” to the right. To remedy this, we use collared tiles, which keep track of the letters adjacent to the given letter. The point is that these collared tiles do force the border and so the machinery of Anderson-Putnam tells us that a certain projective limit converges to the tiling space of any admissible tiling\footnote{The Fibonacci tiling is determined by a primitive substitution, hence any admissible tiling will yield isomorphic tiling space [1, Th. 1.5]{sudan}}. There are a total of four admissible three letter patches, namely:

$A = (b)a (b) ,\quad B = (a)b(b), \quad C = (b)b(a) \quad D = (a)b(a).$

with substitution rules (see [1, p. 28]):

$\begin{align*} \sigma(A) &= (ab) b (ab) = C \sigma(B) &= (b) ab (ab) = AD \sigma(C) &= (ab) ab (b) = AB \sigma(D) &= (b) ab (b) = AB. \end{align*}$

The Anderson-Putnam complex ${\tilde{\Gamma}_0 = \Gamma_1}$ of an admissible collared tiling with the above substitution rule is depicted by the CW-complex below (shamelessly copied from [1, Fig. 2.4])

The substitution rules determine a continuous map ${\sigma: \Gamma_1 \to \Gamma_1}$ (uniquely up to homotopy). In particular, if we denote by ${AD}$ denote the path obtained by first tracing ${A}$ and then the ${D}$ in the direction of the arrows, then ${\sigma(AD)}$ is the triangle ${CAB}$ . Similarly the upper triangle ${ABC}$ is mapped to ${\sigma(ABC) = CADAB}$ which is a path similar to how a regular person would write the figure eight by hand.

By a theorem of Anderson and Putnam the tiling space ${\Omega}$ of the Fibonacci tiling is homeomorphic to the projective limit

$\Omega = \varprojlim_{n} ~ (\Gamma_1, ~\sigma).$

Thus

$K^i(\Omega) = \varinjlim_{n} ~ (K^i(\Gamma_1), ~ \sigma^*)$

where ${K^i}$ denotes complex ${K}$ -theory. Since ${\Gamma_1}$ is homotopic by to the wedge sum ${S^1\vee S^1}$ we have the following (reduced) K-theory groups

$K^0(\Gamma_1) = K^0(S^1\vee S^1) = K^0(S^1) \oplus K^0(S^1) = 0.$

Similarly

$K^1(\Gamma_1) = K^1(S^1) \oplus K^1(S^1) = \Z\oplus \Z.$

1. Cycles in ${K^1}$ : A short digression

Let ${X }$ be a compact Hausdorff space. Cycles in ${K^{1}(X)}$ are represented by homotopy classes of continuous maps

$f: X \to U_n$

for some ${n}$ where ${U_n}$ is the unitary group of ${n\times n}$ -matrices. Two maps ${f: X\to U_n}$ and ${h: X\to U_m}$ are said to be equivalent if they are homotopic after imbedding ${U_n}$ and ${U_m}$ into ${U_{n+m}}$ . We can add two cycles by adding their images in a block diagonal matrix of the form\footnote{By a result called Whiteheads lemma this group operation is the same the usual matrix multiplication

$[f] \oplus [h] = [f \cdot h]$

where ${(f\cdot h)(x) = f(x) h(x)}$ and the left hand side is ordinary matrix multiplication in a sufficiently large ambient unitary group, say ${U_{n +m}}$ .}:

$f\oplus h : X \to U_n\oplus U_m \subset U_{n+m}.$

Given a continuous map ${T: X \to Y}$ between compact Hausdorff spaces, the induced map on the ${K^1}$ -groups is given by pre-composition by ${T}$ , i.e.

$T^*[f] = [f\circ T] \in K^1(X) \qquad [f] \in K^1(Y).$

The most trivial setting is where ${X = S^1}$ . Then homotopy class of map ${f: S^1 \to U_n}$ are clearly in bijection with the elements in ${\pi_1(U_n) = \Z}$ . Since the imbedding ${\iota: U_n \to U_{n + 1}}$ determines an isomorphism on the level of fundamental groups, we deduce that two cycles ${f: S^1 \to U_n}$ and ${h: S^1 \to U_m}$ are equivalent, if and only if they determine the same map on fundamental groups

Thus for any ${n}$ we have

$K^1(S^1) = \pi_1(U_n) = \Z$

with cycles represented by homotopy classes of maps ${f: X\to U_n}$ .

2. The K-theory of the tiling space of a Fibonacci tiling

Returning back to our space ${\Gamma_1}$ we can, in a similar vain, show that ${K^1(S^1\vee S^1) = \Z\oplus \Z}$ is the free abelian group generated by two cycles represented by maps

$f_i : S^1\vee S^1 \to U_n \qquad (i= 1, 2)$

which are constant one copy of ${S^1}$ and restricts to a generator of ${\pi_1(U_n) = \Z}$ on the other copy of ${S^1}$ . Let us choose ${[f_1] = (1, 0) \in \Z\oplus \Z}$ and ${[f_2] =(0,1) \in \Z\oplus \Z}$ with ${f_1}$ is constant on the loop ${ABC}$ while ${f_2}$ is constant on ${AD}$ .

To determine the map ${\sigma^*: K^1(\Gamma_1) \to K^1(\Gamma_1)}$ we simply need to check how it behaves on these two loops.

For our ${\sigma}$ we have seen that it sends the loop ${AB}$ (the lower loop) to the loop ${CAB}$ . Thus

$\sigma^*([f_1]) = [f_1 \circ \sigma] = [f_2]$

while, since ${ABC}$ is sent to ${CADAB}$ , we get

$\sigma^*([f_2]) = [f_2 \circ \sigma] = [f_1] + [f_2].$

Thus the map ${\sigma^*: \Z\oplus \Z \to \Z\oplus \Z}$ has the following matrix realization

$\sigma^* = 1 & 1 0 & 1$

Using the Anderson Putnam theorem [1, Th. 2.1] we get

$K^1(\Omega) = \varinjlim_{n\in \N} \left(\Z\oplus \Z, 1 & 1 0 & 1 \right) = \bigsqcup_{n\in \N} \Z\oplus \Z/\sim$

The equivalence relation is given as follows: Assume ${(r,s)}$ is in the n’th copy of ${\Z\oplus \Z}$ while ${(t,u)}$ is the the m’th copy with ${m\geq n}$ . Then ${(r,s) \sim (t, u) }$ if and only if ${(\sigma^*)^k(r,s) = (t,u)}$ for ${k = m-n}$ . Since ${\sigma^*}$ is invertible this means every equivalence class of ${\sim}$ is uniquely represented by a single copy of ${\Z\oplus \Z}$ and so

$K^1(\Omega) = \Z\oplus \Z.$

Now we can easily compute the K-theory of the transformation groupoid given by the ${\R}$ -action on ${\Omega}$ using Connes-Thom isomorphism:

$K^i(\Omega\rtimes \R) = K_i(C(\Omega)\rtimes \R) \simeq K^{i+1}(\Omega) = \begin{cases}\Z\oplus \Z & i = 0 \\ 0 & i = 1. \end{cases}$

[1] – Sadun, Lorenzo Adlai. Topology of tiling spaces. Vol. 46. American Mathematical Soc., 2008.

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