Decomposing X w 0

Here we decompose thewhite 0-tileX_w⁰ into white trees.

Consider the white 1-tiles in X_w⁰. We assume in the next lemma that they are all connected at all 1-verticesvin theinteriorofX_w⁰, anddisconnectedat all 1-vertices onC.

The resulting white connection graph may not be connected.

Lemma 7.6. The white connection graph in X_w⁰ as above has exactly one (white) cluster that intersects all sides (0-edges).

Proof. LetK be a (white) cluster inX_w⁰ as above. Consider one componentB (in the standard topological sense) of X_w⁰\K. We call the set a:=∂B∩K a boundary arc ofK.

Claim 1. Every boundary arc aas above is contained in a single black 1-tile.

Clearlya is a union of 1-edges. Eithera starts and ends at two distinct 1-vertices v, w∈C, or a is a closed curve. Let E, E⁰3v be two 1-edges contained in a which are consecutive in∂B, wherev /∈C is a 1-vertex. Note that by construction all white 1-tiles Xj3v are connected atv. ThusE and E⁰ are contained in the same black 1-tile. The claim follows.

Assume now thatais a closed curve. Thenais a Jordan curve in the boundary of a single black 1-tile. Thus the corresponding componentB is the interior of a single black 1-tile. Henceadoes not separateK from any other distinct white clusterK⁰ inX_w⁰.

We call a black 1-tileY⊂X_w⁰ non-trivial ifY∩Ccontains at least two 1-vertices. A complementary component ofY is the closure of a componentX_w⁰\Y.

Claim2. LetX, X⁰⊂X_w⁰ be two distinct white 1-tiles. ThenXandX⁰are contained in distinct white clusters K, K⁰⊂X_w⁰ if and only if there is a black 1-tile Y⊂X_w⁰ such thatX andX⁰ are contained in complementary components ofY.

The implication (⇐) is clear. To see the other implication we note that if X⁰ is contained in a cluster distinct from the cluster K⊃X, then X⁰ has to be contained in the closure of one component ofX_w⁰\K. Such a component is separated from K by a boundary arca. However, if adoes not contain two 1-vertices v, w∈C, this component is a single black 1-tile, meaning that it does not containX⁰. OtherwiseX⁰ is separated fromX by the black 1-tileY containinga, proving the claim.

Recall that we assumed that no 1-tile joins opposite sides ofC(see Lemma7.5). Thus for every non-trivial black 1-tile Y there is a complementary component ofY, denoted byKY, that intersects all 0-edges.

We now defineK:=T

Y K_Y, where the intersection is taken over all non-trivial black 1-tilesY⊂X_w⁰. Since two non-trivial black 1-tilesY, Y⁰⊂X_w⁰ do not cross, it follows that Kintersects all 0-edges.

By Claim 2 it follows that all white 1-tiles contained inKare connected, i.e., belong to the same cluster denoted byK.

Assume thatKintersects a given 0-edgeE⁰in a 1-edgeE. This cannot happen ifE is contained in a black 1-tileY⊂X_w⁰, sinceY would be non-trivial, and the corresponding setKY does not containE. ThusE is contained in a white 1-tile, which is inK.

IfKintersectsE⁰only in a 1-vertexv, there is a boundary arca⊂∂K containingv.

LetY⊂X_w⁰ be the corresponding non-trivial black 1-tile containing a. LetE⊂abe the 1-edge containingv. SinceE is not inC, the white 1-tile containingE is in K.

This means that there is a white 1-tile in Kthat intersectsE⁰.

In each white cluster inX_w⁰ define a spanning tree (see Definition6.11). The spanning tree in the cluster from Lemma 7.6 is called the main tree K_M, the spanning trees in the other clusters are called thesecondary trees inX_w⁰. Theconnections at all 1-vertices v∈X_w⁰\C are thusdefined, they will not be changed any more in the construction.

Let E be the boundary circuit of the main treeKM (see Definition6.21 and Lem-ma6.23). Letv0, ..., vN−1be the 1-vertices onC thatE visits (in this order). Note that a 1-vertexv may appear several times in this list.

Notation. Given points v, w∈C, denote by [v, w] (resp. (v, w)), the closed (resp.

open) positively oriented arc onC fromv tow. Note that (v, v)=∅.

Lemma 7.7. The points {vi}^N_i=0⁻¹ satisfy the following conditions (indices are taken modN here):

(1) Each (open) arc (vi, vi+1) contains no point vl. This means that the points {vi}^N−1_i=0 are positively oriented on C.

(2) The points v_i and v_i+1 are not contained in disjoint 0-edges,in particular each 0-edge contains at least one point v_i.

(3) For all i there is a black 1-tile Y3vi, vi+1.

(4) Let K be a secondary tree in X_w⁰. Then there is an arc [vi, vi+1]such that K∩C ⊂[vi, vi+1].

Proof. (1) LetK_M,εbe a geometric representation ofK_M as in Lemma6.23(3). The pathγionE betweenvi andvi+1 is then represented by a Jordan arcγi,εwith endpoints vi,ε andvi+1,εsuch that|vi−vi,ε|and|vi+1−vi+1,ε|are arbitrarily small. Since all white 1-tiles are disconnected at every 1-vertexv∈C, we may assume thatvi,ε∈Candγi,ε⊂X_w⁰ for alli.

The arcsγi,ε are non-crossing, and thus the points{vi,ε}^N_i=0⁻¹ are ordered cyclically or anti-cyclically onC. Hence the points{vi}^N−1_i=0 are ordered cyclically or anti-cyclically onC.

The winding number ofE aroundx /∈Eis 1 if and only ifxis in the interior of a white 1-tile of the main tree. This follows from an inductive argument as in Corollary6.20.

Assume that the points {vi}^N_i=0⁻¹ are ordered anti-cyclically on C. Let Ci be the (positively oriented) arc onC between vi and vi+1. Then γi+Ci has winding number 0 around any pointx in the interior of a 1-tile of the main tree. ThusE+C has winding number 0 around such anx. This is a contradiction.

(3) Considervi andvi+1. Then

• eithervi=vi+1 in which case the statement is trivial;

• or [v_i, v_i+1] is a 1-edge, property (3) is then clear again;

• orv_i andv_i+1 are the boundary points of a boundary arcaofK_M, as in Claim 1 from the proof of Lemma7.6. In this case there is a black 1-tile Y⊃a.

(2) This follows immediately from (3) and the assumption that no 1-tile intersects disjoint 0-edges. Furthermore KM intersects a 0-edge E if and only if it intersects it in some 1-vertex. The set of all 1-vertices in which KM intersects C is equal to the set {vi}^N_i=0⁻¹. Thus, sinceKM intersects each 0-edge, it follows that each 0-edge contains one pointvi.

(4) The reader is reminded of Claims 1 and 2 in the proof of Lemma7.6. For every secondary componentKthere is an arcacontained in a (non-trivial) black 1-tileY such that intKis in the component ofX_w⁰\anot intersecting all 0-edges. Letvi andvi+1be the endpoints ofa(see the discussion from (3)), then

K∩C ⊂[vi, vi+1].