Draft of a Proof of Wester's Theorem 15mar13 Wester [2006 Structural Morphology ... ] refers to an approach by Baglivo&Graver [1983 Incidence ...] for the "rectangular case". And, Wester sketches his theorem for the "Penrose case". I have been unable to locate more from Wester elsewhere. However, I found a very pretty sketch of the "rectangular case" here: [http://new.math.uiuc.edu/quasicrystals/hyperlinks/DaveRicheson.html]. (I relinked his article on my space because the original was difficult to find.) I will now try to reword and amplify Richeson's argument in a way that should apply to the Penrose case. At times, I might be pedantic, but it's so easy to be fooled by pictures in this stuff. GIVEN a rectangular "rods and pinions" lattice of rhombi in the plane. I'll call this a (rhombic) _carpet_ for short. FIND a minimal set of rhombi to brace, so that all rhombi have to be squares. In other words, the braced carpet is _rigid_, it moves (turns, translates) in the plane as a whole. That is, no piece of the carpet can rotate about another piece. SOLUTION Construct the Wester graph for the ribbons in this carpet and brace a maximal subtree of this graph. DEFINITIONS: Every rhombic carpet, whether it is a distortion of a rectangular one or not, decomposes into a multigrid of ribbons as follows. Every rhombus has a two pairs of parallel sides. Each pair has a direction in the plane. This is an unoriented direction, and every rhombus has two of them. For an undistorted rectangular carpet we take these to be literally horizontal and vertical. A rhombus is a square if and only if its two directions are perpendicular. A maximal succession of adjacent rhombi (in both directions), with a common direction is called a (Amman) _ribbon_, denoted by a lower case letter. Of course, a ribbon could consist of a single rhombus when there are no adjacent ones with that direction. We shall now assume that the carpet has an additional property, namely that two ribbons have at most one rhombus in common. If there is none, the two ribbons are said to be parallel. Thus every rhombus in the carpet is uniquely identified by pair of ribbons (ab)=(ba). And a ribbon, a, is a succession of rhombi: a = (ay)...(af)(ad)(ab)(ac)(ae)...(az) so that one of the two directions associated with each rhombus in the ribbon is common to all the rhombi in the ribbon. We denote this direction by ^(a). The shape of a rhombus (ab) is completely determined by ^(a) and ^(b). In particular, if ^(a) _|_ ^(b) then (ab) is a square, but it might be tilted. In particular, if ^(a) is horizontal, and ^(b) is vertical, then rhombus (ab) has to be a "cardinal" square (i.e. sides aligned with the axes.) NOTE Two ribbons, a,b, in a carpet either cross on a rhombus (ab) or they do not cross, in which case we call them parallel ribbons, a||b . What makes a carpet rectangular is that for any of its ribbons a=(ab)(ac)....(az), the crossing ribbons are all parallel, b||c||d||...||z||. This need not be the case in general. It is clearly the case when the carpet is in its final, normal position. And distorting it cannot create new "intersections" of formerly parallel ribbons. And when we deform an initially rectangular carpet, the two classes of mutually parallel ribbons can continue be called the "horizontal ribbons" and the "vertical ribbons" from their original directions. CONJECTURE: I feel certain that two ribbons cannot cross each other twice, at different rhombi, even if we don't make the rectangular assumption. I can't prove it right now. But since it is a hypothesis in the rectangular case, and there is an analogous hypotheses in the Penrose case, thanks to DeBruijn, I don't care right now. So, for the rest of the discussion, the carpets will be Euclidean (i.e. 2 lines don't cross twice). We need Euclidean carpets or the symbol (ab) could be ambiguous because we don't know which rhombus we mean. WESTER GRAPH: This is the graph whose vertices are the ribbons of a carpet, V={a,b,c,....}, and its edges are the rhombi of the carpet E={(ab), etc}.. The Euclidean condition makes sure that two vertices have at most a single edge between them. We shall also assume that the carpet is such that its Wester graph is connected. Otherwise the carpet comes in at least two pieces such that bracing each one of them is enough to make them stable separately, there still would be no way of stabilizing all the pieces relative to each other. TREES: A subgraph T of a graph G is a tree if it has no circuits, i.e. there is a unique edge path connecting each pair of vertices in the tree. A tree is maximal if it cannot be extended. It is a spanning tree if every vertex in G is also in a vertex of T. The two concepts (maximal and spanning) are equivalent (an exercise in graph theory and induction) and the length (number of edges) in a maximal tree is one less than the number vertices in the graph (a second exercise in graph theory.) WESTER'S THEOREM says that we must brace exactly the rhombi in maximal tree in the Wester graph of the carpet. CAVEAT Actually, that's not necessary but may be sufficient in a general carpet. One tacit assumption of Wester seems to be that the carpet is not only connected, but also simply connected, i.e. no "holes". For instance, take a 3x4 square carpet, but remove the middle rod, leaving a 1x2 hole. This has 6 vertical and 4 horizontal ribbons. (You can't make ribbons go through the hole.) But bracing the four base squares, and 3 additonal squares in the vertical walls, stabilizes the holey carpet. The reason being that the rhombi all have the sadme side lenghth. See figure. RICHESON. Finally we get to Richeson's proof for the rectangular case. Note that here, since every horizontal ribbon crosses every vertical ribbon and vice versa, taking one horizontal ribbon and one vertical ribbon, and bracing every rhombus in them, clearly stabilizes the whole carpet. Since there are m+n-1 squares being braced, this gives strong evidence for Wester's theorem. The Wester Graph for the distortion of an initially rectangular, connected and simply connected carpet is a complete, bipartite graph. That is the vertices are either horizontal or vertical. No two vertices of the same class are connected by an edge, but every pair from differing classes are connected by an edge. The bracing described in the previous paragraph corresponds to the maximal tree of one (horizontal) vertex with all of its edges, plus all of the edges through one remote (vertical) vertex. Of course, there are other maximal trees. LEMMA 1. If we brace a maximal tree T, so that all of is edges are squares, the carpect will be rigid. PROOF. Consider an edge=rhombus (az). Since T is spanning, both ribbon=vertices a and z are in T. Being a tree, there is a unique edge path (ab)(bc)(cd)....(yz) from a to z in the tree. Thus all of these rhombi (ab), (bc), ...,(yz) are braced to be squares. But the squares could still be tilted relative to each other. Therefore, ^(a)_|_^(b) and ^(b)_|_^(c), so ^(a) || ^(c). Depending on the parity of edges between them, one side of the square (ab) is parallel or perpendicular to a side of the square (yz). So all the squares (ab), (bc), .... (yz) are aligned (parallel to each other.) So once one of our braced squares is "normal" (horizontal-vertical), all the braced squares are normal, and every rhombus is a normal square as well. The carpet is rigid, and not distorted after all. CONVERSELY? Well, I don't know if the converse is true. Namely, if we unbrace even one of the edges in some Wester tree, the whole carpet becomes unstable. But for the moment I would rather go on to the sufficiency of bracing a Wester tree in a Penrose Carpet. PENROSE CARPET. This is a connected and simply connected piece of a planar tiling by fat (72-108) and thin (36-144) rhombi which comes from a DeBruijn pentagrid projection method. This implies that the ribbons are paritioned into five families of mutually parallel ribbons. Each ribbon r has a direction ^(r) which matches the points on a standard five-pointed star (pentagram). We'll call these the 5 cardinal directions. I believe DeBruijn is credited with proving that a Penrose carpet arrived at in any other way could have been made from a pentagrid. So the ribbons have directions aligned with the pentagram: if ^(a)=^(b) then the two ribbons are parallel (don't intersect, i.e. (ab) is not an edge in the Westergraph.) If they are not paralle, the two ribbons intersect on on fat or skinny rhombus and both ^(a) and ^(b) are cardinal, either 72 or 144 degrees apart. Hence (ab) is a skinny or fat rhombus. Instead of a bipartite Westergraph, we now have a penta1partite Westergraph G. In particular G is also Euclidean. Lets brace a maximal tree T in G. So for each rhombus in T, is either a fat or skinny one, but, a priori, two of them might be tilted relative to each other. We have to show that every rhombus is fat or skinny, and its two directions are cardinal. Let's rotate one of the braced rhombi, call it (ab), to be cardinal in that its sides are aligned with two of the five cardinal directions. For every other edge (yz) in T, there is a succession (ab)(bc)(cd)...(xy)(yz) of braced rhombi (either fat or skinny) connecting vertex a to z. The ribbon b transmits one of the cardinal directions, namely ^(b), from the cardinal rhombus (ab) to the fat or skinny rhombus (bc). So one of the sides of (bc) is cardial. But since it is fat or skinny (being braced!), the whole rhombus (bc) has to be cardinal (not sure here). If this holds, then the whole tree is cardinal (maybe we shoul call it a college of cardinals in honor of current events;-). Finally, we pick any rhombus (pq) which might not be skinny or fat, in a distorted Penrose but with a braced maximal Westertree. Or, even if it is fat or skinny, it might not be cardinal (it might be tilted relative to the cardinal rhombus (ab).) Since its a maximal tree both vertices p and q are in the tree, and there is a unique edgepath from a to p and one from a to q. These two edgpaths might start out to share and initial path of edges, and then fork, or reach p first but continue to q, or vice versa. Consider the edgepath (au)(uv)...(xp) of cardinal rhombi in the Westertree. Since (au) is cardinal, ^(u) is cardinal. Since (uv) is braced it is skinny or fat. Because ^(u) is cardinal, so is ^(v), since (uv) is skinny or fat. So we see that ^(p) is cardinal. Ditto ^(q). Hence the rhombus (pq) must be skinny or fat, and cardinal. I'm not entirely convinced of this last argument. It seems too simple. I'm worrid about the other five cardinal directions you get by a rotation of 36 degrees. But I'm sure that the trick is to use the fact that for rhombi, a single angle with cardinal sides (i.e. 36, 58, 72 or 144 degrees) identifies the rhombus as a fat or a skinny one. Is it enough to know that just one edge of a fat or skinny rhombus is cardinal to predict the whole rhombus to be cardinal. Or could it be "upside down" ?? But I'm tired and will just end here. Maybe someone else will find the hole, and know how to patch it. QUASICRSTALS. I'm pretty sure that we have to use the condition that the quasicrystal frame begins by coming from a hexagrid of parallel planes a la DeBruijn. That makes things "Euclidean" regarding the intersection of ribbons, worms, or carpets in the 3D frame. The worms are contiguous rhombohedra (rhomboids?, bricks?) each sharing a rhombus with the next, these rhombi having parallel faces separating each from the next, define the plane of the worm ^(a). I'm not clear how much machinery needs to be defined here to support the 3D analog of the above argument. On the other hand, the analog of a ribbon in a Penrose may not be a worm in the quasicrystal at all! I might be _sheets_ of rhombohedra. As the rod-and-pinions frameworks sags, and we start bracing, we have a lot of apriori properties that the "worms", or "sheets" will have, in particular non-parallel worms cross on a single rhomboid, non-parallel sheets cross along a worm. Maybe we need to first "Westerize" a cubical lattice to get the drift of the argument?