Penrose tilings and noncommutative geometry - MathOverflow
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2013-05-19T16:19:35Z
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http://mathoverflow.net/questions/42582/penrose-tilings-and-noncommutative-geometry
Penrose tilings and noncommutative geometry
John Mangual
2010-10-18T04:49:42Z
2010-10-25T15:25:35Z
<p>Are there "elementary" resources on <a href="http://en.wikipedia.org/wiki/Penrose_tiling" rel="nofollow">Penrose Tilings</a> in relation to <a href="http://www.alainconnes.org/docs/book94bigpdf.pdf" rel="nofollow">noncommutative geometry</a>? It's all a big blur to me. There are two transformations S and T that can grow the tilings and every tiling corresponds to a sequence of S's and T's. Somehow, there is a C* algebra related to this.</p>
<p>Can the elements of this algebra be interpreted as observables in a quantum mechanical system? What is the "geometry" of this noncommutative space?</p>
<p><img src="http://www.jdawiseman.com/papers/trivia/penrose_tiling.png" alt="Penrose Tiling"></p>
http://mathoverflow.net/questions/42582/penrose-tilings-and-noncommutative-geometry/42602#42602
Answer by Thomas Riepe for Penrose tilings and noncommutative geometry
Thomas Riepe
2010-10-18T09:05:56Z
2010-10-20T19:24:11Z
<p>These (<a href="http://arxiv.org/abs/hep-th/9802129" rel="nofollow" title="arxiv">1 </a>, <a href="http://arxiv.org/abs/math-ph/0204022" rel="nofollow" title="arxiv">2 </a>, <a href="http://arxiv.org/abs/hep-th/9701078" rel="nofollow" title="arxiv">3 </a>) look to me elementary, but I only browsed them very superficially. EdiT: A recent article: <a href="http://www.sciencedirect.com/science?_ob=ArticleURL&_udi=B6TJ8-516669D-3&_user=10&_coverDate=10%2F08%2F2010&_rdoc=1&_fmt=high&_orig=search&_origin=search&_sort=d&_docanchor=&view=c&_acct=C000050221&_version=1&_urlVersion=0&_userid=10&md5=2bcf6754510fa9c0b3cf76dc0cf9a039&searchtype=a" rel="nofollow" title="link">"C*-algebras of Penrose hyperbolic tilings"</a>.</p>
http://mathoverflow.net/questions/42582/penrose-tilings-and-noncommutative-geometry/42676#42676
Answer by Nick S for Penrose tilings and noncommutative geometry
Nick S
2010-10-18T17:51:35Z
2010-10-18T17:51:35Z
<p>I don't know about the relation to non-commutative geometry, but for the C*-algebra you should take a look at this paper:</p>
<p>J.E. Anderson and I.F. Putnam, Topological invariants for substitution tilings and
their associated C*-algebras, Ergodic Theory & Dynamical Systems 18 (1998), 509-
537.</p>
http://mathoverflow.net/questions/42582/penrose-tilings-and-noncommutative-geometry/42677#42677
Answer by José Figueroa-O'Farrill for Penrose tilings and noncommutative geometry
José Figueroa-O'Farrill
2010-10-18T18:00:37Z
2010-10-18T18:00:37Z
<p>Much of <a href="http://arxiv.org/find/all/1/au%3A+kellendonk/0/1/0/all/0/1" rel="nofollow">the work of Johannes Kellendonk</a> deals with physical applications of Penrose tilings. Not sure I'd call this "elementary", though -- but your mileage might vary.</p>
http://mathoverflow.net/questions/42582/penrose-tilings-and-noncommutative-geometry/43008#43008
Answer by Dan for Penrose tilings and noncommutative geometry
Dan
2010-10-21T04:53:02Z
2010-10-21T04:53:02Z
<p>Chp 9 in "A WALK IN THE NONCOMMUTATIVE GARDEN" by ALAIN CONNES AND MATILDE MARCOLLI, and you can find the references there.</p>
http://mathoverflow.net/questions/42582/penrose-tilings-and-noncommutative-geometry/43045#43045
Answer by Paul Siegel for Penrose tilings and noncommutative geometry
Paul Siegel
2010-10-21T12:33:13Z
2010-10-21T12:39:03Z
<p>I don't have any magical references for you, nor do I understand the NCG point of view on the Penrose tiling all that well. I learned just enough about this to convince myself that I didn't need to learn more, and I'll try to convey the solace that I achieved.</p>
<p>First, let me say a few words about the philosophy of Connes' NCG. The basic premise is that sometimes when you have a space equipped with an equivalence relation it is not necessarily a good idea to pass to the space of equivalence classes. The best-advertised justifications for this premise are examples where the space is nice and the equivalence relation is nice but the quotient is miserable, but I want to remark before going on that the tools of NCG are still extremely useful when the quotient is nice (such as when a manifold is viewed as the quotient of its universal cover by its fundamental group).</p>
<p>Here is how the philosophy plays out for Penrose tilings. One regards a Penrose tiling as a tiling of the plane by isometric copies of two specific triangles that are only allowed to connect in a few very specific ways. See page 181 of <a href="http://www.alainconnes.org/docs/book94bigpdf.pdf%20Connes%27%20book" rel="nofollow">Connes' Book</a> for pictures (and a more detailed, but elementary, description of what I am about to say). We declare two Penrose tilings to be equivalent if there is an isometry of the plane which carries one to the other. There is a more down to earth way to express the space of Penrose tilings modulo this equivalence relation by interpreting it combinatorially. It is the space of sequences $(a_n)$ of $0$'s and $1$'s with the property that $a_{n+1} = 0$ whenever $a_n = 1$ modulo the equivalence relation of eventual equality: $a_n = b_n$ for $n$ sufficiently large. What isn't necessarily obvious at the outset is that this space has absolutely no sensible local structure. In terms of the Penrose tilings, this can be expressed by observing that any finite patch in one tiling appears infinitely often in any other tiling by the same tiles.</p>
<p>There is by now a standard way of fitting such a setup into the machinery of NCG. Any equivalence relation on a space gives rise to a certain groupoid whose objects are points in the space and whose morphisms are determined by the equivalence relation. The idea is supposed to be that the groupoid keeps track of which points in the space are equivalent as well as the reason why they are equivalent, rather than violently collapsing each equivalence class down to a single point. In most cases it is possible to equip the groupoid with a compatible system of measures and then use integration with respect to that system to define a convolution product on a suitable space of functions on the groupoid (generally the original space and the groupoid have a topology and the space of functions is the space of continuous functions). The C* algebra of the groupoid is defined as a certain completion of this convolution algebra.</p>
<p>Notice that none of that last paragraph involves the details of the Penrose tiling construction in any essential way: it is a purely mechanical procedure which starts with an equivalence relation and spits out a C* algebra. One can think of this as analogous to the procedure of replacing a function with an operator (in this case convolution against the function) which flies under the moniker "quantization" in physics. Indeed, various sorts of quantization in physics can be realized via convolution algebras on groupoids - though the groupoids are generally more complicated than those which arise from an equivalence relation. As far as I know, the relationship between Penrose tilings and physics ends with this analogy. I could be tragically wrong.</p>
<p>So when you ask about the "geometry" of the noncommutative space of Penrose tilings, you are really asking about the structure of the C* algebra spat out by the machine described above. What is the structure of this C* algebra? I don't really know. Connes claims that it has trivial center and a unique trace, which has consequences on its "measure theoretic" structure. It was at this point in the story that I noticed that I'm more interested in the geometry of manifolds and decided to move on. Still, it wouldn't surprise me if there turns out to be way more to this story than meets the eye.</p>
http://mathoverflow.net/questions/42582/penrose-tilings-and-noncommutative-geometry/43530#43530
Answer by Jon Bannon for Penrose tilings and noncommutative geometry
Jon Bannon
2010-10-25T15:25:35Z
2010-10-25T15:25:35Z
<p>There is a concise description of this in section 1.1 of this nice paper <em>Trees, Ultrametrics, and Noncommutative Geometry</em> by Bruce Hughes: <a href="http://arxiv.org/abs/arXiv:math/0605131" rel="nofollow">http://arxiv.org/abs/arXiv:math/0605131</a>.</p>
<p>This isn't particularly elementary, I guess, but it is a line of investigation that embarks from Connes's example. </p>
<p>I hope it is helpful!</p>