lattice-theory &laquo; WordPress.com Tag Feed http://en.wordpress.com/tag/lattice-theory/ Feed of posts on WordPress.com tagged "lattice-theory" Fri, 24 May 2013 21:19:32 +0000 http://en.wordpress.com/tags/ en <![CDATA[Some reasons to teach lattice theory to undergraduates]]> http://theorylunch.wordpress.com/2013/03/14/some-reasons-to-teach-lattice-theory-to-undergraduates/ Thu, 14 Mar 2013 14:54:46 +0000 Silvio http://theorylunch.wordpress.com/2013/03/14/some-reasons-to-teach-lattice-theory-to-undergraduates/ First-year students in mathematics and computer science are often troubled with the Schröder-Bernstein theorem, which proves that the natural ordering between cardinal numbers is in fact a partial order, but has a lengthy and convoluted proof. A more accurate study of order structures (often neglected by basic course) would, however, allow to see this fact as an almost immediate consequence of a much simpler, and very powerful, theorem due to Bronislaw Knaster and Alfred Tarski.

Recall that a partially ordered set (shortened as poset) is a pair (X,\leq) where X is a set and \leq is a reflexive, transitive, and antisymmetric binary relation on X. A directed acyclic graph, where x \leq y if and only if there is a directed path from x to y, is a prime example of a poset.

If (X,\leq) is a poset and if \geq is defined as x \geq y if and only if y \leq x, then (X,\geq) is a poset too: just think of the original poset, turned upside down—or, equivalently, to the order graph with all the edges reversed.

Given two posets (X,\leq_X), (Y,\leq_Y), a function F : X \to Y is said to be monotone if a \leq_X b implies F(a) \leq_Y F(b). If F is monotone as a function from (X,\leq_X) to (Y,\leq_Y), then it is also monotone as a function from (X,\geq_X) to (Y,\geq_Y): this simple observation often allows to make proofs in only one direction.

Let (X,\leq) be a poset and let U \subseteq X. An upper bound for U is an element m \in X such that x \leq m for every x \in U. A subset U of X may have a least upper bound, that is, an upper bound m_U such that, if x is any upper bound for U, then m_u \leq x: in this case, such least upper bound is usually indicated as \bigvee U. Dually, a lower bound for U is an \ell \in X such that \ell \leq x for every x \in U, and a greatest lower bound is an element \ell_U = \bigwedge U such that, if x is a lower bound for U, then x \leq \ell_U.

A partially ordered set (X,\leq) such that every subset U of X have both a greatest lower bound \bigwedge U and a least upper bound \bigvee U, is called a complete lattice.

It follows from the definition that a complete lattice (X,\leq) is nonempty. In fact, it must contain at least an element \top = \bigvee X and an element \bot = \bigwedge X, although these two may coincide.

Power sets, ordered by inclusion, are complete lattices: the greatest lower bound of a family of subsets is the intersection of the subsets, and the least upper bound is the union. The real line is not a complete lattice: hovewer, any closed and bounded interval of the real line is a complete lattice. Moreover, if (X,\leq) is a complete lattice, and a,b \in X with a \leq b then the interval [a,b] = \{x \in X \mid a \leq x \leq b\} is a complete lattice: if U \subseteq [a,b], then a \leq x \leq b for every x \in U, thus a \leq \bigwedge U \leq \bigvee U \leq b too.

Finite boolean algebras are complete lattices. Infinite boolean algebra are only guaranteed to have finite meets and joins: indeed, the boolean algebra of the recursive subsets of \mathbb{N} is not a complete lattice. To see this, let X be a recursively enumerable subset of \mathbb{N} which is not recursive: then for a suitable total recursive function f : \mathbb{N} \to \mathbb{N} we have X = f(\mathbb{N}) = \bigvee \left\{\{f(m) \mid 0 \leq m \leq n\} \mid n \geq 0\right\}.

The following statement was proved by Knaster in 1928 for power sets, then by Tarski in 1955 for arbitrary complete lattices.

Knaster-Tarski fixed point theorem. Let (X,\leq) be a complete lattice with top element \top and bottom element \bot; let F : X \to X be a monotone function, and let \mathrm{Fix}(F) = \{x \in X \mid F(x) = x\}. Then:

  1. (\mathrm{Fix}(F), \leq) is a complete lattice.
  2. The smallest fixed point x_m of F satisfies
    x_m = \bigwedge \{x \in X \mid F(x) \leq x\} \,,
    and the largest fixed point x_M of F satisfies
    x_M = \bigvee \{x \in X \mid x \leq F(x)\} \,.
  3. If X is finite, then the sequence defined by x_0 = \bot and x_{n+1} = F(x_n) for n \geq 0 converges to x_m in finitely many iterations, and the sequence defined by x_0 = \top and x_{n+1} = F(x_n) for n \geq 0 converges to x_M in finitely many iterations.

Observe that (\mathrm{Fix}(F), \leq) is not, in general, a full sublattice of (X,\leq): the least upper bound of U in \mathrm{Fix}(F) may not coincide with the least upper bound of U in X. To understand why, let X = \mathbb{N} \sqcup \{a,b\} where the natural ordering of \mathbb{N} is extended by putting a < b and x < a for every x \in \mathbb{N}. Let F : X \to X be defined by F(a) = b and F(x) = x if x \neq a: then the greatest upper bound of \mathbb{N} \subseteq \mathrm{Fix}(F) is a in X, but b in \mathrm{Fix}(F).

Proof. We first prove point 2. Call \mathrm{Fix}^+(F) = \{x \in X \mid x \leq F(x)\} the set of post-fixed points, and x^+_M its least upper bound. Then for every x \in \mathrm{Fix}^+(F) we have x \leq x^+_M, thus also x \leq F(x) \leq F(x^+_M), i.e., F(x^+_M) is an upper bound for \mathrm{Fix}^+(F): then x^+_M \leq F(x^+_M), so that the greatest lower bound of the set of post-fixed point is itself a post-fixed point! But if x^+_M is post-fixed, then F(x^+_M) is post-fixed as well as F is monotone: as x^+_M is by construction the least upper bound of the set of the post-fixed points, F(x^+_M) \leq x^+_M. By antisymmetry of the order relation, F(x^+_M) = x^+_M, i.e., x^+_M is actually a fixed point: but if the least upper bound of the overfixed points is a fixed point, then it cannot help but be the greatest fixed point as well! Dually, F has a least fixed point, which is the greatest lower bound of the set of pre-fixed points.

We now use point 2 to prove point 1. Let U \subseteq \mathrm{Fix}(F) and let \overline{x} = \bigvee U \in X. By monotonicity of F, for every z \in [\overline{x},\top] and x \in U we have x = F(x) \leq F(z), so that \overline{x} \leq F(z) too: as clearly F(z) \leq \top for z \in [\overline{x},\top], F is also a monotone function from [\overline{x}, \top] to itself. As the latter is a complete lattice, F has a least fixed point y in it: if z \in \mathrm{Fix}(F) satisfies x \leq z for every x \in U, then it also satisfies z \in [\overline{x},\top], thus also y \leq z. Hence, this y is the least upper bound of U in \mathrm{Fix}(F), although it might happen that \overline{x} < y. Dually, U has a greatest lower bound in \mathrm{Fix}(F), which might be strictly greater than the greatest lower bound of U in X.

Finally, suppose X is finite. Then the sequence \{x_n\}_{n \geq 0} where x_0 = \top and x_{n+1} = F(x_n) also has finitely many elements, so there exist n \geq 0 and k > 0 such that x_{n+k} = x_n. But x_{n+k} \leq x_{n+k-1} \leq \ldots \leq x_{n+1} \leq x_n by construction, so x_{n+1} = F(x_n) = x_n and x_n \in \mathrm{Fix}(F). If x \in \mathrm{Fix}(F) too, then x = F^{(n)}(x) \leq F^{(n)}(\top) = x_n: then x_n is a fixed point and is greater than every fixed point, so ti must be the greatest fixed point x_M. Dually, if x_0 = \bot and x_{n+1} = F(x_n), then x_n = x_m, the least fixed point, for some n \geq 0. \Box

The Knaster-Tarski theorem can be used to define specific constructions as least or greatest fixed points of monotone functions over complete lattices: if the latter are finite, then it also provides an algorithm for their construction. An example of this is bisimilarity. Recall that a bisimulation on a family \mathrm{Proc} of processes is a binary relation \mathcal{R} \subseteq \mathrm{Proc} \times \mathrm{Proc} such that, for every (P,Q) \in \mathcal{R}:

  1. if P \stackrel{a}{\rightarrow} P' then there exists Q' \in \mathrm{Proc} such that (P',Q') \in \mathcal{R} and Q \stackrel{a}{\rightarrow} Q'; and
  2. if Q \stackrel{a}{\rightarrow} Q' then there exists P' \in \mathrm{Proc} such that (P',Q') \in \mathcal{R} and P \stackrel{a}{\rightarrow} P'.

Two processes are bisimilar if there is a bisimulation that contains them. As any union of bisimulations is a bisimulation, whatever \mathrm{Proc} is, there exists a largest bisimulation on \mathrm{Proc}, called bisimilarity, and usually indicated by \sim.

If \mathrm{Proc} is finite, then bisimilarity can be constructed via the Knaster-Tarski fixed point theorem. In fact, consider the powerset 2^{\mathrm{Proc} \times \mathrm{Proc}} as a complete lattice according to inclusion, and define a function F : 2^{\mathrm{Proc} \times \mathrm{Proc}} \to 2^{\mathrm{Proc} \times \mathrm{Proc}} by saying that (P,Q) \in F(\mathcal{R}) if and only if conditions 1 and 2 of bisimulation are satisfied for (P,Q). Then not only F is monotone, but bisimulations are precisely those \mathcal{R} \subseteq \mathrm{Proc} \times \mathrm{Proc} such that \mathcal{R} \subseteq F(\mathcal{R}): hence, bisimilarity, which is the least upper bound of the family of bisimulations, is precisely the greatest fixed point of F!

Coming back to the Schröder-Bernstein theorem: let f : X \to Y and g : Y \to X be injective functions. Then

F(U) = X \setminus g \left( Y \setminus f(U) \right)

is a monotone function on the complete lattice (2^X, \subseteq). Surely f : U \to f(U) is a bijection. By the Knaster-Tarski theorem, there exists U \subseteq X such that U = X \setminus g(Y \setminus f(U)): but for such U we have g(Y \setminus f(U)) = X \setminus U, so that g: Y \setminus f(U) \to X \setminus U is a bijection too! Define then a bijection h = h_U : X \to Y as follows:

  1. if x \in U, let h(x) = f(x);
  2. otherwise, let h(x) be the unique y \in Y such that g(y) = x.

Observe that the standard proof is obtained by explicitly constructing a fixed point for F.

We conclude with a very interesting converse to the Knaster-Tarski theorem, proved by Anne Davis in 1955, in an article that appeared on the same issue as Tarski’s—actually, immediately after it! Recall that a lattice is a poset where every finite subset has a least upper bound and a greatest lower bound.

Davis’ characterization of complete lattices. Let (X,\leq) be a lattice. Suppose that every monotone function F : X \to X has a fixed point. Then (X,\leq) is a complete lattice.

An interesting exercise, is to construct a proof of Davis’ theorem by showing that meets and joins of arbitrary sets can be seen as fixed points of suitable monotone functions.

]]> <![CDATA[for the video game players, philosophers, science fiction fans, programmers, sceptics, and so on...]]> http://malibrary.wordpress.com/2012/12/18/for-the-video-game-players-philosophers-science-fiction-fans-programmers-sceptics-and-so-on/ Tue, 18 Dec 2012 00:10:32 +0000 Trevor http://malibrary.wordpress.com/2012/12/18/for-the-video-game-players-philosophers-science-fiction-fans-programmers-sceptics-and-so-on/ Hologram image from the film Prometheus.

image from the film Prometheus.

Recently a few articles have come out which posit, seriously, that we may be living in a simulated universe (ala The Matrix). Here’s a little from the University of Washington researchers:

Observable consequences of the hypothesis that the observed universe is a numerical simulation performed on a cubic space-time lattice or grid are explored. The simulation scenario is first motivated by extrapolating current trends in computational resource requirements for lattice QCD into the future. Using the historical development of lattice gauge theory technology as a guide, we assume that our universe is an early numerical simulation with unimproved Wilson fermion discretization and investigate potentially-observable consequences. Among the observables that are considered are the muon g-2 and the current differences between determinations of alpha, but the most stringent bound on the inverse lattice spacing of the universe, b^(-1) >~ 10^(11) GeV, is derived from the high-energy cut off of the cosmic ray spectrum. The numerical simulation scenario could reveal itself in the distributions of the highest energy cosmic rays exhibiting a degree of rotational symmetry breaking that reflects the structure of the underlying lattice. (retrieved from: http://arxiv.org/abs/1210.1847)

Basically, it posist that were cosmic rays to not travel out in all directions with equal force, but instead to travel more readily diagonally across a lattice, then there is at least enough evidence that we need to investigate further the notion that we are simulated. Pretty crazy. And it gets a little crazier because were this to be true, then there are probably many “layers” of simulation, as whenever a culture becomes advanced enough to model a universe, they will do so and create a new “layer.”

If you want to read more, you can read a good explanation of the lattice theory here: http://mashable.com/2012/12/17/the-lattice/.

and a good philosophical argument against it here: http://philosophynow.org/issues/75/The_Simulated_Universe.

]]> <![CDATA[Zero-divisor graphs of nilpotent-free semigroups]]> http://tonbak.wordpress.com/2011/12/02/zero-divisor-graphs-of-nilpotent-free-semigroups/ Fri, 02 Dec 2011 09:02:17 +0000 Dr. Peyman Nasehpour http://tonbak.wordpress.com/2011/12/02/zero-divisor-graphs-of-nilpotent-free-semigroups/ <![CDATA[QCD Phases on the Lattice and Quantum Gravity]]> http://blog.vixra.org/2010/07/24/qcd-phases-on-the-lattice-and-quantum-gravity/ Sat, 24 Jul 2010 14:23:03 +0000 Philip Gibbs http://blog.vixra.org/2010/07/24/qcd-phases-on-the-lattice-and-quantum-gravity/ Yesterday there were some sessions on Lattice techniques aimed at non-specialists attending the ICHEP conference. Apparently the attendance was disappointing. That is not very surprising given the competition from other parallel sessions where new physics could be announced. Lattice theory has been around for a long time and mostly looks at QCD which is far from new.

As an ex-lattice gauge theorist myself I think there are some aspects of it that people working on more sexy subjects such as quantum gravity would benefit from understanding better. In particular they should understand how the phase diagram of QCD at high temperature and density is being charted using these non-perturbative methods. The reason they need to know this is that a similar phase structure should exist in quantum gravity and there is likely to be a strong (but approximate) correspondence through AdS/CFT duality that relates quantum gravity to a QCD-like theory.

In the QCD theory of the strong interactions there is believed to be a temperature known as the Hadgdorn temperature above which nuclear matter breaks down into a quark gluon plasma. This happens at around 10 billion degrees Kelvin. In quantum gravity according to string theory (if you don’t like string theory dont switch off, this is just a short diversion) there is another Hagdorn temperature at around the Planck scale. That’s about 1032 degrees Kelvin. What happens there?

According to string theory the length of strings becomes very large and effectively the concept of the string breaks down. Sometimes string theorists call this the topological phase of string theory because they think that spacetime loses its geometry in the hotter phase. The truth is that not much is known about what really happens because most of string theory is based on perturbative calculations and phase transitions are very non-perturbative. What might happen is that not only geometry of space-time is lost but topology too. In that case it should be called the non-topological phase, or pregeometric phase. To put it another way, spacetime evaporates. Even if you don’t believe in string theory you might still consider this possibility. Some non-string theorists talk about geometrogenesis which is the process of cooling from the high temperature pregeometric phase to the more familiar geometric phase at the start of the big bang.

For now we can get some feel for the phase structure of quantum gravity by looking at the phase structure of QCD which brings me to one of the ICHEP talks from yesterday. However I’ll do that in a separate post in case people get confused and think it was about quantum gravity.

]]> <![CDATA[Complete lattice generated by a partitioning - finite meets]]> http://portonmath.wordpress.com/2009/10/20/generated-lattice-finite-meets/ Tue, 20 Oct 2009 15:14:04 +0000 porton http://portonmath.wordpress.com/2009/10/20/generated-lattice-finite-meets/ I conjectured certain formula for the complete lattice generated by a strong partitioning of an element of complete lattice. Now I have found a beautiful proof of a weaker statement than this conjecture. (Well, my proof works only in the case of distributive lattices, but the case of non-distributive lattices is outside of my research area.)

Let’s denote R = \left\{ \bigcup{}^{\mathfrak{A}}X | X\in\mathscr{P}S \right\} where S is a strong partitioning an element of the complete lattice \mathfrak{A}. Our conjecture is trivially equivalent to the statement that R is closed under arbitrary meets and joins.

That R is closed regarding any joins is obvious. To finish proving the conjecture we need to show that R is closed under arbitrary meets. In this post I prove weaker result that R is closed under finite meets.

I hope this finite case may serve as a model for the general infinite case. However it seems that generalizing it to infinite case is non-trivial.

Theorem Let \mathfrak{A} is a distributive complete lattice and S is a strong partitioning of some element of this lattice. Then R is closed under finite meets.

Proof Let X,Y\in\mathscr{P}S.

Then \bigcup^{\mathfrak{A}} X \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y = \bigcup^{\mathfrak{A}} ((X \cap Y) \cup (X \setminus Y)) \cap^{\mathfrak{A}}\bigcup^{\mathfrak{A}} Y = ( \bigcup^{\mathfrak{A}} (X \cap Y)\cup^{\mathfrak{A}} \bigcup^{\mathfrak{A}} (X \setminus Y))\cap^{\mathfrak{A}} \bigcup Y = ( \bigcup^{\mathfrak{A}} (X \cap Y)\cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y) \cup ( \bigcup^{\mathfrak{A}} (X\setminus Y) \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y) = (\bigcup^{\mathfrak{A}} (X \cap Y) \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}}Y) \cup^{\mathfrak{A}} 0 = \bigcup^{\mathfrak{A}} (X \cap Y)\cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y.

Applying the formula \bigcup^{\mathfrak{A}} X \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y = \bigcup^{\mathfrak{A}} (X \cap Y) \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y twice we get

\bigcup^{\mathfrak{A}} X \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y = \bigcup^{\mathfrak{A}} (X \cap Y) \cap^{\mathfrak{A}} \bigcup (Y \cap (X \cap Y)) = \bigcup^{\mathfrak{A}} (X \cap Y) \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} (X \cap Y) = \bigcup^{\mathfrak{A}} (X \cap Y).

But for any A,B\in R exist X,Y\in\mathscr{P}S such that A=\bigcup^{\mathfrak{A}}X and B=\bigcup^{\mathfrak{A}}Y. So A\cap^{\mathfrak{A}}B = \bigcup^{\mathfrak{A}} X \cap^{\mathfrak{A}} \bigcup^{\mathfrak{A}} Y = \bigcup^{\mathfrak{A}} (X \cap Y) \in R.

]]> <![CDATA[Complete lattice generated by a partitioning of a lattice element]]> http://portonmath.wordpress.com/2009/10/20/complete-lattice-generated-by-a-partitioning-of-a-lattice-element/ Mon, 19 Oct 2009 23:02:25 +0000 porton http://portonmath.wordpress.com/2009/10/20/complete-lattice-generated-by-a-partitioning-of-a-lattice-element/ In this post I defined strong partitioning of an element of a complete lattice. For me it was seeming obvious that the complete lattice generated by the set S where S is a strong partitioning is equal to \left\{ \bigcup{}^{\mathfrak{A}}X | X\in\mathscr{P}S \right\}. But when I actually tried to write down the proof of this statement I found that it is not obvious to prove. So I present this to you as a conjecture:

Conjecture The complete lattice generated by a strong partitioning S of an element of a complete lattice \mathfrak{A} is equal to \left\{ \bigcup{}^{\mathfrak{A}}X | X\in\mathscr{P}S \right\}.

Proposition Provided that this conjecture is true, we can prove that the complete lattice [S] generated by a strong partitioning S of an element of a complete lattice is a complete atomic boolean lattice with the set of its atoms being S (Note: So [S] is completely distributive).

Proof Completeness of [S] is obvious. Let A\in[S]. Then exists X\in\mathscr{P}S such that A=\bigcup{}^{\mathfrak{A}}X. Let B=\bigcup{}^{\mathfrak{A}}(S\setminus X). Then B\in[S] and A\cap B=0. A\cup B=\bigcup{}^{\mathfrak{A}}S is the biggest element of [S]. So we have proved that [S] is a boolean lattice.

Now let prove that [S] is atomic with the set of atoms being S. Let z\in S and A\in[S]. If A\neq z then either A=0 or x\in X where A=\bigcup^{\mathfrak{A}}X, X\in\mathscr{P}S and x\neq z. Because S is a strong partitioning, \bigcup^{\mathfrak{A}}(X\setminus\{z\})\cap^{\mathfrak{A}}z=0 and \bigcup^{\mathfrak{A}}(X\setminus\{z\})\neq 0. So A=\bigcup^{\mathfrak{A}}X=\bigcup^{\mathfrak{A}}(X\setminus\{z\})\cup^{\mathfrak{A}}z\nsubseteq z.

Finally we will prove that elements of [S]\setminus S are not atoms. Let A\in[S]\setminus S and A\neq 0. Then A\supseteq x\cup^{\mathfrak{A}}y where x,y\in S and x\neq y. If A is an atom then A=x=y what is impossible. QED

The above conjecture as a step to solution to the original conjecture may also be considered for the polymath research problem. Or maybe we should research both these two problems in a single polymath set, as the solution of one of them may inspire the solution of the other of these two problems.

]]> <![CDATA[Partitioning of a lattice element: a conjecture]]> http://portonmath.wordpress.com/2009/10/17/partitioning-lattice-element/ Sat, 17 Oct 2009 16:49:54 +0000 porton http://portonmath.wordpress.com/2009/10/17/partitioning-lattice-element/ Let \mathfrak{A} is a complete lattice. Let a\in\mathfrak{A}.

I will call weak partitioning of a a set S\in\mathscr{P}\mathfrak{A}\setminus\{0\} such that

\bigcup{}^{\mathfrak{A}}S = a \text{ and } \forall x\in S: x\cap^{\mathfrak{A}}\bigcup{}^{\mathfrak{A}}(S\setminus\{x\}) = 0.

I will call strong partitioning of a a set S\in\mathscr{P}\mathfrak{A}\setminus\{0\} such that

\bigcup{}^{\mathfrak{A}}S = a \text{ and } \forall A,B\in\mathscr{P}S: (A\cap B=0\Rightarrow \bigcup{}^{\mathfrak{A}}A\cap{}^{\mathfrak{A}}\bigcup{}^{\mathfrak{A}}B = 0).

Question: Do exist complete lattices for which weak partitioning and strong partitioning are not the same?

If this conjecture (that it is the same for arbitrary complete lattices) is indeed false for arbitrary complete lattices, we must find the cases when it is true. (I strongly suspect that it is true for distributive complete lattices.)

]]>