Product-Decomposition of distributive lattices - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T20:39:54Z http://mathoverflow.net/feeds/question/97844 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/97844/product-decomposition-of-distributive-lattices Product-Decomposition of distributive lattices Sebastian 2012-05-24T15:06:48Z 2012-07-10T14:03:32Z <h2><strong>EDIT</strong></h2> <p>I now (strongly) believe that the following claim answers my question (see the text below). However, if it does, then I am sure that it is known. It is not difficult to prove and the question must have arised before. Thus, can anybody tell me a reference where the following is shown or mentioned? </p> <blockquote> <p><strong>Claim</strong> For each $n \in \mathbb{N}$, a bounded distributive lattice $\mathbf{D}$ can be written as the direct product of $n$ product-irreducible bounded distributive lattices if the there exist $n$ (but not more) elements $a_1,\ldots,a_n \in D \setminus \{0\}$ such that $a_i \wedge a_j = 0$ for all $i \neq j$ and $\bigvee a_i = 1$.</p> </blockquote> <p>I would like to use this fact for a paper, but since I assume its known, I would like to give a reference instead of a proof (in particular since it only appears as an example).</p> <hr> <p><em>I am sorry for having a question that is probably very basic for lattice theorists. However, I was not able to find the answer by looking into textbooks or by using the magic google mashine.</em></p> <p>Let $\mathbf{D} = \langle D,0,1,\vee,\wedge \rangle$ be a bounded (not necessarily finite) distributive lattice and let $n$ be an integer such that $\mathbf{D}$ can be decomposed into the (direct) product of $n$ product-irreducible distributive lattices. </p> <p>Is there any inner property of the lattice that characterizes the number $n$ in a different, preferrably easy, way?</p> <p>What I would like to have is a statement similar to the following one holding (I think) for Boolean algebras:</p> <blockquote> <p><strong>Proposition.</strong> For each $n \in \mathbb{N}$, a Boolean algebra $\mathbf{B}$ can be written as the direct product of $n$ product-irreducible Boolean algebras if and only if $\mathbf{B}$ has $2^n$ elements.</p> </blockquote> <p>I believe the truth of this proposition can best be seen by using the duality between Boolean algebras and Stone spaces and the simple observation that any Stone space is the coproduct of $n$ coproduct-irreducible Stone spaces if and only if it contains exactly $n$ elements (a consequence of the fact that the coproduct in the category of Stone spaces is the disjoint union).</p> <p>I think that the following statement is also true:</p> <blockquote> <p><strong>Proposition.</strong> For each $n \in \mathbb{N}$, a <em>finite</em> bounded distributive lattice $\mathbf{D}$ can be written as the direct product of $n$ product-irreducible bounded distributive lattices if the graph given by the partial order restricted to the nonzero join-irreducible elements among $\mathbf{D}$ has exactly $n$ connected components.</p> </blockquote> <p>However, if the distributive lattice is infinite, then I am not so sure what could said about the number $n$ (if such a characterization is possible at all). Does anybody know?</p> http://mathoverflow.net/questions/97844/product-decomposition-of-distributive-lattices/97847#97847 Answer by Patricia Hersh for Product-Decomposition of distributive lattices Patricia Hersh 2012-05-24T15:54:07Z 2012-05-25T10:11:46Z <p>Edit: You seem to have edited your question in such a way that my answer no longer fits your new question. In particular, what I call your claim'' below is now a proposition, and you have a new claim. My answer does not address the new revised question.</p> <p>Original answer: For what it is worth, your claim seems to follow in the case of <em>finite</em> distributive lattices from the Fundamental Theorem of Finite Distributive Lattices (FTFDL), which I quote from p. 290 of Enumerative Combinatorics, Volume I, by Richard Stanley:</p> <p>Theorem 3.4.1 (quoted from ECI, proved originally by Birkhoff): Let $L$ be a finite distributive lattice. Then there is a unique (up to isomorphism) poset $P$ for which $L = J(P)$. </p> <p>Given a finite poset $P$, then $J(P)$ is the poset of order ideals'' in $P$, ordered by containment, where an order ideal is a set $S$ such that $v\in S$ and $u\le v$ implies $u\in S$.</p> <p>In the finite case, one recovers $P$ from $J(P)$ by taking the subposet of join irreducibles (cf. Proposition 3.4.2 in ECI), so it seems interesting that you are also using that operation. Stanley also mentions in the notes at the end of chapter 3 of ECI that there are generalizations to the non-finite case of FTFDL by M.H. Stone and by H.A. Priestley, so maybe these papers could help.</p> http://mathoverflow.net/questions/97844/product-decomposition-of-distributive-lattices/100720#100720 Answer by NN for Product-Decomposition of distributive lattices NN 2012-06-26T21:04:51Z 2012-06-26T21:04:51Z <p>The following can surely be found in Birkoff, lattice theory. Almost surely also in Gratzer.</p> <p>The decompositions of a poset in finite direct products are the same as the "partitions of unity" in a certain Boolean algebra. When the poset has universal bounds (a mimimum element 0 and a maximum element 1) the Boolean algebra is the Boolean algebra of central elements of the poset. When the poset is a distibutive lattice with 0 and 1, the central elements are exactly the complemented elements of the lattice, and so the center is the largest Boolean subalgebra of the lattice. The poset is a direct product of n directly indecomposable components iff the center is a Boolean algebra with N atoms. So your claims are correct, with the N atoms of the center being the N elements of the distributive lattice which are complemented but are not <em>disjoint</em> union of two smaller nonzero elements. However, also in the finite case, an element can be directly irreducible without being join irreducible; consider the following Hasse diagram:</p> <p>1 / \ a b \ / c | 0</p> <p>i.e. the length 3 distributive lattice with two co-atoms a,b and one atom c. In the dual of the above lattice, join irreducible elements are indecomposable. In a finite distributive lattice, coincidence of join irreducible elements with directly indecomposable elements, plus the dual condition, happens iff the lattice is a direct product of chains. </p> <p>So I do not see real advantages in the use of join irreducible elements in comparation with indecomposable ones to describe direct decompositions. (But note that I never have been interested in combinatorics, so that you might see things differently for your specific application) </p> <p>As you note, to express the results in terms of the poset of join-irreducible elements (instead of the center of the lattice) you can use the Bikhoff transform. If you want to use Birkhoff transform (categorical dual equivalence between finite posets and finite distributive lattices, so that disjoint unions of posets [of join-irreducible elements] correspond to direct products for the lattices), the infinite case is the following (and in particular note that it does not apply to all distributive lattices, only to special ones): </p> <p>posets are dually equivalent to algebraic and dually algebraic distributive lattices and also to Alexandroff discrete topological spaces: the poset is the poset of points with the specialization order; the lattice is the lattice of open sets. Also, the lattice is the lattice of order ideals in the poset, and the poset is the poset of (completely) join-irreducible elements of the lattice. Central elements of the lattice correspond to clopen sets of the topological space.</p>