5
$\begingroup$

Is there a difference between the join operator, $\wedge$, and the union of a set?

In particular, what is the join of $a \wedge b $ and $b \wedge c$? Is it $a\wedge b \wedge c$ or is it $0$?

I seem to have read both answers (in physics textbooks where they have skimmed over the details of how they define their operators).

The answer $0$ comes from a geometric algebra book studying projective geometry, where they identify the geometric exterior product (= Grassmann's exterior product) with the join operator. Since the exterior product is anti-commutative and associative it follows that for vectors $a$, $b$, $c$,

$a\wedge a=0 \implies (a\wedge b)\wedge(b\wedge c) = 0$.

They went on to define the meet in terms of the exterior product

$(a\vee b)^* = a^*\wedge b^*$

where the star denotes the dual of the (in this case) vectors $a$ and $b$. (see for example Universal Geometric Algebra by David Hestenes)

The set union answer comes from a discussion of lattices and probabilities (different book) where they identify join and meet with set union and intersection. So for example, (since my terminology might be wrong), they drew a lattice such as follows,

    {a,b}    /    \  {a}    {b}    \    /      {}  

So in this case the join is $a \cup b$.

Does join/meet have a strict definition that is distinct from union/intersection - or can you define it however you like given the circumstances? If its the latter, which is the more usual definition?

  • 0
    @CarlMummert Ba! The book I was using was `Geometric Algebra for physicists`. Just noticed that in the article I linked they were a little more careful of their definition of the join, stating that it matches the exterior product only when the exterior product is not equal to zero! I guess that solves my problem!2011-09-29

4 Answers 4

10

Join is a lattice-theoretic concept that need not have anything to do with unions. For instance, the positive integers partially ordered by divisibility are a lattice in which the join of two integers is their least common multiple and the meet is their greatest common divisor. Another example is $\mathbb{R}^2$ partially ordered so that $(a,b) \preceq (c,d)\text{ iff }a\le c\text{ and }b\le d;$ in that lattice $(a,b)\lor (c,d) = (\max\{a,c\},\max\{b,d\}),$ and $(a,b)\land(c,d) = (\min\{a,c\},\min\{b,d\}).$

For yet another example, if $X$ is a set, and $\mathbb{T}$ is the set of all topologies on $X$, $\langle \mathbb{T},\subseteq\rangle$ is a lattice in which the meet of two topologies is their intersection, but the join of two topologies generally is not their union: rather, it’s the topology generated by taking their union as a subbase.

  • 0
    @Tom: Yes, it depends on the lattice. You might want to take a look at [this article](http://en.wikipedia.org/wiki/Lattice_%28order%29).2011-09-29
3

There is a particular type of lattice - the lattice of subsets - where each element of the lattice can labelled by subsets of some set maximal set $S$, and the join (respectively, meet) of any two elements of the lattice can be evaluated by looking for the element of the lattice labelled by the set union (resp. intersection) of the two given elements' labelling sets. However, just because the join is designated to be the union in this example does not mean that the join is defined that way or even carries the same structure generally. The typical definitions go like this:

join: unique supremum, or least upper bound; meet: unique infimum, or greatest lower bound

For example, you can take partitions of a set instead of subsets and the join and meet will no longer be evaluable in terms of simple unions or intersections. There is a diagram of this on the Wikipedia page for lattices. Of course, lattices don't need elements that are necessarily sets - for example you could alternatively define a lattice of Young diagrams (which, admittedly, represent partitions), or vertices of a hypercube (which, admittedly, are isomorphic to lattices of subsets, but hopefully you get the idea).

  • 0
    yes your right, I looked at a $f$e$w$ of the original references (rather than my physics book $w$hich was derived from them), and they add the caviat that the exterior product must not vanish for it to be considered a meet. I have copied the appropriate paragraph as a separate answer.2011-09-29
2

Note that $\wedge$ is just a symbol. This symbol can have different interpretations. The exterior product is very different from anything coming from lattices. So you have to know from the context what the interpretation of $\wedge$ is.

Let us forget about exterior products for now.
If we are talking about lattices (i.e., partially ordered sets where any two elements $a$ and $b$ have a least upper bound and a greatest lower bound, often denoted by $a\vee b$ and $a\wedge b$, respectively), I am not sure why you call $a\wedge b$ the join of $a$ and $b$. I would rather call $a\vee b$ the join of $a$ and $b$.

In any case, unions ($\cup$) and intersections ($\cap$) enter the picture as follows: Namely, if the lattice you are looking at is the partial order of subsets of a set, ordered by $\subseteq$, then the intersection of two sets is the greatest lower bound of the two sets in that partial order and the union is the least upper bound.

2

After being prompted by Carl I tried to find the original references I found the answer. My question, it turns out, was explicitly dealt with in Projective Geometry with Clifford Algebra, by Hestenes and Ziegler, where it says:

We define the join $J$ of blades $A$ and $B$ (to within a scalar factor) as a common dividend of lowest step. The support of $J$ is then the usual ‘lattice join’ of the supports of $A$ and $B$. We define $J$ explicitly by

$J =A\wedge B$ (3.4)

when $A\wedge B$ does not vanish.... When $Step A + Step B \ge n$ and the join of $A$ and $B$ can be identified with the unit pseudoscalar $I$.

This entirely answers my question - for I was worried about the case when $A\wedge B = 0$, where the join is identified as the pseudoscalar (as a special case).