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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?

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I thought the join and meet spitted out a single value. That is, they are defined on a partially ordered set, and one gives the supremum of the sets, while the other the infimum...So they are entirely different things from the union (which gives a set)! –  user1729 Sep 29 '11 at 11:33
    
@Swlabr - my terminology might be wrong, I have edited my question to show the lattice's they were drawing ( a lattice of sets - if that makes sence) –  Tom Sep 29 '11 at 11:37
    
I think the exterior product is something completely different from the join in lattice & poset theory. So I don't understand this remark –  Thomas Rot Sep 29 '11 at 11:44
    
@anon, they do go on to define the meet in terms of the duals (as edited) (i.e. it obeys the de Morgan rule) –  Tom Sep 29 '11 at 11:48
2  
The symbol $\wedge$ is the "meet". It stands for the greatest lower bound of two elements in a poset. The symbol $\vee$ is the "join". It stands for the least upper bound of two elements in a poset. I have downvoted because the text does not match the symbol in the question, or in the Hasse diagram. See en.wikipedia.org/wiki/Lattice_%28order%29#Lattices_as_posets –  Carl Mummert Sep 29 '11 at 11:54

4 Answers 4

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.

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Thank you for you answer. So would I be write in saying that the join and meet are defined on a lattice by lattice basis (so long as the meet certain rules such as associativity?). So that the answer to my question is it depends how you define you lattice? –  Tom Sep 29 '11 at 11:55
2  
On a lattice, the join of two elements $a,b$ is their supremum, denoted $a \vee b$, and the meet of the two elements is their infimum, denoted $a \wedge b$. The definition of a lattice says that it is a poset with least and greatest elements such that the join and meet are defined for all pairs of elements. –  Carl Mummert Sep 29 '11 at 12:05
    
@Tom: Yes, it depends on the lattice. You might want to take a look at this article. –  Brian M. Scott Sep 29 '11 at 12:07

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).

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So the least upper bound of the 2 areas defined by the (non-parallel) vectors $a,b$; $b,c$ is the volume? (or does it depend upon exactly how I have defined my algebra?) –  Tom Sep 29 '11 at 12:11
    
@Tom: I've never come across exterior algebras presented as lattices - and technically, under the wedge product, they aren't (due to $e_i\wedge e_i=0$), but they do have lattice structure outside of the special vanishing cases. But not only do exterior algebras only have partial lattice structure, they have much more extraneous structure due to the scalar multiplicative factors, anticommutativity, and addition. The "nilpotence" in differential forms is a key geometric feature, and it serves as an important distinction between exterior algebras and bona fide lattices. –  anon Sep 29 '11 at 12:56
    
yes your right, I looked at a few of the original references (rather than my physics book which 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. –  Tom Sep 29 '11 at 13:04

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.

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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).

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