Why is addition not commutative under PM's notion of relation number?

Question

Quoting Bertrand Russell's "The Principles of Mathematics" p468 §299:

299. It is worth while to repeat the definitions of general notions involved in terms of what may be called relation-arithmetic*. If P, Q be two relations such that there is a one-one relation S whose domain is the field of P and which is such that Q = (˘S)PS, then P and Q are said to be like. The class of relations like P, which I denote by λP, is called P’s relation-number. If the fields of P and Q have no common terms, P + Q is defined to be P or Q or the relation which holds between any term of the field of P and any term of the field of Q, and between no other terms. Thus P+Q is not equal to Q+P. Again λP+λQ is defined as λ(P + Q). *Cf. Part IV, Chap. xxix, §231.

Where Principia Mathematica symbols and terminology use:

• the symbol ˘ to denote the converse relation.
• the word "field" to denote the union between the domain and co-domain of the relation.

First, a point of confusion is the tendency to assume binary relations in calling out "domain" and "co-domain". Is it not the case that n-place relations may have up to n distinct "domains", each of which supplies a value for a given place in a particular relationship of the relation's extension?

Second, as Russell specifies the two (n-ary) relations have "no terms in common" between their fields (as n-way unions of their respective n-domains), it seems the only compositional intension would be the disjunction (or) of their respective extensions. To bring this down to simple relation-tables: One takes the rows of P and the rows of Q and puts them in the same relation table, regardless of row-order. This is what one would expect of the + operator's logical analogue of or which, also, is commutative.

What am I missing?

Answer 1

Re: (1), I question whether relations in this context are $n$-ary or specifically binary - your question suggests the former, but relation algebra is usually set up for the latter (and I don't know how to make sense of the converse of a more-than-$2$-ary relation).

Re: (2), I believe the key point is the clause

"or the relation which holds between any term of the field of P and any term of the field of Q

which distinguishes $P+Q$ from the mere union of $P$ and $Q$ (which is indeed commutative) - the idea is that if $p$ is in the field of $P$ and $q$ is in the field of $Q$, then $(p,q)\in P+Q$ but $(q,p)\not\in P+Q$ and conversely $(q,p)\in Q+P$ but $(p,q)\not\in Q+P$.

Answer 2

It makes sense that addition between relations is not commutative. Take this easy example: let P be the series of natural number in order of magnitude and let Q the ordered couple (-1;-2). Then the series P+Q is just the series that you get if, metaphorically, you "put" the couple Q at the "end" of P; while Q+P is just the series that you get by putting Q at the beginning of P. The first series P+Q is a well ordered series but is not a progression (i.e: is not isomorphic to the series of natural number in order of magnitude; that term is used in Principia Mathematica *263); the second one, i.e, Q+P, is still a progression.

When P and Q are well ordered finite series, then it's true that P+Q=Q+P; but that result isn't true in general.