A note on the correct definition of a proportion space

Recently I’ve been learning about torsors. A torsor is, in informal terms, an algebraic structure that’s like a group with the identity element “forgotten”. The precise definition (whose connection to the informal notion is not meant to be obvious!) is: a torsor is a set equipped with a group acting on such that for any two elements and of , there is a unique with .

My interest in torsors comes from the idea that the set of all possible values of a dimensional physical quantity, such as mass, is intrinsically a torsor, because the choice of the unit of measurement is arbitrary. Therefore it might make sense to formalize dimensional analysis using the notion of a torsor.

I might have more to say about torsors themselves, and the connection to dimensional analysis, in a future post. But for now, I just want to write about one alternative way of capturing the idea of a “group with the identity element forgotten”, which was described in a blog post by Matt Baker.

Baker introduces the notion of a proportion space, which is a non-empty set equipped with an equivalence relation on such that:

1. (PS1) For any two elements , of , we have .
2. (PS2) For any three elements , , of , there is a unique such that .

The idea here is that represents a proportion statement, or four-part analogy: for any four elements , , , of , the relation is thought of as having the meaning “ is to as is to ”. Another notation for this meaning which is sometimes seen is

For example, in elementary mathematics, where , , , are numbers, this may be used to express that the ratio of to is the same as the ratio of to , or in symbols . In linguistics, the same notation may be used with words in place of , , , to represent an analogy, e.g.

Even if you heard about torsors for the first time when reading this article, you likely correctly inferred that “torsors” was the plural of “torsor” by way of the above analogy. Note however that the “existence” part of (PS2) does not seem to be appropriate in a linguistic context: I can easily think of triples of words where there isn’t a natural fourth word to complete the analogy, such as

Anyway, Baker claims that proportion spaces, as defined above, are essentially the same as torsors. More precisely, he sketches a proof of the following theorem:

For every set , there is a natural bijection between the proportion spaces on and the equivalence classes of torsors on . (See Baker’s post for details on how equivalence of torsors is defined.)

Unfortunately, this theorem appears to be false for the chosen definition of proportion spaces. The flaw is in the second part of the proof, where a torsor is constructed from a proportion space. In order to construct the group the torsor is to be over, Baker takes the quotient set and defines a product operation on it by the rule that for any four elements , , , of , we have

where is the unique such that given by (PS2). But he omits the verification that this operation is well-defined, i.e. that for any eight elements , , , , , , , of , such that and , we have

It turns out that for a general proportion space, this can’t be verified, because it’s a false statement!

Here’s a counterexample. Let , and partition the 8 elements of into equivalence classes as follows:

The resulting equivalence relation clearly satisfies (PS1), since , , and are all in the same equivalence class. As for (PS2), you can scan the above equations and verify that for each , each of the three equivalence classes , , has exactly one element with as its second projection. Since each belongs to exactly one equivalence class, this implies (PS2). So we may regard as a proportion space. However, if we take

then we have

so for the product operation to be well-defined, we would have to have

and this is false.

By the way, I didn’t find this counterexample by myself. I used Mace4 to do it, using the following input:

formulas(assumptions).

R(x,y,x,y). % reflexivity
R(x,y,w,z) -> R(w,z,x,y). % symmetry
R(x,y,w,z) & R(w,z,u,v) -> R(x,y,u,v). R(x,x,y,y). % transitivity
R(x,y,w1,z) & R(x,y,w2,z) -> w1 = w2. % PS2 (uniqueness)
exists w R(x,y,w,z). % PS2 (existence)

end_of_list.

formulas(goals).

  R(x1,y1,x2,y2) & R(z1,w1,z2,w2)
& R(x1,y1,u1,z1) & R(x2,y2,u2,z2)
  -> R(u1,w1,u2,w2).

end_of_list.

Anyway, what this shows is that the definition of proportion space given by Baker is too weak: to get the desired equivalence with torsors, we need to add an additional axiom to the definition, or strengthen one of the existing axioms. My preference is to strengthen (PS1) to the following statement:

(PS1′) For any four elements , , , of , if , then .

This implies (PS1), since we may take and , and then the antecedent holds by the reflexivity of while the succedent is (PS1). And it implies that the product operation given by (1) is well-defined, because for any ten elements , , , , , , , , , of such that

we have by symmetry and transitivity applied to (4), (2) and (5), and hence by (PS1′) we have . Applying (PS1′) to (3) as well gives us , and putting these conclusions together using transitivity gives

as desired.

Edit (9 June 2024): It turns out that I made a mistake too, in the opposite direction! Axiom (PS1′) is in fact too strong. For example, if you consider a group as a torsor over itself, then when turning the torsor into a proportion space, (PS1′) becomes

but this is only true when the group is Abelian (in general, only implies ).

In terms of the operation, (PS1′) is equivalent to the identity

There is a well-known alternative characterization of torsors, which says that a torsor is equivalent to a heap, which is a set equipped with a ternary operation on satisfying the identities

Mace4 readily finds a counterexample where the heap axioms are satisified, but not (2).

OK, here’s my second attempt at amending the definition: I suggest keeping (PS1) in its original form, which says that , and adding a third axiom

This can be understood as saying that the binary relation on defined by

is transitive (while (PS1) says it’s reflexive; one can also easily prove that it’s then symmetric). Checking this with Prover9/Mace4, the conjunction of (PS1), (PS2) and (PS3) (along with the stipulation that is an equivalence relation) does seem to both imply, and be implied by the heap axioms.

Here’s another proof that the product defined by (1) is well-defined, using (PS3) instead of (PS1′). We are given ten elements , , , , , , , , , of such that

and we need to prove that . As before, by symmetry and transitivity we obtain . Then we just put this together with (3), using (PS3), and we’re done.

Let’s also check that if we have a -torsor , then the relation on defined by

satisfies (PS3). (Here denotes the unique such that , which exists by the characteristic property of a torsor.) Indeed, if , and , then

using the identity , which is related to the third heap axiom.