[FOM] Foundationalist introduction to the field with one element, part I

[Andrius Kulikauskas]

Mario, Harvey, Colin,

Thank you for your responses to my question about the possible relevance 
of the one-element field F1 to the Foundations of Mathematics.

I will add two other interpretation of the Gaussian binomial 
coefficients.  Richard Stanley's "Enumerative Combinatorics" includes 
the counting of k-dimensional subspaces of an n-dimensional vector space 
over a field of characteristic q.  He shows that these coefficients also 
count the number of Young tableaux which fit within a  k x (n-k) 
rectangle where each tableaux cell has weight q.

But they also count the k-simplexes in an n-simplex where the vertices 
of the n-simplex are given weights 1, q, q2, q3...qn-1 and the edges are 
all given weights 1/q and the weight of a k-simplex is the product of 
the weights of its vertices and edges.  In a k-simplex, all of the 
vertices are linked to each other.  Thus we can think of the k-simplexes 
as suborders within the total order defined by the n-simplex.  So as q 
goes to 1 we are shifting from an ordered set to an unordered set.  Note 
that generally at the heart of the counting of the elements of a finite 
field is a notion of a canonical basis e1, e2...en, that is, a basis 
that has a definite order.

In this context it is interesting to think about the -1 simplex. It is 
implied by the leftmost diagonal of Pascal's triangle:
https://en.wikipedia.org/wiki/Simplex
For example, the tetrahedron has:
1  ?     4 vertices     6 edges   4 faces   1 volume

My presentation
http://www.ms.lt/sodas/Book/DiscoveryInMathematics
shows my interpretation of the -1 simplex which I believe is relevant 
for interpreting the field with one element.

Andrius

Andrius Kulikauskas
ms at ms.lt
+370 607 27 665


2016.06.20 02:13, Colin McLarty rašė:
> I will describe one typical but very limited issue about the 
> one-element field F1 that is fairly easy to see, and I will compare 
> Harvey's pseudo fields on these issues.  In a following post I will 
> describe a more general issue which may be the key issue but is 
> naturally more arcane.  That more general issue connects the 
> one-element field to modern methods in math where the point is not to 
> look at structures on sets but to look at structural relations between 
> objects (even if the logical foundation is taken to be ZFC so that 
> every object is actually a set).
>
> Let q be any positive power of a prime, q=p^n, n>0. Then up to 
> isomorphism there is a unique field Fq with q elements, and every 
> finite (conventional) field is of this form.  This has been an 
> extremely productive idea in algebraic geometry and combinatorics the 
> past 70 years.
>
> From this point of view the one-element field F1 is the special case n=0.
>
> >From this point of view F1 should be a subfield of every finite 
> field, which is impossible with the standard definition of field.  And 
> further considerations suggest F1 should be a subfield of the ring of 
> integers Z, which is also impossible by standard definitions.  To 
> avoid overloading the notation I will not use the expression p^n 
> below, but much of the motivation is that 1 is the special case of a 
> prime power p^n where n=0.
>
> As to the logical foundations of the idea: A lot of work on the idea 
> of a one-element field F1 considers only finite fields as motivation, 
> and that is all strictly finite combinatorics.  Considering the ring Z 
> of integers of course is still finitary in some way but it is not just 
> finite combinatorics.  Further considerations can link F1 to the 
> complex numbers and more, so the logical level of the idea of F1 will 
> depend on things that are not entirely settled in the literature today.
>
> The issue for this post is to count the points in the n-dimensional 
> projective space P^n(F1) over  F1.
>
> For any conventional field k the  points of the n-dimensional 
> projective space
> P^n(k) over  k are equivalence classes of non-zero points in the n+1 
> dimensional vector space k^(n+1) where two points of k^(n+1) are 
> treated as identical when they are scalar multiples of each other.  If 
> k is a finite field Fq then the number of points is q^(n+1)-1 divided 
> by q-1.  That is just the number of non-zero points in k^(n+1), 
> divided by the number of scalars in k.
>
> When k is a conventional finite field Fq this quotient can be written 
> as a finite sum q^n+q^(n-1)+...+1.
>
> But when q=1 the quotient is 0/0, while the sum is simply n. The sum 
> and the quotient do not agree.
>
> A correct treatment of the one-element field  F1 should make the 
> number of points of P^n(F1) equal n.  (More precisely, experts agree 
> this should be so, but the point of this discussion is that there is 
> not yet any consensus on precisely how to define the one-element field.)
>
> But if we define the one-element field as the one-element pseudo field 
> by Harvey's axioms, and construct the projective spaces P^n(F1) as the 
> usual equivalence classes of non-zero points in the vector space 
> F1^(n+1) then  P^n(F1) has no points, since  F1^(n+1)  has no non-zero 
> points.
>
> Many more sophisticated combinatoric questions can be taken as 
> algebraic-geometric questions over finite fields Fq for q a positive 
> power of a prime.  Often one approach to such a question has a 
> naturally appealing, and significant, extension for q=1.  In our 
> example the approach to counting points of P^n(Fq) as a sum has an 
> attractive generalization to q=1.  But other approaches which are 
> equivalent in the conventional case of q>1 are not equivalent when 
> q=1.  In our example counting points of  P^n(Fq) as a quotient is 
> undefined when q=1.
>
> And in these more sophisticated questions, as in our example here, we 
> can make the case q=1 well defined by using the pseudo field axioms.  
> But that well defined answer is not the intuitively correct answer in 
> our example -- according to the intuition of people like Tits.  And 
> the same is likely to happen for other issues besides counting points 
> of projective spaces.
>
> Colin
>
>
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