Point-free affine real line?

Point-free affine real line?

[Steve Vickers]

Algebraic geometry defines the affine line over a field k as an affine scheme, the spectrum of k[X]. It includes a copy of k, each element a being present as the irreducible polynomial X-a, with local ring the ring of fractions got by inverting polynomials f(X) such that f(a) is non-zero.

You can carry this out for the real line R, but it is very much R as a set, and the copy of R in the underlying space of the spectrum has the discrete topology.

Does algebraic geometry provide an analogous construction that could lead to  the point-free R? Can the locally ringed space be topologized (point-free) so that the copy of R has its usual topology?

I've run into various problems.

1. It is not obvious to me that R[X] exists point-free. By that I mean that,  without presupposing a set R[X], or using non-geometric constructions, I can't see how to define a geometric theory whose models are the polynomials. The problem comes with trying to pin down the requirement that all but finitely many of the coefficients of a polynomial must be zero. You cannot continuously define the degree of a polynomial, because the function R -> N, a |-> degree(aX + 1), is not continuous.

That suggests the construction as Spec(R[X]) might have to be adjusted. Is there still some locally ringed space that does the trick?

2. The "structure sheaf" cannot be a sheaf. We hope its fibres are point-free local rings, but, whatever they are, they must be R-algebras and so cannot  have the discrete topology. The space is locally ringed by some bundle other than a sheaf (local homeomorphism).

3. The usual local rings, got as rings of fractions as described above, may be problematic point-free in the same way as R[X] is. I don't know what would  do instead. The power series ring R[[X]]? (At least as fibre over 0.) It does have the property of inverting those polynomials f for which f(0) is non-zero. And it can be defined point-free, as R^N. (However, the finitely presented approximations R[X|X^n = 0] happily exist point-free.)

Thanks for any references you can provide,

Steve.




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Re: Point-free affine real line?

[Steve Vickers]

Dear Johannes,

"Point-free" is as opposed to "point-set" - there is no assumption that
the real numbers form a set. Instead, they are taken to be the models of
a (geometric) theory of Dedekind sections of the rationals (which do
form a set). More generally, a point-free space is one where the points
are defined as models of a propositional geometric theory. The topology
is then defined intrinsically, opens being the geometric propositions.

My motivation for this comes from topos theory. In any elementary topos,
each point-free space does in fact have a set (object) of points.
Topologically, it amounts to approximating bundles by local
homeomorphisms. However, that construction is not geometric - preserved
by inverse image functors, and by pullback of bundles -, so the
point-set version is not robust under change of base. There are
advantages to staying within the geometric fragment of elementary topos
logic, and I am exploring how far that can be taken. I also now have
other semantics using arithmetic universes, where the point-set versions
simply don't exist.

If it helps, the point-free real line in an elementary topos E is the
localic geometric morphism p: F -> E got from F as topos of sheaves of
the internal locale of formal reals in E. If E is an S-topos, then you
can do this generically over S to get R as the S-classifier for Dedekind
sections, and then p is just the bipullback E x_S R. A point-set R would
be some local homeomorphism over E (then equipped separately with a
topology), but in general it is not got as a bipullback of a local
homeomorphism over S.

The question about the affine real line represents a challenge to this
geometric approach, and I'd like to form a better idea of whether it is
simply a difficult problem, or a fundamental limitation to my approach.
To put it another way, am I following in Grothendieck's footsteps in the
way I think of toposes, or am I mishandling his ideas in ways that have
no bearing on what he was trying to do?

All the best,

Steve.

On 01/06/2018 10:22, huebschm@math.univ-lille1.fr wrote:
> Dear Steve
>
> I am not sure whether I understand.
> What precisely do you mean by "point-free"?
>
>
>
> On Thu, 31 May 2018, Steve Vickers wrote:
>
>> Algebraic geometry defines the affine line over a field k as an
>> affine scheme,
> the spectrum of k[X]. It includes a copy of k, each element a being
> present as the irreducible polynomial X-a,
> with local ring the ring of fractions got by inverting polynomials
> f(X) such that f(a) is non-zero.
>>
>> You can carry this out for the real line R, but it is very much R as
>> a set, and the copy of R in the underlying space of the spectrum has
>> the discrete topology.
>
>
>
>
> The standard approach leads to the Zariski topology.
>
>
>
>> Does algebraic geometry provide an analogous construction that could
>> lead to  the point-free R? Can the locally ringed space be
>> topologized (point-free) so that the copy of R has its usual topology?
>>
>> I've run into various problems.
>>
>> 1. It is not obvious to me that R[X] exists point-free. By that I
>> mean that,  without presupposing a set R[X], or using non-geometric
>> constructions, I can't see how to define a geometric theory whose
>> models are the polynomials. The problem comes with trying to pin down
>> the requirement that all but finitely many of the coefficients of a
>> polynomial must be zero. You cannot continuously define the degree of
>> a polynomial, because the function R -> N, a |-> degree(aX + 1), is
>> not continuous.
>>
>> That suggests the construction as Spec(R[X]) might have to be
>> adjusted. Is there still some locally ringed space that does the trick?
>>
>> 2. The "structure sheaf" cannot be a sheaf. We hope its fibres are
>> point-free local rings, but, whatever they are, they must be
>> R-algebras and so cannot  have the discrete topology. The space is
>> locally ringed by some bundle other than a sheaf (local homeomorphism).
>>
>> 3. The usual local rings, got as rings of fractions as described
>> above, may be problematic point-free in the same way as R[X] is.
>> I don't know what would  do instead. The power series ring R[[X]]?
>
>
>
> The power series ring recovers a single point ("formal geometry" in
> the sense of Grothendieck, Gelfand-Kazhdan, Kontsevich, etc.).
>
>
> Best Johannes
>
>
>
>
>> (At least as fibre over 0.)
>> It does have the property of inverting those polynomials f for which
>> f(0)
> is non-zero. And it can be defined point-free, as R^N. (However, the
> finitely presented approximations R[X|X^n = 0] happily exist point-free.)
>>
>> Thanks for any references you can provide,
>>
>> Steve.
>>
>>



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Re: Point-free affine real line?

[Graham Manuell]

Dear Steve,

I have been working on very similar ideas for my PhD thesis.

If R is the real line, then R[X] has at least 3 reasonable topologies: the
compact-open topology, the Whitney topology and the coarsest topology
making the coefficient maps continuous. None of these gives a localic ring
since they are not complete. Their completions are, of course, the ring of
continuous functions on R, the ring of smooth functions on R and the ring
of formal power series R[[X]], respectively. Perhaps this indicates that
these are more appropriate than the polynomial ring when R has a
non-discrete topology.

It seems to me that the classical Zariski spectrum is only correct for
discrete rings. The most obvious example where it gives pathological
results is for the ring of continuous functions on a compact Hausdorff
space. For Gelfand duality we must consider not all the ideals, but only
the closed ideals. The Zariski spectrum is given by taking the localic
reflection of the quantale of ideals of a discrete ring. For a localic
ring, I believe we should instead take the localic reflection of the
quantale of overt ideals. This should give the expected results for both
discrete rings and for rings of continuous functions. For the localic rings
above, I expect R to appear inside the spectrum with its usual topology,
but I've only checked this in the first two cases.

I don't know if the construction of the quantale of overt ideals from a
localic ring is geometric. I think it might not be. There is a geometric
theory that gives the elements of this quantale as its points. This
tentatively suggests to me that the resulting spectrum will not be a
locale, but a colocale, but you know much more about these than I do.

I have thought less about the corresponding 'sheaf'. In particular, I do
not know what the correct notion of localic local ring is. I found some
work on localisation of topological rings (see here
<https://www.sciencedirect.com/science/article/pii/0166864194900353>), but
I haven't thought how this would work in the pointfree setting.

Best regards,

Graham

On Fri, 1 Jun 2018 at 01:02 Steve Vickers <s.j.vickers@cs.bham.ac.uk> wrote:

> Algebraic geometry defines the affine line over a field k as an affine
> scheme, the spectrum of k[X]. It includes a copy of k, each element a being
> present as the irreducible polynomial X-a, with local ring the ring of
> fractions got by inverting polynomials f(X) such that f(a) is non-zero.
>
> You can carry this out for the real line R, but it is very much R as a
> set, and the copy of R in the underlying space of the spectrum has the
> discrete topology.
>
> Does algebraic geometry provide an analogous construction that could lead
> to  the point-free R? Can the locally ringed space be topologized
> (point-free) so that the copy of R has its usual topology?
>
> I've run into various problems.
>
> 1. It is not obvious to me that R[X] exists point-free. By that I mean
> that,  without presupposing a set R[X], or using non-geometric
> constructions, I can't see how to define a geometric theory whose models
> are the polynomials. The problem comes with trying to pin down the
> requirement that all but finitely many of the coefficients of a polynomial
> must be zero. You cannot continuously define the degree of a polynomial,
> because the function R -> N, a |-> degree(aX + 1), is not continuous.
>
> That suggests the construction as Spec(R[X]) might have to be adjusted. Is
> there still some locally ringed space that does the trick?
>
> 2. The "structure sheaf" cannot be a sheaf. We hope its fibres are
> point-free local rings, but, whatever they are, they must be R-algebras and
> so cannot  have the discrete topology. The space is locally ringed by some
> bundle other than a sheaf (local homeomorphism).
>
> 3. The usual local rings, got as rings of fractions as described above,
> may be problematic point-free in the same way as R[X] is. I don't know what
> would  do instead. The power series ring R[[X]]? (At least as fibre over
> 0.) It does have the property of inverting those polynomials f for which
> f(0) is non-zero. And it can be defined point-free, as R^N. (However, the
> finitely presented approximations R[X|X^n = 0] happily exist point-free.)
>
> Thanks for any references you can provide,
>
> Steve.
>

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Re: Point-free affine real line?

[Ingo Blechschmidt]

Dear Steve,

On Thu, May 31, 2018 at 10:40:41AM +0100, Steve Vickers wrote:
> Does algebraic geometry provide an analogous construction that could
> lead to  the point-free R? Can the locally ringed space be topologized
> (point-free) so that the copy of R has its usual topology?

very interesting question. I feel that the following remarks have some
relevance.

* Algebraic geometers do consider an analogue of the usual spectrum
   construction for topological rings: the "formal spectrum".
   Unfortunately, the theory is only developed for actual topological
   rings, not point-free versions of them.

* One can define a point-free notion of a spectrum for apartness rings
   (ring objects in the category of sets-equipped-with-apartness-relation).
   This secretly comes up in algebraic geometry: Let X be a scheme.
   Inside the topos Sh(X), there is the ring O_X to which we can apply the
   usual point-free spectrum construction. We might hope that this just
   yields the one-point space; but unless X was zero-dimensional to begin
   with, this hope is false. In fact, the externalization of this
   construction is a locally ringed locale which comes equpped with a morphism
   of ringed locales to (X,O_X); but this morphism is not a morphism of
   *locally* ringed locales.

   The solution to this problem is to observe that O_X, being a local
   ring, has canonically the structure of an apartness ring. The variant
   of the spectrum construction for apartness rings applied to it yields
   the one-point space (i.e. (X,O_X) itself from the external point of
   view), as one would expect.

* Both the problem and the solution can be generalized a bit.

   Let X be a scheme. Let A be an O_X-algebra. Then algebraic geometers
   consider the "relative spectrum of A", which is always a locally
   ringed locale over X and will be a scheme if A is quasicoherent.
   As before, it's not true that one can obtain the relative spectrum
   simply by carrying out the usual point-free spectrum construction
   internally in Sh(X). A variant is needed.

A description of these variants of the usual construction can be found in Section 12
of https://rawgit.com/iblech/internal-methods/master/notes.pdf. However,
I don't know yet a natural generalization of these constructions which
could be directly helpful to your project -- they are tailored to
apartness rings and to algebras over local rings. While it's certainly
better to view the reals as an apartness ring instead of an ordinary ring,
information has still been lost from the point-free version of the
reals.

Cheers,
Ingo


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Re: Point-free affine real line?

[Vaughan Pratt]

  > The question about the affine real line represents a challenge to this
> geometric approach, and I'd like to form a better idea of whether it is
> simply a difficult problem, or a fundamental limitation to my approach.

An affine space over any given field differs *k* from a vector space over
*k* only in its algebraic structure, not its topological structure.
Whereas the algebraic operations of a vector space over *k* consist of all
finitary linear combinations with coefficients drawn from *k*, those of an
affine space consist of the subset of those combinations whose coefficients
sum to unity, the barycentric combinations.  Since the former includes the
constant 0 as a linear combination while the latter does not, a consequence
is that 0 is a fixpoint of linear transformations but not of affine
transformations, whence the latter can include the translations.

This is equally true whether *k* is the rationals or the reals.   So
whatever method you use to obtain the real line from the rational line
should also produce the affine real line from the affine rational line.

The relevance of the equational theory I gave in my previous message is
that it is equivalent to the equational theory of affine spaces over the
rationals yet does not require a separate theory of the rationals.  The
insight here is that the polynomials formed from the operations 2b - a and
all finitary centroid operations, one for each positive finite arity, are
precisely the barycentric linear combinations with rational coefficients.
The real line is a model of this theory but lacks the usual topology that
expands its algebraic operations to include linear combinations with all
real coefficients.  I cannot conceive how the absence of the linear
combinations whose coefficients don't sum to unity could be an obstacle for
you.

Vaughan Pratt

On Fri, Jun 1, 2018 at 3:11 AM, Steve Vickers <s.j.vickers@cs.bham.ac.uk>
wrote:

> Dear Johannes,
>
> "Point-free" is as opposed to "point-set" - there is no assumption that
> the real numbers form a set. Instead, they are taken to be the models of
> a (geometric) theory of Dedekind sections of the rationals (which do
> form a set). More generally, a point-free space is one where the points
> are defined as models of a propositional geometric theory. The topology
> is then defined intrinsically, opens being the geometric propositions.
>
> My motivation for this comes from topos theory. In any elementary topos,
> each point-free space does in fact have a set (object) of points.
> Topologically, it amounts to approximating bundles by local
> homeomorphisms. However, that construction is not geometric - preserved
> by inverse image functors, and by pullback of bundles -, so the
> point-set version is not robust under change of base. There are
> advantages to staying within the geometric fragment of elementary topos
> logic, and I am exploring how far that can be taken. I also now have
> other semantics using arithmetic universes, where the point-set versions
> simply don't exist.
>
> If it helps, the point-free real line in an elementary topos E is the
> localic geometric morphism p: F -> E got from F as topos of sheaves of
> the internal locale of formal reals in E. If E is an S-topos, then you
> can do this generically over S to get R as the S-classifier for Dedekind
> sections, and then p is just the bipullback E x_S R. A point-set R would
> be some local homeomorphism over E (then equipped separately with a
> topology), but in general it is not got as a bipullback of a local
> homeomorphism over S.
>
> The question about the affine real line represents a challenge to this
> geometric approach, and I'd like to form a better idea of whether it is
> simply a difficult problem, or a fundamental limitation to my approach.
> To put it another way, am I following in Grothendieck's footsteps in the
> way I think of toposes, or am I mishandling his ideas in ways that have
> no bearing on what he was trying to do?
>
> All the best,
>
> Steve.
>
> On 01/06/2018 10:22, huebschm@math.univ-lille1.fr wrote:
>> Dear Steve
>>
>> I am not sure whether I understand.
>> What precisely do you mean by "point-free"?
>>
>>
>>
>> On Thu, 31 May 2018, Steve Vickers wrote:
>>
>>> Algebraic geometry defines the affine line over a field k as an
>>> affine scheme,
>> the spectrum of k[X]. It includes a copy of k, each element a being
>> present as the irreducible polynomial X-a,
>> with local ring the ring of fractions got by inverting polynomials
>> f(X) such that f(a) is non-zero.
>>>
>>> You can carry this out for the real line R, but it is very much R as
>>> a set, and the copy of R in the underlying space of the spectrum has
>>> the discrete topology.
>>
>>
>>
>>
>> The standard approach leads to the Zariski topology.
>>
>>
>>
>>> Does algebraic geometry provide an analogous construction that could
>>> lead to  the point-free R? Can the locally ringed space be
>>> topologized (point-free) so that the copy of R has its usual topology?
>>>
>>> I've run into various problems.
>>>
>>> 1. It is not obvious to me that R[X] exists point-free. By that I
>>> mean that,  without presupposing a set R[X], or using non-geometric
>>> constructions, I can't see how to define a geometric theory whose
>>> models are the polynomials. The problem comes with trying to pin down
>>> the requirement that all but finitely many of the coefficients of a
>>> polynomial must be zero. You cannot continuously define the degree of
>>> a polynomial, because the function R -> N, a |-> degree(aX + 1), is
>>> not continuous.
>>>
>>> That suggests the construction as Spec(R[X]) might have to be
>>> adjusted. Is there still some locally ringed space that does the trick?
>>>
>>> 2. The "structure sheaf" cannot be a sheaf. We hope its fibres are
>>> point-free local rings, but, whatever they are, they must be
>>> R-algebras and so cannot  have the discrete topology. The space is
>>> locally ringed by some bundle other than a sheaf (local homeomorphism).
>>>
>>> 3. The usual local rings, got as rings of fractions as described
>>> above, may be problematic point-free in the same way as R[X] is.
>>> I don't know what would  do instead. The power series ring R[[X]]?
>>
>>
>>
>> The power series ring recovers a single point ("formal geometry" in
>> the sense of Grothendieck, Gelfand-Kazhdan, Kontsevich, etc.).
>>
>>
>> Best Johannes
>>
>>
>>
>>
>>> (At least as fibre over 0.)
>>> It does have the property of inverting those polynomials f for which
>>> f(0)
>> is non-zero. And it can be defined point-free, as R^N. (However, the
>> finitely presented approximations R[X|X^n = 0] happily exist point-free.)
>>>
>>> Thanks for any references you can provide,
>>>
>>> Steve.
>>>
>>>
>
>

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Re: Point-free affine real line?

[Peter Johnstone]

On Sun, 3 Jun 2018, Vaughan Pratt wrote:

> > The question about the affine real line represents a challenge to this
>> geometric approach, and I'd like to form a better idea of whether it is
>> simply a difficult problem, or a fundamental limitation to my approach.
>
> An affine space over any given field differs *k* from a vector space over
> *k* only in its algebraic structure, not its topological structure.
> Whereas the algebraic operations of a vector space over *k* consist of all
> finitary linear combinations with coefficients drawn from *k*, those of an
> affine space consist of the subset of those combinations whose coefficients
> sum to unity, the barycentric combinations.  Since the former includes the
> constant 0 as a linear combination while the latter does not, a consequence
> is that 0 is a fixpoint of linear transformations but not of affine
> transformations, whence the latter can include the translations.
>
> This is equally true whether *k* is the rationals or the reals.   So
> whatever method you use to obtain the real line from the rational line
> should also produce the affine real line from the affine rational line.
>
Not quite: the affine rational line doesn't have a definable total order,
since it has order-reversing automorphisms, so any definition using
Dedekind sections is problematic. However, it does have a ternary
`betweenness' relation, and it should be possible to rewrite the
geometric theory of Dedekind sections of Q, as presented on p. 1015
of `Sketches of an Elephant', in terms of this relation (but note that
sections will have to be unordered rather than ordered pairs of
subobjects of Q).

Peter Johnstone


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Re: Point-free affine real line?

[Vaughan Pratt]

Peter is right that Q as a field contains order information absent from Q
as either a linear or affine space.  (−*x*² ≤ 0 ≤ *x*².)

> For the order to be definable from the algebraic structure, you need to
consider Q as a field

It certainly suffices, and indeed is traditional.

But is it necessary for the construction of the affine real line?  In
particular do we have to define multiplication on the rationals only to
throw it away at a later stage?

Obviously the ordered dyadic rationals would suffice.  Equally obviously
they don't form a field.

But can they be defined as an ordered affine space with the same degree of
formality and economy as we routinely define the ordered linear space of
rationals?

Claim.  The ordered algebra (D, xy, x+y, ≤) consisting of the set D  of
dyadic rationals under the two binary operations xy denoting 2y - x and x+y
denoting (x + y)/2 can be defined order-algebraically up to isomorphism as
the free ordered algebra on the ordinal 2 satisfying finitely many
equations in the two operations along with the inference rules, from x ≤ y
infer each of

y ≤ xy
yx ≤ x
x ≤ x+y
x+y ≤ y

The Dedekind cuts being cuts in the rational line, the following claim
depends on a distinct name for the corresponding notion of a cut in the
ordered affine line of dyadic rationals, which I suggest calling a dyadic
cut.  (When cutting at a dyadic rational follow a consistent convention as
to which side to associate that rational, as done with the Dedekind cuts,
and always have both sides of the cut nonempty.)

Claim.  The dyadic cuts in the ordered affine space of dyadic rationals are
in order-preserving bijection with the Dedekind cugts in the ordered linear
space of rationals, with the cuts at dyadic rationals correctly matched to
their Dedekind counterparts.

Vaughan Pratt

On Sun, Jun 10, 2018 at 6:54 AM, Peter Johnstone <ptj@dpmms.cam.ac.uk>
wrote:

> Sorry, what I wrote was a bit sloppy. Vaughan is right that the problem
> doesn't arise with the passage from considering Q as a linear space
> to considering it as an affine space, since it already has order-
> reversing linear automorphisms. For the order to be definable from
> the algebraic structure, you need to consider Q as a field, which is
> what the usual Dedekind-section construction does.
>
> Peter Johnstone
>
>
> On Thu, 7 Jun 2018, Vaughan Pratt wrote:
>
> > Not quite: the affine rational line doesn't have a definable total
>> order,
>>
>>> since it has order-reversing automorphisms, so any definition using
>>> Dedekind sections is problematic.
>>>
>>
>> Morphism-wise, since the affine transformations are just the composition
>> of
>> a linear transformation with a translation, and translation of the
>> rational
>> line preserves order, affinity can't be the problem here.
>>
>> Structure-wise, one can equip the rational line with either its linear
>> combinations or its linear order, or both.  Using both eliminates the
>> order-reversing linear transformations.   "Affine" only makes sense in the
>> context of having the linear combinations, as "affine" limits the linear
>> combinations to those whose coefficients sum to one.   If it is ok for the
>> linear combinations and the linear order to coexist, it must be even more
>> ok for the affine combinations and the linear order to coexist.
>>
>> So whether one considers the morphisms or the structure they preserve,
>> affinity (affineness?) must be a red herring here: any problem for the
>> rational line as an affine space is surely also a problem for it as a
>> vector space.
>>
>> Vaughan Pratt
>>
>>
>>

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Re: Point-free affine real line?

[Peter Johnstone]

Sorry, what I wrote was a bit sloppy. Vaughan is right that the problem
doesn't arise with the passage from considering Q as a linear space
to considering it as an affine space, since it already has order-
reversing linear automorphisms. For the order to be definable from
the algebraic structure, you need to consider Q as a field, which is
what the usual Dedekind-section construction does.

Peter Johnstone

On Thu, 7 Jun 2018, Vaughan Pratt wrote:

> > Not quite: the affine rational line doesn't have a definable total order,
>> since it has order-reversing automorphisms, so any definition using
>> Dedekind sections is problematic.
>
> Morphism-wise, since the affine transformations are just the composition of
> a linear transformation with a translation, and translation of the rational
> line preserves order, affinity can't be the problem here.
>
> Structure-wise, one can equip the rational line with either its linear
> combinations or its linear order, or both.  Using both eliminates the
> order-reversing linear transformations.   "Affine" only makes sense in the
> context of having the linear combinations, as "affine" limits the linear
> combinations to those whose coefficients sum to one.   If it is ok for the
> linear combinations and the linear order to coexist, it must be even more
> ok for the affine combinations and the linear order to coexist.
>
> So whether one considers the morphisms or the structure they preserve,
> affinity (affineness?) must be a red herring here: any problem for the
> rational line as an affine space is surely also a problem for it as a
> vector space.
>
> Vaughan Pratt
>
>
>
> On Tue, Jun 5, 2018 at 3:55 AM, Peter Johnstone <ptj@dpmms.cam.ac.uk> wrote:
>
>> On Sun, 3 Jun 2018, Vaughan Pratt wrote:
>>
>>> The question about the affine real line represents a challenge to this
>>>
>>>> geometric approach, and I'd like to form a better idea of whether it is
>>>> simply a difficult problem, or a fundamental limitation to my approach.
>>>>
>>>
>>> An affine space over any given field differs *k* from a vector space over
>>> *k* only in its algebraic structure, not its topological structure.
>>> Whereas the algebraic operations of a vector space over *k* consist of all
>>> finitary linear combinations with coefficients drawn from *k*, those of an
>>> affine space consist of the subset of those combinations whose
>>> coefficients
>>> sum to unity, the barycentric combinations.  Since the former includes the
>>> constant 0 as a linear combination while the latter does not, a
>>> consequence
>>> is that 0 is a fixpoint of linear transformations but not of affine
>>> transformations, whence the latter can include the translations.
>>>
>>> This is equally true whether *k* is the rationals or the reals.   So
>>> whatever method you use to obtain the real line from the rational line
>>> should also produce the affine real line from the affine rational line.
>>>
>>> Not quite: the affine rational line doesn't have a definable total order,
>> since it has order-reversing automorphisms, so any definition using
>> Dedekind sections is problematic. However, it does have a ternary
>> `betweenness' relation, and it should be possible to rewrite the
>> geometric theory of Dedekind sections of Q, as presented on p. 1015
>> of `Sketches of an Elephant', in terms of this relation (but note that
>> sections will have to be unordered rather than ordered pairs of
>> subobjects of Q).
>>
>> Peter Johnstone
>>
>
>
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>


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Re: Point-free affine real line?

[Vaughan Pratt]

  > Not quite: the affine rational line doesn't have a definable total order,
> since it has order-reversing automorphisms, so any definition using
> Dedekind sections is problematic.

Morphism-wise, since the affine transformations are just the composition of
a linear transformation with a translation, and translation of the rational
line preserves order, affinity can't be the problem here.

Structure-wise, one can equip the rational line with either its linear
combinations or its linear order, or both.  Using both eliminates the
order-reversing linear transformations.   "Affine" only makes sense in the
context of having the linear combinations, as "affine" limits the linear
combinations to those whose coefficients sum to one.   If it is ok for the
linear combinations and the linear order to coexist, it must be even more
ok for the affine combinations and the linear order to coexist.

So whether one considers the morphisms or the structure they preserve,
affinity (affineness?) must be a red herring here: any problem for the
rational line as an affine space is surely also a problem for it as a
vector space.

Vaughan Pratt



On Tue, Jun 5, 2018 at 3:55 AM, Peter Johnstone <ptj@dpmms.cam.ac.uk> wrote:

> On Sun, 3 Jun 2018, Vaughan Pratt wrote:
>
> > The question about the affine real line represents a challenge to this
>>
>>> geometric approach, and I'd like to form a better idea of whether it is
>>> simply a difficult problem, or a fundamental limitation to my approach.
>>>
>>
>> An affine space over any given field differs *k* from a vector space over
>> *k* only in its algebraic structure, not its topological structure.
>> Whereas the algebraic operations of a vector space over *k* consist of all
>> finitary linear combinations with coefficients drawn from *k*, those of an
>> affine space consist of the subset of those combinations whose
>> coefficients
>> sum to unity, the barycentric combinations.  Since the former includes the
>> constant 0 as a linear combination while the latter does not, a
>> consequence
>> is that 0 is a fixpoint of linear transformations but not of affine
>> transformations, whence the latter can include the translations.
>>
>> This is equally true whether *k* is the rationals or the reals.   So
>> whatever method you use to obtain the real line from the rational line
>> should also produce the affine real line from the affine rational line.
>>
>> Not quite: the affine rational line doesn't have a definable total order,
> since it has order-reversing automorphisms, so any definition using
> Dedekind sections is problematic. However, it does have a ternary
> `betweenness' relation, and it should be possible to rewrite the
> geometric theory of Dedekind sections of Q, as presented on p. 1015
> of `Sketches of an Elephant', in terms of this relation (but note that
> sections will have to be unordered rather than ordered pairs of
> subobjects of Q).
>
> Peter Johnstone
>


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Re: Point-free affine real line?

[Vaughan Pratt]

(FMCS and BLAST attendees will have seen this before.)

Apropos of affine spaces over a field (the affine real line being a
one-dimensional example), here are some axioms for a language with one
binary operation ab, with (ab)c abbreviated to abc (left-associative
convention as for combinatory algebra etc.), satisfying

A1.  aa = a
A2.  abb = a
A3.  abc = ac(bc)
A4.  abcd = adcb

(Side remark: Axiom Ai requires i variables.)

The intended model of ab is 2b - a over an arbitrary abelian group, e.g.
the integers.  (A4 forces abelian.

Obviously the free algebra on n generators for n < 2 has n elements.

Identify the free algebra F2 on {a,b} with the integers ..., -3, -2, -1, 0,
1, 2, 3, 4, ... as follows.
... baba, aba, ba, a, b, ab, bab, abab, ...

(Exercise: This are the only elements of F2.)

As usual the elements of F2 can be understood as binary operations on F2.
Abbreviate the binary operation identified with n as a[n]b; thus a[0]b = a,
a[1]b = b, a[2]b = ab, a[-1]b = ba, etc.  Interpreting ab as 2b - a over
the integers, a[n]b = a + n(b - a), and 0[n]1 = n, making F2 yet another
binary notation for the integers, albeit exponentially longer.

Now expand this one-operation language with a family c1, c2, c3, c4, ... of
operations of respective finite arities 1, 2, 3, 4, ... satisfying

c1(a) = a
cm(a1, ..., am)[n]cn(a1, ..., an) = an  for all n > 1 where m = n-1
cn(a1, ..., am, cm(a1, ..., am))[n]b = b      ditto

The intended model of cn(a1, ..., an) is the centroid or mean of a1, ...,
an over an arbitrary field, i.e. (a1 + ... + an)/n.

Claim.  The variety defined by this equational theory, including its
homomorphisms as standardly defined, is equivalent to the category of
affine spaces over the rationals.  (Affine transformations over the
rationals are linear combinations whose coefficients sum to 1, e.g.
centroid.  The idea is that other models such as the real line can be
pulled apart by homomorphisms into uncountably many copies of the rational
line because the operations of the theory can only make rational
connections between points of the line.)

I mention this in case it has any bearing on Steve's question about a
point-free version of the real affine line.  Not that I see one myself.

Vaughan Pratt


On Thu, May 31, 2018 at 2:40 AM, Steve Vickers <s.j.vickers@cs.bham.ac.uk>
wrote:

> Algebraic geometry defines the affine line over a field k as an affine
> scheme, the spectrum of k[X]. It includes a copy of k, each element a being
> present as the irreducible polynomial X-a, with local ring the ring of
> fractions got by inverting polynomials f(X) such that f(a) is non-zero.
>
> You can carry this out for the real line R, but it is very much R as a
> set, and the copy of R in the underlying space of the spectrum has the
> discrete topology.
>
> Does algebraic geometry provide an analogous construction that could lead
> to  the point-free R? Can the locally ringed space be topologized
> (point-free) so that the copy of R has its usual topology?
>
> I've run into various problems.
>
> 1. It is not obvious to me that R[X] exists point-free. By that I mean
> that,  without presupposing a set R[X], or using non-geometric
> constructions, I can't see how to define a geometric theory whose models
> are the polynomials. The problem comes with trying to pin down the
> requirement that all but finitely many of the coefficients of a polynomial
> must be zero. You cannot continuously define the degree of a polynomial,
> because the function R -> N, a |-> degree(aX + 1), is not continuous.
>
> That suggests the construction as Spec(R[X]) might have to be adjusted. Is
> there still some locally ringed space that does the trick?
>
> 2. The "structure sheaf" cannot be a sheaf. We hope its fibres are
> point-free local rings, but, whatever they are, they must be R-algebras and
> so cannot  have the discrete topology. The space is locally ringed by some
> bundle other than a sheaf (local homeomorphism).
>
> 3. The usual local rings, got as rings of fractions as described above,
> may be problematic point-free in the same way as R[X] is. I don't know what
> would  do instead. The power series ring R[[X]]? (At least as fibre over
> 0.) It does have the property of inverting those polynomials f for which
> f(0) is non-zero. And it can be defined point-free, as R^N. (However, the
> finitely presented approximations R[X|X^n = 0] happily exist point-free.)
>
> Thanks for any references you can provide,
>
> Steve.
>
>

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