Is space a union of circles?
Andrej Bauer
MathOverflow, see
His question is in the context of classical set theory. But do we even
know that the decompositions he talks about cannot be achieved
constructively?
Specifically, can there be a constructive proof that the
three-dimensional space R^3 is a union of pairwise disjoint circles?
(Note that Theorem 1.1 in http://www.mscand.dk/article.php?id=77 is
_not_ constructive, at least the way I understnand it).
With kind regards,
Andrej
Thomas Streicher
toposes where all functions on R are continuous, e.g. function
realizability. But also not in the topos sSet of simplicial sets since
it admits a finite limit preserving functor to Sp having a right adjoint
(Geometric Realization).
Thomas S.
Andrej Bauer
> toposes where all functions on R are continuous, e.g. function
> realizability.
Excellent. What is a "continuous decomposition", concretely speaking?
I presume it would be a continuous map I x Circle -> R^3 for some
indexing space I. Then we're left with stating a concrete reason why
such a map does not exist. I can almost drive it home: given such a
map, we would have a continuous action of SU(2) on R^3 without fixed
points. By such-and-such fixed point theorem this is impossible. But
which fixed-point theorem?
With kind regards,
Andrej
Thomas Streicher
A bijection I x Circle -> R^3 would give rise to a decomposition of R^3
into a sum of circles. But that's too naive since I needn't be discrete.
Of course, we can get R \setminus {0} isomorphic to I x circle. Then
it should be a homotopy argument, isn't it?
Thomas
Steve Vickers
map R^3 -> I whose fibres are all circles. (Think of the Clifford
bundle, a.k.a. Hopf fibration, of the 3-sphere S^3 over S^2. I don't
believe you get that as I x S^1 -> S^3.)
But then maybe the argument comes down to the same thing. In the
topos of sheaves over I (whatever I is) you will be looking for a
torsor for the circle group SU(1). (Did you really mean SU(2)? Maybe
I've misunderstood something.) That will give an SU(1) action on the
bundle space R^3, without fixpoints.
Regards,
Steve.
Fred Richman
Suppose that's what it means. Consider the circle C that contains the origin. Given a point in space, we can ask what circle it is on, so we can find out whether or not it is on C. That property is enough to derive LPO, so we get a traditional Brouwerian counterexample.
--Fred
Steve Vickers
circles, the notion "any two circles are either disjoint or equal"
amounts to having decidable equality in I and so is rather strong.
Steve.
Fred Richman
--Fred
Han Geurdes
A line is a union of disjoint points. Can I map disjoint circles to disjoint points to get a plane when points give me the line? I think that is possible. Is 3 space a conjunction of disjoint planes like a line is of disjoint points? If yes then indeed 3 space can be build out of disjoint circles.
Don't laugh at this idiotic attempt with simple geometry. :-)
Fred Richman
--Fred
----- Original Message -----
From: Han Geurdes <han.g...@gmail.com>
To: construc...@googlegroups.com
Sent: Tue, 10 Apr 2012 14:06:30 -0400 (EDT)
Subject: Re: Is space a union of circles?
Thomas Streicher
"disjoint union of circles". I understood it as X being isomorphic to
Y x C for some Y (where C is the circle). There arises also the
problem to say what is C in a topos say or a ccc. The answer to that
is not clear.
But there is a precise question in the category Sp of spaces, namely
whether R^3 with its Euclidean topology is isomorphic to X \times C
for some space X (where C is the circle with its Euclidean topology).
When I mentioned function realizability I though about the above
question in the category QCB_0 of T_0 quotients of subspaces of Baire
space.
In case of SSet I thought of C = \partial \Delta[2] (border of the
triangle). But what is R there? Well, presumably Sing(R) = Sp(|-|,R)
where I : \Delta -> Sp is geoemtric realization.
Thomas S.
Thomas Streicher
some space I. The homotopy group of R^3 is trivial. The homotopy group
of C is \Z. Thus the homotopy group of I x C is G x \Z which never
is trivial. (Of course, for I the homotopy group G depends on the choice
of base point of I but this doesn't affect the argument).
Thomas
han.g...@gmail.com
Han
Verzonden van mijn HTC
Paul Taylor
Thomas has pointed out the homotopical reason why R^3 is not
homeomorphic to the product of the circle S^1 with something else.
However, I would observe that if it had been, such an important fact
would have been fully exploited long ago in geometry, cartography
and astronomy!
So the question must be whether there is a continuous and bijective
function whose pointwise inverse is not continuous
f: I x S^1 --> R^3
This means that it must be crucial that I and R^3 be not compact,
so any attempt to extend this function to compact spaces must
be seriously pathological.
Also, surely the set theorists asked about R^3 because the similar
question for R^2 is trivially false.
An interesting thing about R^3 is that circles may be linked,
maybe multiply so, whereas we have the Jordan curve theorem in R^2.
This suggests that the function, if it exists, must involve linking
of circles, to arbitrarily complicated extents.
With excluded middle we can certainly say how many times two disjoint
circles link in R^3. I think this can be done constructively too.
(At least, if we identify R^2 with C, we can say how often a
closed curve - continuous image of S^1 - that doesn't pass through
0 circles it, by integrating 1/z around the curve.)
OK, try this. For i neq j in I, the images f(i,S^1) and f(j,S^1)
are disjoint circles, which link some (positive, zero or negative)
integer n(i,j) times.
This suggests that the complement of the diagonal in IxI must have
a lot of components. This is bijective with some subspace of R^6.
Can anybody take the argument on from there?
I am inclined to think that this all has to be a simple application
of homotopy theory, and therefore that the "categories" list would
be a more productive forum in which to ask the question.
Paul
Fred Richman
Andrej Bauer
1) The easy one is whether there is a bijection f : I x C -> R^3 for
some I, circle C and space R^3.
2) The seemingly less easy one is whether there is g : R^3 -> J such
that its fibers are circles.
Regarding 1), I almost believe Thomas, except for a small detail. As
Thomas observes, it is consistent to assume that "all maps are
continuous" (by passing to some topos). But then to use this fact to
conclude that f is continuous, we need some topology or metric on the
indexing set I. Where does it come from? "Synthetic" topology? Do we
really know that the usual homotopy-theoretic arguments will "just
work"? We could also look at this model-theoretically, but then I
could be an arbitrary object of some topos, and I don't know what it
would mean for I to be a space in a traditional sense of the word.
The same question can be asked for 2). Suppose we had a constructive
argument that such a g exists. If we want to presume that g is
continuous, we must first make J into a space. If we do so by
assumption then we don't need any fancy stuff, I am sure there will be
a homotopy-theoretic argument why such a g does not exist.
With kind regards,
Andrej
Andrej Bauer
These are two different eyes, of course.
Steve Vickers
>>>
I'm thinking you ought to formulate it more generally than this, as a
map R^3 -> I whose fibres are all circles. (Think of the Clifford
bundle, a.k.a. Hopf fibration, of the 3-sphere S^3 over S^2. I don't
believe you get that as I x S^1 -> S^3.)
But then maybe the argument comes down to the same thing. In the
topos of sheaves over I (whatever I is) you will be looking for a
torsor for the circle group SU(1). (Did you really mean SU(2)? Maybe
I've misunderstood something.) That will give an SU(1) action on the
bundle space R^3, without fixpoints.
Regards,
Steve.
Thomas Streicher
> 1) The easy one is whether there is a bijection f : I x C -> R^3 for
> some I, circle C and space R^3.
>
> 2) The seemingly less easy one is whether there is g : R^3 -> J such
> that its fibers are circles.
I must say only 1) is a fair translation of the original problem
because in the set theoretic framework it's a question about
cardinality, isn't it? But, of course, one may always ask more general
questions :-) BTW it is easy in ZFC to show that C is isomorphic to R^3
because both C and R have the same cardinality and R^3 has the same
cardinality as R^3 (every infinite set S is isomorphic so S x S under AC
as is shown in most set theory text books at the beginning).
You are right that with the function realizability topos RT(K_2) there
is the question what topology to take on I when we assume I x C \cong R^3.
Well, what I had in mind was the fact that QCB_0 is a full reflective
subcat of the ambient topos RT(K_2). Since QCB_0 is cc the reflector R
preserves finite products. Suppose i : I x C -> R^3 is an isomorphism
then the reflection map r_{I x C} \cong r_I x r_C is an iso, too, and
thus r_I is an iso because r_C is.
Best regards,
Thomas
Steve Vickers
(If we're talking topos theory, then they could be internal frames in
the topos, but it's generally more convenient to use the equivalent
description as localic geometric morphisms into the topos ("Localic
Bundle Theorem").) That seems to be the only way the mathematics
stays sane. All hell breaks loose as soon as you try to ignore the
topology and deal with the spaces as objects in a topos, the "sets of
points".
From that point of view, in the hypothesized map g, J comes - point-
free - with its topology already assigned. It may be discrete, but
that's a special case.
Given g, we then have to ask what it means to say all its fibres are
circles. We wouldn't want to quantify over the global points of J,
since J need not be spatial. In fact, if it involves the reals, then
constructively it's probably not spatial. Instead we should work
generically and ask that, in the internal logic of Sh(J), the bundle
(as internal locale) is a circle. But, non-classically, we get non-
standard circles, the torsors for the circle group: they are locally
but not globally isomorphic to the standard circle. As bundles, I
believe these are going to correspond to what are usually called
principal U(1)-bundles. (The standard circle is the trivial bundle J
x S^1 -> J.)
This then should come down to Andrej's original suggestion, that what
we need to rule out is a fixpoint-free action of U(1) on R^3 - but
all continuous, of course.
Steve.
Steve Vickers
calls it) S^3 -> S^2. I would say it does display S^3 as a disjoint
union of circles, but I believe (I could be wrong) it doesn't come
from a continuous S^2 x S^1 -> S^3.
S^3 embeds in C^2 (C here is the complex plane) as the space of those
(z,w) such that |z|^2 + |w|^2 = 1. Each ray (1-dimensional subspace)
in C^2 meets that S^3 in a circle, and they are disjoint. But the
rays are the points of the complex projective line, which - as the
Riemann sphere - is S^2.
Given a point (z,w) in S^3, we have either z or w apart from 0, and
we get either w/z or z/w in C. Using some stereographic projection we
find a corresponding point on S^2; and if z and w are both apart from
zero we get the same point of S^2 either way. This gives the map S^3 -
> S^2.
It seems to me the argument does not rely on classical principles.
Steve.
Thomas Streicher
the Clifford bundle is certainly an interesting object (thanks for the
detailed description). Much more interesting than the usual
set-theoretic argument that S^1 \cong R^3.
But either the map itself or it's inverse has to be non-continuous for
sake of the simple homotopy argument of yesterday.
So your argument shows that one can prove the claim without
choice. But my argument shows that it can't be proved in IZF.
Thomas
han.g...@gmail.com
Now the line connecting the origins of the parallel planes created out of the circles is arbitrary. I can construe a sphere and have the line of origins outside that sphere.Then the diameter of the sphere goes to infinity.
>
Steve Vickers
If you puncture S^3 you are left with R^3. One of the circles is
punctured and becomes a copy of R. The rest, RxRxR\R, becomes a
disjoint union of circles.
I got this from Penrose "The Road to Reality" fig 33.15.
Steve.