From: rusin@washington.math.niu.edu (Dave Rusin)
Newsgroups: sci.math.research
Subject: Contractions of infinite-dimensional spheres
Date: 16 Sep 1995 15:32:45 GMT
In an article which has since rolled off my site, [name deleted] presented
several spaces, each of which could arguably be called the infinite-dimensional
sphere. He asked which were contractible and asked for the homotopy and
homology of their quotients by Z_2.
Homotopy theorists take as "S-infinity" the union of the S^n ,
[the poster]'s second space. This set is contractible. Although
I can't remember an explicit contraction, I offer instead an
interesting construction which gives a homeomorphic space on which
an explicit contraction can be described. It's a useful construct
for general classifying spaces.
Here's the new space. Let X be the set of all integrable
functions f : [0,1] --> {+1, -1} which have only a finite number of
discontinuities. That sounds pretty high-falutin', but the graphs of
such functions f are just pieces of straight lines at y=1 or y=-1.
We give X the metric it inherits from L^1(R). Note that elements of
X are really _equivalence classes_ of functions (equal almost everywhere).
This X contains a family of interesting subsets X_n, the
collection of such functions with exactly n points of discontinuity.
X_0 is the set of two constant functions. X_1 contains the functions which,
for some t, equal 1 on [0,t] and -1 on [t,1]. (X1 also contains the
negatives of these functions.) In general, an element of X_n which equals
y=+1 near x=0 is describe by a sequence of numbers (t1, t2, ..., tn), each
positive and summing to 1: this n-tuple stands for the function which is
y=+1 on [0,t1], y=-1 on [t1,t1+t2], y=+1 on [t1+t2, t1+t2+t3], and so on.
(The elements of X_n which equal y=-1 near x=0 may be similarly
parameterized.) In this way it's easy to see that X_n is homeomorphic
to the union of two n-cells, each of whose boundaries are exactly
X_(n-1), so that X_n is homeomorphic to the sphere S^n (with a
homeomorphism extending the one X_(n-1) -> S^(n-1).) Since X is
just the union of the X_n, it follows that this X is homeomorphic
to the sphere S-infinity.
OK, _now_ I can give the explicit contraction. Let F : X x I --> X
be the map F(f, t) = fbar, where fbar is the function in X given
by
/ f(x), if x > t
fbar(x) = <
\ 1, if x <= t.
Use the metric in X to show that F is continuous (in fact, uniformly so).
Note F(f,0) = f and F(f,1) is the constant function 1 in X, so
F is a contraction.
Incidentally, it is easy to show that F(X_n x I) is contained in X_(n+1),
so this contraction is no more complicated than you might think: each
sphere S^n contracts within the next one S^(n+1).
There is an action of Z/2Z on X consistent with all this, namely
pointwise multiplication f |--> -f. Since X is contractible, the
quotient X/(Z/2Z) is a classifying space for Z/2Z, and its cohomology
is the algebraic cohomology of the group Z/2Z, namely the ring
Z[T]/2T (T is in dimension 2).
You can construct a similar space X by allowing the functions to take
values in any other discrete group G; then you have the CW complex EG,
and its quotient by the action of G is the classifying space BG.
While this is in principle just a geometric realization of the bar
construction of E(Z/2Z), I am not aware of any printed references to
such a concrete realization of this space.
dave
==============================================================================
Date: Sun, 17 Sep 1995 13:19:38 -0700
From: [Permission pending]
To: rusin@washington.math.niu.edu
Subject: Re: Contractions of infinite-dimensional spheres
Thanks for an interesting an informative response to my query!
BTW, I was pretty sure that at least S^oo is contractible, since I can
imagine shrinking S^k to a point inside of S^(k+1) suring the time interval
[1 - 1/2^(k-1), 1 - 1/2^k], and extending to all of S^oo by the homotopy
extension property.
This seems to lead to points moving with an unbounded set of velocities as
t -> 1.
So now I wonder: Can S^oo be contracted using a homotopy
H: S^oo x [0,1] -> S^oo
of "bounded speed", in the sense that the set
{ (||H(x,b)-H(x,a)||) / (b-a) : x is in S^oo, and 0 <= a < b <= 1 }
is bounded???
[sig deleted]
==============================================================================
From: Michael.Murray@Adelaide.Edu.Au (Michael K. Murray)
Newsgroups: sci.math.research
Subject: Re: Topology of Infinite-Dimensional Spheres and Projective Spaces
Date: Sun, 24 Sep 1995 14:51:49 +0930
In article <[identifier deleted]>, [original poster] wrote:
>Let H denote the Hilbert space of square-summable sequences of real numbers.
>
>Let R^oo denote the countable product of R with itself, as a vector space,
>(i.e., all sequences of real numbers).
>
>Consider the following infinite-dimensional spheres:
>
>1. S_H, defined as { x in H : ||x|| = 1}
>
>
>a) Which, if any, of the spheres is contractible, and for those that are,
> what is an explicit homotopy of the identity map to a constant???
>
S_H is certainly contractible and you proceed as follows. Define T(x) =
(0, x_1, x_2, ...)
if x = (x_1, x_2 ... ) \in S_H. Let S' \subset S_H be the image of S_H
under T. This is the equator and, of course, a copy of S_H! Now to
contract you notice that for any point x, T(x) is not its antipode, so you can
slide x along the line joining x to T(x) in a well defined manner. In fact
you can write down the explicit map. This shows that
the identity is homotopic to T. Now you can find a family of maps
g_t: S' \to S_H where g_0 is the inclusion and g_1 is the
constant map onto the south pole and g_t maps a point on the
equator a certain distance down the line of longitude
towards the south pole. Then g_t o T defines a homotopy of T to the
constant map onto the south pole. Again you can write out g_t
explicitely.
I would be interested if you learn of a method that avoids
the two-step procedure.
BTW any cts map from S^n to S_H must have compact image
and as S_H is not compact cannot be onto. But S_H minus
a point can be stereographically projected onto H so
is contractible. Hence S_H has all homotopy groups
vanishing. I don't know if this helps in the other cases.
Michael
Michael Murray Fax: 61+ 8 232 5670
Department of Pure Mathematics Phone: 61+ 8 303 4174
University of Adelaide Email: mmurray@maths.adelaide.edu.au
Australia 5005 http://macpure.maths.adelaide.edu.au/mmurray/mmurray.html
==============================================================================
Date: Mon, 25 Sep 95 14:37:32 CDT
From: rusin (Dave Rusin)
To: [Permission pending]
Subject: Re: Contractions of infinite-dimensional spheres
>BTW, I was pretty sure that at least S^oo is contractible, since I can
>imagine shrinking S^k to a point inside of S^(k+1) suring the time interval
>[1 - 1/2^(k-1), 1 - 1/2^k], and extending to all of S^oo by the homotopy
>extension property.
>
>This seems to lead to points moving with an unbounded set of velocities as
>t -> 1.
>
>So now I wonder: Can S^oo be contracted using a homotopy
>
> H: S^oo x [0,1] -> S^oo
>
>of "bounded speed", in the sense that the set
>
> { (||H(x,b)-H(x,a)||) / (b-a) : x is in S^oo, and 0 <= a < b <= 1 }
>
>is bounded???
I wish I had the time to sit down and write out the homeomorphism between
the space I described and the usual S^oo. Clearly the contraction I
proposed moves at low speed, in fact, (recall H(f,a) agrees with f
to the right (I think) of t=a and is a constant on the other side)
dist( H(f,a), H(f, b) ) = integral |H(f,a)(x)-H(f,b)(x)| dx
is the integral of zero except over [a,b], where it has value at most 2,
so the quotient at the end of your quote is at most 2.
What I didn't do was demonstrate an _isometry_ between my space and S^oo.
I don't think they are isometric (well, I know they're _not_ because
distances in my space are bounded by 1, not 2 as in S^oo), but I
think the distortion is not too bad.
However, I think even in the usual metric you can find a contraction with
the properties you desire. Rather than shrinking each sphere one at a time,
shrink all at once (this is what I did in my space). You define maps
S^n x I --> S^(n+1) with H(x,0)=x and H(x,1)={North pole} by induction,
extending the given map from S^(n-1) x I --> S^n. If it's important for
some reason I could probably work out some formulas to justify this, but
I think you can extend the situation for n=0, which goes as follows.
You already know how to contract S^0 within S^1: H(1,t)=1, and
H(-1,t)= (cos(u), sin(u)), where u can be any function of t with
u(0)=pi, u(1)=0.
You also know how to contract S^1 within S^2: pin down a rubber band
at one point on the equator, with the rest of the rubber band running the
equator. Paint a dot on the rubber band at the point opposite the pinned
point. Then raise the band by the dot so the circle slips off the sphere
and contracts to the pinned point.
As you do this, you get circles of diminishing radius. You can adjust the
speed of the raising so that the circular distance from the pointed point
to the pinned point equals u(t). If for example u is linear in t,
then you shouldn't need the motion of the circle on the sphere to
approach a high velocity as t--> 1 since no point is moving faster than
the painted point, and its velocity is (say) 1.
OK, so maybe it's not very convincing, but I think this approach could be
made to work if one were patient or clever enough. It's just necessary to
start moving all spheres at the same time.
dave
(Sorry about the delay in responding. Mailbox runneth over.)
==============================================================================
Date: Mon, 25 Sep 1995 12:44:06 -0700
From: [Permission pending]
To: rusin@math.niu.edu (Dave Rusin)
Subject: Re: Contractions of infinite-dimensional spheres
No, it's pretty convincing. Thanks for your comments.
I'm disappointed to learn that bounding the average velocities
probably won't exclude this contraction. At the moment I'm
very much hoping that there will be a way to define the idea
of an infinite dimensional hole, despite all these striking
results that indicate not. (This contrctibility of S^oo, as
well as th fact that H - {0} is homeomorphic to H (= Hilbert space).
Thanks,
[sig deleted]
==============================================================================
Date: Mon, 25 Sep 95 15:11:34 CDT
From: rusin (Dave Rusin)
To: [Permission pending]
Subject: Re: Contractions of infinite-dimensional spheres
Ah, so _that's_ your goal --
One may think of the finite homotopy groups as measuring 'holes', so the
question you're asking is more or less, is there anything else? The answer
is roughly 'no', although it depends on the category. Whitehead proved
that among CW complexes, a space is contractible iff all homotopy groups
vanish (more generally, if X and Y are CW spaces and f:X--> Y is
a map whch induces an isomorphism on each pi_n, then f is a homotopy
equivalence.) So this more or less rules out infinite dimensional holes.
On the other hand, this category, which certainly larger than the category
of finite complexes (e.g. it includes S^oo and other complexes with
finite skeleta, but there's even more), does not exhaust the category of
topological spaces. I'm pretty sure there are non-contractible spaces
with vanishing homotopy, although I can't remember a reference offhand.
Possibly these would be your choice of "infinite dimensional holes".
For that matter, it may be that the first space in your post is a candidate
for this, since I don't really know how to show the sphere in hilbert space
is contractible; it's not a CW complex in any way that's obvious to me.
I'll try to remember a nontrivial space with no homotopy, but it might be
faster to post again to s.m.r.
dave
==============================================================================
Date: Mon, 25 Sep 1995 14:09:27 -0700
From: [Permission pending]
To: rusin@math.niu.edu (Dave Rusin)
Subject: Re: Contractions of infinite-dimensional spheres
>Ah, so _that's_ your goal --
>One may think of the finite homotopy groups as measuring 'holes', so the
>question you're asking is more or less, is there anything else? The answer
>is roughly 'no', although it depends on the category. Whitehead proved
>that among CW complexes, a space is contractible iff all homotopy groups
>vanish (more generally, if X and Y are CW spaces and f:X--> Y is
>a map whch induces an isomorphism on each pi_n, then f is a homotopy
>equivalence.) So this more or less rules out infinite dimensional holes.
>
>On the other hand, this category, which certainly larger than the category
>of finite complexes (e.g. it includes S^oo and other complexes with
>finite skeleta, but there's even more), does not exhaust the category of
>topological spaces. I'm pretty sure there are non-contractible spaces
>with vanishing homotopy, although I can't remember a reference offhand.
>Possibly these would be your choice of "infinite dimensional holes".
>For that matter, it may be that the first space in your post is a candidate
>for this, since I don't really know how to show the sphere in hilbert space
>is contractible; it's not a CW complex in any way that's obvious to me.
>
>I'll try to remember a nontrivial space with no homotopy, but it might be
>faster to post again to s.m.r.
>
>dave
----------------------------------------------------------------
Although your "roughly 'no'" answer MAY be ultimately correct, I am
not willing to give up this easily.
(I am well aware of Whitehead's theorem.)
But I don't yet know that, if the category of maps is suitably
restricted, one might not find a finer classication which includes
distinguishing "infinite-dimensional holes" from non-holes.
And I am not thinking of weird pathological spaces, but things as simple
as H - {0} vs. H.
[sig deleted -- djr]
==============================================================================
Date: Tue, 26 Sep 95 14:27:01 CDT
From: rusin (Dave Rusin)
To: [Permission pending]
Subject: Re: Contractions of infinite-dimensional spheres
Just to follow up: I knew there were non-contractible spaces with no
homotopy. They're called "weakly contractible", and figure prominently
in the Sullivan conjecture (proved about 10 yr ago by Haynes Miller),
which asserts (more or less) that the space of maps from any BG to
a finite-dimensional complex is weakly contractible.
Thus in principle you could measure infinite-dimensional holes in a space
Y by the set of homotopy classes [map(BG,X), Y] , although you
have multiple choices for G and X -- I don't know what would consitute
the most natural choice, although I would imagine it makes a difference.
I frankly admit this is not pretty; you're looking at a "double exponential"
of spaces -- Y^(X^BG) -- which does not lend itself to convenient
visualization.
Sorry about blanking out on the instances of these spaces. You can get
some pointers from MathSciDisc with a search on "weakly contractible" or
"phantom maps" (\pi_n(f) = 0 for all n).
dave
==============================================================================
From: baez@guitar.ucr.edu (john baez)
Newsgroups: sci.math.research
Subject: Re: Topology of Infinite-Dimensional Spheres and Projective Spaces
Date: 26 Sep 1995 01:59:08 GMT
In article Michael.Murray@Adelaide.Edu.Au (Michael K. Murray) writes:
>In article <[identifier deleted]>, [original poster] wrote:
>>Let H denote the Hilbert space of square-summable sequences of real numbers.
and consider
>> S_H, defined as { x in H : ||x|| = 1}
>S_H is certainly contractible and you proceed as follows.
[procedure omitted]
>I would be interested if you learn of a method that avoids
>the two-step procedure.
The way I've seen this done is as follows. Think of H as L^2[0,1]
rather than as square-summable sequences --- after all, all countable-
dimensional Hilbert spaces are isomorphic (use the Fourier transform
if you doubt me). Suppose f is in S_H. Define f_t as follows. For
x in [0,t] let f_t(x) = 1. For x in [t,1], let f_t(x) = c_t f(x)
where c_t is a positive constant chosen so that f_t is in S_H.
Note that f_0 = f and f_1 = 1, and note that f_t depends continously
on t (wrt the topology on S_H induced from the norm topology on H).
Note with a bit more work that this really defines a contraction of
H.
==============================================================================
From: [Permission pending]
Newsgroups: sci.math.research
Subject: Infinite dimensional holes
Date: 26 Sep 1995 19:15:47 GMT
A number of people have responded to my query about various infinite-
dimensional spheres, confirming that they are in fact all contractible.
[They were, in my notation, as follows:
S_F = the union of finite-dimensional unit spheres,
S_H = the unit sphere in Hilbert space
S^oo = the space of linear rays in R^oo (the real vector space of
all countable-tuples of real numbers).]
Particularly helpful replies were received from Michael Murray, Dave Rusin,
Michael Hirsch and Ross Geoghegan. Ross Geoghegan suggested the following
references, which I mention in case others are interested:
* R.D. Anderson and R.H. Bing, A complete elementary proof that Hilbert
space is homeomorphic to the countable infinite product of lines, Bull.
AMS, vol. 74, 1968, pp. 771-192.
* book: Infinite Dimensional Topology, J. van Mill, North-Holland, 1989.
* book: Selected Topics in Infinite-Dimensional Topology, C. Bessaga
and A. Pelczynski, Polish Scientific Publishers, 1975.
* various articles by H. Torunczyk.
to which I'll add
* Open problems in infinite-dimensional topology, R. Geoghegan (ed.),
Topology Proc. vol. 4 (1979) no. 1, pp. 287 - 338 (1980). [I haven't
looked this up yet, so I don't understand the presence of both 1979 and
1980 in the reference.]
* various papers by Geoghegan, T.A. Chapman, R.D. Edwards, N. Kuiper,
and D. Burghelea.
-----------------------------------------------------------------------------
Given the striking results that infinite-dimensional spheres are contractible,
and that in fact Hilbert space minus one point is homeomorphic to Hilbert space
itself (Can this homeomorphism be made real analytic?)...
QUESTION: Are there any known ways to rigorously capture the idea of an
infinite-dimensional hole?
Perhaps by suitably restricting the class of allowable mappings?
[sig deleted -- djr]
==============================================================================
From: Gavin Wraith
Newsgroups: sci.math.research
Subject: Re: Contractions of infinite-dimensional spheres
Date: Tue, 26 Sep 1995 08:23:44 +0100
Dave Rusin, and others on this thread, might be interested
in my paper "Using the Generic Interval", in Cahiers de Topologie
et Geometrie Differentielle Categoriques, Vol XXXIV-4,
pp 259-266, (1993), where some of the things Rusin mentions
appear. If he sends me his address I will mail him a reprint.
This paper describes the Boolean completion of the unit interval,
thought of as a topological distributive lattice, as the infinite
sphere (i.e. direct limit of finite dimensional spheres with
equatorial embeddings). I also give another description in terms
of functions with discontinuities, which is related to a functor
invented by Ronnie Brown and S.Morris - videte "Embeddings in
Contractible or Compact Objects", Colloquium Mathematicum, 38,
pp 213-222, (1978).
-- Gavin Wraith
==============================================================================
From: Gavin Wraith
Date: Thu, 28 Sep 1995 08:50:45 +0100
To: rusin@math.niu.edu
Subject: Infinite-dim spheres etc
Thanks for your mail, and for the tip about http://www.ams.org. It
is only in the last few weeks that I have got a web browser up and
running. The article is on its way - it was really a write-up of a
snippet of conversation on the question "What is the Boolean completion
of the unit interval?". The answer is the infinite-dim sphere, with
negation given by antipodal involution; the unit interval embeds as
a longditude from pole to pole.
-- Gavin Wraith
==============================================================================
From: palais@math.brandeis.edu (Richard S. Palais)
Newsgroups: sci.math.research
Subject: Re: Topology of Infinite-Dimensional Spheres and Projective Spaces
Date: Mon, 02 Oct 1995 13:21:48 -0400
>> In article <[identifier deleted]>, [original poster] wrote:
>>
>> >Let H denote the Hilbert space of square-summable sequences of real
numbers.
>> >
>> >Let R^oo denote the countable product of R with itself, as a vector space,
>> >(i.e., all sequences of real numbers).
>> >
>> >Consider the following infinite-dimensional spheres:
>> >
>> >1. S_H, defined as { x in H : ||x|| = 1}
>> >
>> >
>> >a) Which, if any, of the spheres is contractible, and for those
>> > that are,what is an explicit homotopy of the identity map to
>> > a constant???
In a paper called "Homotopy Theory of Infinite Dimensional Manifolds",
Topology, vol. 5, pp.1--16 (1966), I proved a very general result that
handles most situatuions of this type. Namely, the Corollary of Theorem 17
of that paper states:
Theorem. Let V be a locally convex topological vector space and let {E_n}
be an increasing sequence of finite dimensional subspaces of V whose union
is dense in V. Given O open in V, let O_n denote the intesection of O with
E_n, and let O_oo denote the union of the O_n with the inductive limit
topology. If V is metrizeable, or more generally if O is paracompact,
then the inclusion of O_oo into O is a homotopy equivalence.
This can be applied to answer questions like that of [original poster]. If M is a
smooth infinite dimensional submanifold in V, then in reasonable cases
(e.g., in the Hilbert space case) we can replace M by a tubular neighborhood
O that has M as a deformation retract (so M --> O is a homotopy equivalence)
and then apply the above theorem to O.
For typical applications see my paper "On the homotopy type of certain groups
of operators", Topology, 3, (1965), in which I show various infinite dimensional
linear groups have the homotopy type of the inductive limits of their finite
dimensional counterparts.
--
Richard S. Palais Dept. of Mathematics Brandeis Univ.
palais@math.brandeis.edu
Home Page: http://rsp.math.brandeis.edu
==============================================================================
Date: Wed, 4 Oct 1995 07:37:03 -0700
From: [Permission pending]
To: rusin@math.niu.edu (Dave Rusin)
Subject: Re: spheres question
Thanks for the clarification. I apologize for reading your message
carelessly. Your point was a very interesting one.
[sig deleted -- djr]
P.S. Now I am thinking about how to detect SOME kind of infinite
dimensional holes in Hilbert space. I believe that at least in the
extremely limited situation of two oo-codimension Hilbert subspaces
H_1 and H_2 of Hilbert space H (and by Hilbert space or subspace I always
mean here something isomorphic to square-summable sequences of reals), then
if we assume that H_1 intersect H_2 is a 1-dimensional subspace L,
and each contains a geometric sphere S_1 and S_2 (resp.) such that S_1 and S_2
are disjoint from each other and intersect L in an alternating pattern
of 4 points (just as two planar geometric circles can line in R^3),
THEN any 1-parameter family of rigid motions of H, starting from the identity,
that are applied only to H_2, say, with the contraint that all images of
H2 never intersect H_1, will preserve the linking property. (Meaning,
H_1 and H_2 will never lie in disjoint halfspaces.) Clearly this is equivalent
to also subjecting H_1 to a family of its own rigid motions of H restricted to
H_1 s.t. the images of H_1 and H_2 at time t are always disjoint.)
Admittedly this is pretty lame, but at least it's something. Now I'd like
to know how far it can be extended (the class of spheres and the class of
motions)...
[sig deleted]