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The Bohr compactification of the Reals

  1. Feb 16, 2004 #1

    marcus

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    the Bohr compactification of the real line
    RBohr
    is essential (it seems) to Loop Quantum Cosmology.

    Here are some links illustrating that:
    https://www.physicsforums.com/showthread.php?s=&postid=147432#post147432

    I want to understand the Bohr compactification better.

    It is putting a different topology on the Reals
    (or the reals embedded as a dense subset of a larger
    collection of numbers) so that they become compact.
    But it is not the "one point" compactification which
    is a widely known trick of adjoining a "point at infinity".
    It is a different and probably more cool compactification.
    The younger brother of Niels thought of it. His name was Harald.

    Well, why shouldnt we just keep on using the line of real numbers
    that we know and love, but have a special topology handy for
    use on Sundays and other special occasions that makes them compact.
    Sounds OK to me.

    To understand the construction we probably need to know the idea of the "dual group" or (a droll synonym) the "group of characters" of a commutative group G.
    The characters of a commut. topol. group G are just the continuous homomorphisms from G to the Unit Circle---the complex numbers with |z|=1, under multiplication.
    [tex]\chi : G \longrightarrow S^1[/tex]

    [tex]\chi(gh) = \chi(g)\chi(h)[/tex]

    a familiar example would be χ(x) = exp(ix)

    notice (please) that the characters form a group themselves
    you can multiply two of them and it is still continuous
    and it still satisfies that multiplicativity condition

    I'll post this and get back to it later.

    Oh, the bohr cmptfn of the Reals is the dual group of the Reals equipped with a wacko topology called the "discrete" topology in which the real line is totally atomized as if by a giant sneeze and no point is near any other point.


    Viqar Husain and Oliver Winkler
    "On singularity resolution in quantum gravity"
    http://arxiv.org/gr-qc/0312094

    Abhay Ashtekar, Martin Bojowald, Jerzy Lewandowski
    "Mathematical Structure of Loop Quantum Cosmology"
    http://arxiv.org/gr-qc/0304074

    I see the Ashtekar/Bojowald/Lewandowski paper came out
    in the 2003 edition of "Advances in Theoretical and Mathematical Physics" 7, 233-268
     
    Last edited: Feb 18, 2004
  2. jcsd
  3. Feb 18, 2004 #2

    marcus

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    Wikipedia
    http://en.wikipedia.org/wiki/Pontryagin_duality#Bohr_compactification_and_almost-periodicity

    the first reference the article gives is Walter Rudin
    Fourier Analysis on Groups (Interscience 1962)

    the idea is that duality is reflexive so you expect G'' = G
    the dual of the dual of a group G is the same group back again
    But suppose you take the dual of G and throw away the usual topology and give it the discrete topology (the power set, same as saying that singletons are open)
    so you say H is equal to G' as a group but atomized as a topological space and all functions from it are continuous.
    THEN you take the dual of H, and that is the bohr compactification of G.

    so it is almost like taking G'' and getting G back
    except you intervene in midstream and change the topology.
     
    Last edited: Feb 18, 2004
  4. Feb 18, 2004 #3

    marcus

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    Now what Wikipedia means by the map H->G'
    is the identity map and of course it is
    continuous (as Wiki notes) because it is coming from the discrete topology from whence all maps are continuous.

    so the identity map H-> G' is a continuous homomorphism and Wiki
    says look at the "dual morphism" to that
    (morphisms come in pairs like Mounds bars, the chocolate coated
    cocoanut treat)
     
  5. Feb 18, 2004 #4

    marcus

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    we are going to "lift" the identity map to its dual
    B(G) is the dual of H, so put it upstairs over H

    and G'' is the same as G so G is the dual of G'
    and lets put G upstairs over G'
    and draw this little mapping diagram such as the people
    in the chalk-dust clouds on the third floor of the math building like to draw.

    [tex]B(G) \leftarrow G[/tex]
    [tex]H \rightarrow G'[/tex]

    notice that the dual morphism arrow points backwards
     
    Last edited: Feb 18, 2004
  6. Feb 18, 2004 #5

    marcus

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    to "lift" the identity map to its dual morphism
    (what Wiki calls it)
    I have to say what g, an element of G, gets mapped to

    it has to be mapped to something in B(G)
    which is defined as the dual of H

    you think i get a kick out of this?
    it is trivial
    g gets mapped to the simple act of evaluating
    a χ at g.

    remember H consists of homomorphisms χ from G to the unit circle S.
    [tex]\chi : G \rightarrow S ; g \rightarrow \chi (g)[/tex]

    so a group element g can actually behave as a homomorphism from H to the unit circle

    [tex]g : H \rightarrow S ; \chi \rightarrow \chi (g)[/tex]

    [tex]B(G) \leftarrow G[/tex]
    [tex]H \rightarrow G'[/tex]

    you are supposed to check that it is in fact a homomorphism but
    the adept graduate student will just drop a ringing name like
    "Pontryagin" and not trouble him or her self with this.

    now we begin to see why the discrete topology was used, because
    otherwise something might not be continuous and therefore would
    be unkosher.
     
  7. Feb 18, 2004 #6

    marcus

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    http://en.wikipedia.org/wiki/Pontryagin_duality

    "the duality interchanges the categories of discrete groups and compact groups"

    OK this must be the meat
    we blew the real line to bits and gave it discrete topology just so that we can come home to a dual version of the real line that is compact

    here is what Wiki says:
    I am still having difficulty imagining what the bohr compactification of the real line is like. can anyone provide some intuition, imagery, anything helpful to understanding it.
    how can the real line (locally compact and abelian but not compact) be mapped into a topological group that is somehow like the real line but compact?
     
    Last edited: Feb 18, 2004
  8. Feb 18, 2004 #7

    marcus

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    the motivation for this is worth recalling
    when general relativity is quantized either quantizing
    the metric (geometrodynamics a la Wheeler/Dewitt) or
    quantizing the connection (a la Rovelli/Ashtekar/Smolin)
    and applied to cosmology then when they use the bohr
    compactification to build the quantum state space
    then the big bang singularity goes away.
    somehow it is a ticket to smoothness at time zero.

    Viqar Husain and Oliver Winkler
    "On singularity resolution in quantum gravity"
    http://arxiv.org/gr-qc/0312094

    Abhay Ashtekar, Martin Bojowald, Jerzy Lewandowski
    "Mathematical Structure of Loop Quantum Cosmology"
    http://arxiv.org/gr-qc/0304074
     
  9. Feb 19, 2004 #8

    Hurkyl

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    All right, this is a mess. :smile:

    All groups in this post will be additive groups. Some of these theorems I've only sketched the proof for, but I'm fairly confident in them. I've omitted the proofs, but I can fill them in if you like.

    If you just want the answer, skip to the bottom. :smile:

    (If someone more knowledgable can check my work, it would be great! I'm fairly confident in everything, but I feel like I've glossed over a few details)


    Let us denote elements of [itex]\mathbb{R}'[/itex], the dual group to [itex]\mathbb{R}[/itex] by [itex]\chi_r : \mathbb{R} \rightarrow \mathbb{R} / \mathbb{Z} : x \rightarrow rx + \mathbb{Z}[/itex], where [itex]r \in \mathbb{R}[/itex]. In other words, [itex]\chi_r(x) = rx (\mathrm{mod} 1)[/itex].

    (I've chosen to map into [itex]\mathbb{R} / \mathbb{Z}[/itex] instead of [itex]S^1[/itex] so that the dual group will be an additive group)


    Now, let's let [itex]H[/itex] denote [itex]\mathbb{R}'[/itex] with the discrete topology.

    Theorem 1: If [itex]G[/itex] is a group with the discrete topology, then [itex]G'[/itex] is the set of all group homomorphisms [itex]\chi : G \rightarrow \mathbb{R} / \mathbb{Z}[/itex], and a convegence in [itex]G'[/itex] coincides with pointwise convergence in [itex]\mathbb{R} / \mathbb{Z}[/itex].


    Now, the question is "What do group homomorphisms from [itex]\mathbb{R}[/itex] to [itex]\mathbb{R} / \mathbb{Z}[/itex] look like?" Allow me to emphasize that these homomorphisms do not need to be continuous.


    Lemma 2: If [itex]\varphi : \mathbb{Q} \rightarrow \mathbb{R} / \mathbb{Z}[/itex] is a group homomorphism, then it may be written as [itex]\varphi : \mathbb{Q} \rightarrow \mathbb{R} / \mathbb{Z} : x \rightarrow rx + \mathbb{Z}[/itex] for some [itex]r \in \mathbb{R}[/itex].


    Theorem 3: If an additive group [itex]V[/itex] is a vector space over [itex]\mathbb{Q}[/itex], and if If [itex]\varphi : V \rightarrow \mathbb{R} / \mathbb{Z}[/itex] is a group homomorphism, then it may be written as [itex]\varphi : V \rightarrow \mathbb{R} / \mathbb{Z} : x \rightarrow vx + \mathbb{Z}[/itex] where [itex]v[/itex] is a real valued linear functional on the vector space [itex]V[/itex].


    Now, it so happens that [itex]\mathbb{R}[/itex] is a vector space (of uncountably infinite dimension) over [itex]\mathbb{Q}[/itex], so applying theorems 1 and 3:

    The dual group of [itex]H[/itex], [itex]B(\mathbb{R})[/itex], can be parametrized by the set of all real valued [itex]\mathbb{Q}[/itex]-linear functions on [itex]\mathbb{R}[/itex]. For any such function [itex]v[/itex], the corresponding element of [itex]B(\mathbb{R})[/itex] is [itex]\varphi_v : \mathbb{R} \rightarrow \mathbb{R}/\mathbb{Z} : x \rightarrow vx + \mathbb{Z}[/itex].

    Convergence of elements in [itex]B(\mathbb{R})[/itex] is given by pointwise convergence over [itex]\mathbb{R} / \mathbb{Z}[/itex].

    Notice that the additive group [itex]\mathbb{R}'[/itex] is a subgroup of [itex]B(\mathbb{R})[/itex]; thus, we can use the isomorphism between [itex]\mathbb{R}[/itex] and [itex]\mathbb{R}'[/itex] to embed [itex]\mathbb{R}[/itex] into [itex]B(\mathbb{R})[/itex].


    Theorem 4: If [itex]\varphi_v = \varphi_w[/itex], then [itex]v = w[/itex].


    So we can simplify the formulation somewhat.

    -----------------------------------------------------------------------------

    The Bohr compactification of [itex]\mathbb{R}[/itex], [itex]B(\mathbb{R})[/itex], is the [itex]\mathbb{Q}[/itex]-vector space of all real valued [itex]\mathbb{Q}[/itex]-linear functions on [itex]\mathbb{R}[/itex]. (Equivalently, the set of all additive group homomorphisms from [itex]\mathbb{R}[/itex] to itself)Convergence in [itex]B(\mathbb{R})[/itex] is pointwise.

    An element [itex]r \in \mathbb{R}[/itex] can be identified with the function [itex]\chi_r : \mathbb{R} \rightarrow \mathbb{R} : x \rightarrow rx[/itex] in [itex]B(\mathbb{R})[/itex].
     
    Last edited: Feb 19, 2004
  10. Feb 19, 2004 #9

    marcus

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    portion about typo deleted since was fixed


    Hurkyl thanks for the help, have been wanting
    help with bohr compactification
     
    Last edited: Feb 19, 2004
  11. Feb 19, 2004 #10

    Hurkyl

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    Good catch. I cut-pasted the theorem from lemma 2, didn't fix all the definitions!

    The end result seems (comparatively) so simple to anything you've quoted from other sources on it; I wonder why this special case isn't mentioned? Have I just made a horrible blunder?
     
  12. Feb 19, 2004 #11

    marcus

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    this is the information i believe i needed most, that the real line is an infinite dimensional vectorspace over the field of rational numbers. never occurred to me to think of it that way.
    any favorite sourcebook about this?

    whoah! sounds from what you just posted that this is a Hurkyl proof!
    bravo!
    so not cribbed from Walter Rudin or some such venerable

    now its after 11PM here so I will not inspect the proof
    any more until tomorrow. It seems all right.
    and also a good thing to use R/Z instead of unit circle
     
    Last edited: Feb 19, 2004
  13. Feb 19, 2004 #12

    Hurkyl

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    Erm... Wikipedia? The only mentions of "Bohr compactification" I've seen have actually came from you or sources linked by you. Never understood it until I poured through that Wikipedia link. It was a fun exercise, IMHO. And I got to learn just what a Haar measure is too, which is nice.

    I'm trying to decide if I should come up with a proof that the set I mentioned is, indeed, compact, before I go to bed. I think bed is winning, but I'll know I'll think about it while trying to go to sleep! :smile:


    I had seen the reals as a rational vector space before, but just as a curiousity that was introduced by my professors. It's a terrible thing to try and wrap your brain around! Countably infinite dimensional vector spaces are bad enough!
     
  14. Feb 19, 2004 #13

    Hurkyl

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    Grr, Lemma 2 is wrong, methinks, and I need it for Theorem 3.

    At the moment, I'm fairly convinced what I mentioned is only a subspace of [itex]B(\mathbb{R})[/itex]...

    There's more work to be done on just what a group homomorphism from [itex]\mathbb{Q}[/itex] to [itex]\mathbb{R}/\mathbb{Z}[/itex] looks like. :frown:
     
  15. Feb 19, 2004 #14

    Hurkyl

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    All right, I got it this time.


    The mistake I made was assuming that if [itex]f : \mathbb{Q} \rightarrow \mathbb{R} / \mathbb{Z}[/itex] is a homomorphism, then [itex]f(r) \rightarrow 0[/itex] as [itex]r \rightarrow 0[/itex].


    However, we can still catalog these homomorphisms!

    Define [itex][f]_0 := f(1) \quad (0 \leq [f]_0 < 1)[/itex]

    Now, consider the possible values of [itex]f(1/2)[/itex]. Because we know the value of [itex]f(1)[/itex], there are only two possibilities for [itex]f(1/2)[/itex]. We define the next statistic implicitly by [itex]f(1/2) = (1/2) (f(1) + [f]_1)[/itex].

    Then, given [itex]f(1/2)[/itex], there are three possibilities for [itex]f(1/6)[/itex]. We define the next statistic by [itex]f(1/6) = (1/3) (f(1/2) + [f]_2)[/itex].

    In general, [itex]f(1/n!) = (1/n) (f(1/(n-1)!) + [f]_{n-1})[/itex].


    Allow me to emphasize that [itex][f]_0 \in \mathbb{R}[/itex], but [itex][f]_i \in \mathbb{N}_i[/itex] for [itex]i > 0[/itex].


    Given these statistics [itex][f]_i[/itex], we can recover the function through the formula:

    [tex]
    f(\frac{p}{q!}) = \frac{p}{q!} \sum_{i=0}^{q-1} (i! [f]_i)
    [/tex]

    Notice that higher order terms ([itex]i >= q[/itex]) of the sum are irrelevant, because we're working mod 1.


    With a little effort, we can see that each sequence of statistics describes a unique homomorphism, and each homomorphism has a sequence of statistics, so we have a catalog of all homomorphisms from [itex]\mathbb{Q}[/itex] to [itex]\mathbb{R}/\mathbb{Z}[/itex].


    So what is the topology of this space? Well, as per my previous post, convergence in this space is given by pointwise convergence in [itex]\mathbb{R}/\mathbb{Z}[/itex], so what does this mean?

    If [itex]f_i \rightarrow f[/itex], then for any [itex]q \in \mathbb{Q}[/itex], we have that [itex]f_i(q) \rightarrow f(q)[/itex].

    Plugging in the formula, this means (after a little rearranging, and applying the homomorphism property) for all positive integers q, the sum

    [tex]\frac{1}{q!} \sum_{i=0}^{q-1} ([f_n]_i - [f]_i)[/tex]

    must converge to 0 as n increases.


    This usually means that [itex][f_n]_0 \rightarrow [f]_0[/itex] as [itex]n \rightarrow \infty[/itex], and that each [itex][f_n]_i[/itex] is eventually equal to [itex][f]_i[/itex], for [itex]i > 0[/itex]. However, things get messy when [itex][f]_0 = 0[/itex]...


    To see what is going on, allow me to comment on this representation. There is something called (IIRC) the "factorial base" representation of nonnegative integers. Recall that the decimal representation of an integer is given by:

    [tex]
    n = \sum_i 10^i n_i
    [/tex]

    If you're careful, on ranges, one can use other things instead of [itex]10^i[/itex] for scaling; the factorial base uses [itex]i![/itex] as the scale factor. With a little playing around, you can see that given the right bounds on the terms, each positive integer can be written uniquely as

    [tex]
    n = \sum_i i! n_i
    [/tex]

    You can see how this relates closely to the formula given above. If we add a fractional term to the factorial base representation, then we can uniquely write any nonnegative real number!

    For example, we can write the integer 37 as: 1201
    And you can check that 37 = 1 * 1! + 0 * 2! + 2 * 3! + 1 * 4!

    Notice that the ordinary addition algorithm works in this representation. For example, 1201 + 1 = 1210 == 38. (Note that because you exceeded the limit of 1 in the 1!'s place, you have to carry to the 2!'s place)


    However, the sequence of interest doesn't have to be eventually zero, so this is somewhat of a transfinite generalization of the factorial base representation. The important thing to note, though, is that how incrementing and decrementing work just like in ordinary arithmetic; you have to be sure to borrow and carry when appropriate.


    Now, back to the messy situation. When [itex][f]_0 = 0[/itex], we lie on a boundary. If we move slightly forward, all of the other [itex][f]_i[/itex] remain unchanged, but if we move slightly backward, we have to "borrow". For example:

    1201 + .1, 1201 + .01, 1201 + .001

    aproaches 1201 + 0

    But going the other way:

    1200 + .9, 1200 + .99, 1200 + .999

    also approaches 1201 + 0.


    Now, there's still more of a mess! Consider this sequence:

    11201 + 0, 101201 + 0, 1001201 + 0, 10001201 + 0, ...

    Note that this also converges to 1201 + 0!!!

    (and I haven't even tried to write any of the messes we can have with a string of digits that is not eventually 0)


    Anyways, the point is that we now have a catalog of all homomorphisms [itex]f\phi: \mathbb{Q} \rightarrow \mathbb{R} / \mathbb{Z}[/itex]. The spirit of my treatment carries on unchanged. We want to consider [itex]\mathbb{R}[/itex] as a vector space over [itex]\mathbb{Q}[/itex], and use the decomposition into [itex]\mathbb{Q}[/itex]-linear combinations of basis elements, and then we apply one of the ugly things we discovered in this post to each of the dimensions of [itex]\mathbb{R}[/itex], and in this way we generate all homomorphisms from [itex]\mathbb{R}[/itex] to [itex]\mathbb{R} / \mathbb{Z}[/itex].

    Yuck!
     
    Last edited: Feb 19, 2004
  16. Feb 21, 2004 #15

    Hurkyl

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    Ok, let me formally present this representation of [itex]\mathrm{Hom}(\mathbb{Q}, \mathbb{R}/\mathbb{Z})[/itex], complete with the appropriate topology.


    Let us define the set [itex]M[/itex] to be the set of all sequences [itex]\{b_i\}_0^{\infty}[/itex] where [itex]b_0 \in \mathbb{R}[/itex], [itex]0 \leq b_0 < 1[/itex], and for [itex]i>0[/itex], [itex]b_i \in \mathbb{N}[/itex], [itex]0 \leq b_i \leq i[/itex]

    We will write elements of [itex]M[/itex] in the following way:

    [tex]\ldots ,b_3,b_2,b_1.b_0[/tex]

    where the commas are optional symbols used for clarity. So, for instance, the sequence [itex]b_0 = 1/2, b_1 = 1, b_2 = 0, b_3 = 1, b_i = 0 (i > 3)[/itex] will be written as

    [tex]\ldots00101.5[/tex]

    Or, in the case where the sequence is eventually zero, we use the shorthand

    [tex]101.5[/tex]

    We can define addition and subtraction on these things by analogy with the way we define addition and subtraction on the usual decimal (or n-ary) representation of real numbers. So, for example:

    [tex]
    \begin{array}{r}
    \ldots 54310.5 \\
    + 1.5 \\
    \hline
    \ldots 54321.0 \\
    \end{array}
    [/tex]

    Note that we had to carry twice; first because [itex].5 + .5 = 1[/itex] in the 0th place, and second because [itex]0 + 1 + 1 = 2[/itex] in the 1st place.


    The set [itex]M[/itex] with this structure for addition forms an additive group.


    There is a natural embedding of [itex]\mathbb{R}[/itex] in [itex]M[/itex], which I will give via the reverse map:

    If [itex]\{ b_i\} \in M[/itex] is an eventually zero sequence, then we can map it to a real number by

    [tex]
    \{ b_i \} \rightarrow \sum_{i=0}^{\infty} b_i i!
    [/tex]

    This gives us any nonnegative real number. We can get the negative real numbers via taking additive inverses of eventually zero sequences. (alternatively, you could use the analog of two's complement notation to give them explicitly)


    So, for instance:

    [tex]
    101.5_M = 0.5 * 0! + 1 * 1!+ 0 * 2! + 1 * 3! = 7.5_{\mathbb{R}}
    [/tex]

    [tex]
    \ldots,4321.0_M = -1_{\mathbb{R}}
    [/tex]

    (check by adding 1 to [itex]\ldots,4321.0[/itex])


    We also want to consider quotient spaces of [itex]M[/itex] by truncating it after a finite number of digits. Note that if we truncate elements of [itex]M[/itex] to n digits long, then [itex]M[/itex] is isomorphic to [itex]\mathbb{R} (\mathrm{mod} n!)[/itex] (via the truncated form of the mapping given above)


    Now elements of [itex]M[/itex] are supposed to represent homomorphisms from [itex]\mathbb{Q}[/itex] to [itex]\mathbb{R}/\mathbb{Z}[/itex]. They do this by "multiplication"; if we take a rational number written in the form [itex]p / q![/itex] and [itex]\{ b_i \} \in M[/itex], we can use the above mapping to formally multiply:

    [tex]
    \{ b_i \} \left( \frac{p}{q!} \right)
    = \left( \sum_{i=0}^{\infty} b_i i!\right) \frac{p}{q!}
    [/tex]


    Note that because the range is [itex]\mathbb{R}/\mathbb{Z}[/itex], the terms with [itex]i \geq q[/itex] don't contribute anything to this sum. Thus, we can rewrite as

    [tex]
    \{ b_i \} \left( \frac{p}{q!} \right)
    = \left( \sum_{i=0}^{q-1} b_i i!\right) \frac{p}{q!}
    [/tex]

    and this works for any element of [itex]M[/itex], not just eventually zero ones.


    (Notice that the fact the domain of the homomorphism is [itex]\mathbb{Q}[/itex] is essential!)

    (Also note that if we add elements in [itex]M[/itex] then convert the result into a homomorphism, the result is the same as if we converted the elements into homomorphisms first then added them)


    Finally, I can describe the topology of [itex]M[/itex]. Recall that we desire convergence of a sequence in [itex]\mathrm{Hom}(\mathbb{Q}, \mathbb{R}/\mathbb{Z})[/itex] to be pointwise convergence.

    This is the same as saying that a sequence converges in [itex]M[/itex] iff it converges in every truncation of [itex]M[/itex].

    And finally, I can describe what neighborhoods look like:

    To construct a neighborhood of [itex]m \in M[/itex] we first construct the set [itex]S_{\epsilon} = \{ m + x | -\epsilon < x < \epsilon\}[/itex] for some real number [itex]\epsilon > 0[/itex]. Then, we construct the set [itex]U_{N, \epsilon}[/itex] which consists of all elements of [itex]M[/itex] that agree with elements of [itex]S_{\epsilon}[/itex] on the first [itex]N[/itex] digits.


    To demonstrate the two "orthogonal" directions of convergence, both of these sequences converge to 0:

    2.5, 1.1, 0.01, 0.001, ...
    1.0, 10.0, 100.0, 1000.0, ...


    One can imagine [itex]M[/itex] being constructed by taking infinitely many copies of the real line and attaching them end to end wrapping them around a torus. Sequences have to converge both by their position in the real line and have to converge to the right copy of the real line!


    Now, I can say explicitly what the Bohr compactification of the real line ([itex]B(\mathbb{R})[/itex]) looks like.

    Choose a basis [itex]B[/itex] of the [itex]\mathbb{Q}[/itex]-vector space of the real numbers. Elements of [itex]B(\mathbb{R})[/itex] are functions from [itex]B[/itex] to [itex]M[/itex].

    If [itex]\sum_{b \in B} q_b b = r[/itex] is an element of [itex]\mathbb{R}[/itex] (the [itex]q_b[/itex] are rational, and only a finite number of them are nonzero), and we're given [itex]f \in B(\mathbb{R})[/itex], then we compute [itex]f(r)[/itex] by:

    [tex]
    f(r) = \sum_{b \in B} f(b)(q_b)
    [/tex]

    (Notice this is well-defined since the summand can only be nonzero a finite number of times)

    (So basically, for each dimension of [itex]\mathbb{R}[/itex] we choose a way in [itex]M[/itex] for that dimension to transform, and we combine dimensions in an additive way)


    Convergence in [itex]B(\mathbb{R})[/itex] is given by pointwise convergence. Or, alternatively, you can say [itex]B(\mathbb{R})[/itex] has the topology it inherits from being the cartesian product of uncountably many copies of [itex]M[/itex].


    Finally, there is a natural embedding of [itex]\mathbb{R}[/itex] in [itex]B(\mathbb{R})[/itex] given by [itex]x \rightarrow R_x[/itex] where [itex]R_x : B \rightarrow M : b \rightarrow bx[/itex]. (The "multiply by x" map)



    So there you have it, this is what the Bohr compactification of the real numbers looks like. It's an awful terrible thing, and I think the only usefulness of this exercise is to appreciate how horrible it really is. :smile: I imagine for any practical purpose you can just say that elements of [itex]B(\mathbb{R})[/itex] are additive homomorphisms from [itex]\mathbb{R}[/itex] to [itex]\mathbb{R}/\mathbb{Z}[/itex] and be done with it!



    edit: Remembered to mention how the negative real numbers fit into [itex]M[/itex]

    removed a description of neighborhoods of [itex]M[/itex] that was missing a detail.
     
    Last edited: Feb 21, 2004
  17. Feb 21, 2004 #16

    marcus

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    I'm actually a little awed
    thanks
    nice to have a definite construction
    instead of just the abstract version
     
  18. Feb 22, 2004 #17

    marcus

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    still a bit swamped----Almost Periodic fcns?

    I'm still not quite afloat here. Wondering about the
    connection to the almost periodic functions.

    the AP functions on the reals are supposed to be
    the same as the continuous functions on the bohr compactification

    AP(R) = C(RBohr)

    for instance right at the top of page 6 of
    Husain/Winkler they say

    "It is well-known that AP(R) is naturally isomorphic to C(RBohr)."

    http://arxiv.org/gr-qc/0312094

    and this is actually not well-known to me and I would like to be able to picture the natural isomorphism. Is it obvious to you Hurkyl from what you already said?
     
  19. Feb 23, 2004 #18
    Every integer maps to a unique wave, with primes being sine waves

    I'm quite out of my depth here, but it occurred to me recently that every prime p could be associated with a sine wave with period p or 1/p and every integer could be represented as a unique wave by taking the sum of the sine waves of it's prime factors.

    If you sum the waves for two integers, you get the wave of their product. I was amateurishly toying with ideas to quickly factorize large primes.
     
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