Support of a function and k-vertices

Discussion in 'Physics & Math' started by arfa brane, Feb 5, 2017.

  1. arfa brane call me arf Valued Senior Member

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    Define a k-vertex as a point embedded in the centre of a k+1 dimensional space.

    A 0-vertex is defined explicitly by the intersection of two lines = 1-surfaces. The centre is the 0-vertex.
    A 1-vertex is a point embedded in the centre of a line which is bounded by 0-vertices, a 2-vertex is in the centre of a region bounded by lines, etc.

    A function has support if it has nonzero values for some set of parameters which is closed under the function. One way to think about this is with the sinc(x) function, which has a singular point at x = 0, but this point has nonzero values either side, hence arbitrarily setting sinc(0) = 1 means assuming we can always get arbitrarily close to x = 0, even infinitely close. We say sinc(x) has support at {0} since there is a local neighbourhood of {0} where the function is not zero.

    A problem I'm looking at is about finding a continuous analytic function which tiles \( \mathbb {R} \) with \( \mathbb {Z} \). The clue here is that the delta function acts as a translate of a set in \( \mathbb {R}^d \). A tiling exists when the translates have empty intersections, or more technically, zero Lebesgue measure.

    Concretely, in the case of a tiling of polygons, such as squares, you would need to define the empty intersections over the set of interiors of each square, since the boundaries of each square tile are nonempty intersections in an edge-to-edge tiling.
     
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  3. arfa brane call me arf Valued Senior Member

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    Suppose we define a delta function as a 'unit' point mass, since the delta function has the restriction that:

    \( \int_{-\infty}^{\infty} \delta (x) dx = 1 \),​

    the "1" is the unit of mass at the point x.

    Then we have convolution of some function f, with this delta "mass" function, in the Fourier domain (!).
     
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  5. arfa brane call me arf Valued Senior Member

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    This arxiv paper from two people at Budapest and Crete universities certainly covers a bit of abstract ground. It's kind of ironic that given one of them is Greek, the ideas they present appear to require a tiling in a Euclidean domain, where we can talk about the interiors of polygons in a tiling as a set of discrete objects, and the pointwise translates as another discrete set, both in the same domain of points.

    The paper is titled Tilings by Translation. Tilings in the plane (and other surfaces) can include reflections and rotations along with translations (viz Escher). The authors exclude all but translations.

    Of course, the notion of a point is an axiom, because you can distinguish them (a tiling is in some sense, a concrete expression of this axiom, as is a pair of intersecting lines). In calculus, it's more about whether points are limits of functions, or intuitively, how close you can get to a point. The idea they take from the Euclidean plane to the Fourier domain and convolution of tiling functions is just that we can assume there is a function f that tiles space with translation.

    A planar graph has one face which is the complement of (the disjoint union of) the faces bounded by edges, i.e. the rest of the plane, hence the centre (a 2-vertex) of this "unbounded" face is at infinity.
    Gosh, really? Is that the same place the y value of the sinc function goes to at zero (i.e. when technically, division by zero occurs in the function: sin(x)/x)?

    So this idea of a function f having support at {0}, or that {0} "is" the support of f, means that at say, x = 0, f(x) can have any nonzero, but not infinite value, this means we use the calculus on the singular point to in some mathematical sense, remove the problem (by showing x = 0 isn't necessarily "there", so to say).
     
    Last edited: Feb 9, 2017
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  7. arfa brane call me arf Valued Senior Member

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    So, what about graphs with a face whose centre can be anywhere outside the interior of the finite graph, which is to say, anywhere on a surface with no boundary

    We adjust what we mean by a centre, and put the 2-vertex "near" its set of bounding edges, in fact we can do this with all the 2-vertices in the planar graph, leaving it topologically the same graph. Just stretch the plane here and there, type of thing or adjust the notion of a rigid distance (to a geometric centre).

    In that case, we give some function that takes a graph to a graph, support (it's a topological function) at a discrete point (the one which can be at infinity on the Euclidean plane, and is a 2-vertex which is a 0-vertex in the dual). We could also notate such graphs with edges extended to this point, without the point being "notated".
     
    Last edited: Feb 9, 2017
  8. arfa brane call me arf Valued Senior Member

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    And towards my own ideas, so far, about starting with the integers to tile the integers, this map of the Mobius function over the first 100 x 100 integers modulo 100.
    (The white squares are where the function is nonzero, black squares where it is zero)

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    The way the domain is mapped to the square matrix (of squares, like a square tiling) reveals patterns. You can see the parallel sets of black squares, some look "continuous", but intersections of subsets are all over the place. So what do the 'coordinates' in this matrix do to the Mobius function's values? The patterns must be saying something about the sets of divisors of a number, and about prime factors of numbers. The squarefree/not squarefree tiling (?) How to consider the translation via a delta function?

    And the above map of the first 10,000 integers is also a map to the same integers on the negative side of zero because of the relation of congruence between numbers (modulo another number). The function that reveals some periodicity and regularity also reveals some underlying randomness--there are no repeating patterns anywhere, this randomness and regularity appears to be a property of numbers themselves, the Mobius function is regular too.

    Why the above is not a tiling is made more apparent with a fuller map that identifies negative (red)or positive (blue) values of the function:

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    Last edited: Feb 9, 2017
  9. arfa brane call me arf Valued Senior Member

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    I'll just go over what a squarefree number is in number theory:

    An integer number is squarefree if it can be written as a product of distinct prime factors with no repeating factors. Hence any prime number is squarefree since its factorization contains one distinct prime. 4 is the first positive non-squarefree number because \( 4 = 2^2 \), so any multiple of 4 is non-squarefree (which is why there are vertical black lines at n = 4,8,12, ...; \( 9 = 3^2 \), so the same applies to any multiple of \( 3^2 \), to any multiple of \( 5^2 \), and so on (although these multiples don't have corresponding vertical lines, but why not?).

    Otherwise a squarefree number has an even or odd number of prime factors and this is represented in the k x k tables above as blue and red respectively. Hence all primes are red. The lowest row in the table is the first 100 integers (i.e. k = 100), since \( 0 \equiv k (mod\; k) \), the 100th and 0th values appear in the leftmost position.
     
    Last edited: Feb 9, 2017
  10. arfa brane call me arf Valued Senior Member

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    So it should be obvious that the above 100 x 100 tables have vertical black lines (for multiples of 4, for instance) because (in the integers) 4 divides 100, so does 25 (sorry, mistake in post 6, and you didn't spot it!). The multiples of 9 aren't vertical lines because 9 doesn't divide 100.

    Then the overall patterns are an artefact of splitting the integers into lots of 100, with a pair of functions parametrised by n, namely, \( f(n) = n\; (mod\; 100) \) along the horizontal, and \( g(n) = \left \lfloor \frac{n}{100} \right \rfloor \) along the vertical. But instead of 100, a general value could be used, so the pattern is arbitrary but exists because of some property of numbers. It emerges when you try to tile the integers and apply the Mobius function at each 'point'.
    Moreover, tiling is something that has a pre-existing pattern as an argument of some tiling function, not a 'search' for periodicity.

    So I could say, conjecture that there is no repeating pattern in the first n = 10,000 integers when these are arguments of the Mobius function (at least I can't find one with visual inspection, and I know humans are pretty good at seeing patterns).

    However, that doesn't mean there isn't a pattern for some other value of n. Proving this I think would be a hard problem . . .
     
    Last edited: Feb 10, 2017
  11. arfa brane call me arf Valued Senior Member

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    Here is a diagram representing the notion of a geometric tiling in \( \mathbb {R}^2 \), with hexagons. We have points \( \omega \in \Omega \) the interior of a hexagon, and a set \( \Lambda \) of points (0-vertices) on the boundary, note that this set includes every second vertex, or half the number of vertices in the boundary of \( \Omega \).

    So the translate of \( \Omega \) is another vertex which is a minimum of two vertices distance along the boundary.

    The authors want to generalise this notion of tiling. Are there algebraic kinds of tiling (of \(\mathbb R\) with \(\mathbb Z\)), or complex kinds? Apparently yes.
    If a tile can be an algebraic pattern, i.e. 123123123... then the translate could be the Euclidean division algorithm, along with multiplication by multiples of 10 (if that's the base you're working in) so no "decimal point" exists. Then there is a (somewhat trivial) way to tile \(\mathbb R\) with \(\mathbb Z\).
     
    Last edited: Feb 13, 2017

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