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Homework Help: Define a pseudometric on unit sphere

  1. Mar 23, 2012 #1
    1. The problem statement, all variables and given/known data

    Define a pseudometric on the unit sphere [itex]S^{n}[/itex] in [itex]ℝ^{n+1}[/itex] that makes the set of lines in [itex]ℝ^{n+1}[/itex] through the origin into a metric space.

    2. Relevant equations
    Pseudometric [itex](X,\rho)[/itex] is where [itex]\rho[/itex] satisfies all conditions of a metric except that [itex]\rho(x,y)[/itex] = 0 need not imply x=y

    3. The attempt at a solution
    I am totally clueless... But I tried the following...

    I was thinking in the [itex]ℝ^{2}[/itex]
    I know that a unit sphere can be described by a set of lines through the origin by specifying the two points that intersect with the unit circle.

    Since it is a unit sphere, I suppose we need d(x,y) = 1 for all when x is the origin and y is the intersection point??

    Help :cry:
  2. jcsd
  3. Mar 23, 2012 #2
    You want to define a metric ON THE SPHERE, so x and y are both on the unit circle for R^2.
  4. Mar 23, 2012 #3
    Does this means that d(x,y) will be the arc length of the unit circle for [itex]ℝ^{2}[/itex]?
  5. Mar 23, 2012 #4
    That's not pseudometric. You want to define one such that d(x,y)=0 if x=y or if x and y are on the same line passing origin.
  6. Mar 23, 2012 #5
    I was thinking when d(x,y) is the angle between the 2 points.... when θ=0 or 180°, d(x,y)=0...

    Also thought about when d(x,y) is the shortest distance between the two points measured in the normal euclidean space. So if x=(a,b) and y=(c,d), then if [itex]\sqrt{(c-a)^{2}+(d-b)^{2}}=0[/itex] or [itex]\sqrt{(c-a)^{2}+(d-b)^{2}}=1[/itex], then d(x,y)=0.... But it's like using the distance function before defining it????
  7. Mar 23, 2012 #6
    The metric defined like you did here probably violates other rules such as the triangle inequality. (Or prove me wrong.) Try thinking about the simplest metric on an arbitrary set, which is almost trivial but it works.
  8. Mar 23, 2012 #7
    Simple metric on an arbitrary set.... Does this means the simplest way to measure distance between 2 elements in the set?

    So it'll be like |y-x|?
  9. Mar 23, 2012 #8
    "Distance," "difference," and "absolute value" all don't make sense on arbitrary sets. Unless the set has some extra structure, the only thing we know how to do with its elements are to check if they are the equal. Does that help?
  10. Mar 23, 2012 #9
    Well, what if there's no "-" or abs defined on the set, such as the set of all students in your class?
  11. Mar 23, 2012 #10
    I still don't understand... Why are we thinking about metric on arbitrary set when we have a set which is a set of points? Sorry, I'm quite slow...

    If the only way we can do anything with the elements in the arbitrary set is to check if they are equal, does it mean the simplest metric on a arbitrary set will be.... erm... I don't know... so sorry
  12. Mar 24, 2012 #11
    No no no that's a very good question: in this case we do have some structure (specifically, points can have the property of lying on the same line) and we will need it. We were simply setting up a simpler example that would be a helpful starting point.

    On an arbitrary set, we can usually only define a very very simple metric, since we don't have additional structure. Think about what that is, and then how we can augment it in to a pseudometric by using our structure (lines through the origin).
  13. Mar 24, 2012 #12
    Is it something like

    For all x,y in the arbitrary set
    d(x,x)=0 and d(x,y)=1 if x≠y?
  14. Mar 24, 2012 #13
    Yes exactly!

    Now think about what you're doing: you've turned "=" into a metric. An equivalence relation is kind-of a metric, but it can only measure two distances: 0 and 1. Furthermore, since different things can be equivalent, equivalence relations are actually pseudometrics (with the one exception you just showed).

    So now you just need to pick the right equivalence relation, and then do what you just did! While considering what relation to use, think about what I mentioned earlier about the structure of your space.
  15. Mar 24, 2012 #14
    Does that means now our x any y are points on the line through the origin and i'll need to find an expression for d(x,y) such that d(x,x)=0 and d(x,y)=1 if x≠y?
  16. Mar 24, 2012 #15
    Two things:

    First of all, you're on the right track, but not quite there. x and y are not necessarily on the same line through the origin - the question we're interested in here is whether they are. In the trivial metric we were interested in whether they were equal, and now we're interested in whether they lie on the same line. Does that help?

    Secondly, who said you need an expression? You didn't have one when you created the trivial metric earlier, and the way you expressed that was just fine. You just need to describe its values: it doesn't really matter if you can write it down in terms of symbols.
  17. Mar 24, 2012 #16
    Will it be something like that?

    For all [itex]x,y \in ℝ^{n+1}[/itex]

    1. d(x,y)≥0
    2. If x=y, d(x,y) = 0.
    3. If x≠y and x,y lies on the same line through the origin, then d(x,y)=0
    4. If x≠y and x,y does not lie on the same line through the origin, then d(x,y)=1

    Is this correct?
  18. Mar 24, 2012 #17
    Yes, exactly!

    Just so you know, 1. can be left out because it follows from the others. Also, 2. and 3. can be merged into "if x and y lie on the same line through the origin, then d(x,y)=0" because if x=y then they definitely lie on the same line. Finally, 4. can be shortened to "If x and y do not lie on the same line through the origin, then d(x,y)=1" because if they don't lie on the same line then they're not equal.

    Even though it can be simplified, however, it is absolutely correct the way you wrote it. it's important to note that this is not the only answer, however. Any function that is zero when x,y lie on the same line, positive otherwise, and satisfies triangle will suffice, but this is the simplest.

    Another option is this: define f(x)=x if x1 is nonnegative and f(x)=-x (reflected across the origin) otherwise. Then d(x,y)=d'(f(x),f(y)) (where d' is normal Euclidean distance) is also a correct answer. Basically what we did was take the points which lie on the same line (which are always a point x and its "reflection" -x) and map them together. We "picked half of the sphere" so that each line only goes through one point, then measure distance between those. Triangle inequality holds because no matter how we flip or rotate, no path that passes through a sphere at three points will be the shortest path between its endpoints.
  19. Mar 24, 2012 #18
    But the question states that to define on a unit sphere [itex]S^{n}[/itex]. From the way I define it, it didn't use the fact that it is on the unit sphere.

    And do we need to prove that it is a pseudometric on the unit sphere but a metric for the lines in [itex]ℝ^{n+1}[/itex] through the origin?
  20. Mar 24, 2012 #19
    In some sense you did: d is a function from SnxSn to ℝ. You "used" it because your distance function is only defined there (although you could've used the whole space). You're right to think that there's nothing special about the sphere, though. The metric you described works for any subset of ℝn, and if that subset happens to be a closed surface about the origin then it also forms a metric for lines in a way I'm about to describe.

    Suppose we have two lines L and K through the origin. Since we're talking about a closed surface around the origin, they intersect that surface in at least one point (in fact, at least two, and exactly two in our case). Pick any point of intersection for each, and call them xL and xK, and then measure the distance between them according to your pseudometric. Because all the points of intersection for the same line are "0 distance apart," it doesn't matter which one we choose: we'll get the same distance no matter which intersection we use.

    This also illustrates a more general idea: every pseudometric on a set can be made into a metric on a smaller set. Think of a pseudometric space as the same as a metric space, except each point is replaced by a collection of points that are all at the same "spot" and are the same distance from any other point. We can just measure distance between those "groups" of points, instead of between individual points, and we have a regular old metric space.

    In this case, we can chop our sphere in half and "reflect" one half, then superimpose it on the other half. We get a half sphere, except that "every point has two points:" the original, and its reflection. Since lines through the origin are totally determined by where they cross the sphere, defining a metric on those points is the same as defining one on the lines.
  21. Mar 25, 2012 #20
    Even in the pseudometric that I define?
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