Define a pseudometric on unit sphere

[Lily@pie]

Homework Statement

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

Homework Equations

Pseudometric $(X,\rho)$ is where $\rho$ satisfies all conditions of a metric except that $\rho(x,y)$ = 0 need not imply x=y

The Attempt at a Solution

I am totally clueless... But I tried the following...

I was thinking in the $ℝ^{2}$
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

 

[sunjin09]

You want to define a metric ON THE SPHERE, so x and y are both on the unit circle for R^2.

 

[Lily@pie]

Does this means that d(x,y) will be the arc length of the unit circle for $ℝ^{2}$?

 

[sunjin09]

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.

 

[Lily@pie]

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 $\sqrt{(c-a)^{2}+(d-b)^{2}}=0$ or $\sqrt{(c-a)^{2}+(d-b)^{2}}=1$, then d(x,y)=0... But it's like using the distance function before defining it?

 

[sunjin09]

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 $\sqrt{(c-a)^{2}+(d-b)^{2}}=0$ or $\sqrt{(c-a)^{2}+(d-b)^{2}}=1$, then d(x,y)=0... But it's like using the distance function before defining it?

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.

 

[Lily@pie]

Try thinking about the simplest metric on an arbitrary set, which is almost trivial but it works.

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|?

 

[alexfloo]

"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?

 

[sunjin09]

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|?

Well, what if there's no "-" or abs defined on the set, such as the set of all students in your class?

 

[Lily@pie]

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

 

[alexfloo]

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 into a pseudometric by using our structure (lines through the origin).

 

[Lily@pie]

Is it something like

For all x,y in the arbitrary set
d(x,x)=0 and d(x,y)=1 if x≠y?

 

[alexfloo]

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.

 

[Lily@pie]

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?

 

[alexfloo]

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.

 

[Lily@pie]

Will it be something like that?

For all $x,y \in ℝ^{n+1}$

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?

 

[alexfloo]

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.

 

[Lily@pie]

But the question states that to define on a unit sphere $S^{n}$. 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 $ℝ^{n+1}$ through the origin?

 

[alexfloo]

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 x_L and x_K, 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.

 

[Lily@pie]

Even in the pseudometric that I define?

 

[alexfloo]

To which part, exactly?

 

[Lily@pie]

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 x_L and x_K, 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.

This part? Because I didn't use a function but just use the concept of the points on a line.

 

[alexfloo]

Actually, you did use a function! The word "function" has nothing to do with a formula, a function can be thought of as a little machine that takes inputs and gives outputs. (Also, it has to be consistent; f(x) can't be 3 one time and 4 the next time, for the same input.)

That's exactly what your d(x,y) is: it takes two points in space, and produces a nonnegative real number. It doesn't matter "how" it does that. It doesn't matter if you have a (good, clear) English description or a mathematical formula. In fact, many of the most interesting functions can't be written down either.

 

[alexfloo]

In fact, the first paragraph of your quote describes a function that takes "lines L and K through the origin" and produce a distance. I described my little machine in words as well, and my description even used your machine.

 

[Lily@pie]

I asked my lecturer on this solution, but he say the metric should be something that is defined along the unit sphere... Not quite what we do here...

 

[sunjin09]

I asked my lecturer on this solution, but he say the metric should be something that is defined along the unit sphere... Not quite what we do here...

What I suggested and alexfloo clarified is valid, but I understand it's different from what your lecturer had in mind, which is probably 1-cos(angle between two lines) or some equivalent variant. I consider this extreme example would help you understand metric better, because this metric meets the MINIMAL requirements and provides no more information about the geometric structure of the unit sphere.

 

[Lily@pie]

So how should I go about defining it on the unit sphere??

 

[Lily@pie]

I was thinking about measuring the distance between two points with the euclidean distance between the points?

 

[sunjin09]

I was thinking about measuring the distance between two points with the euclidean distance between the points?

d(x,y)= min{arc(x,y), arc{x,y'}} where arc(x,y) is the arc length between x and y on the sphere and y' is the point opposite to y on the sphere.