# [bi|sur|in]jective coordinates

Discussion in 'Physics & Math' started by RJBeery, Sep 29, 2010.

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1. ### RJBeeryNatural PhilosopherValued Senior Member

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This is my understanding, please correct as needed:

A bijective function has a one-to-one correspondence from x in X to y in Y, meaning that every y has exactly one x mapped to it and every x has a mapping to exactly one y.

A surjective function is a many-to-one correspondence, meaning that every y has at least one x mapped to it, but that x is not necessarily unique.

An injective function maps at most one x to a given value y.

To simplify:
Surjective: At least one mapping from x to any and all y
Injective: At most, one mapping from x to any and all y
Bijective: Exactly one mapping from x to any and all y

Looking at it in the above manner it's easy to see that surjective + injective implies bijective (at least 1, and at most 1, necessitates exactly 1).

Now when we speak of coordinate systems in Physics I simply think of it in terms of functions mapping numbers to the real world. This might be wrong, or this might be the very definition of coordinate systems, I don't know; I'm just explaining how I "see it in my head". With that in mind I've always made the presumption that the real world, in its entirety, could and should be represented by a single coordinate system. Specifically, a bijective or surjective representation is valid, while an injective representation suggests to me that the coordinate system fails. Is this wrong? The answer will resolve some other questions I have...(yes, prometheus/Guest/AN, they are related to our previous discussions)

3. ### temurman of no wordsRegistered Senior Member

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Basically correct, the only thing is that injectivity does not require "all y", and you don't need "at most one x" because by definition a map must send every x to something. Injection fails to be bijection only because the range does not cover the target set. Injection is a bijection if you restrict the target set to its range. You can think of an injection as putting the domain inside the target set as its subset. Another, bit more suggestive term for injection is "one to one", for surjection is "onto", and for bijection is "one to one and onto", or "invertible".

5. ### RJBeeryNatural PhilosopherValued Senior Member

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temur: perfect, thanks for that clarification. To me, "all y" not being required is actually equivalent to saying "at most, one x" (because the absence of x is technically at most, one) but it's good to make that point explicitly.

Here's where I'm headed with this thread, and maybe you can help:
...A sheet of paper can be represented bijectively by a Cartesian coordinate system. No one can deny the "physicality" of a sheet of paper.
...A sphere can be represented surjectively by a polar coordinate system. No one can deny the "physicality" of a sphere.
1) Are there objects, whose "physicality" we cannot deny (toroids, mobius strips, etc), which cannot be so represented by bijective/surjective coordinates?
2) Are the Schwarzschild coordinates injective on the mathematical structure of a black hole?

Note that I'm note asking you directly how you feel about black holes. I'm also not asking you how you feel about my "presumption that the real world, in its entirety, could and should be represented by a single coordinate system." I'm just asking for some lending of your mathematical knowledge.

7. ### temurman of no wordsRegistered Senior Member

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What you are talking about is the concept of manifolds. By definition, manifold is an object that can be studied with a collection of (overlapping) coordinate systems on it. An example would be the surface of a sphere, where you have to use at least two coordinate patches to cover it totally. The coordinate system you call "polar" does not count as covering by single coordinate patch because it has singularities and you have to put extra patches on these singularities to get nice smooth stitching.

Schwarzschild coordinates are injective but not surjective. They have singularities at the horizon and miss the part of the manifold inside the horizon. So you have to put an extra patch to see the whole picture just like on sphere.

8. ### RJBeeryNatural PhilosopherValued Senior Member

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OK, but polar coordinates are still surjective when describing a sphere, correct? (r = constant, Theta = 0, Phi = "anything") for the North Pole does not invalidate the coordinate system in its ability to represent the sphere in its entirety; it may certainly cause problems when attempting a transformation to another coordinate system but that's not what I'm asking.

9. ### RJBeeryNatural PhilosopherValued Senior Member

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Also, we're now getting to why I started this thread (specifically because Guest246 and Prometheus demand that I admit I was wrong about something; I am willing to do so but I must first verify, in fact, that I'm in error). Now, when you say
Why is it that when I post the following
prometheus responds with the wonderfully condescending
and AlphaNumeric responds with
It sounds to me like either I'm wrong and you are wrong, or AlphaNumeric is wrong. (It's safe to say that Prometheus is a douchebag regardless...J/K!

) I don't mean to drag you into the middle of this but, frankly, I had assumed that there was something that I "just wasn't getting" and now you have me questioning those doubts...

10. ### prometheusviva voce!Registered Senior Member

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FYI, the facepalm in your last thread (not the first, and not even the second where the discussion had been going on,) was because you'd been asking the same question over and over in the hope that someone would give in and validate your ideas that black holes don't exist. Over the course of that thread Guest, AN, Ben, DH and JamesR (and probabaly others I haven't mentioned) and myself spend a long time trying to explain why you were wrong but it didn't seem to sink in, and we are still talking about it. It's kind of exasperating, don't you think?

Basically, your misunderstanding comes because you assume the usual Schwarzschild metric; $ds^2 = -\left(1-\frac{2M}{r}\right) dt^2 + \frac{dr^2}{\left(1-\frac{2M}{r}\right) } + r^2 d\Omega_2^2$ covers the whole space. As pointed out many times and most recently by temur above, they do not. Physical reasoning tells you that an observer far away from the black hole (or collapsing shell) sees the same geometry by Gauss' law. Mathematical reasoning tells you the coordinate system you are using to describe the geometry breaks when you get to the horizon. Either way, the conclusion is the same - you need to use a coordinate system that doesn't have that problem like kruskal coordinates.

How's your degree coming along BTW?

11. ### Guest254Valued Senior Member

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You "see it" wrong. Coordinate systems are defined via homeomorphisms, which are mappings with special properties. In particular, the map needs to be continuous and have continuous inverse.

If all you want is a way of labelling points, and no other conditions, then you're not talking about coordinate systems. It's obviously possible to map an N-dimensional manifold bijectively to Euclidean N-space, since both of these sets have the same cardinality.

12. ### temurman of no wordsRegistered Senior Member

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Yes. But to connect to what you had in mind, and to speak in your language, you can cover the sphere in its entirety by one coordinate system does not mean that every conceivable coordinate system must cover it. Consider the stereographical projection of the sphere from the North Pole onto the plane. Then the North Pole will not be represented by any point on the plane. In other words, the coordinate system defined by the inverse projection is *not* surjective. The Schwarzschild coordinates are an example of this type.

13. ### QuarkHeadRemedial Math StudentValued Senior Member

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Well I feel defrauded! When I opened this thread just now I thought "here is someone with a genuine question for once", only to find there is a hidden agenda. And an unseemly one at that.

By edit, I see others got there before me!
Well first you need to explain what you mean by "coordinates are surjective".

The terms injective, surjective and bijective are reserved for mappings from one space to another (which may, or may not, be the same space). In fact these terms are generally reserved for set functions, but that is neither here nor there.

So what spaces do your "coordinates" map from and to? More accurately, what are the domain, codomain and range of these coordinates? Given a point $x$ in the domain, what is its image set in the codomain?

But to answer your main query, try this pencil and paper exercise.

Sit the 2-sphere $S^2$ on the 2-plane $R^2$ and, for each point $p \in S^2$ project its tangents down to $R^2$. Each point $p \in S^2$ has a unique image in $R^2$ EXCEPT for the uppermost point, call it the North Pole, which has two tangents, each running off to infinity.

Call this mapping $f:S^2 \to R^2$ the function that assigns a coordinate system to $S^2$ This so called function is by the above, undefined at the North pole. This is a problem, and there is more than one possible solution to it; here is one.

"Delete" the North Pole, which we can call $NP$. So we have the situation that, for every point $p \in S^2\setminus NP$ there is a corresponding image point $f(p) \in R^2$, and a preimage $f^{-1}(f(p)) = p \in S^2 \setminus NP$. This is certainly a bijection, then.

Trouble is $S^2 \setminus NP \ne S^2$ , and they are not even homeomorphic (points missing!) from which I (at least) conclude that there is no invertible continuous mapping $S^2 \to R^2$. But as Guest has said, the definition of a coordinate system for any object $X^n$ is that there is a continuous invertible mapping from it to some subset of $R^n$

So there can be no GLOBAL coordinate system for any $S^n$.

There can be LOCAL ones, however, as others have told you. Which leads to the concept of a manifold.

14. ### RJBeeryNatural PhilosopherValued Senior Member

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If by "hidden agenda" you mean I was trying to determine how my understanding of injective coordinates was wrong, you are correct. It's also completely unfair to label it such. I've had multiple demands that I admit I was wrong about this very topic, so I started this thread to understand what I was missing (I even announced I would start the thread with that intent). It turns out, apparently, that the issue AlphaNumeric was demanding an admission of error on didn't even deserve one. Shall I expect one from him now? Shall I demand it?

The truth is, I don't want one and I couldn't care less. AlphaNumeric is one of only a handful of posters who gets my respect for having an education AND a keen intellect. I would say he lacks "big picture" vision occasionally, but that's not a requirement to be a good instructor or experimental physicist.

You though, I don't know. Prom and Guest, in my opinion and regardless of their education, simply lower the signal-to-noise ratio of the thread.

Last edited: Sep 29, 2010
15. ### funkstarratsknufValued Senior Member

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1,390
Incidentally, there's a lovely yellowbook which I remember enjoying back in the day: Elementary Differential Geometry, by Pressley.

So if someone is, you know, a bit confused about, say, what a coordinate system is, there's help to be had.

16. ### prometheusviva voce!Registered Senior Member

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Well, from my point of view there isn't a coherent signal from you at all. You can at least spell reasonably well which is more than can be said for many armchair crackpots, but it's really all just noise.

My pleasure.

17. ### RJBeeryNatural PhilosopherValued Senior Member

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Are you saying my links don't work? They work for me...hmm. My intent was to link to the posts in the threads to provide readers context.

Anyway Prom I'm letting your other comments slide (including those in #7). If you want to press it I guess we can continue the discussion, but I think we both know you're just "running interference" now...(American Football :bravo

18. ### Guest254Valued Senior Member

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You've inferred from the posts in this thread that your comments regarding the global validity of spherical coordinates were correct?!?!

19. ### prometheusviva voce!Registered Senior Member

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Your links take me to the thread but not the post (not even the right page in fact). I was serious in post 7 about your degree. You must be about a year in now right? How's it going?

20. ### RJBeeryNatural PhilosopherValued Senior Member

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No, Guest, in fairness you're right; I got fixated on AlphaNumeric's demand. Looking at your recent post you brought up a separate issue. Let's talk about it.

What does it mean mathematically for coordinate mappings to be "valid"? (sincere question) As I mentioned in the OP, surjective mappings, when describing physically undeniable objects (sheets of paper, spheres, toroids, etc), contain no "invalid coordinates" as I define them. When I say that I mean there is no point of indeterminability on the physical object being described. The redundancy of coordinate mappings to a specific point (i.e. the North Pole of a 2-sphere) does not "invalidate" that point, as it is clearly able to be described perfectly well. The fact that it's able to be described by any one of an infinite set of coordinates is irrelevant.

Injective mappings are different, however, because they leave open the possibility that points exist on the physical object which are indeterminate (e.g. ill-defined or require the value of infinity). My belief is that if a physical object is described by injective coordinates then those coordinates are improper. If there is no other way to describe the physical object then the object's existence itself is called into question. Despite my layman's terminology, and my possible inability to articulate my point, there is a guiding principle for me believing this which is related to infinity being nonphysical.

Now...refuting this theory is as simple as coming up with a counter example. Can anyone provide a physically undeniable object that can only be represented by an injective coordinate mapping?

21. ### RJBeeryNatural PhilosopherValued Senior Member

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Ahh crap. My links work for me because I have paging turned off (I hate paging). I guess I'll go fix the links, thanks for the heads-up.

Regarding my education, I wish I was a year along! Here's some math I learned though:

Large mortgage + 3 kids = full time job

Full time job + family time + part-time schooling > hours in the day

Family time - part-time schooling = divorce threats

Anyway I'm in 2nd year Physics...

22. ### RJBeeryNatural PhilosopherValued Senior Member

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The links in post 11 work now (hopefully). Someone please follow them so the nontrivial effort of fixing them was not in vain.

23. ### prometheusviva voce!Registered Senior Member

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This maybe a slightly naive answer, but I believe any manifold will do, for example, a 2 sphere

The links now work (the ones I checked at least).