Coordinate functions of a many-to-1 function

  • #1
Stephen Tashi
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Main Question or Discussion Point

How many coordinate functions of a many-to-1 function must also be many-to-1 ?

Let ##F## be a function from ##\mathbb{R}_n## into ##\mathbb{R}_n##. Represented as an ##n##-tuple in a particular (not necessarily Cartesian) coordinate system ##h##, ##F## is given by ##n## coordinate functions as ##F(X) = (f_1(X),f_2(X),...f_n(X))## where each##f_i(X)## is a mapping from ##n##-tuples ##X## of real numbers to single real numbers.

Given a free choice of the coordinate system ##h##, there is some maximum number ##M## of coordinate functions that we can make 1-to-1.##\ ## For a particular ##F##, is the value of ##M## given by some theorem in topology?

I suppose that we'd have to restrict the choice of coordinate systems to those that are continuous mappings in order for topology to tell us anything.
 

Answers and Replies

  • #2
andrewkirk
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One could start with linear functions, ie endomorphisms of ##\mathbb R^n##. We represent an endomorphism ##f## by a ##n\times n## matrix. To be many-to-one, the rank of the matrix must be ##r\in \{0,...,n-1\}##. We can write ##\mathbb R^n## as a direct sum of the matrix's nullspace K (kernel of the function) and another space U and these have dimensions (n-r) and r respectively. The function ##f|_U## will be injective and an isomorphism from U to a r-dimensional subspace of ##\mathbb R^n##. If we choose a coordinate system using a basis that is the union of orthogonal bases of U and K with the r basis vectors of U making up the first r coordinate positions then the first r coordinate functions of ##f## in that basis will be one-to-one. It is not possible to have more than r coordinate functions one-to-one.

If the function is differentiable but non-linear I expect the maximum number of one-to-one coordinate functions will be equal to the lowest rank of the Jacobian matrix anywhere in ##\mathbb R^n##, because the function will approximate a linear function locally. But I wouldnot be surprised if the maximum could be lower than that for non-linear cases.
 
  • #3
Stephen Tashi
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One could start with linear functions, ie endomorphisms of ##\mathbb R^n##.
This raises the interesting topic of how we will define ##\mathbb{R}_n## to be a vector space. By default, we define addition ##X + Y## as component-by-component addition. If we have a 1-to-1 mapping ##h(x): \mathbb{R}_n \rightarrow \mathbb{R}_n## then we can define a different addition operation ##+^h## on ##\mathbb{R}_n## by ##X +^h Y = h^{-1}( h(X) + h(Y))## and a different scalar multiplication operation ##*^h## by ##\lambda *^h X = h^{-1} ( \lambda h(X))##.
 
  • #4
andrewkirk
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This raises the interesting topic of how we will define Rn" role="presentation">Rn to be a vector space
Interesting! At first I thought it would not be a vector space under those operations. But I played around trying to break the VS axioms and couldn't. Then I realised that actually the vector space is identical (isomorphic?*) to the usual ##\mathbb R^n##, but just uses different labels for the elements. In the new vector space the components no longer relate to an orthonormal basis of the space and the zero element may not be (0,0,....,0). For instance if the function h just adds 1 to the first component then (-1,0) is the zero element of the new space, rather than (0,0).

A couple of transformations I was playing with:

(1) h takes ##\vec v## to ##\vec v \sin(|\vec v|)/|\vec v|##. This function is many to 1 on all cartesian coordinates (I think regardless of change of Cartesian basis) but if we use a hyper-polar system then it will only be many-to-one on the radial coordinate.

(2) h acts on the modulus as per (1) but also doubles every angular coordinate and deducts or adds the range-width of that angular coordinate as needed to get it back in the allowable range. Then I think this will be two-to-one on all angular coordinates and many to one on the modulus. I wonder if there is any coordinate system in which any of the coordinate functions of this h are one to one.

* Actually, the map is an isomorphism. It just doesn't look like one because of the funny labelling.
 
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  • #5
Stephen Tashi
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(1) h takes ##\vec v## to ##\vec v \sin(|\vec v|)/|\vec v|##. This function is many to 1 on all cartesian coordinates (I think regardless of change of Cartesian basis) but if we use a hyper-polar system then it will only be many-to-one on the radial coordinate.
Generalizing that, we can imagine a coordinate system for ##\mathbb{R}_n## that is defined by the level surfaces of a function ##H(X)## of ##n-1## real variables together with the parameter ##r##. Further imagine that the level surfaces ##H(X) = r## are topologically like nested spheres and that we have some coordinate system defined on each level surface that represents points on that surface.

A particular type of many-to-one function ##F(Y):\mathbb{R}_n \rightarrow \mathbb{R}_n## can be defined by letting f(r) be a many-to-one mapping ##\mathbb{R} \rightarrow \mathbb{R}##, and for each pair of level surfaces ##H(X) = r_1, H(X) = r_2## choose a paticular 1-1 continuous mapping ##\phi_{(r_1, r_2)}## between them, which is guaranteed to exist because of their topological equivalence. Then let ##F(Y) = (f(Y_1),\phi_{(Y_1, f(Y_1))}(Y_2,Y_3,..Y_n) ) ##
 
  • #6
Infrared
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Sorry if I'm missing something obvious, but the question confuses me. No continuous function from [itex]\mathbb{R}^n[/itex] to [itex]\mathbb{R}[/itex] for [itex]n>1[/itex] can be one-to-one, and this is what your coordinate functions [itex]f_i[/itex] are.

Proof of the above claim: Let [itex]f:\mathbb{R}^n\to\mathbb{R}[/itex] be continuous. Since [itex]\mathbb{R}^n[/itex] is connected, the image of [itex]f[/itex] must be an interval, which we may assume to have more than one point. Removing a point [itex]x[/itex] from the interior of this interval disconnects it and so [itex]\mathbb{R}^n\setminus f^{-1}(\{x\})[/itex] must also be disconnected. This can't happen if [itex]f^{-1}(\{x\})[/itex] is a singleton, so [itex]f[/itex] isn't injective.
 
  • #7
andrewkirk
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@Infrared Good observation! I see that I meant something different by 'coordinate function' from what you, quite reasonably, are taking it to mean. Further I suspect yours is the usual meaning. Let me try to explain what I meant by coordinate function.

What I had in mind for the ##j##th coordinate function was a function that mapped the ##j##th coordinate of a point ##\vec x\in \mathbb R^n## to the ##j##th coordinate of ##f(\vec x)##, where the coordinates are in a chosen coordinate system. That function would be a map from a subset of ##\mathbb R## to itself and so is not barred from being injective.

I'll try to formalise that as follows:

Let C be a coordinate system for ##\mathbb R^n##. That means there are subsets ##A_1,...,A_n## of ##\mathbb R##, projection functions ##p^C_j:\mathbb R_n\to A_j## for ##j=1,...,n## and a bijective coordinate map ##p^C:\mathbb R^n\to A_1\times....\times A_n## such that ##p^C(\vec x)=(p_1^C(\vec x), ...,p_n^C(\vec x))##.

Let ##S_j^C## be the Cartesian product of the ##A_1,...,A_n##, excluding ##A_j##, so it's a subset of ##\mathbb R^{n-1}##.
Then for ##\vec u\in S_j^C## define the function ##f^{C,j}_{\vec u}:A_j\to A_j## such that, for ##r\in A_j## we have ##f^{C,j}_{\vec u} (r)= p^C_j\circ f\circ \left(p^C\right)^{-1}\left((u_1,u_2,...,u_{j-1},r,u_j,....u_{n-1})\right)##.

With that background, what I meant by saying the ##j##th coordinate function of ##f## in coordinate system C is one-to-one is that ##f^{C,j}_{\vec u}## is one-to-one for all ##\vec u\in S_j##. To avoid confusion with the usual meaning, let's say that the '##f##-rank' of a coordinate system C is the number of dimensions ##j## such that ##f^{C,j}_\vec u## is injective for all ##\vec u\in S^C_j##

The question is then, for a given ##f##, what is the maximum ##f##-rank of any coordinate system for ##\mathbb R^n##, and which coordinate systems have that ##f##-rank?

We could then define the 'coordinate rank' of any function ##f:\mathbb R^n\to\mathbb R^n## as the maximum ##f##-rank of any coordinate system.

If ##f## is linear of rank ##n## and injective, I think any Cartesian coordinate system will have ##f##-rank equal ##n##, so the coordinate rank of ##f## is ##n##. I suspect that for injective, nonlinear functions, some coordinate systems may have ##f##-rank less than ##n## but there will be at least one whose ##f##-rank is ##n## so that the coordinate rank of ##f## is ##n##. But That's just a hunch. It may be that if we require continuity, for all injective ##f## all coordinate systems will have ##f##-rank ##n##.

Consider the function that maps every point P in ##\mathbb R^n## to the point where the ray OP (O is the origin) intersects an oblate spheroid surrounding the origin. In a Cartesian coordinate system C every ##f^{C,j}_\vec u## will be non-injective. But if C is a hyper-polar coordinate system, ##f^{C,j}_\vec u## will be injective for every ##j## other than 1, where we take dimension 1 to be the radial coordinate. So C will have ##f##-rank ##n-1## and ##f## will have coordinate rank ##n-1##.

Another one to consider is where ##f## is the projection function onto a ##(n-1)##-dimensional hyperplane through the origin. Most Cartesian coordinate systems will have ##f##-rank zero. But the coordinate system whose basis vectors are an orthogonal basis for the hyperplane together with a vector normal to it, will have ##f##-rank ##n-1##, as the functions ##f_\vec u^{C,j}## are all the identity function on ##\mathbb R##, where ##j## is the dimension of one of the basis vectors of the hyperplane.

The coordinate rank of a function would be a measure of how much it reduces the dimensionality of its domain. For nonlinear functions, the reduction in dimensionality may vary between different parts of the domain.
 
  • #8
Stephen Tashi
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@Infrared Good observation! I see that I meant something different by 'coordinate function' from what you, quite reasonably, are taking it to mean.
Me, too!

Before analyzing mappings of ##\mathbb{R}_n## into ##\mathbb{R}_n## that map coordinates to coordinates, there is the matter of how points in ##\mathbb{R}_n## are mapped to m-tuples in a particular coordinate system of m-tuples. Obviously the individual coordinate functions of that mapping are not 1-to-1 because, for example, lots of points can have the same first coordinate.

It's simplest to regard points in ##\mathbb{R}_n## as being defined by cartesian coordinates and their distances apart defined by the usual Euclidean metric.

We can think of coordinate functions in a different (possibly non-cartesian) system as being real valued functions of n-real variables and define continuity of the coordinate functions in the usual way - i.e. we accept the absolute value function ##|a - b|## as the metric used in the range of the coordinate functions.

That still leaves the problem of what to do about systems like polar coordinates - how to avoid a discontinuity at ##\theta = 2\pi "=" 0 ## I think "charts and atlases" solves that problem, but I don't know the technical details.

(One guess is that ##\lim_{X \rightarrow P} c_i(X) = r## is defined by choosing a local coordinate patch that includes an open interval about ##P## in ##\mathbb{R}_n## and an open interval about ##r## in ##\mathbb{R}##.)

Like @andrewkirk , I want to consider a modified definition of "1 to 1" that would let mappings like ##(x,y,z) \rightarrow x## or ##(x,y,z) \rightarrow y ## satisfy that definition. The intuitive idea is that a mapping like ##(x,y,z) \rightarrow x ## "does not depend" on ##y## and ##z##. How to formulate that idea precisely is an interesting problem. We have to distinguish between "does not depend on" and "is not a function of". If ##y## is defined to be in the domain of a mapping, we can can't really say the mapping "is not a function of" ##y##.
 

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