Dear William, sure, it is possible.

An idea is revealed in Peter´s answer, a complete excellent solution is here: https://math.stackexchange.com/questions/183361/examples-of-bijective-map-from-mathbbr3-rightarrow-mathbbr

@peter ; essentially its a bijectiction F from a real interval could be [0,1] , does not have to be, as the dom(F) but does not have to be to the set of all probability triples

F: [0,1] \toM  M= {_i|, \forall i,st, (x,y,z}\in( 0,1) x+y+z=1}

(the set of all probability triples,  three positive number which sum to one i am not sure what the set of all probability triples have a name, a kind of pythagoraen triple) hence

the open interval if positive in  . for each combination or set  of three values which are positive (x,y,z)>0,(x,y,z)

Dear VIerra, given its a bijection I presume I can take the inverse function of its, to get what I want. Although the only difference between my questioon and that (it may make it easier is that its the set of all probability triples, the set of values which in addition sum to one)

@peter and vierra is there name for set of 3 outcome probability triple, by which I mean 3-tuples of positive or negative numbers, 0

When i meant orthogonal triples, I meant with regard to unit vectors or probability triples. It was a bit of a slip of the tongue, I had gleasons theorem on my mind at the time. I am just thinking about set of all n tuples of positive numbers (3 numbers in (0,1) or if need be non negative in [0,1], which are unit vectors I suppose in 1-norm ie x+y+z =1, for each such vector all of which sum to one)

so i presume in the case of probability , classically speaking  the unit 3d sphere in 1 norm,with the restriction to the absolute value which can be easily handled I presume

Thank you, Peter :)

I think that we cannot close our eyes and put the head in the sand, if we see a trolling, bullying etc.

I was aware that it was risky! I will delete my message - and will still keep logic since I wrote my appeal as an implication :D

I hope that William read it already ... maybe also the addressee.

I presume that for any set of three real numbers is there which sum to one and are positive and smaller then 1, there is something unnique to those parameters such as x1-x2, x2-x3 so if x1-x2= 0.1 x1=x2+0.1, x2-x3=0.4, x2=x3+0.4, so x1=x3+0.5

so x1+x2+x3=1, 3x3+0.9=1, x3=0.03333. whese appear to be always unique , if it were is there some function of (x1-x2,x2-x3)\in [0,1]^2 onto a single number such as (x1-x2)^2 +(x2-x3)^2\in [0,1], where i can ignore say the negative entries if necessary so that is uniquely designates those two difference and those vectors for all vectors. I think one needs for four degnerataes and the interval form [-,1,1] ie to getl

AH gees i spend some time trying to clear up the question in any case I think that if its a the set of all triple three outcome probability vectors should it be the n=2 standard simplex. the the probability tetrahedron seems to the set of all quadruples: ie https://en.wikipedia.org/wiki/Categorical_distribution ; the tetrahedron simplex may how one forms it, but it appears to be the set of all quadruples as a standard probability simplex. I think i will attach documents as lyx ones so that they readable from now on, I had put in some of the contraints. Include the symmetric aspects that to holds, ie that no combinations presnt in the diagonal on distinct vectors which sum to 1 or2,3  cannot contain a set of 2,3,4vaalues or sums of values which are not present on one singular vector in the system somewhere I presume that maybe my updated question was noticed and I accidentally cut out the details

@ peter, do you much about barycentric coordinates, and ternary representations of the unit simplex built over an equalitaterial triangle as a function of x,y, in cartesian coordinates.

Apparently they are hardly homogeneous in that only finitely many vectors with positive entires are present. Or at least their cardinality is finite in contrast to the points on the edges of the triangle which are primarily  contains zeros and ones,  Its said to be convex but I doubt that it is

I presume there was some reason why this kind of representationo was not used in q a quantum like system, ie which uses three coordinates, and a spherical design

does it define a pt or the entire simplex, the whole pt is that every set of three pts are supposed to sum to one in my system. The question however, is, do some triples in the probability simplex non negative numbers that sum to one, go missing, is is a bijection onto the unit probability triangle. barycentric coordinates looka  bit ungainly, and I am wondering if all convex combinations of three non negative numbers 0

Dear followers,

If  the unit probability simple  S   is   the triangle simplex of all triples of non-negative numbers

Thanks for this. I presume the homeomorphism between open unit disc and the plane R^2 is the real analytic bijection

he function

f ( z ) = z 1 − | z | 2 {\displaystyle f(z)={\frac {z}{1-|z|^{2}}}}https://en.wikipedia.org/wiki/Unit_disk

yes, the bary-centric coordinates or three dimensional coordinates, just are weights/probability values.

I am wondering, Whether the map F(x,y)from all of the two dimensional points within the  closed equilateral traingle with unit side length in 2 norm in the euclidean plane F(x,y)  dom (F)=F([0,1]\times[0,srt(3)/2].to  the standard 2 probability  simplex: ie is bi-jective and analytic function to said bary-centric three dimensional simplex (the standard 2 probability simplex). Where

F: [0,1]\times[0,srt(3)/2]. to the standard 2 probability  simplex  i, \delta^2= {p_i=_i\in R^3| \forall and only(pi\in[0,1]^3; p1i+p2i+p3_i=1, 0

Dear @ William Balthes, ·

I am agree that  the homeomorphism between open unit disc and the plane R^2 is the real analytic bijection

given by 

f ( z ) = z/( 1 − | z | 2) {\displaystyle f(z)={\frac {z}{1-|z|^{2}}}}https://en.wikipedia.org/wiki/Unit_disk

Do you know if any of the triples go missing or in these equilateral representations, I have been told that one had to be careful . Certain elements of the simplex do not appear, in the Marshak/Machina isoceles represntations: http://www.econport.org/econport/request?page=man_ru_advanced_icurves

I know that if one uses an isoceles representation that certan triples in the simplex will not be mapped to. ( I think that maybe because in those representations, they use a function of one variable)

Dear @ William Balthes, ·

The projection defined $p(x_1,x_2,x_3)= (x_1,x_2)$ maps S0   on planar tringle in $x_1x_2$-plane.

Dear professor Mateljevic, where you say  mean the projection, do you mean a bijective projective map, from some closed domain in R^2  that is isomorphic to the unit probability simplex (projective transformation, or bijective projective map) or the

  the projection of a single point in (x,y) to the a single vector in the probability simplex. For example in p.5  the attached document they give an example but it seems to be the identity function. If that works and is easy to plo

t I suppose I use that. In two dimensions, ie use ie F(x)= where dom(F)=[0,1], which is just really two functions 1-x and x, are b-ijective partition of unity (of course when there three dimensions one cannot do it this way; and every probability double is covered, a bijection from the unit interval to the set of all probability doubles.

  defined do you mean the one defined as above, or the identity projection, (ie where the cartesian coordinates of a probability triples, are in effect one and the same as the probability coordinates) .

It has to be projective- transformation or a bijective function from the x,y cartesian plane to the set of all and only non negative 3-tuples whose entries sum to one.

whereby the first two Cartesian,x_1, y_1,  coordinates just are the first two probability values, p1, p2; ie the cartesian coordinates just are the probability coordinates

(\forall x1, x2 in [0,1]^2; x1+x2

I presume that it may not be the IM(F)=S, the probability 2 simplex, in that it may not homogeneous with regard to the third variable. Ie some combinations of probability triples where p1+p2 +p3=1, 1>=p1, p2, p3>=0 may not be present, (one would require that x1, x2, x3,  and x1+x2, x2+x3, x1+x3, all assume each value in [0,1] infinitely many times, ie such that every element of the domain maps to and only to a point in the simplex ( a set of three non negative numbers that sum to one) and conversely every triple is present, (surjective over every combination), and injective, No triple can go missing.

unless that is equivalent to for all x1, x2, x3 in [0,1]^3, x1+x2+x3=1, F(x1,x2, x3)= which I presume would work (its just the identity function with regard to the simplex, the triples and the probabilty values themselves) ;

ie the dom(F) is the unit 2 probability standard simplex simplex, the set of all and only the, non negative cartesian coordinates,(real values) whose sum is 1,

being mapped direcltly onto the unit 3 probability simplex. the set of all non negative numbers which sum to one,

where  if  if the Cartesian coordinates of some probability vector,   in the probability standard 2 simplex S  , are   then, =

F(x1,x2,x3)==

I was wondering if there are pascal simplex ie akin to the sample space of the n=2 dirichlet distribution which gives only entries with all entries positive (symmetric uniform) but with the edges (the zero entries (0, 1/2 ,1/2)  such that it also uniformly gives the values for the corresponding completing boolean algebra ie for all triples probability triples a+b+c=1 0

I have been told that as dimensions increase the euclidean projection using an equilaterial or some other convex regular solid from a lower dimension,  is particularly uniformly and most values are approximately zero and on the edges where some pts go missing ie some n-tupes. I presume isosceles represntations would be like that as well. I presume then that a multi linear reprsetentation just as just absolute barycentric coordinates is  one the only way to ensure this does not happen (which is really jus the simplex of iteslf.  F(p1. p2)=(p1, p2, 1-p1-p2) although is this particularly uniform?

I presume one could have  a whole of recursive linear Bernstein or multi line equations, where for each point in one probability axis,  p1 as function of x, you have further two bernstein linear polynomials projecting over entire of z, x= contant plane,s where p1 is constant,

such that along every pt on a line segment for x1 , you interleave the axis for z, and define an entire two new linear  functions  on the z axis , over a new range (possibility) for z, on the [0, 1-p1] times {x=p1}   plane, one for p2, another p3, which sum to 1-x1 (partition of 1-x1, for that particular x value) and mapped onto some kind of y axis. and you continually just build lines and from pts in x, until every pt is attained. Otherwise I presume one just takes the simplex as one domain.

, so that you have another two line segments  or functions for p1 ,p2 defined, at every pt on p1 line. Ma

incribing a rectangle.

\forall z in [0 1-z] times {p1}on another axis, which range over all z in [0,1-p1] times x in {p1} and are a partition of 1-z rather than unity, along the y axis

(sum to

where you flot the first function forall

x in [0,1]  p1(x)=x for p1, and then for value of given x =p1value, and for all x in [0,1]. compute recursively plot

compute sum(x)=1-x, and plot and further two curves on the Z=[0,z=p1]\times {x=p1} plane  \forall z in [0,1-[x=p1]], and x={p1} p2(z,x=p1)= z,  p3( z, x=p1)=[1-p1]-z  p, sum(z,x=p1}=1-p1 p1(z, x=p1)=p1

constantly plotting a whole bunch of two curves or a plane of

,; its alright so long as no pts go missing. Otherwise one could have a multidimension bernstein represntation for the n=2 (trople simplex is for the first entry  forall x in[0,1]] F(p1)=x and on the other two axis  have when F(p)=p, two axis x, z. such that for every specific p1, there is a two dimensional face, z,x coordintes [0,1]^2, where

\forall x=p1 forall y in [0,1] F(p2)=(1-p1)y  and and forall x=p1 forall y in [0,1] F(p3)=(1-p1)- (1-p)z so if the the first parameter p1 = 0.3 then this entries p1=0.3 in the p2, +p3 =0.7 rectangle the  x,=0.3 /p1,=0.3  pt, features in all tuples dsecribed by the entire  set of  all non negative pts 0>=p2, p3

Dear Professor Mateljevic,

When you say that the unit disk is a real analytic bijection to R2, and has a bijective analytic function rto R^2, does it have one ; is it b-ijective to the simplex; are all pts in (p1, p2, p3)  in IM(F) of the unit disk function.  Do any go missing. I presumed that there were certain issues, with the surface area of the sphere

F: dom (unit disk) to . I was wondering if a quadratic function of the northern surface of the unit sphere, would be bijective as defined by F:[0, 2pi times [0, pi]in R^2\to < [p1=sinpi* cosphi]^2, p2=[sin pi * sin phi]^2, p3=cos^2phi> in R^3where  p1+p2+p3=1, and should be non-negative due to the squares.

Whilst the triangular functions appear to be the other way around and maximize surface area

Dear Peter I was just wondering whether there is any canonical form of the canonical 2 probability simplex ; considered as a probability vector space,  other than the convex hull of the three vertices of the euclidean triangle (1,0,0) or (0,0,1) , (0,1,0); as a function of three variables; apart from just itself. In the same sense that surface of the 2- sphere can be parameterized. Does it happen any other form, other than F(x,y,z)=x(1,0,0)+y(1,0,0)+z(0,0,1); which appears to be just F(x,y,z)= for all and only x,y,z>=0, x+y+z=1

for all x,y, z x+y+z=1, 1=0; and in particular is there name of the union 2- plex   ie the entire 2 simplex where each probability space, (x,y,z, x v z,=x+z, y v z= y+z, zux=z+x, emptyset=0, sigma=1, x+y+z=1, 1=0,

there is then going to be another probability function puts onto the entire entity as well subject to certain constaints, but I want to know whether the canonical 2 probability simplex is actually reliable. Is it compact,  does points go missing, by which I mean in the canonical probability 2 simplex in R3

ie see the standard 2-simplex in R3; for instance is there any functional closed form of that entity which plots that to begin with. I have been tolds that these euclidean triangle probability simplex can unreliable unless their norm >1

by which I mean this https://en.wikipedia.org/wiki/Simplex

The function has infinitely many equivalence classes, for instances; all of the pi in [0,1]  contained in that image, and each of the 8 elements of the vector has a local rank,  given by the values themselves denoted by another probability G, subject to similar axioms, and then there is a global function Gw, between all of the individual 8 elements in all of the uncountable many probability space, which are tied together due to a form of conditional logic. But before I go into this, I need to know whether the canonical 2 probability plex is itself a reliable domain, overal global frame  function kind of thing that  , that