Note: this example session was done with a previous version of the
package, and under mathematica 1.2.  It should be checked for accuracy
and updated.  For example, I think the shortcut illustrated in In[5]
below is not be supported anymore, and the line

In[26]:= Line[{p1,p2,p3,p1}] /. Render[PoincareBall]

is certainly obsolete -- in the current version you would say 
SetOptions[Convert,Model->PoincareBall] at the beginning, and then
and Line[{p1,p2,p3,p1}] will convert itself into arcs of circle
at rendering time.

Also some capabilities (such as 3-D line plotting) are not
illustrated here.

carrot:; math
<<Hyperbolic.m
Mathematica (sun3.68881) 1.2 (August 3, 1989) [With pre-loaded data]
by S. Wolfram, D. Grayson, R. Maeder, H. Cejtin,
   S. Omohundro, D. Ballman and J. Keiper
with I. Rivin and D. Withoff
Copyright 1988,1989 Wolfram Research Inc.
Temporary license: expiration date is 31 Dec 1989.
 -- Terminal graphics initialized -- 

In[1]:= 
In[2]:= Convert[PoincareBall[{.3,.4}],Projective]

Out[2]= Projective[{0.48, 0.64}]

In[3]:= Convert[%,PoincareBall]

Out[3]= PoincareBall[{0.3, 0.4}]

In[5]:= Convert[{.3,.4},PoincareBall,Projective]

(* not recommended except where speed is of the essence *)
(* doesn't actually work: do we want it to ? OG. *)

Out[5]= {0.48, 0.64}

In[6]:= Convert[PoincareBall[{.3,.4}],Minkowski]

Out[6]= Minkowski[{0.8, 1.06667, 1.66667}]

In[7]:= Convert[Minkowski[{1,2,3}],PoincareBall]

(* this is garbage, because {1,2,3} was not normalized. *)
(* Minkowski points generated by Convert are automatically normalized
   unless they're at infinity. *)

                      1  1
Out[7]= PoincareBall[{-, -}]
                      4  2

In[8]:= Convert[Normalize[Minkowski[{1,2,3}]],PoincareBall]

(* this will work *)

                      1  2
Out[8]= PoincareBall[{-, -}]
                      5  5

In[9]:= Convert[%,Minkowski]

                   1     3
Out[9]= Minkowski[{-, 1, -}]
                   2     2

In[11]:= v1=PoincareBall[Vector[{0,0},{0,1}]]

Out[11]= PoincareBall[Vector[{0, 0}, {0, 1}]]

In[12]:= v2=Convert[v1,Projective]

Out[12]= Projective[Vector[{0, 0}, {0, 2}]]

In[14]:= VectorLength /@ {v1,v2}

(* they're the same vector, so should have the same length *)

Out[14]= {2, 2}

In[15]:= p={a,a Sqrt[3]}

(* we'll solve the (2,3,7) triangle.  Let the Pi/3 angle
be at the origin, and the short edge along the x-axis.  In the
projective mode, the right angle will be at {a,0} and the Pi/7
angle will be at {a,a Sqrt[3]}. *)

Out[15]= {a, Sqrt[3] a}

In[19]:= Angle[Projective[Vector[p,{0,-1}]],Projective[Vector[p,-p]]] //
  Short
         

                                                  3
                                      -4 Sqrt[3] a
                                      ------------- + <<2>>
                                               2 3
                                       (1 - 4 a )
Out[19]//Short= ArcCos[----------------------------------------------------]


                                   2                          4
                               -3 a                      -16 a
                       Sqrt[----------- + <<2>>] Sqrt[------------ + <<2>>]
                                    2 3                          3
                            (1 - 4 a )                (1 + <<1>>)

In[20]:= Cos[%] // Simplify

            Sqrt[3]
Out[20]= --------------
                     2
         2 Sqrt[1 - a ]

In[21]:= Solve[N[%==Cos[Pi/7]],a]

Out[21]= {{a -> 0.275798}, {a -> -0.275798}}

In[23]:= p1 = Projective[N[p /. %21[[1]]]]

Out[23]= Projective[{0.275798, 0.477696}]

In[24]:= p2= Projective[{%[[1,1]],0}]

Out[24]= Projective[{0.275798, 0}]

In[25]:= p3=Projective[{0,0}]

Out[25]= Projective[{0, 0}]

In[26]:= Line[{p1,p2,p3,p1}] /. Render[PoincareBall]

(* we get two straight lines and an arc of circle; the result
   can be passed to Graphics *)

                                     -16
Out[26]= {Circle[{3.62585, 3.15217 10   }, 3.48522, {3.06679, 3.14159}], 

>    Line[{{0.140626, 0.}, {0., 0.}}], Line[{{0., 0.}, {0.150371, 0.260451}}]}

In[27]:= Show[Graphics[%],AspectRatio->Automatic]
                                                                             
                                                                            
                                                      #                       
                                                     ##                       
                                                    # #                       
                                                   #  #                       
                                                  #  #                        
                                                ##   #                        
                                               #     #                        
                                              #      #                        
                                             #       #                        
                                            #        #                        
                                           #         #                        
                                          #          #                        
                                         #           #                        
                                        #            #                        
                                      ##             #                        
                                     #              #                         
                                    #               #                         
                                   #                #                         
                                  #                 #                         
                                 #                  #                         
                                #                   #                         
                               #                    #                         
                              #                     #                         
                             #                      #                         
                           ##                       #                         
                          #                         #                         
                         #                          #                         
                        #                           #                         
                       #                            #                         
                      ###############################                         
                                                                              
                                                                              

Out[27]= -Graphics-

In[28]:= Distance[p1,p2]

Out[28]= 0.545275

In[29]:= {q1,q2,q3} = Convert[#,PoincareBall]& /@ {p1,p2,p3}

Out[29]= {PoincareBall[{0.150371, 0.260451}], PoincareBall[{0.140626, 0}], 
 
>    PoincareBall[{0, 0}]}

In[30]:= Distance[q1,q2]

Out[30]= 0.545275 (* better be the same... *)

In[31]:= CompleteFrame[PoincareBall[Vector[{0,0},{1,0}]]]

(* we'll build up the reflection in each of the three sides *)
(* the reflection in side p2-p3 takes a positive frame including *)
(* Vector[{0,0},{1,0}] to a negative frame including the same vector *)

Out[31]= Minkowski[Frame[{{0, -1, 0}, {1, 0, 0}, {0, 0, 1}}]]

In[32]:= CompleteFrame[PoincareBall[Vector[{0,0},{1,0}]],-1]

Out[32]= Minkowski[Frame[{{0, 1, 0}, {1, 0, 0}, {0, 0, 1}}]]

In[33]:= r23=Isometry[%%,%]

Out[33]= Isometry[{{1, 0, 0}, {0, -1, 0}, {0, 0, 1}}]

In[34]:= %.PoincareBall[{.3,.4}]

Out[34]= PoincareBall[{0.3, -0.4}]

In[35]:= CompleteFrame[PoincareBall[Vector[Peel[q2],{0,1}]]]

(* Peel[q2] peels off the label, PoincareBall.  First would also work *)

Out[35]= Minkowski[Frame[{{1.04035, -1.62076 10   , 0.286926}, {0, 1., 0}, 
 
>      {0.286926, 0, 1.04035}}]]

In[36]:= CompleteFrame[PoincareBall[Vector[Peel[q2],{0,1}]],-1]

                                               -17
Out[36]= Minkowski[Frame[{{-1.04035, 1.62076 10   , -0.286926}, {0, 1., 0}, 
 
>      {0.286926, 0, 1.04035}}]]

In[37]:= r12=Isometry[%%,%] //Chop

Out[37]= Isometry[{{-1.16465, 0, 0.597007}, {0, 1., 0}, 
 
>     {-0.597007, 0, 1.16465}}]

In[38]:= Inverse[%]

(* we expect the same, since this is a reflection *)

Out[38]= Isometry[{{-1.16465, 0, 0.597007}, {0, 1., 0}, 
 
>     {-0.597007, 0, 1.16465}}]

In[39]:= %.PoincareBall[{0,0}]

Out[39]= PoincareBall[{0.275798, 0}]

In[40]:= {Distance[q2,q3],Distance[q2,%]}

Out[40]= {0.283128, 0.283128}

(* reflection preserves distances *)

In[41]:= CompleteFrame[PoincareBall[Vector[Peel[q1],-Peel[q1]]]]

                                                     -17
Out[41]= Minkowski[Frame[{{-0.866025, 0.5, 9.59344 10   }, 
 
>      {-0.59944, -1.03826, -0.661297}, {0.330648, 0.5727, 1.19888}}]]

In[42]:= CompleteFrame[PoincareBall[Vector[Peel[q1],-Peel[q1]]],-1]

                                                      -17
Out[42]= Minkowski[Frame[{{0.866025, -0.5, -9.59344 10   }, 
 
>      {-0.59944, -1.03826, -0.661297}, {0.330648, 0.5727, 1.19888}}]]

In[43]:= r13=Isometry[%%,%] // Chop

Out[43]= Isometry[{{-0.5, 0.866025, 0}, {0.866025, 0.5, 0}, {0, 0, 1.}}]

In[45]:= r13.r23.r13.r23.r13.r23 // Chop

Out[45]= Isometry[{{1., 0, 0}, {0, 1., 0}, {0, 0, 1.}}]

In[46]:= r23.r12.r23.r12 // Chop

Out[46]= Isometry[{{1., 0, 0}, {0, 1., 0}, {0, 0, 1.}}]

In[47]:= r12.r13.r12.r13.r12.r13.r12.r13.r12.r13.r12.r13.r12.r13 // Chop

Out[48]= Isometry[{{1., 0, 0}, {0, 1., 0}, {0, 0, 1.}}]

In[50]:= Drag[PoincareBall[Vector[Peel[q2],{0,1}]],100] //N

(* find where the geodesic determined by this vector intersects the sphere: *)
(* by walking 100 units along it, we get pretty close to the boundary; *)
(* in the future there may be a special function to do this right *)

Out[50]= PoincareBall[Vector[{0.275798, 0.961216}, 
 
                 -16            -43
>     {1.02577 10   , 1.40258 10   }]]

In[51]:= Distance[q2,First /@ %]

(* this should be 100, but because of roundoff the point *)
(* defined in Out[50] is actually slightly outside the ball, *)
(* with unpredictable consequences *)

Out[51]= 0.27942 I