74 ANALYTICAL TRLATM ENT OF THE POLYTOPES REGULARLY 



can do so by means of the same set of expansions, (P) s must be 

 central symmetric. So we have to solve first the new question if 

 the simplex S^u-^-l) admits a space 8 n _t reflecting it into aeon- 

 centric simplex S'(?i-{-l) oppositely orientated to S(n-\-l). We once 

 more suppose that there is such a space S n _ ± and we examine the 

 consequences of this supposition. Let A X A 2 . . . . A n be the given 

 simplex and A^ A 2 ' . . . . A' n+i the concentric simplex of opposite orien- 

 tation, the common centre O being at the same time centre of the 

 n -j- 1 segments A- t Al, (i = 1, 2, . . ., n -f- 1) Then the image of A 1 

 wit respect to S n __ i must be either A' t or one of the other verti- 

 ces of S'(n-\-l) y say A' 2 - We consider these two different cases each 

 for itself. 



If A\ is the mirror image of A x , the space S n _ ± is normal to 

 the line A x O joining in S(n-\-l), if produced, the vertex A t with 

 the centre M x of the opposite limit S(n), which line A l M i may 

 be called a "first transversal" of S (n -{-■!); this S(n) with the ver- 

 tices A 2 A s ...A n + 1 is contained in a space S n _ ± parallel to S n _ if 

 whilst the mirror image of it is the limit S' (n) of S' {n -j- 1) oppo- 

 site to A' ± , with the vertices A' 2 , A'%. . . ., A' n+i . So, as a whole, 

 these S(?i) and S' (n) have to be at the same time equipollent and opposi- 

 tely orientated to each other, equipollent as reflections of figures 

 lying in spaces S n _ i and S' n _ i parallel to the mirror S n _ i , oppo- 

 sitely orientated as corresponding parts of the oppositely orientated 

 simplexes S(n~{-1) and S'(n-\-l). This is impossible for n^>2, 

 e. g. two triangles lying in parallel planes cannot be equipollent and 

 oppositely orientated at the same time. Now the case n === 1 , meaning- 

 less in itself, leads to two conciding simplexes, i. e. to a point of 

 symmetry of the simplex of the linear domain, the line segment. 

 So the case n = 2 of the triangle A v A. 2 A% with the lines through 

 O parallel to the sides is the only remaining one. 



If A' 2 (fig. 12) is the mirror image of A lt we consider the tri- 

 angle A x A 2 A' ' 2 with a right angle in A 1} as the centre O of A 2 A' 2 is 

 at equal distance from the three vertices; if M i2 is the centre of A 1 A 2% 

 the line M i2 0, parallel to A X A' 2 , passes, if produced, through the 

 centre of the limit S(n — 1) of S(n-\-l) containing A- à ,A^ . . ., A n+i as 

 vertices and may therefore be called a "second transversal" of S{n -f- 1). 

 Now the mirror S n _ i bisects A X A 2 orthogonally and is therefore 

 normal to the second transversal M X2 0, while the limits S(n — 1) 

 with the vertices A 3 A^, . . .,A n+i and S' (n — 1) with the vertices 

 A' 3 A\ . . . A' n+i lie in parallel spaces S n _ 2 and #' n _ 2 . Here too these 

 limits have to be at the same time equipollent and oppositely orientated 

 to each other, which is impossible for n — 1 > 2. So we find here 



