This page in Ukrainian.
First we choose a convenient embedding of the tetrahedron
in R^3. With i,j,k the standard unit vectors (these may also
be interpreted correctly as unit quaternions), we consider
the regular tetrahedron with vertices A =
-i-j-k, B = -i+j+k,
C = i-j+k and D = i+j-k.
The symmetries of order 3 of the tetrahedron
(in terms of permutations of the vertices) are generated by
the symmetries of order two are
The corresponding SU(2) elements of order six are
+/- a, +/- a^2,
+/- b, +/- b^2,
+/- c +/- c^2, +/- d, +/- d^2
a = (1/2) ( 1- i- j- k),
(And thus
The corresponding elements of order four are
+/- i, +/- j, +/- k;
1 and - 1 complete the list of 24 elements.
The Multiplication Table for
the binary tetrahedral group may be conveniently calculated
and written out in this notation.
(BCD) = 120-degree rotation about A,
(ADC) = 120-degree rotation about B,
(ABD) = 120-degree rotation about C, and
(ACB) = 120-degree rotation about D;
(AB)(CD) = 180-degree rotation about i,
(AC)(BD) = 180-degree rotation about j, and
(AD)(BC) = 180-degree rotation about k.
where a = (1/2)(1+A), and similarly for b, c, d. This yields
b = (1/2) ( 1- i+ j+ k),
c = (1/2) ( 1+ i- j+ k),
d = (1/2) ( 1+ i+ j- k).
a^2 = (1/2) (- 1- i- j- k),
b^2 = (1/2) (- 1- i+ j+ k),
c^2 = (1/2) (- 1+ i- j+ k),
d^2 = (1/2) (- 1+ i+ j- k).)
Corrected September 26 2004.
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