Path: utzoo!attcan!uunet!mcsun!i2unix!inria!seti!bora.inria.fr!hussein
From: hussein@bora.inria.fr (Hussein Yahia)
Newsgroups: comp.graphics
Subject: Re: 4-Space Basis Rotation Matrices
Message-ID: <1538@seti.inria.fr>
Date: 13 Aug 90 12:31:24 GMT
References: <2498@ryn.esg.dec.com>
Sender: news@seti.inria.fr
Reply-To: hussein@bora.inria.fr (Hussein Yahia)
Organization: INRIA
Lines: 55



<1279@enuxha.eas.asu.edu>, hollasch@enuxha.eas.asu.edu (Steve Hollasch)
writes...

>>    I'm writing a 4D wireframe-viewer and have run into the problem of
>>generating rotation matrices for the 4D viewpoint.  What are the basis
>>rotation matrices in 4D?


  Let f be an isometry of the n dimensional space. Then, according to a 
theorem of algebra, there exists subspaces E1, E2, ..., Eq, F1, ..., Fr
orthogonal each others, such that the sum

       E1 + E2 + ... + F1 + ... + Fr is direct and such that:


    - dimension of Ei = 2, dimension of Fj = 1,
    - the restriction of f to Ei is a 2 - dimensional rotation,
    - the restriction of f to Fj is either Identity or a central symmetry.



  So, if n = 3 we get the standard basis for a 3 - dimensionnal rotation:

                 cos(a)      -sin(a)        0
                 sin(a)       cos(a)        0
                   0            0           1


  If n = 4, the theorem implies that EVERY rotation matrix in 4-space can
be put in the following form:


                 cos(a)       -sin(a)       0         0
                 sin(a)        cos(a)       0         0
                   0             0         cos(b)    -sin(b)
                   0             0         sin(b)     cos(b)

in a suitable basis.

Hence a 4-rotational matrix does not have an axis, and it has 2 angles, and
we see from the preceding form that is is merely the commutative product of 
two 2-dimensionnal rotations acting on two orthogonal planes in 4 dimensional
space.



  Hussein Yahia
  INRIA
  Syntim Project
  BP 105 78153 le Chesnay Cedex
  France.

  E-mail: hussein@bora.inria.fr