> as opposed to what the a sibling comment implies, the dimensionality of Clifford algebras is 2^N for an N-dimensional carrier. Indeed you are correct -- I sh

> as opposed to what the a sibling comment implies, the dimensionality of Clifford algebras is 2^N for an N-dimensional carrier. Indeed you are correct -- I should have used the word "degree of freedom" to describe it more accurately in the sense of dimensional constraints, as opposed to "dimensions" which are the individual members. To elaborate, the 3D space have the following possible members: scalar component (3 choose 0): () vector components (3 choose 1): x | y | z bivector components (3 choose 2): xx | yy | zz | xy=-yx | yz=-zy | zx=-xz trivector components (3 choose 3): xyz=yzx=zxy=-zyx=-xzy=-yxz which is 1+3+3+1 = 8 = 2^3 different members