MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Are there any three dimensional subalgebras of GA(n) where GA(n) is the geometric algebra corresponding to $R^n$? If yes, what about for GA(2)?

Edit: a geometric algebra is a Clifford algebra.

share|cite|improve this question
up vote 1 down vote accepted

I see from Wikipedia that "geometric algebra" is another name for a Clifford algebra, which I suppose explains the clifford-algebras tag. Since you are working over the real numbers, you need to specify what the signature of the quadratic form is, as the isomorphism type of the Clifford algebra strongly depends on the signature.

When $n=2$ and you take a quadratic form of Euclidean signature (i.e. a positive-definite symmetric bilinear form), for instance, the Clifford algebra is isomorphic to $M_2(\mathbb{R})$, and there is the subalgebra consisting of all matrices of the form $$ \begin{pmatrix} * & * \\\ 0 & * \end{pmatrix}, $$ or anything conjugate to that.

For forms with other signatures, there are tables written down which list the isomorphism type of the associated Clifford algebras. For instance, there is this Wikipedia page. Also there is a reasonably good explanation of the classification in Chapter 5 of the book Elements of Noncommutative Geometry, by Gracia-Bondia, Varilly, and Figueroa.

In general, you end up with either a matrix algebra or a direct sum of two matrix algebras, with coefficients in $\mathbb{R},\mathbb{C}$, or $\mathbb{H}$. In any of those cases, you should be able to find lots of three-dimensional subalgebras.

If you want a more intrinsic description, it might help to say why you're looking for these things in the first place.

share|cite|improve this answer
Thanks! The motivation comes from looking for 3d analogues of the Mandelbrot set. – antianticamper May 27 '12 at 21:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.