0
$\begingroup$

Many kinds of Clifford algebras have corresponding sub-algebras of matrix algebras in sense of isomorphism. Say, quaternion, spacetime algebra and also Dirac algebra. Generally, Clifford algebra has matrix representation. Is the converse still true? Is there any sub-algebra of Clifford algebra isomorphic to matrix algebra of $n$ dimension?

$\endgroup$
  • $\begingroup$ What is $n$? Any Clifford algebra contains a subalgebra isomorphic to $1 \times 1$ matrices. $\endgroup$ – S. Carnahan Apr 27 '14 at 3:16
  • $\begingroup$ @S.Carnahan $n$ is dimension of matrices. Or I want to find subalgebra of Clifford algebra isomorphic to all $n$ by $n$ matrices. $\endgroup$ – Shuchang Apr 27 '14 at 3:19
  • $\begingroup$ How is $n$ connected to the Clifford algebra? Are you taking the Clifford algebra of a quadratic space of dimension $n$? $\endgroup$ – S. Carnahan Apr 27 '14 at 3:20
  • $\begingroup$ @S.Carnahan I'm not sure. $n$ is best but maybe the real case is not. $\endgroup$ – Shuchang Apr 27 '14 at 3:23
  • $\begingroup$ The Chapter on Clifford algebras in Werner Greub's book "Multilinear Algebra" (Springer, 1978) might be helpful. $\endgroup$ – UwF Apr 27 '14 at 7:31
1
$\begingroup$

It depends to your Clifford algebra. The Clifford algebra of a hyperbolic quadratic form $q$ of dimension $2n$ over a field $k$ is isomorphic to the full matrix algebra $M_{2^{2n}}(k)$. By contrast it may happen that a Clifford algebra be a division algebra, so it cannot contain a full matrix subalgebra.It also depends on your base field, the Clifford algebra of the form $q=-x^2-y^2$ over the field of real numbers is isomorphic to the division algebra of hamiltonian quaternion. The Clifford algebra of the same form considered over the field of complex numbers is isomorphic to the full matrix algebra $M_2(\mathbb{C})$. Also if you are just interested to investigate the Clifford algebra of of the standard quadratic forms $q=x_1^2+\cdots+x_n^2$ (or $q=-x_1^2-\cdots-x_n^2$) over $\mathbb{R}$ you can use Bott periodicity.

$\endgroup$
0
$\begingroup$

The conventional real Clifford algebras are matrix algebras over the real numbers, complex numbera, or quaternions, and are simple algebras in the sense of ring theory. Any subalgebra that is a simple algebra will also be such a matrix algebra. And conversely: those types of rings are all the simple algebras, up to isomorphism, that are finite dimensional over the reals.

$\endgroup$
  • 1
    $\begingroup$ Some real Clifford algebras are direct sums of matrix rings. $\endgroup$ – S. Carnahan Apr 27 '14 at 4:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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