I'm studying the "Clifford Lie Algebra" (see http://arxiv.org/pdf/1007.2481.pdf page 30). It's basically a way to look at Clifford algebras and their properties in a Lie algebraic setting (which I find appealing). I'm looking for a reference that looks at this in more detail; one that discusses the lie subalgebras even better. Thanks.
A little bit of what you want can be found in Chapter 5 of GraciaBondia, Varilly, and Figueroa's book Elements of Noncommutative Geometry. They don't say much about subalgebras, I think, but they do prove (Lemma 5.7, page 182) the result that bivectors in the Clifford algebra $Cl(V)$ are closed under taking commutators, and that the adjoint action of bivectors on vectors in $V$ induces an isomorphism of the Lie algebra of bivectors with $\mathfrak{so}(V)$. That might be a good place to start.

3$\begingroup$ Thanks I'll try to track this down. Finding $so(n)$ in $cl(n)$ is, as expected, easy to do; here $cl(n)$ is the Clifford lie algebra; it has dimension $2^n1$ (I think this can also be extended to be $2^n$). $so(n+1)$ also shows up in $cl(n)$ if I did my calculations correctly. $\endgroup$ – Y Macdisi Jan 17 '13 at 6:54

1$\begingroup$ I do not understand the statement about the dimension. The Clifford algebra $Cl(V)$ is an associative algebra of dimension $2^n$ when $\mathrm{dim}V=n$. Then it acquires a Lie algebra structure where the Lie bracket is the (scaled) commutator. This does not change the dimension. Also, upon examining these notes a little more closely, I would say that if you want to learn about Clifford algebras and spinors from a mathematical viewpoint, you would be better off consulting another source. These notes are coming more from a physics perspective (which can be useful if you're a physicist)... $\endgroup$ – MTS Jan 17 '13 at 16:33

$\begingroup$ but the presentation here is kind of ad hoc. For instance the Clifford algebra of $V$ is defined not as a quotient of the tensor algebra of $V$ but as a new multiplication on the exterior algebra $\Lambda(V)$. So if you're going to be rigorous about things, you need to check that this new product is associative. On the other hand you get that for free if you define it as a quotient. In addition to the book I recommended above, another good (and classic) source is Chevalley's monograph The algebraic theory of spinors and Clifford algebras. A different viewpoint is given in... $\endgroup$ – MTS Jan 17 '13 at 16:41

1$\begingroup$ Clifford algebras and spinor norms over a commutative ring, by Hyman Bass. But that one is probably much more general than what you're looking for. $\endgroup$ – MTS Jan 17 '13 at 16:42

1$\begingroup$ I agree with you on the dimension, it should be $2^n$; there is an element in the lie algebra that commutes with everything; my guess is that the authors did not want to include it. Clifford algebras are a large field and my guess is that angle is just not very popular. One motivation for me is the ability to look at the adjoint rep of $cl(n)$ as a rep of $so(n)$; for example $cl(4)$ $\to 1 \oplus 4 \oplus 6 \oplus 4 \oplus 1$ as an $so(4)$ rep... $\endgroup$ – Y Macdisi Jan 17 '13 at 21:06
I don't know anyone else who calls this the "Clifford Lie Algebra". It is just one of the basic applications of Clifford algebras. Given the Clifford algebra of a quadratic form, the quadratic elements of the Clifford algebra give you the Lie algebra of the orthogonal group of that quadratic form.
There are many places to read about this, one of them would be Chapter 1.6 of "Spin Geometry" by Lawson and Michelson. I've written up some notes for a graduate course that include this, see here:
http://www.math.columbia.edu/%7Ewoit/LieGroups2012/cliffalgsandspingroups.pdf

3$\begingroup$ Thanks for the response. It is actually the other elements (nonquadratic) that make things more interesting. Altogether these give a $2^n$ dimensional lie algebra which can stand on its own as an abstract lie algebra that includes $so(n)$ as a lie subalgebra. The multivectors correspond to invariant subspaces of this subalgebra (adjoint action). Also $so(n+1)$ and I believe $so(n+2)$ also occur as lie subalgebras...all this motivated the question. $\endgroup$ – Y Macdisi Jan 24 '13 at 2:34