# Clifford Lie Algebras

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.

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A little bit of what you want can be found in Chapter 5 of Gracia-Bondia, 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.

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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 $2^n-1$ (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. –  Y Macdisi Jan 17 at 6:54
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)... –  MTS Jan 17 at 16:33
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... –  MTS Jan 17 at 16:41
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. –  MTS Jan 17 at 16:42
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... –  Y Macdisi Jan 17 at 21:06
Thanks for the response. It is actually the other elements (non-quadratic) 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. –  Y Macdisi Jan 24 at 2:34