# Clifford Algebra is driving me crazy

How do we determine the Clifford Algebra of a quadratic form $q$? Thanks!

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The downvotes are probably because it looks like you are pretty lost, and haven't added much of your own thoughts. I think I know where you are coming from, though. Without altering the original question, it would be good if you typed in specific things that you are unsure about, or would like to know more about. –  rschwieb Jan 15 '13 at 18:30
It is also probably too broad. Specific requests about the algebra might help to turn that around. People considering voting to close might also want to take into account the author has not had much time to enact changes. (Right now, at least.) –  rschwieb Jan 15 '13 at 18:39

I don't want to reprove the existence of Clifford algebras in general here. You will have to go look that up, for a rigorous explanation.

I can give a heuristic explanation, though, on how to think of it.

The quadratic form $Q:V\rightarrow F$ gives rise to a bilinear form $B:V\times V\rightarrow F$ that works like an inner product on $V$. From $Q$ and $V$, one can prove the existence of the algebra $C\ell(V,Q)$.

To get a grip on what it looks like, begin with any orthonormal basis $b_1\dots b_n$ for $V$. (It doesn't have to be finite dimensional, and it doesn't have to be orthonormal, but I'll assume so for convenience.) By the way the algebra was built, you are guaranteed that the multiplication works this way on the basis: $b_i\otimes b_i=Q(b_i)$ and $b_i\otimes b_j=-b_j\otimes b_i$ for $i\neq j$.

As you probably know, $C\ell(V,Q)$ has dimension $2^n$, but we only have $n$ things in the set of $b_i$, so it is clearly not a basis. To get a complete basis you form every product of the $b_i$ possible with ascending indices. By doing this, we ensure that the set of $2^n$ elements formed is a linearly independent set. We don't want to include $b_2\otimes b_1\otimes b_3$ for example, because that is the same as $-b_1\otimes b_2\otimes b_3$, and that is linearly dependent with $b_1\otimes b_2\otimes b_3$. Similarly you don't want to include $b_2\otimes b_3\otimes b_3$, because that is the same as $b_2\cdot Q(b_3)$ which is linearly dependent with $b_2$.

Using the anticommutativity and squaring rules, along with the associative and distributive property of multiplication, you can multiply any two elements of the Clifford algebra.

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There could be something special said about fields of characteristic two, but I had really written thinking of $char(F)\neq 2$. –  rschwieb Jan 15 '13 at 18:38