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I am reading a book named "An Introduction to Clifford Algebras and Spinors" by J. Vaz Jr. and R. da Rocha. In page 78, I met a confusing formula (3.89), written as:

$$\gamma(\mathbf{v})\gamma(\mathbf{u})=\gamma[\gamma(\mathbf{v})(\mathbf{u})]$$

where $\gamma:V\to\mathcal{C}\ell(V,g)$ is a Clifford mapping and $\mathbf{v},\mathbf{u}\in V$. The book said this formula is due to "the associativity of Clifford algebra".

I really don't understand this, could anyone give some hints of why?

Context:

In the preceding section the authors constructed Clifford algebra using creation operators $\mathbf{E}:V\to\mathrm{End}(\bigwedge(V))$ where $\mathbf{E}(\mathbf{v})(A)=\mathbf{v}\wedge A$, and annihilation operators $\mathbf{I}:V^*\to\mathrm{End}(\bigwedge(V))$ where $\mathbf{I}(\alpha)(A)=\alpha\rfloor A$. Using notation $\flat:V\to V^*$ as the index lowering operator, the Clifford mapping is defined as

$$\gamma=\gamma_+=\mathbf{E}+\mathbf{I}\circ\flat$$

After that confusing formula, the authors used above formula to get ($g$ is the associated bilinear quadratic form)

$$\begin{eqnarray} \gamma(\mathbf{v})(\mathbf{u})&=&(\mathbf{E}+\mathbf{I}\circ\flat)(\mathbf{v})(\mathbf{u})\\ &=&\mathbf{v}\wedge\mathbf{u}+\mathbf{v}_\flat\rfloor\mathbf{u}\\ &=&\mathbf{v}\wedge\mathbf{u}+g(\mathbf{v},\mathbf{u}) \end{eqnarray}$$

And, thus

$$\begin{eqnarray} \gamma(\mathbf{v})\gamma(\mathbf{u})&=&\gamma[\gamma(\mathbf{v})(\mathbf{u})]\\ &=&\gamma[\mathbf{v}\wedge\mathbf{u}+g(\mathbf{v},\mathbf{u})]\\ &=&\gamma(\mathbf{v}\wedge\mathbf{u})+g(\mathbf{v},\mathbf{u})\mathbf{1} \end{eqnarray}$$

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  • $\begingroup$ I'm confused by part of the definition of $\mathbf E$ and $\mathbf I$ but I'm probably making incorrect assumptions. I assume $V$ is a vector space and $\text{End}(V)$ is the set of endomorphisms on $V$, i.e. linear functions $V\to V$. But then $\mathbf E(\mathbf v)(A)\notin V$, similarly for $\mathbf I$, at least for my usual understanding of $\wedge$ and $\rfloor$, but that also probably doesn't match the definitions being used here. $\endgroup$ Commented Feb 1, 2018 at 8:29
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    $\begingroup$ Despite the above confusions, it does seem to me that $\gamma(\mathbf u)(\mathbf v)$ is a round-about way to write the geometric product and so if $(\gamma(\mathbf u)\gamma(\mathbf v))(\mathbf w)$ means $\gamma(\mathbf u)(\gamma(\mathbf v)(\mathbf w))$, then the formula is indeed associativity of the geometric product. If we write $\gamma(\mathbf u)(\mathbf v)$ as simply $\mathbf u\mathbf v$, then the formula (applied to $\mathbf w$) is $\mathbf u(\mathbf v\mathbf w)=(\mathbf u\mathbf v)\mathbf w$. $\endgroup$ Commented Feb 1, 2018 at 8:36
  • $\begingroup$ My mistake, $\mathrm{End}(V)$ should be $\mathrm{End}(\bigwedge(V))$. $\endgroup$
    – AmFCG
    Commented Feb 1, 2018 at 9:31
  • $\begingroup$ Ah, that makes a lot more sense. $\endgroup$ Commented Feb 1, 2018 at 9:35
  • $\begingroup$ So, do you have any idea about that formula? I cannot understand how that could hold. @DerekElkins $\endgroup$
    – AmFCG
    Commented Feb 1, 2018 at 9:37

1 Answer 1

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I asked the original author via email and finally understand the formula. So let me share my understanding below.

Suppose we have an associative operator $\vartriangle$, satisfying $$(\mathbf{v}\vartriangle\mathbf{u})\vartriangle\mathbf{w}=\mathbf{v}\vartriangle\mathbf{u}\vartriangle\mathbf{w}=\mathbf{v}\vartriangle(\mathbf{u}\vartriangle\mathbf{w})$$

So here if $\mathbf{v}\vartriangle\mathbf{u}=\gamma(\mathbf{v})(\mathbf{u})$, then we have $$\begin{eqnarray} \gamma[\gamma(\mathbf{v})(\mathbf{u})](\mathbf{w})&=&(\mathbf{v}\vartriangle\mathbf{u})\vartriangle\mathbf{w}\\ &=&\mathbf{v}\vartriangle(\mathbf{u}\vartriangle\mathbf{w})\\ &=&\gamma(\mathbf{v})(\gamma(\mathbf{u})(\mathbf{w}))\\ &=&\gamma(\mathbf{v})\gamma(\mathbf{u})(\mathbf{w}) \end{eqnarray}$$

Removing $\mathbf{w}$ from both sides, we finally get $$\gamma(\mathbf{v})\gamma(\mathbf{u})=\gamma[\gamma(\mathbf{v})(\mathbf{u})]$$

And the associativity is due to Clifford algebra.

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