Let $C = \operatorname{Cl}(V,q)$ be a Clifford algebra where $V$ is an $N$-dimensional space with basis $B = \{e_1,e_2, \dotsc, e_N\}$. I'm looking for a way to invert elements.

What I've already worked out for myself is that if $x = \sum_{I \subseteq B} \lambda_I \hat{I}$ where $\hat{I}$ is the (ordered) product of the elements of $I$ then since left multiplication is linear there is a matrix $M^x$ such that $xy = M^xy$ for $y$ in $C$ (given an ordering on $\mathcal{P}(B)$). Then $x$ is a zero divisor if $\operatorname{Det}(M^x) = 0$ and invertible if $\operatorname{Det}(M^x)$ is invertible, in fact $x^{-1} = (M^x)^{-1}\emptyset$.

So far so basic linear algebra, but there's a problem. The matrix $M^x$ grows faster than Jack's beanstalk. By the time $N=3$ it's already $8 \times 8$ which I could maybe work out if I was sufficiently determined, but certainly nothing bigger. Maybe some people could do bigger space $V$ if they had access to Maple or similar but there's still going to be some fairly low upper bound.

So my question is `is there a nice formula for $\operatorname{Det}(M^x)$ and $x^{-1}$ of the form $x^{-1} = \sum_{I \subseteq B} \mu_I \hat{I}$ where the $\mu_I$ are rational functions in the $\lambda_I$ and $q_i = q(e_i)$?'

I've got a funny feeling that $\operatorname{Det}(M^x) = \sum (-1)^{\sigma_I} \lambda_I^2$ where $\sigma_I = \left(\begin{array}{c}|I| \\ 2 \end{array}\right)$, $|I|$ choose $2$, but I'm far from sure. Certainly I can't prove this. I also suspect that any formula for the coefficients of $x^{-1}$ will have $\operatorname{Det}(M^x)$ as a denominator but I wouldn't put money on it.

This seems a fairly natural thing to want, but I haven't found it mentioned in any of the places that I've looked. On the other hand I am quite new to Clifford algebras so this may be well known to everybody else. I tried to work it out myself because that's a good way to learn but now it's taking a while and I think I've got as much out of it as I'm likely to, so is this a well known thing or shall I keep plugging away at it myself?

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    $\begingroup$ This seems to be an open problem in general per the answer in: math.stackexchange.com/q/3154032/347489 $\endgroup$
    – user347489
    May 21 at 3:50
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    $\begingroup$ Do you really mean $(M^x)^{-1}\{\emptyset\}$? I'm having trouble parsing that notation. $\endgroup$
    – LSpice
    May 21 at 4:16
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    $\begingroup$ Yes, I'm pretty sure that's right. $V$ has a basis $B = \{e_1,\cdots e_N\}$ and so $C$ (as a vector space) has a basis the collection of subsets of $B$ - including $\emptyset$. The set $\lambda \emptyset$ for some $\lambda$ in the ground field are the scalars of the algebra and so $\emptyset$ is the multiplicative identity. , Maybe I shouldn't have put brackets round it, I'll change that now. $\endgroup$
    – Stevie H
    May 21 at 7:07

You should have a look at F. Reese Harvey's book "Spinors and Calibrations", where your question is answered for Clifford algebras of every signature.

The main point is that, when the ground field is $\mathbb{R}$, each Clifford algebra is actually isomorphic to a classical matrix algebra (or sum of two matrix algebras), and, once you identify the isomorphism the answer is obvious.

For example, if $q$ is positive definite and $V$ has dimension $3$, then the $8$-dimensional algebra $Cl(V,q)$ is isomorphic to $\mathbb{H}\oplus\mathbb{H}$, i.e., to pairs of quaternions. Thus, an element $(p_1,p_2)\in \mathbb{H}\oplus\mathbb{H}$ is invertible if and only if neither $p_1$ nor $p_2$ is zero.

As another example, if $q$ is positive definite and $V$ has dimension $6$, then the $64$-dimensional algebra $Cl(V,q)$ is isomorphic to $M_8(\mathbb{R})$, and hence an element is invertible if and only if its determinant as an $8$-by-$8$ matrix is nonzero. Thus, there is a polynomial $\Delta$ of degree $8$ on $Cl(V,q)$ such that $\Delta(x)\not=0$ if and only if $x$ is invertible. It turns out that the determinant of $M_x$ is just the $8$-th power of $\Delta$.

For the full story, see the book referenced above.


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