# Finding inverses in Clifford Algebras

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?

• This seems to be an open problem in general per the answer in: math.stackexchange.com/q/3154032/347489 May 21 at 3:50
• Do you really mean $(M^x)^{-1}\{\emptyset\}$? I'm having trouble parsing that notation. May 21 at 4:16
• 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. May 21 at 7:07

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 $$\Delta$$.