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Let $V$ be a finite dimensional vector space over $\mathbb{R}$ or $\mathbb{C}$, $Q$ a quadratic form on $V$ and $\mathrm{Cl}(V,Q)$ the corresponding Clifford algebra. Now consider the multiplicative group of units of this Clifford algebra, namely:

$$ \mathrm{Cl}^\times(V,Q):=\{\phi\in\mathrm{Cl}(V,Q):\exists\psi\in\mathrm{Cl}(V,Q)\quad\phi\psi=\psi\phi=1\} $$

I would like to prove that this group is a Lie group, so anybody with an answer is welcome!

My idea is the following. Equip the Clifford algebra with the topology given by the quadratic form. Then the unit ball is included in $\mathrm{Cl}^\times(V,Q)$, so we can prove that $\mathrm{Cl}^\times(V,Q)$ is an open set of $\mathrm{Cl}(V,Q)$ using the same method used to prove that the set of isomorphisms between to Banach spaces is an open set, if I am not wrong. Therefore, $\mathrm{Cl}^\times(V,Q)$ is a manifold. The inverse operation is smooth because

$$ \forall v\in V\quad v^{-1}=-v/Q(v) $$

and the multiplication is certainly smooth because it comes basically from the map $v\mapsto v\otimes\cdot$ which is linear. Therefore we have a Lie group.

Are these arguments correct? At least, I would like to know if there is a generalization of this, like the group of units of a Banach algebra is always a Lie group (or at least a topological one)?

Many thanks!

t.b.
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Benjamin
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    Yes, the group of invertible elements of a Banach algebra $A$ is a Banach Lie group. It is an open subset of $A$, multiplication is smooth because it is the restriction of a bilinear map and inversion is smooth because it can be written using the geometric series in a neighborhood of the identity. I haven't quite followed the details of your argument concerning the topology (I'd probably try to use the one coming from the identification of the Clifford algebra with the exterior algebra as a vector space but I haven't really thought about it). BTW it's prove not proove but that's a detail. – t.b. Aug 25 '11 at 13:21

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Real Clifford algebras Cl(n) are algebra isomorphic to matrix algebras M(p,K)or direct sums M(p,K) + M(p,K), where p is an integer that depends on n and K = R,C or H. Hence the units of Cl(n) are the units of these matrix algebras and they form a group that is open and dense in Cl(n), at least in the case that Cl(n) = M(p,K), where K = R or C. See for example the book by Lawson and Michelsohn on Spin Geometry for details.

Patrick Eberlein
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  • Can you please give some reference for this? I cant find a proof for Cl(n) to be isomerphic to some M(p, R) (a full real matrix algebra). – Prajwal Samal Jun 23 '21 at 03:30