# Group of units of a Clifford algebra

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!

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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