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I am reading a text by Jean Gallier on Clifford algebras, Pin and Spin groups. I have a problem with one little innocent-looking paragraph establishing Pin and Spin as Lie groups on the page 37. I don't know what he uses in the arguments. I'm gonna go through the difficulties I have and add some of my reasoning. I would be glad for any light shed on them.

What is the topology? Let $C_{p,q}$ denote the real Clifford algebra associated with a non-degenerate quadratic form of signature $(p,q)$. Its underlying vector space $V$ is isomorphic to a certain Euclidean space ($V \cong \mathbb{R}^{2^{p+q}}$). I assume, the euclidean topology from this isomorphism is THE topology for $C_{p,q}$, but I'm not really sure.

Why is the group of units a topological group? How do we know, that multiplication and inverses are even continuous? Since the multiplication is inherited from tensor algebra, I don't see how it all plays together.

Continuity of the map $C_{p,q}\rightarrow \mathrm{End}(V)$ given by $x \mapsto (y\mapsto xy)$. Maybe If I understood the multiplication (previous point), I would see how is this continuous.

If I would understood these, then it would be clear, why the group of units $C_{p,q}^\times$ is an open subset. But why is it a Lie group? It is certainly locally euclidean, hence by a fundamental theorem of Gleason, Montgomery and Zippin, it is a Lie group. Is there an easier way to see that? I guess the author would at least mentioned that he uses such a big gun. What we effectively need is just to show that multiplication is smooth. Am I right? I'm rather unfamiliar with the Lie theory.

Then Pin and Spin groups are closed subgroups of a Lie group and hence both Lie groups themselves. The fact, that they are given by closed condition isn't crystal clear to me - but I believe if I've understood the multiplication's continuity, I would be able to supply a formal proof of continuity of the norm.

I'm sorry that this is a somewhat open question. Can anyone direct me from the abyss of Lie ignorance?

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When you work in finite dimension, all these questions are basically trivial (this adresses mainly your first two questions).

There is only one topology on any finite-dimensional vector space that makes it into a separated locally convex topological space. This is most likely the only topology you will ever see on any such space. And of course this is the euclidian topology.

And for this topology, everything linear is automatically continuous. So in particular any finite-dimensional real algebra is a topological algebra for this topology.

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  • $\begingroup$ Thanks! I've suspected the euclidean bit, but it is better to ask. Alright, obviously multiplication by some fixed element is linear, hence continuous. I haven't done the extra step to see that multiplication itself is bilinear, and hence given by a matrix anyway. That settles the first two. The map in the third is linear, so that's done as well. Still not sure about the Lie stuff though. Thanks again! $\endgroup$ – liczman Apr 17 '16 at 21:28

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