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It is known that

$$ \exp : \mathbb{M}(n,\mathbb{C}) \to \operatorname{GL}(n,\mathbb{C}) $$

This relationship only works over the complex field.

My question is:

$$ \exp : \mathcal{G}_n(\mathbb{R})\to ? $$

where $\mathcal{G}_n(\mathbb{R})$ is the set of all real multivectors of a Clifford algebra of dimension $n$.


I know that taking the exponential of a multivector produces an invertible multivector due to $\exp V \exp -V = I$. I wonder if it is group isomorphic to $\operatorname{GL}(n,\mathbb{R})$:

$$ \exp ( \mathcal{G}_n(\mathbb{R}) )\cong \operatorname{GL}(n,\mathbb{R}) ? $$

Anon21
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  • "This relationship" -- what do you mean? That the exp map is surjective? That's true for many Lie groups and not true for many more, go through the links in the comments of https://math.stackexchange.com/q/1591689/96384. – Torsten Schoeneberg Sep 27 '20 at 16:17

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