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Let $A$ be a finite dimensional real algebra, and let $F:A\times A\to A$ be the multiplication map $F(a,b)=ab$. Is it true that the map $F$ is smooth? (Here I am considering the canonical smooth structure of $A$: choose a real linear basis $v_1,\dots,v_n$ of $A$ and use this basis to linearly identify $A$ with $\Bbb R^n$; this gives a smooth structure on $A$, and it is independent of the choice of the basis.)

P.S. Actually I want the result when $A$ is the Clifford algebra $Cl(V)$ of a finite dimensional inner product space $V$.

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Write $m(v_i,v_j) = \sum_k C_{ij}^k v_k$ with constants $C_{ij}^k$ (structure constants). Since $m$ is bilinear, then then $m(\sum_i a_i v_i,\sum_j b_j v_j) = \sum_{i,j,k} a_i b_j C_{ij}^k v_k$. So the isomorphism of vector spaces $\mathbb{R}^n \to A$, $e_i \mapsto v_i$ transports this to the map $m' : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n$, $m'(a,b) = (\sum_{i,j} a_i b_j C_{ij}^k)_{k=1,\dotsc,n}$, which is a polynomial in each entry and hence smooth.

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