# Multiplication of a real algebra is a smooth map

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 $$V$$.

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.