Convention on Clifford Product When studying the Clifford Algebra associated to some $(V,Q)$, one finds two basic identities differing by a sign:


*

*$vv=Q(v)$ (see, for instance, Wikipedia)

*$vv=-Q(v)$ (see, for instance, MathWorld or Lawson-Michelson Spin Geometry)
I suppose it is not a great problem this uncertainty in the product, but I wonder
Are there any reasons to prefer any of the possibilities? At first sight, $vv=Q(v)$ may seem more 'canonical', but I suppose there may be good reasons to prefer the other sign.
Any suggestion or idea is welcome.
 A: You obtain different sequence of algebras with $+$ and different with minus. Let $C_k$ denote algebra generated by $k$ anticommuting letters which square to $-1$. Then $C_k$=$\mathbb C$, $\mathbb H$, $\mathbb H+\mathbb H$, $M_2\mathbb H$, $M_4 \mathbb C$, $M_8 \mathbb R$, $M_8 \mathbb R+M_8\mathbb R$, $M_{16}\mathbb R$.
Let $ C_k^{\prime}$ denote algebra generated by $k$ anticommuting letters which square to $1$. Then $C_k^{\prime}$=$\mathbb R+\mathbb R$, $M_2\mathbb R$, $M_2\mathbb C$, $M_2\mathbb H$, $M_2 \mathbb H+M_2\mathbb H$, $M_4\mathbb H$, $M_8\mathbb C$, $M_{16}\mathbb R$.
For k>8 we need to take tensor product of $M_{16}\mathbb R$ (possibly few copies) with one of the above i.e. $C_{k+8}=C_k\otimes M_{16}\mathbb R$ and similar for $C_k^{\prime}$. Additionally we have $C_{k+2}=C_k^{\prime}\otimes \mathbb H$ and $C_{k+2}^{\prime}=C_k\otimes M_2\mathbb R$ which should help to recreate the algebras above from memory.
One may ask what algebra will be obtained by taking $k$ letters which square to $-1$ and $l$ letters which square to $1$ all anticommuting. Unfortunately or luckily we obtain one of the above.
In my opinion Clifford algebras and octonions is neglected subject on mathematics study - presented during excercises. For me it was quite big "discovery" that e.g. $M_4 \mathbb R$=$\mathbb H\otimes \mathbb H$.
EDIT
I do not see advantage of using quadratic form $Q$ in definition of Clifford algebra. The definition on MathWorld says that this is "general case". But we do not obtain any new algebras by using quadratic form than by using just $k$ letters $e_i$ which anticommute and square to $1$ or $-1$. The only advantage - maybe - is that we can obtain exterior algebra when assuming that base letters square to $0$. In my opinion using quadratic form just confuses, complicates beautiful simple definition used by Clifford ( I imagine, because I never read any Clifford work).
