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

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.

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    $\begingroup$ I think most books use the natural convention $v^2=Q(v)$. I guess the only argument in favor of the "$-$" convention is that the standard quadratic form $q_0$ over $\mathbb{R}$ gives a more interesting algebra, e.g. the quaternions when $\dim V=2$. I do not find this convincing because it suffices to consider $C(\mathbb{R}^n,-q_0)$, while the "$-$" convention does not make any sense over $\mathbb{C}$ for instance. $\endgroup$
    – abx
    Commented Mar 9, 2015 at 10:59
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    $\begingroup$ @abx: Perhaps if one is always working in a differential geometry setting where only the compact form of the Spin group is of interest it gets annoying to keep writing the sign every time. $\endgroup$
    – user74230
    Commented Mar 9, 2015 at 17:09
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    $\begingroup$ Clifford himself used the $-$ convention, so it's perhaps partially tradition. The motivation for Clifford algebras was to generalise the sequence $\mathbb{R}$, $\mathbb{C}$ and $\mathbb{H}$ and this requires the $-$ sign. $\endgroup$
    – José Figueroa-O'Farrill
    Commented Mar 10, 2015 at 1:47
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    $\begingroup$ See Henriques' meta-answer. Another way to see which preference you have is what is your preference between $Pin_+$ and $Pin_-$. $\endgroup$
    – AHusain
    Commented Jun 29, 2016 at 11:35

1 Answer 1

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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).

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    $\begingroup$ Clifford algebras also exist (and are very useful!) over other fields than $\Bbb{R}$. $\endgroup$
    – abx
    Commented Jun 29, 2016 at 12:57
  • $\begingroup$ OK, but $1$ and $-1$ exist in any field, aren't they ? Therefore simple definition used by Clifford and me :) still works. $\endgroup$
    – user21230
    Commented Jun 29, 2016 at 13:27
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    $\begingroup$ I was referring to your edit: yes, the quadratic form $Q$ is definitely useful. $\endgroup$
    – abx
    Commented Jun 29, 2016 at 13:34
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    $\begingroup$ The first advantage that comes to my mind for the $Q$ version of the definition is that it makes obvious the fact that the orthogonal group of $(V,Q)$ acts on the Clifford algebra by algebra-automorphisms. (This is a special case of the fairly general phenomenon that, in linear algebra, things tend to be easier to see when you don't make them depend on a particular basis.) $\endgroup$
    – Andreas Blass
    Commented Jun 29, 2016 at 15:43

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