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for a symmetric bilinearform $\beta$ on a $\mathbb{K}$-vectorspace $V$ the associated Clifford Algebra $Cl(\beta)$ is the associative algebra with unit subject to the relations $$v\cdot v=\beta(v,v)\cdot 1\qquad\forall v\in V.$$It is then often said that $Cl(-\beta)$ is isomorphic to the opposite Algebra $Cl(\beta)^\text{op}$. Why is that?

Cheers, Robert

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The opposite algebra of an algebra is build by the same elements and the same addition, and the multiplication is the same, but in swapped order. – Ben Jul 25 '12 at 12:14
I know what the opposite of an algebra is. I want to know why $Cl(-\beta)\cong Cl(\beta)^\text{op}$. – Robert Jul 25 '12 at 13:01
I don't have an answer, so I'll just make the observation that in the case of an $\mathbb{R}$-vector space where $\beta$ is positive definite, the isomorphism can't take vectors to vectors, because the square of a vector is negative in $Cl(-\beta)$ but positive in $Cl(\beta)^{op}$. – Ted Jul 26 '12 at 8:03
I agree on that. In a book i use quite a lot (Budinich, Trautman: "The spinorial chessboard"), which i believe doesn't do stupid mistakes it says the following: The injection $\iota:V\hookrightarrow Cl(\beta)^\text{op}$ is a Clifford map of $(V,−\beta)$ and then extends to the desired Isomorphism $Cl(-\beta)\cong Cl(\beta)^\text{op}$. However, $\iota(v)^2=\beta(v,v)$ so $\iota$ is a Clifford map of $(V,\beta)$, not $(V,-\beta)$... i don't get it. – Robert Jul 26 '12 at 10:08
Now i'm really starting to believe the claim to be wrong. For instance, if $V=\mathbb{R}$ and $\beta(x,y)=-xy$ we see that $Cl(\beta)^\text{op}=\mathbb{C}^\text{op}=\mathbb{C}$ is not isomorphic to $Cl(-\beta)$, which is isomorphic to $\mathbb{R}^2$ with the product $$(a,b)\cdot(c,d)=(ac+bd,ad+bc).$$ – Robert Jul 27 '12 at 13:23
up vote 2 down vote accepted

First, care is needed! We need the right notion of "op", for behold $Cl(1) \simeq \mathbb{R}^2 \not\simeq \mathbb{C} \simeq Cl(-1)$, which as algebras are both commutative!

Over $\mathbb{R}$, the Clifford algebras arise naturally in the braided symmetric monoidal category of $\mathbb{Z}/2$-graded vector spaces, or "super-vector spaces" (alternatively...), as (super-)tensor products of the basic algebras $\mathbb{R}[\varepsilon]$ and $\mathbb{R}[i]$, where $\varepsilon$ and $i$ denote odd roots of $\pm 1$.

As with ordinary vector spaces, there is an isomorphism $$ \tau : A \otimes B \to B \otimes A, $$ and you need something like this to make an algebra of $A\otimes B$, given the algebra structures on $A$ and $B$ separately: $$ (A\otimes B) \otimes (A\otimes B) \overset{A\otimes \tau\otimes B}\longrightarrow (A\otimes A)\otimes(B\otimes B) \overset{m_A\otimes m_B}\longrightarrow A \otimes B .$$ Figuring out different ways to define $\tau$ can be confusing at first, but you'll get used to it.

Now, in ordinary vector spaces, the "opposite" product $m^{op}$ is derived from $m$ by pre-composition with the transpose on $A\otimes A$: $$ m^{op} = A\otimes A \overset{\tau}\to A\otimes A \overset{m}\to A, $$ and the same expression is the "right way" to define the opposite multiplication on a superalgebra; only the $\tau$ means something slightly different for superspaces vs. for ordinary vector spaces.

So, the exercises to start with are to understand $\tau$, and then to verify that $m_{-1}^{op} = m_1 $ and vice-versa --- or, understanding $\tau$, to carry-out the calculation over clifford algebras in full generality.

My favourite reference for superalgebras is Trimble's notes, but now you've read the words "symmetric" and "braided" and "monoidal" together, you can go on digging. Have fun!

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Your favorite reference for superalgebras is...what? – Noah Stein Aug 9 '12 at 18:05
Sorry, Trimble's notes on superdivision algebras; I can't remember why, but it made an impression. (when I wrote the answer I didn't have 10 rep yet, so it wouldn't let me include a third link) – Jesse C. McKeown Aug 10 '12 at 3:42
No problem. I took the liberty of editing it into your response (should be visible once the edit is "peer reviewed"). Thanks! – Noah Stein Aug 10 '12 at 14:36
Do I unterstand you correctly that the reason for my confusion is that i am using the "wrong" definition of $Cl(\beta)^\text{op}$? – Robert Aug 21 '12 at 13:56

There are different conventions for Clifford algebras, which is confusing. If I understand your question correctly. If the convention for the bilinear form is

$ v . v=\beta(v,v)\cdot 1\qquad\forall v\in V $

then the algebra is denoted as $\mathcal{Cl}_{p,q}$.

In the opposite convention

$v . v= - \beta(v,v)\cdot 1\qquad\forall v\in V $

it is equivalent to the algebra $\mathcal{Cl}_{q,p}$.

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