# Quaternions as Clifford algebra: confusion over grade

If the quaternions are defined as the even grade multivectors in $Cl_{3,0}(\mathbb{R})$ then $i,j,k$ are all even. If they are defined as the Clifford algebra $Cl_{0,2}(\mathbb{R})$, then $i,j$ are odd grade, $k=ij$ is even grade. So in what sense is grade a well-defined concept, if isomorphisms do not preserve it? Or does this mean that the two definitions of quaternions are not truly isomorphic as Clifford algebras? (in the sense that $i\wedge j=0$ in the first, but $i\wedge j=k$ in the second) If so, is any one preferable?

• An algebra can have different gradings. – Qiaochu Yuan Jun 20 '16 at 7:41
• @Yuan Can a Clifford algebra be defined in a grade-free manner? Your answer seems to imply that. I guess one can always look at the set of vectors in an algebra such that $v^2\in\mathbb{R}$ and check if it's a subspace that generates the rest of the algebra in a 'free' way. – Chrystomath Jun 20 '16 at 16:40
• It depends on the definition of the wedge product. – user48672 Jun 21 '16 at 14:49
• $\Bbb R$ and $\Bbb C$ are isomorphic as rational vector spaces, that doesn't mean fields aren't well-defined. – arctic tern Dec 7 '16 at 15:30

Can a Clifford algebra be defined in a grade-free manner?

The definition of a Clifford algebra (at least, every one I have seen) is totally independent of the notion of "grade." Given a Clifford algebra $A=\mathcal{C}(V,q)$ where $V$ is a quadratic space with form $q$, you can then define "grade."

So in what sense is grade a well-defined concept, if isomorphisms do not preserve it?

Grade is determined by the choice of $V$. You could, in principle, find an isometric copy of $V$ inside $A$ (call it $V'$), and the Clifford algebra for that quadratic space would coincide with $A$, but the grading would be totally different if $V'\neq V$, since the grade 1 elements all live in the quadratic space. Or you can take a sub-quadratic space of $V$ and generate its Clifford algebra inside $A$, and it will have its own grading etc.

For this reason and others, you can see there's no good reason to expect that the grading of an algebra is important to its subalgebras, or even to algebra-isomorphisms.

Or does this mean that the two definitions of quaternions are not truly isomorphic as Clifford algebras?

The reason for this problem is the failure to take into account the importance of the specificiation of $(V,q)$ in the definition of a Clifford algebra. If $A=\mathcal{C}(V,q)$ and $A'=\mathcal{C}(V',q')$ are two Clifford algebras, then a Clifford algebra morphism $f:A\to A'$ should be an algebra morphism from $A\to A'$ such that when it's restricted to $V$, it becomes a morphism of quadratic spaces $(V,q)\to (V',q')$.

In particular, it has to map $V\to V'$, and that's something that isn't happening with the map that you're describing from $Cl_{0,2}(\mathbb R)\to Cl_{3,0}(\mathbb R)$. While the algebra isomorphism does preserve the geometric product and addition properly, the link with the quadratic form has been broken. So the fact that the wedge product (which can be defined in terms of $V,q$ and the geometric product) is also broken comes as no surprise.

So the answer to the highlighted question here is: yes, the two algebras are not isomorphic as Clifford algebras. They are isomorphic as $\mathbb R$ algebras, though, and that says a lot. Really when talking about the quaternions, though, we don't often think of the underlying quadratic space that generates it as a Clifford algebra. That's why we can still talk about it and be interested in it as an $\Bbb R$ algebra without these nice things automatically happening for it as a Clifford algebra.