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Some context first: one can construct $\mathbb{H}$ as the even subalgebra of the Clifford algebra $Cl_{0,3}(\mathbb{R})$ using the Clifford product, which states that given $u,v\in \mathbb{R}^3$ then $uv+vu=-2(u\cdot v)$. The basic property of the quaternions $$i^2=j^2=k^2=ijk=-1$$ is then shown to follow from the Clifford product.

Recall the Clifford algebra is the freest until associative algebra and as such and element of $Cl_{0,3}(\mathbb{R})$ looks like $$u= a_0 +a_1e_1+a_2e_2+a_3e+3+a+4e_1e_2+a_5e_1e_3+a_6e_2e_3+a_7e_1e_2e_3,$$ where $e_1,e_2,e_3$ is an orthogonal basis for $\mathbb{R}^3$. The quaternions are identified with even degree elements via: $$i=e_2e_3, j=e_3e_1, k=e_1e_2,$$ from which the basic property stated above follows, using the Clifford product.

My question is: what can be done with the elements of odd degrees? What do they form? anything interesting? If they do, what is the structure called?

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    $\begingroup$ The set of odd elements does not form a subalgebra, since it's not closed under multiplication. $\endgroup$ Commented Nov 26, 2021 at 15:29
  • $\begingroup$ @HansLundmark Thanks for that comment Hans, they indeed can't form a subalgebra $\endgroup$ Commented Nov 26, 2021 at 16:38
  • $\begingroup$ I think you meant to write $$u= a_0 +a_1e_1+a_2e_2+a_3e+a_4e_1e_2+a_5e_1e_3+a_6e_2e_3+a_7e_1e_2e_3,$$ insteead of your $$u= a_0 +a_1e_1+a_2e_2+a_3e+3+a+4e_1e_2+a_5e_1e_3+a_6e_2e_3+a_7e_1e_2e_3,$$. $\endgroup$
    – mma
    Commented Dec 9, 2023 at 6:46

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I got the answer to my question (albeit trivial but thanks HansLundmark) but found something better along the way, so I'll post it here for posteriority. it turns out that the Clifford Algebra $Cl(V,Q)$ where $V$ is a vector space and $Q$ is the quadratic form, can be given a structure of a $\mathbb{Z}_2$-graded algebra. What this means is the following:

Consider the linear map $f:V\to V$ given by $f(v)=-v$, it preserves the quadratic form $Q$. This is a reflection. By the universal property of Clifford Algebras, this map $f$ can be extended to a map $g$ which is an automorphism of algebras, given by $$g:Cl(V,Q)\to Cl(V,Q)$$ Notice that $g^2=Id$ hence we have a positive eigenspace of $g$ and a negative eigenspace of $g$ given by: $Cl^0(V,Q)$ and $Cl^1(V,Q)$. We then have that $Cl(V,Q)= Cl^0(V,Q) \bigoplus Cl^1(V,Q)$, and $$Cl^i(V,Q)=\{x\in Cl(V,Q): g(x)=(-1)^i x\}$$

$Cl^0(V,Q)$ is the even subalgebra and $Cl^1(V,Q)$ is not a subalgebra.

Correction: $Cl_{0,2}(\mathbb{R})$ is isomorphic to $\mathbb{H}$, not $Cl_{0,3}(\mathbb{R})$. this seems to depend on the "grading". The even subalgebra of $Cl_n(\mathbb{C})$ is isomorphic to $Cl_{n-1}(\mathbb{C})$. $SU(2)$ is isomorphic rather to the quaternions of norm 1! So: what are the quaternions of norm 1 in $Cl_{0,2}(\mathbb{R})$? A specific subgroup of the Clifford algebra, any name to this specific subgroup?

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  • $\begingroup$ "The even subalgebra is always isomorphic to the Clifford algebra itself." Are you saying that $Cl^0(V,Q)\simeq Cl(V,Q)$? Because they are not of the same dimension, so that's going to be difficult. Also, $SU(2)$ is not an algebra but a group. $\endgroup$ Commented Nov 27, 2021 at 11:43
  • $\begingroup$ @CaptainLama thank you for this clarification. There's a few things I need to correct up there: $Cl_{0,2}(\mathbb{R})$ is isomorphic to $\mathbb{H}$, not $Cl_{0,3}(\mathbb{R})$. Seems like it depends on the "grading". The even subalgebra of $Cl_n(\mathbb{C})$ is isomorphic to $Cl_{n-1}(\mathbb{C})$. And sorry yes, $SU(2)$ is isomorphic rather to the quaternions of norm 1! So: what are the quaternions of norm 1 in $Cl_{0,2}(\mathbb{R})$? A specific subgroup of the Clifford algebra, any name to this specific subgroup? $\endgroup$ Commented Nov 27, 2021 at 16:00

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