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I am looking for a explicit isomorphism between $Cl(8)$ (Clifford algebra over $\mathbb{R}^8$ with standard Euclidean metric) and $\mathbb{R}(16)$ (algebra of $16\times 16$ matrices over $\mathbb{R}$). More concretely, it would be very useful to know:

  • The image of the basis elements $e_k$

  • The image of the volume element $v=e_1e_2e_3e_4e_5e_6e_7e_8$

  • The equations of the $\pm$-eigespaces of $v$, $\Delta_{\pm}$

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2 Answers 2

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Here's a standard explicit formula: Let $\mathbb{O}\simeq\mathbb{R}^8$ denote the algebra of octonions, and for $x\in\mathbb{O}$, let $L_x$ (respectively $R_x)$ denote the linear map from $\mathbb{O}$ to itself generated by left (respectively, right) multiplication by $x$ and let $C:\mathbb{O}\to\mathbb{O}$ be conjugation in the octonions. Now define $\rho(x):\mathbb{O}\oplus\mathbb{O}\to\mathbb{O}\oplus\mathbb{O}\simeq\mathbb{R}^{16}$ to be the matrix in $\mathbb{R}(16)$ $$ \rho(x) = \begin{pmatrix} 0 & CR_x\\ - CL_x & 0\end{pmatrix}. $$ Then one easily computes that $\rho(x)^2 = -|x|^2 \mathrm{Id}_{\mathbb{O}\oplus\mathbb{O}}$, so $\rho$ extends to a homomorphism $\rho:Cl(8)\to \mathbb{R}({16})$, which is necessarily an isomorphism.

I should also remark that the two eigenspaces of $v$ are the two given $\mathbb{O}$-summands in $\mathbb{O}\oplus\mathbb{O}$; which is $\Delta_+$ depends on which orientation of $\mathbb{O}$ you choose.

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  • $\begingroup$ Thanks a lot! And, under this isomorphism, have the images described in the two first points of the question nice associated matrices? $\endgroup$
    – Jjm
    Jan 19, 2015 at 19:12
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    $\begingroup$ @Jjm: Well, yes, but they are the matrices of multiplication by an orthonormal basis of $\mathbb{O}$, so 'nice' depends on how 'nice' you think the octonions are (they are basically signed permutation matrices, though, in the usual basis). As for the matrix of $v$, in the above form, depending on your orientation, it's either a diagonal matrix with $+1$s in the first 8 slots and $-1$s in the last 8 slots or it's the negative of that matrix. $\endgroup$ Jan 19, 2015 at 19:16
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    $\begingroup$ Another formula that works is $$\rho(x) = \left(\begin{matrix} 0 & L_x\\-L_{\overline{x}} & 0 \end{matrix}\right).$$ $\endgroup$ Jan 19, 2015 at 20:42
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    $\begingroup$ @Jjm: All you need to know is that $Cl(8)$ has no nontrivial $2$-sided ideals, for this will imply that $\rho$ is injective (since it clearly isn't zero). The fact about no nontrivial $2$-sided ideals is not hard to prove directly, but I admit that it's not completely obvious. However, it does follow from the general fact that $Cl(2n,\mathbb{C})=\mathbb{C}(2^n)$ is irreducible, specialized to the case that $n=4$. $\endgroup$ Feb 23, 2015 at 19:28
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    $\begingroup$ @Jjm: No, because $Cl(7)\simeq \mathbb{R}(8)\oplus \mathbb{R}(8)$, so there can't be an isomorphism $Cl(7)\longrightarrow \mathbb{R}(8)$. $\endgroup$ Nov 26, 2015 at 14:09
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In order to establish isomorphism of Clifford $C_8$ with $M_{16}\mathbb R$ it is enough to define 8 letters generators which square to $-1$ and anticommute. Define $e_k$=$$\begin{pmatrix} L_k & \\ & R_k \end{pmatrix} $$ where $L_k$ and $R_k$ are left and right multiplication by imaginary base octonions, $k$=1..7. Last letter $e_8$=$$\begin{pmatrix} & C \\ -C & \end{pmatrix} $$. In this presentation we can see that first 7 letters generate $C_7$=$M_8\mathbb R$+ $M_8\mathbb R$. Alternatively you can use $\bigl( \begin{smallmatrix} L & \\ & -L \end{smallmatrix} \bigr)$ and $\bigl( \begin{smallmatrix} & -I \\ I & \end{smallmatrix} \bigr)$.

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