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A simple Lie group $𝑆𝑝𝑖𝑛(10)$ has a spinor representations of 16 dimensions, which is distinct from the vector representation of 10 dimensions (coming from standard vector representation of SO(10)).

My question concerns what are the matrix matrix representations of the 45 Lie algebra generators for the rank-16 spinor representations?

  1. The vector representation of $Spin(10)$ is also the vector representation of $SO(10)$. We know the 45 of rank-10 matrices can be obtained by taking any two basis vectors $v_i$ and $v_j$ of 10-dimensional vector space, and assign $\pm 1$ along the off diagonal component with this matrix: $$\begin{bmatrix} 0&-1\\1&0\end{bmatrix}$$ namely along the wedge product of the two basis vectors: $$v_i \wedge v_j = v_i \otimes v_j - v_j \otimes v_i.$$ Include this matrix into the rank-8 matrix, we get $\frac{10 \cdot 9}{2}=45$ such set of matrix $$\begin{bmatrix} 0& \cdots & 0 & 0 \\ 0& \cdots &-1 &\vdots\\ \vdots & 1& \ddots & \vdots \\ 0& \cdots & \cdots & 0 \end{bmatrix}$$ namely along the wedge product of the two basis vectors: $$v_i \wedge v_j = v_i \otimes v_j - v_j \otimes v_i.$$ for any $i,j\in \{1,2,\dots,8,9,10\}$ with $i \neq j$.

Above we derive the explicit Lie algebra matrix representations of $Spin(10)$ also $SO(10)$: vector representation, with $\frac{10 \cdot 9}{2}=45$ Lie algebra generators of rank-10.

  1. Question --- How do we derive the explicit Lie algebra matrix representations of $Spin(10)$ with 45 Lie algebra generators of rank-16?
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    $\begingroup$ You've asked a lot of questions about finding explicit matrix representations for various Lie algebra representations. It might be a good idea instead to understand how to construct these in general. As an example the Wikipedia page on spin representations demonstrates a few different ways to construct the spin representations. Converting these to a matrix representation is just choosing (nice) bases on the vector space and the Lie algebra. $\endgroup$
    – Callum
    Aug 10, 2021 at 15:11

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