I am new to spin geometry and I am trying to understand spinor bundles in dimension 6 for something else I am reading. $\mathcal{Cl}(6,0)$ has a unique irreducible real 8-dimensional representation $S$. Given any $s \in S$ non zero, I think that a basis of $S$ can be given by $\{e_i \cdot s\}_{i=1}^6 \cup \{s, \omega \cdot s\}$, where $\{e_i\}$ is an orthonormal basis of $\mathbb{R}^6$ and $\omega$ is the Clifford volume form. I will sketch why.
Clearly, for dimension arguments, it suffices to show that the given elements are linearly independent. Now, if one writes $$ \sum_{i=1}^6 \lambda_ie_i \cdot s + \lambda_7 s + \lambda_8 \omega \cdot s = 0 $$ taking the inner product on both sides with $s, \omega \cdot s$ and $e_i \cdot s$, and using that both $\omega$ and $e_i$ act skew-symmetrically, one has the thesis.
Now I believe that if a two form $\theta$ that is given by $\iota_v \Omega$, where $v$ is a vector $\iota$ denotes the interior product and $\Omega$ is a negative stable three form (i.e. is equivalent to the real part of the standard complex volume form on $\mathbb{C}^3$), and moreover $s$ is fixed by the stabilizer of $\Omega$, then there is a one form $\zeta$ such that $$ \theta \cdot s = \zeta \cdot s $$ that is to say, the $s$ and $\omega \cdot s$ components of $\theta \cdot s$ are $0$. It is easy to see that the $s$ component is $0$ by taking the inner product with $s$, but I can't manage to show that the $\omega \cdot s$ component vanishes: taking the inner product with $\omega \cdot s$ dos not make me conclude despite seeming the natural thing to do.