Let $Spin(9,1)$ denote the universal (double) cover of $SO(9,1)$. $Spin(9,1)$ acts linearly on $\mathbb{R}^{16}$ (see e.g. p.29 here <https://arxiv.org/pdf/math/0105155v4.pdf> ).


Consider the induced action of $Spin(9,1)$ in the space of symmetric quadratic forms on $\mathbb{R}^{16}$, i.e. in $Sym^2((\mathbb{R}^{16})^\*)$.


**How to decompose (the complexification of) this space into irreducible components? In particular, how many irreducible components there exist?**


Remark. In the answer to this post [Decomposition into irreducible components of a representation of $Spin(9)$](https://mathoverflow.net/questions/243886/decomposition-into-irreducible-components-of-a-representation-of-spin9) it was stated that the decomposition of the above space under the action of the smaller group $Spin(9)\subset Spin(9,1)$ contains exactly 3 irreducible components which are pairwise non-isomorphic.



A representation of Spin(9,1)
In this case if your 16-dimensional $\mathrm{Spin}(9,1)$-representation is the one of highest weight $(0,0,0,0,1)$, then the $\mathrm{Spin}(9,1)$-irreducible decomposition of its symmetric square is just into two pieces: The 10-dimensional piece $\mathbb{R}^{9,1}$ isomorphic to the standard (Lorentzian) representation of $\mathrm{SO}(9,1)$ (with highest weight $(1,0,0,0,0)$) and the 126-dimensional representation of highest weight $(0,0,0,0,2)$. This latter representation can be shown to be isomorphic to the space of *self-dual* $5$-forms on $\mathbb{R}^{9,1}$, i.e., $\Lambda^5\_+(\mathbb{R}^{9,1})\subset \Lambda^5(\mathbb{R}^{9,1})$.


