My question concerns writing the $Z(SO(n))=\mathbb{Z}/2$ center subgroup inside the spinor representation of $Spin(n)$ which has $Z(Spin(n))=\mathbb{Z}/4$ for $n = 2 \mod 4$ and $n>2$. Let us take $n=10$ as an example.
The center subgroup $\mathbb{Z}/2$of the $SO(10)$ written as the rank-10 matrix under the 10-dimensional vector representation of $SO(10)$ is given by a negative diagonal identity matrix: $$ g=-\text{diagonal}(1,1,1,1,1,1,1,1,1,1) $$ for $$ Z(SO(10))=\mathbb{Z}/2. $$
I am curious how this $Z(SO(10))=\mathbb{Z}/2$ is written as in the spinor representation of $Spin(10)$. Note that the spinor representation of $Spin(10)$ is a 32-dimensional irreducible representation, or a 16-dimensional irreducible representation. This means this element $g$ may be expressed as a rank-32 or rank-16 matrix.
The $Spin(10)$ has a center $Z(Spin(n))=\mathbb{Z}/4$ for $n = 2 \mod 4$ and $n>2$.
My speculation: We cannot express this center $Z(SO(10))=\mathbb{Z}/2$ inside the spinor representation of $Spin(10)$ into a rank-32 or rank-16 matrix. Can we prove or disprove my statement?