Let $\Omega\subset \mathbb{R}^d$ be a polyhedron. Denote the faces of $\partial \Omega$ as $\{e_i\}_{i=1}^n$ for some $n$. Define $u_i\colon \partial\Omega \to \mathbb{R}$ as
$$u_i(x) = \left\{\begin{array}{cl} 1 & x\in \mathrm{interior}(e_i)\\ 0 &\mathrm{otherwise} \end{array} \right. .$$
My question is, is $u_i\in H^{1/2}(\partial\Omega)$? I am interested in knowing the answer, of course, but more interested in a proof.
No, piecewise constants are not in $H^{1/2}$. Begin with $d=2$ case: then $u$ is a piecewise constant function on the circle. Its Fourier coefficients $\hat u(n)$ decay like $1/n$ and no faster (either by direct calculation, or considering that $u'$ is a combination of Dirac deltas). Hence, $$ \sum_n |n| |\hat u(n)|^2 = \infty $$ A narrow miss; the function is in $H^s$ for every $s<1/2$.
For the higher dimensional case, one can either use a similar approach replacing the $(d-1)$-dimensional boundary with the torus $\mathbb{T}^{d-1}$, or use a Fubini-type reasoning: if $u$ was in a Sobolev class, its restrictions to almost every line segment crossing a face of $e_i$ would be in the same class. So the answer stays negative.