It is well known that if $M$ is a smooth $(n-1)$-dimensional surface in $\mathbb R^n$ (e.g. a subspace) then there is a continuous trace operator $W^{s,p}(\mathbb R^n)\to W^{s-1/p,p}(M)$. Now suppose that $M$ is a manifold with boundary, e.g. $M=\{0\}\times\mathbb R^+\times \mathbb R^{n-2}\subset\mathbb R^n$, and in this case call the boundary $\Sigma:=\partial M=\{(0,0)\}\times\mathbb R^{n-2}$.

The question (it's more a poll) is how to define the trace on $\Sigma$ (and I'd expect a function $W^{s,p}(\mathbb R^n)\to W^{s-2/p,p}(\Sigma)$) in such a way that the composed trace $$W^{s,p}(\mathbb R^n)\to W^{s-1/p,p}(M)\to W^{s-2/p,p}(\Sigma)$$ is independent of the (smooth) manifold $M$.

Another (i.e. the more challenging) question is if you can (or prove it impossible to) find an example of two *different* compositions of linear continuous functionals as above (with two $M, M'$ smooth and with boundaries both equal to $\Sigma$, which by locality we suppose here to be orthogonal half-hyperplanes), without the requirement that the first functions $W^{s,p}(\mathbb R^n)\to W^{s-1/p,p}(M\:or\:M')$ be the usual traces, but such that then composed with the traces within $M, M'$ they give the codimension $2$ trace on $\Sigma$ of the first question.

The question is inspired from a small book by Tartar ("An intorduction to Sobolev spaces and interpolation spaces", chapter 40) where (codimension one) trace spaces are identified with interpolation spaces, baiscally by parametrizing the "killed" spatial coordinate in $\mathbb R^n$ with the interpolation parameter. In the light of that construction it would appear that the reason why the first question should have a positive answer is because partial derivatives of linear functions commute: if no example could be done for the second question, it would (to me) indicate strongly that *linearity is a die-hard property*, so to speak.