Laplace operator is always closable but it is in general, when the vertex degree is not bounded, not essentially self-adjoint. That is the reason why we will only consider [*rrsc*]{}s of bounded degree. Before we go on, we will summarize some results about the question, when the Laplace operator is essentially self-adjoint. In [@wojciechowski] Wojciechowski proves that the $0$th Laplace operator on a locally-finite graph is essentially self-adjoint. In [@bordenave Proposition 2.2] it is shown that the $0$th Laplace operator for unimodular measures on the space of rooted locally finite graphs $\mathcal{G}_*$ is essentially self-adjoint. In [@anne] Anné and Torki-Hamza defined a property called $\chi$-completness for graphs, which implies that the $1$st-Laplace operator is essentially self-adjoint. This property was extended by Chebbi [@chebbi] to $2$-dimensional simplicial complexes, where it also implies essentially self-adjointness of the $1$st and $2$nd Laplace operator. Chebbi likewise gives an example of a $2$-dimensional simplicial complex with a non self-adjoint Laplace operator. In [@linial-peled] Linial and Peled studied the spectral measures of random simplicial complexes $Y(n,\frac{c}{n})$ and showed that they weakly converge to the spectral measure of a Poisson d-tree, which has a self-adjoint Laplace operator.

Even though the most things hold true in the unbounded 