We know that for a simple graph (i.e. a graph with no self-loops and multiedges), the Graph Laplacian uniquely characterizes it in the sense that if two graphs have the same Graph Laplacian, then the graphs are the same. I was wondering whether a similar property holds for Hodge Laplacians.
By this I mean, if two simplicial complexes have the same sequence of Hodge Laplacians (also known as p-Laplacian if it calculated for an element belonging to chain group of p-simplices), then they must be the same simplicial complex (upto reordering of the vertices or rearrangement of the orientations of the subsimplices in the simplical complex). If the above statement is false, can a counterexample be constructed? While providing counterexamples, it will be appreciated, but not necessary, if the intuition used for constructing it is described.
For relevant definitions, I would request all to refer to http://pi.math.cornell.edu/~goldberg/Papers/CombinatorialLaplacians.pdf.
I am very new to this field, so I apologize if the answer to this question is very obvious.
It would be very helpful if a diagram is also provided while referring to any particular topological space (such as a torus) or simplicial complex.
EDIT: Following Connor Malin's suggestion, I am stating the definitions used.
For each boundary operator $\partial_{d}: C_{d} \rightarrow C_{d-1}$ of $K,$ we let $\mathcal{B}_{d}$ be the matrix representation of this operator relative to the standard bases for $C_{d}$ and $C_{d-1}$ with some orderings given to them. Let $\partial_{d}^{*}$ be the coboundary operator, which for our purpose is the linear adjoint of the boundary operator of $\partial_{d}$. So matrix representation of $\partial_{d}^{*}$ relative to the standard bases for $C_{d}$ and $C_{d-1}$ using the same orderings as for $\partial_{d}$ is $\mathcal{B}_{d}^{T}$.
Let $K$ be a finite oriented simplicial complex, and let $d \geq 0$ be an integer. The $d$ th combinatorial Laplacian is the linear operator $\Delta_{d}: C_{d} \rightarrow C_{d}$ given by $$ \Delta_{d}=\partial_{d+1} \circ \partial_{d+1}^{*}+\partial_{d}^{*} \circ \partial_{d} $$ For convenience, we use the notations $\Delta_{d}^{U P}=\partial_{d+1} \circ \partial_{d+1}^{*}$ and $\Delta_{d}^{D N}=\partial_{d}^{*} \circ \partial_{d},$ so that $\Delta_{d}=\Delta_{d}^{U P}+\Delta_{d}^{D N}$
The $d$ th Laplacian matrix (Hodge Laplacian) of $K$, denoted $\mathcal{L}_{d}$, relative to some orderings of the standard bases for $C_{d}$ and $C_{d-1}$ of $K,$ is the matrix representation of $\Delta_{d}$. Observe that $$ \mathcal{L}_{d}=\mathcal{B}_{d+1} \mathcal{B}_{d+1}^{T}+\mathcal{B}_{d}^{T} \mathcal{B}_{d} $$ As above, for convenience, we use the notations $\mathcal{L}_{d}^{U P}=\mathcal{B}_{d+1} \mathcal{B}_{d+1}^{T}$ and $\mathcal{L}_{d}^{D N}=\mathcal{B}_{d}^{T} \mathcal{B}_{d},$ so that $\mathcal{L}_{d}=\mathcal{L}_{d}^{U P}+\mathcal{L}_{d}^{D N}$
Note: $d = 0$ refers to the (unnormalized) graph Laplacian.
