Characterization of Simplicial Complex using p-Laplacian/Hodge Laplacian 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.
 A: A counter example is given by the torus and the wedge of a sphere and two circles. The attaching maps from the 1-simplices to the 0-simplices are the same in both cases, so the 0th Hodge Laplacian is the same.
The 1st Hodge Laplacian is the same because the attaching map of the 1-simplices to the 0-simplices are the same, again, plus the boundary map from the 2-chains to the 1-chains is 0 in both cases, so the only contribution comes from the boundary map between the 1 and 0 simplices.
The 3rd Hodge Laplacian is the same in both cases because there are no higher simplices and the boundary map from the 2-chains to the 1-chains in both cases are 0.
Maybe an intuitive reason why the 1-dimensional case is different is that 1-dimensional spaces have no interesting cup products, but higher dimensions do. When we use the chain complex to define an invariant, we most likely have no interaction with the cup product which comes from the cochains. Good to note is that these spaces I gave have the same homology and cohomology, but are distinguished by their cup product.
