Determining the Laplacian for a signed graph I was reading the paper: On the Laplacian Eigenvalues of Signed Graphs, when I had this question. When we determine the Laplacian of a signed graph, why do we take the unsigned degree matrix D, but the signed adjacency matrix A? The question is more towards the unsigned degree matrix. Is there some intuition I could follow? Thanks!

One way that the standard graph Laplacian $L$ arises is by assigning each edge of $G$ an orientation, and considering the incidence matrix $M=(M_{ie})$  where $M_{ie} =1$ if node $i$ is the head of $e$ and $M_{ie} =-1$ if node $i$ is the tail  of $e$; otherwise, let $M_{ie}=0$. Then $L=I-A_G=MM^T$. This is useful, e.g., in proving the matrix tree theorem via the Binet-Cauchy expansion of determinants.
One way to obtain a signed Laplacian is to start with a signed incidence matrix $S=(S_{ie})$  where $S_{ie} \in\{\pm 1\}$ if node $i$ is on edge $e$ and $0$ otherwise. Then define
$\widetilde{L}=I-SS^T$. Note that the diagonal elements of $\widetilde{L}$
are the unsigned degrees.
