3

I had the need to calculate the adjacency matrix $L$ of the line graph of a certain planar $k$-regular graphs $G(n,e)$ ( $n$ vertices and $e=\frac k2 n$ edges) given its adjacency matrix $A_G$. Here I went:

Let $N_k$ an "outer-diagonal" indexing matrix, e.g. $N_4=\pmatrix{0&1&2&3\\1&0&4&5\\2&4&0&6\\3&5&6&0}$.

Calculate the Hadamard Dot product $B=N_k\odot A$. I found that $$L_{B(a,b),B(c,d)}=(1-\delta_{ac}\delta_{bd})(\delta_{ac}+\delta_{bd})A(a,b)A(c,d) \tag{$*$}$$ works fine for the examples I checked. We get a matrix in $\mathbb Z_2^{n(n-1) \times n(n-1)}$, but since the line graph operator acts effectively as $L(A): \mathbb Z_2^{n\times n}\mapsto \mathbb Z_2^{e\times e} $, you can remove plenty of rows/columns.

Is $(\text{$*$})$ correct and is there also straight representation of the line graph operator in terms of matrix mapping $A$ to $L(A)$? Maybe superoperators can help...

EDIT: The $A(a,b)A(c,d)$ product points towards something like $A\otimes A$ maybe suited with an according $\Delta$ matrix that is Hadamard-$\odot $-multiplied: $$ L\overset{?}{=}\Delta\circ (A\otimes A) $$ A construction without Hadamard-$\odot $-product would also be nice$\dots$

draks ...
  • 18,449
  • 1
    I don't know if that helps bu you can express the adjacency matrix for the line graph by using the incidence matrix. If $E$ is the incidence matrix of your graph $G$ then $E^T E - 2I = A(L(G))$ – Jernej Jan 28 '16 at 14:06
  • @Jernej yes I knew that, but I didn't follow that road to the end, since it looked as stony as the one took finally. If you could show that it's in fact easier, this would help... – draks ... Jan 28 '16 at 15:27
  • I am sorry I don't understand. You don't know how to prove this identity? – Jernej Jan 28 '16 at 21:15
  • @Jernej sorry. I meant, I thought about using the "incidence matrix" identity (I don't need a proof), but I managed my goal without it. If you could come up with a construction using it, it would be interesting to compare yours with mine... – draks ... Jan 29 '16 at 09:00
  • 1
    Could you clarify why you need the "outer diagonal indexing matrix" here? I mean, couldn't you just work with $L_{a,b,c,d}$ as a $n^2 \times n^2$ matrix (and then delete rows and columns which become zero)? – BharatRam Jun 19 '17 at 08:52
  • @BharatRam our chat room is frozen. The reference I mentioned is M. Kempton, Non-Backtracking Random Walks and a Weighted Ihara’s Theorem. On page 10 he mentions a representation of the DiLineGraph operator, I think. Followed by a crisp proof of Ihara's Theorem... – draks ... Jun 30 '17 at 07:37
  • @BharatRam I had another but similar approach here... – draks ... Jun 30 '17 at 12:51
  • @draks Will have a look. Also, I have posted another question you might be interested in: https://math.stackexchange.com/questions/2345021/imbalance-in-a-signed-graph Do tell me if you know anything about it. Its about non-backtracking and zeta in signed graphs. – BharatRam Jul 03 '17 at 14:16
  • @BharatRam I once had something similar in mind: For planar cubic graphs (this implies 3-edge-colorability) you can assign orientations $\mathfrak O ={o_k}$ to the vertices $v_k$ corresponding to the edges oriented around a specific the vertex. I called them left or right oriented. While walking without back-tracking you can now assign a $\pm 1$ towards your current step, either if you go left or right (since "go back" is not an option) along the local orientation $o_k$... – draks ... Jul 04 '17 at 10:23
  • There was interesting link to Riemann surfaces connected to these graphs (see here)... – draks ... Jul 04 '17 at 10:23

0 Answers0