# Number of spanning trees of a quotient graph

Let $G$ be a finite connected graph on a $2m$-element vertex set $V$. For any graph with vertices $u,v$, let $\mu(u,v)$ denote the number of edges between $u$ and $v$. Suppose that $G$ has an automorphism $f$ that is a fixed-point free involution on the vertices. We can define a quotient graph $G/f$ by letting the vertices of $G/f$ be the orbits $[u]=\left\lbrace u,f(u)\right\rbrace$ of $f$, and setting $$\mu([u],[v]) = \mu(u,v) + \mu(f(u),v).$$ Let $\kappa(H)$ denote the number of spanning trees of a graph $H$. By the Matrix-Tree Theorem and some simple linear algebra, one can show that $2\kappa(G)$ is divisible by $\kappa(G/f)$. My question is whether the factor of 2 is necessary, i.e., is it always true that $\kappa(G)$ is divisible by $\kappa(G/f)$? Similar questions can be asked for more complicated automorphism groups of $G$.

• I think this works in the general case too if we define a quotient according to $\mu([u],[v])=\mu(u,v)+\mu(u,f(v))+\cdots+\mu(u,f^{n-1}(v))$, where $f$ is an automorphism of degree $n$, but I havent checked the details. I wonder if there are any other graph related quantities that behave this way under taking quotients. – Gjergji Zaimi Jun 4 '10 at 14:40

Let us group the vertices as $U=\{u_1,u_2,\dots,u_n\}$ and $V=\{v_1,v_2,\dots, v_n\}$ where $f(u_i)=v_i$. Let $L_0$ be the laplacian of the graph with vertex set $U$ and edges as restricted from $G$, let $L _1 = \operatorname{diag}\left( \sum _{j=1}^n \mu(u _i, v _j)\right)$ and $L=L _0+L _1$, also let $A$ be the symmetric matrix whose $a _{ij}$ is $-\mu(u _i,v _j)$. Clearly the Laplacian of $G$ is $M=\left( \begin {array} {cc} L & A \\\ A & L \end {array} \right)$. Let $M^*$ stand for the matrix $M$ with deleted first row and column. We have $$\kappa(G)=\det \left( \begin {array} {cc} L & A \\\ A & L \end {array} \right) ^ *=\det \left( \begin {array} {cc} B & C \\\ D & (L+A)^ * \end {array} \right)$$ for some block matrices $B,C,D$ of size $n\times n,n\times (n-1),$ and $(n-1)\times n$ where this second matrix was obtained by adding the $i$th row of $M^{\*}$ to it's $n+i$th row for $1\le i\le n-1$ and then adding the first (or last) $n-1$ columns to the $n$th column. So $D$ is the matrix $(L+A)*$ together with a last column of zeros, making $(L+A)^{\*-1}D$ with integer entries. Next we factor it using one of these identities $$\kappa(G)=\det(L+A)^ * \det(B-C(L+A)^{*-1}D)$$ and observe that $L+A$ is the Laplacian of $G/f$ so $\det(L+A)^ *=\kappa(G/f)$, and since the second factor is an integer we get the desired divisibility.

• It seems to me that the statement "Clearly the Laplacian of G is M = [[L,A],[A,L]]" is false. Can the proof be fixed? – Richard Stanley May 29 '10 at 21:33
• Fortunately it was just a typo, I think it's fixed now. – Gjergji Zaimi May 31 '10 at 5:55
• The proof looks o.k. now. Getting the last column of zeros was a key step. Thanks for your answer! – Richard Stanley Jun 21 '10 at 1:12