= 1 {, \dots ,} n$ and any $m = 1 {, \dots ,} k$ one has $$\label{Eq:Deriv}
    \frac{\partial^m}{\partial w_{ii}^m} \langle W \mid
    \det_{n,k}^{\varnothing}\rangle = \langle W \mid
    \det_{n,k-m}^{\varnothing}+ \det^{\{i\}}_{n,k-m}\rangle.$$

See [@Epstein Lemma 86] for a formula similar to (with $m=1$ and a finite difference instead of a derivative).

Main results {#SSec:Main}
------------

Let $G \in \Gamma_{n,k}$, $p \in \{1 {, \dots ,} k\}$ and $i,j \in \{1 {, \dots ,} n\}$. Denote by $R_{ab;p} G \in \Gamma_{n,k}$ the graph obtained from $G$ by replacement of its $p$-th edge by the edge $[ab]$ bearing the same number $p$.

Consider now a linear operator $B_p: \Graph_{n,k} \to \Graph_{n,k}$ acting on every basic element $G \in \Gamma_{n,k}$ as follows: $$B_p(G) =  \begin{cases}
G, &\text{if the $p$-th edge of $G$ is not a loop},\\
-\sum_{b \ne a} R_{ab;p} G, &\text{if the $p$-th edge of $G$ is the loop
$[aa]$}.
 \end{cases}$$ In particular, $B_p = 0$ if $n=1$ (and $k > 0$).

\[Df:Laplace\] The product $\Lapl {\stackrel{\mbox{\scriptsize def}}{=}}B_1 \dots B_k: \Graph_{n,k} \to \Graph_{n,k}$ is called [*Laplace operator*]{}.

If $n = 1$ and $k > 0$ then $\Delta = 0$; also take $\Delta = \name{id}$ by definition if $k = 0$.

The operators $B_p$, $p 