= 1 {, \dots ,} k$, are commuting idempotents: $B_p^2 =
B_p$ and $B_p B_q = B_q B_p$ for all $p, q = 1 {, \dots ,} k$. Therefore, $\Lapl$ is an idempotent, too: $\Lapl^2 = \Lapl$.

Let $W = (w_{ij})_{i,j=1}^n$ be a $n \times n$-matrix, like in Example \[Ex:Det\] and Proposition \[Pp:DetProp\]. Denote by $\widehat{W}$ the corresponding Laplace matrix, i.e. a matrix with nondiagonal elements $w_{ij}$ ($1 \le i \ne j \le n$) and diagonal elements $-\sum_{j \ne i}
w_{ij}$ ($1 \le i \le n$). It follows from Definition \[Df:Laplace\] that $$\langle \widehat{W} \mid X \rangle = \langle W \mid \Lapl(X) \rangle$$ for any $X \in \Graph_{n,k}$. This equation explains the name “Laplace operator” for $\Lapl$. Note that since $\Lapl(X)$ is a sum of graphs containing no loops, one is free to change diagonal entries of $W$ in the right-hand side; in particular, one can use $\widehat{W}$ instead.

The main results of this paper are the following two theorems:

\[Th:Diag\] $$\label{Eq:LaplDiag}
\Lapl(\det_{n,k}^I) = \frac{(-1)^n}{k!} U(\AC_{n,k}^I).$$

and

\[Th:Codim1\] $$\label{Eq:LaplCodim1}
\Lapl(\det_{n,k}^{i/j}) = \frac{(-1)^n}{k!} U(\AC_{n,k}^{\{i\}}).$$

Applying the functional $\langle \widehat{W}\vert\relax$ to equation with $k = n-s$ and to equation with $k = n-1$ and using Examples \[Ex:Det\] and \[Ex:Forest\] one obtains

\[Cr:Diag\] 