 Bosonic trace vertices from a finite fermionic determinant

Integrating finite Grassmann variables out of a quadratic action gives $\det [B-gC(x)]$. If $B$ is invertible, divide by $\det B$ and expand its logarithm to obtain $V_{\mathrm{eff}}={\sum }_{m\ge 1}{g}^m\mathrm{Tr}[B^{-1}C(x){]}^m/m$. Since the bosonic variables commute, an $m$-leg effective Feynman rule uses the fully symmetrized trace tensor, with weight $-g^m(m-1)!\mathrm{Sym}\mathrm{Tr}[B^{-1}{C}^{(k_1)}\cdots B^{-1}{C}^{(k_m)}]$. The minus sign comes from $e^{-V_{\mathrm{eff}}}$. This organizes closed fermion loops into bosonic vertices. It is a formal perturbative identity; a singular $B$ prevents the normalized expansion, even though the unfactored determinant polynomial still exists.