# Betti number of Feynman graph

Let $\mathcal{L}$ be a Lagrangian, which contains polynomials of bosonic fields $\phi$. After Wick's rotation we obtain a perturbation expansion od Green's function. In this expansion there are terms of the form

$$\frac{(-1)^n}{n!}\int\hat\phi(p_1)...\hat\phi(p_N)\prod\limits_{j\in A}\xi_jd\mu,$$

where $\xi_j$ are monomials of the form $\frac{-z}{d!}\phi_E(x)^d$ ($A$ is just a set of indices). These monomials come from decomposition of interaction part of $\mathcal{L}$ onto monomials.

It it well-known that we obtain a Feynman graph from pairings in the integration by parts

$$\int\hat\phi(p_1)...\hat\phi(p_N)\prod\limits_{j\in A}\prod\limits_{i=1}^{d_j}\hat\phi\left(k_i(j)\right) d\mu,$$

where $k_i(j)$ are momentum variables carried by half-line associated to $\xi_j$.

In the Green's (or rather Schwinger - because we done Wick's rotation) function there is so-called unrenormalized value of the graph $V(\Gamma)$- multiple integral obtained from the product of terms given by Feynman rules.

My question is: How to prove that for a connected Feynman graph the number of free integration variables in $V(\Gamma)$ is equal to first Betti number $b_1$ of the geometric realization of $\Gamma$ ? ($b_1=\#\Gamma^{}_{\mathrm{int}}-\#\Gamma^{}+1$, where $\Gamma^{}$ is a set of vertices, $\Gamma^{}_{\mathrm{int}}$ is a set of internal edges. )

1. In the momentum representation, we should integrate over all the internal momentum variables. The number of internal momentum variables is the number $E$ of internal edges in the Feynman diagram.
2. Each vertex obeys momentum conservation. (Note that the external momentum variables in a connected component are assumed to already satisfy momentum conservation.) Hence, the number of constraints is the number $V$ of vertices minus the number $C$ of connected components.
3. Therefore the number of independent momentum integrations is $b_1=V-E+C$. This number is the first Betti number in graph theory, a.k.a. as the cyclomatic number or circuit rank.