**3. LO**

The 1/*N* expansion at the LO accuracy amounts to the following substitutions: *<sup>A</sup>*(*p*) = 0, <sup>Π</sup>(*p*) = *a*/|*p*| and <sup>Γ</sup>*<sup>ν</sup>*(*p*, *k*) = *γ<sup>ν</sup>*, where the fermion mass was neglected in the calculation of <sup>Π</sup>(*p*). The LO diagram contributing to the gap Equation (2a), see Figure 1, reads:

$$\Sigma(p) = \frac{8(2+\xi)a}{N} \int \frac{d^3k}{(2\pi)^3} \frac{\Sigma(k)}{(k^2 + \Sigma^2(k))\left[(p-k)^2 + a\left|p-k\right|\right]}.\tag{4}$$

Following [6], we consider the limit of large *a* and linearize Equation (4) which gives

$$
\Sigma(p) = \frac{8(2+\xi)}{N} \int \frac{d^3k}{(2\pi)^3} \frac{\Sigma(k)}{k^2 \left| p - k \right|}. \tag{5}
$$

The mass function can be parameterized as [6]:

$$
\Sigma(k) = B \left(k^2\right)^{-n},\tag{6}
$$

where *B* is arbitrary and the index *α* must be found in a self-consistent way. Using this ansatz, Equation (5) reads:

$$
\Sigma^{(\rm LO)}(p) = \frac{4(2+\xi)B}{N} \frac{(p^2)^{-a}}{(4\pi)^{3/2}} \frac{2\beta}{\pi^{1/2}},\tag{7}
$$

from which the LO gap equation is obtained:

$$1 = \frac{(2 + \xi)\beta}{L} + O(L^{-2}), \quad \text{or} \quad \beta^{-1} = \frac{(2 + \xi)}{L} + O(L^{-2}), \tag{8}$$

where

$$\beta = \frac{1}{\alpha \left(1/2 - \alpha\right)} \text{ and } L \equiv \pi^2 N. \tag{9}$$

Note that the two equations in (8) are completely equal to each other. Solving them, we obtain:

$$n\_{\pm} = \frac{1}{4} \left( 1 \pm \sqrt{1 - \frac{16\left(2 + \xi\right)}{L}} \right),\tag{10}$$

which reproduces the solution given by Appelquist et al. [6]. The gauge-dependent critical number of fermions: *Nc* ≡ *Nc*(*ξ*) = 16(2 + *ξ*)/*π*2, such that <sup>Σ</sup>(*p*) = 0 for *N* > *Nc* and Σ(0) exp−2*π*/(*Nc*/*<sup>N</sup>* − <sup>1</sup>)1/2, for *N* < *Nc*. Thus, D*χ*SB arises when *α* becomes complex, that is, for *N* < *Nc*.

**Figure 1.** LO diagram to dynamically generated mass <sup>Σ</sup>(*p*). The crossed line indicates a mass insert.

The gauge-dependent fermion wave function can be computed in a similar way. At LO, Equation (2b) simplifies as:

$$A(p)p^2 = -\frac{2a}{N} \text{Tr} \int \frac{d^D k}{(2\pi)^D} \frac{P\_{\mu\nu}^{\mathbb{Z}}(p-k)\not p\gamma^\mu \hat{k}\gamma^\nu}{k^2|p-k|}\,\tag{11}$$

*Particles* **2020**, *3*

where the integral is dimensionally regularized with *D* = 3 − 2*ε*. Taking the trace and calculating the integral on the r.h.s. outputs:

$$A(p) = \frac{\overline{\mu}^{2\varepsilon}}{p^{2\varepsilon}} \mathbb{C}\_1(\xi) + \mathcal{O}(\varepsilon) \,, \ \mathbb{C}\_1(\xi) = + \frac{2}{3\pi^2 N} \left( (2 - 3\xi) \left[ \frac{1}{\varepsilon} - 2\ln 2 \right] + \frac{14}{3} - 6\xi \right) \,, \tag{12}$$

where the *MS* parameter *μ* has the standard form *μ*2 = 4*πe*<sup>−</sup>*γ<sup>E</sup> μ*2 using the Euler constant *γE*. Note that in the *ξ* = 2/3-gauge, the value of *<sup>A</sup>*(*p*) is finite and *<sup>C</sup>*1(*ξ* = 2/3)=+4/(<sup>9</sup>*π*2*N*). From Equation (12), the LO wave-function renormalization constant can be extracted: *λA* = *μ*(*d*/*dμ*)*A*(*p*) = 4(2 − <sup>3</sup>*ξ*)/(<sup>3</sup>*π*2*N*), that matches the result of of [46,47].
