# Reality properties of auxiliary fields after Wick rotation

I was reading the treatment of the large $N$ limit of the Non-Linear Sigma Model (NLSM) in Peskin & Schroeder, Sec. 13.3, and I noticed something strange in the evaluation of the path-integral by steepest descent.

P&S are evaluating the NLSM euclidean partition function: $$Z_E=\intop \mathcal D n \,\delta [n^2 -1]\exp [-N\intop \text d ^d x \frac{1}{2\tau_0}(\partial _\mu n)^2],\tag{13.112}$$ where $n$ is an $N$-component vector subject to the constraint $n^2=1$, represented here by a functional $\delta$, and $\tau_0=g_0^2/N$ is the bare 't Hooft coupling, which is held fixed in the $N\to \infty$ limit.

Next, they get rid of $\delta [n^2-1]$ by integrating-in an auxiliary field $\alpha$: $$Z_E=\intop \mathcal D n \mathcal D \alpha \,\exp [-N\intop \text d ^d x \lbrace \frac{1}{2\tau_0}(\partial _\mu n)^2-\frac{i}{2\tau_0}\alpha(n^2-1)\rbrace ].\tag{13.113}$$

The "$i$" in front of $\alpha(n^2-1)$ correctly reproduces a $\delta$-function, as also Ron Maimon remarked here.

However, as they go on in doing the large $N$ limit by the steepest descent method (I omit for short the passages, see Eqs. 13.114 through 13.115 in P&S), they arrive to the following equation for the saddle point in the $\alpha$ integration: $$\langle x\vert \frac{1}{-\partial ^2 +i\alpha}\vert x \rangle = \frac{1}{\tau_0}.\tag{13.115}$$ They argue that such equation requires $\alpha = -im^2$, where $m^2$ is a positive real number given by: $$\frac{1}{2\pi} \ln \frac{\Lambda}{m} = \frac{1}{\tau_0}.\tag{13.118}$$ ($\Lambda$ is a cut-off).

Now I am confused. The original $\alpha$ was supposedly a real scalar field, but the steepest descent method requires it to be pure imaginary.

So my question: does this mean that

1. The auxiliary field $\alpha$ was really imaginary from the beginning, or
2. The auxiliary field is real, but the expansion is done around a non-physical value of $\alpha$.
3. Neither of these.

Either possibilities (1) or (2) sound equally phony to me, so any suggestion is appreciated.

Option 2 is correct. The Lagrange multiplier $\alpha(x)$ is originally integrated along the real axis $\mathbb{R}$, but the method of steepest descent allows for deformations of the $\alpha$-integration contour into the complex $\alpha$-plane.