# Delta function constraint as Lagrange multiplier in SYK model calculation?

In eq. (112) of these lecture notes the author is introducing a 1 into an integral in the following way

This looks like an integral representation of the delta function $$\delta(x)\sim \int dy\, e^{i x y}$$, but I am confused by the fact that the factor in the exponential function seems to be not $$i$$ but instead $$-\frac{1}{2}$$. The author comments something about $$\Sigma$$ being a Lagrange multiplier enforcing the delta function constraint, which I have never seen being done like that before.

Maybe someone here can explain in a few more details and a bit more rigor how the transformation of delta functions to Lagrange multipliers works?

• Similar question on Math.SE: math.stackexchange.com/q/2829562/11127 – Qmechanic Jan 26 at 17:48
• In the path integral of $\Sigma$ field, one needs to assume a proper contour, e.g. going from -i $\infty$ to +i$\infty$. One puzzling thing is that the physical saddles of $(G,\Sigma)$ are real, but there is no contradiction. – Yingfei Gu Mar 9 at 20:32