Rescale

W \mapsto t W

with

t > 0

. After Berezin integration, the unnormalized holomorphic function insertion is

I_{t} [f] = \int_{C} d^{2} z t^{2} ∣ P^{'} ∣^{2} f (z) e^{- t^{2} ∣ P ∣^{2}} .

\frac{d I _{t} [ f ]}{d t} = 2 t \int_{C} d^{2} z f P^{'} \overline{P^{'}} (1 - t^{2} ∣ P ∣^{2}) e^{- t^{2} ∣ P ∣^{2}} = 2 t \int_{C} d^{2} z \partial_{\overset{z}{ˉ}} (f P^{'} \overline{P} e^{- t^{2} ∣ P ∣^{2}}) = 0.

The last equality again follows from polynomial growth and exponential decay. Thus the insertion is independent of

t

t \to \infty

at the zeros of

P

z_{a}

,

P (z) = P^{'} (z_{a}) (z - z_{a}) + O ((z - z_{a})^{2})

. The local Gaussian integral cancels the factor

∣ P^{'} (z_{a}) ∣^{2}

and contributes

π f (z_{a})

m_{a}

, write

P (z) = a_{a} (z - z_{a})^{m_{a}} + \dots

. Its leading radial integral is

2 π t^{2} m_{a}^{2} ∣ a_{a} ∣^{2} \int_{0}^{\infty} ρ^{2 m_{a} - 1} e^{- t^{2} ∣ a_{a} ∣^{2} ρ^{2 m_{a}}} d ρ = π m_{a} .

The insertion tends to

f (z_{a})

there; contributions away from the zeros vanish. Therefore the full answer, including degenerate critical points, is

⟨ f (z)⟩ = π a : P (z_{a}) = 0 \sum m_{a} f (z_{a}) .

This is an unnormalized expectation. For simple zeros all

m_{a} = 1

Z = π r

would give the normalized average, but is not part of the requested insertion. Taking

f = 1

recovers the partition function, and taking

f = P g

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 Incoming links: Supersymmetric localization