Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 306 / 1 / iv

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- Solution iv

Solution

For the Möbius transformation

z^{'} = (a z + b) / (cz + d)

with

a d - b c = 1

,

z_{j}^{'} - z_{k}^{'} = \frac{z _{j} - z _{k}}{( c z _{j} + d ) ( c z _{k} + d )}, d^{2} z_{i}^{'} = \frac{d ^{2} z _{i}}{∣ c z _{i} + d ∣ ^{4}} .

The power of

∣ c z_{i} + d ∣

contributed by every pair containing

i

is

- α^{'} \sum_{j \neq = i} p_{i} \cdot p_{j} = α^{'} p_{i}^{2} = 4,

where momentum conservation and the mass-shell condition

p_{i}^{2} = 4/ α^{'}

were used. The Koba-Nielsen factor therefore contributes

∣ c z_{i} + d ∣^{4}

at each insertion, exactly cancelling the transformed measure. Hence the remaining integral is

S L (2, C)

invariant, and division by its volume removes the residual conformal-gauge redundancy.

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