38 2. THEORIES, LAGRANGIANS AND COUNTERTERMS Proof. We will prove this by first verifying the result for P = 0, and then checking that both sides satisfy the same differential equation as a function of P . When P = 0, we have seen that W (0,I) = I which of course is the same as log exp( ∂P ) exp(I/ ). Now let us turn to proving the general case. It is easier to consider the exponentiated version: so we will actually verify that exp ( −1 W (P, I) ) = exp( ∂P ) exp(I/ ). We will do this by verifying that, if ε is a parameter of square zero, and P ∈ Sym2 U, exp ( −1 W ( P + εP , I )) = (1 + ε∂P ) exp ( −1 W (P, I) ) . Let a1,...,ak ∈ U, and let us consider ∂ ∂a1 · · · ∂ ∂ak exp ( −1 W (P, I) ) (0). It will suﬃce to prove a similar differential equation for this expression. It follows immediately from the definition of the weight function wγ(P, I) of a graph that ∂ ∂a1 · · · ∂ ∂ak eW (P,I)/ (0) = γ,φ g(γ) |Aut(γ, φ)| wγ,φ(P, I)(a1,...,ak) where the sum is over all possibly disconnected stable graphs γ with an isomorphism φ : {1,...,k} ∼ T (γ). The automorphism group Aut(γ, φ) consists of those automorphisms preserving the ordering φ of the set of tails of γ. Let ε be a parameter of square zero, and let P ∈ Sym2 U. Let us consider varying P to P + εP . We find that d dε ∂ ∂a1 · · · ∂ ∂ak exp ( −1 W ( P + εP , I )) (0) = γ,e,φ g(γ) |Aut(γ, e, φ)| wγ,e,φ(P, I)(a1,...,ak). Here, the sum is over possibly disconnected stable graphs γ with a distin- guished edge e ∈ E(γ). The weight wγ,e,φ is defined in the same way as wγ,φ except that the distinguished edge e is labelled by P , whereas all other edges are labelled by P . The automorphism group considered must preserve the edge e as well as φ. Given any graph γ with a distinguished edge e, we can cut along this edge to get another graph γ with two more tails. These tails can be ordered

Purchased from American Mathematical Society for the exclusive use of nofirst nolast (email unknown) Copyright 2011 American Mathematical Society. Duplication prohibited. Please report unauthorized use to cust-serv@ams.org. Thank You! Your purchase supports the AMS' mission, programs, and services for the mathematical community.