40 2. THEORIES, LAGRANGIANS AND COUNTERTERMS If U is a complex vector space and Φ is a non-degenerate complex linear inner product on U, then the same formula holds, where we integrate over any real slice UR of U. Proof. It suffices to show that, for all functions f ∈ O(U ), x∈U e(2 )−1Φ(x,x) f(x + a) = e ∂P f (where both sides are regarded as functions of a ∈ U). The result is clear when f = 1. Let l ∈ U ∨. Note that e ∂ P (lf) − le ∂ P (lf) = [∂P , l]e ∂ P (lf). In this expression, [∂P , l] is viewed as an order 1 differential operator on O(U ). The quadratic form Φ on U provides an isomorphism U → U ∨ . If u ∈ U, let u∨ ∈ U ∨ be the corresponding element and, dually, if l ∈ U ∨ , let l∨ ∈ U be the corresponding element. Note that [∂P , l] = − ∂ ∂l∨ . It suffices to verify a similar formula for the integral. Thus, we need to check that x∈U e(2 )−1Φ(x,x) l(x)f(x + a) = − ∂ ∂l∨ a x∈U e(2 )−1Φ(x,x) f(x + a). (The subscript in l∨ a indicates we are applying this differential operator to the a variable). Note that ∂ ∂l∨ x e(2 )−1Φ(x,x) = −1 l(x)eΦ(x,x)/ . Thus, x∈U e(2 )−1Φ(x,x) l(x)f(x + a) = x∈U ∂ ∂l∨ x e(2 )−1Φ(x,x) f(x + a) = − x∈U e(2 )−1Φ(x,x) ∂ ∂lx∨ f(x + a) = − x∈U e(2 )−1Φ(x,x) l(x) ∂ ∂l∨ a f(x + a) = − ∂ ∂l∨ a x∈U e(2 )−1Φ(x,x) f(x + a) as desired.
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