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) = −

∂

∂la ∨

x∈U

e(2 )−1Φ(x,x)f(x

+ a).

(The subscript in la

∨

indicates we are applying this differential operator to

the a variable).

Note that

∂

∂lx∨

e(2

)−1Φ(x,x)

=

−1l(x)eΦ(x,x)/

.

Thus,

x∈U

e(2

)−1Φ(x,x)l(x)f(x

+ a) =

x∈U

∂

∂lx∨

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)

∂

∂la∨

f(x + a)

= −

∂

∂la ∨

x∈U

e(2

)−1Φ(x,x)f(x

+ a)

as desired.