ON INTEGRATION IN A SPACE WITH COMPLEX DIMENSION

Pavel Etingof

These are comments on dimensional regularization of Feynman integrals, which was discussed in the lectures of K.Gawedzki and D.Gross. This material can be found in physical literature, but in a slightly different form.

We will first define a certain remarkable operator (the D-dimensional integral), which acts on functions of nonnegative symmetric bilinear forms, and then show how it applies to dimensional regularization.

Acknowledgement: I am grateful to Pierre Deligne, who helped me improve this text.

1. The D-dimensional integral.

Let E be a real vector space of dimension N with a volume element dv_{E}. Denote by S^{2}E^{∗} the space of symmetric bilinear forms on E, and by S_{+}^{2}E^{∗} ⊂ S^{2}E^{∗} the set of nonnegative symmetric bilinear forms. We think of elements of S^{2}E^{∗} as of selfadjoint operators E → E^{∗}. For any such operator A, we can canonically define its determinant det(A) ∈ ℝ.

Let dA be a volume element on S^{2}E^{∗} defined by dv_{E}. That is: identify E^{∗} with E by any positive operator X : E → E^{∗} with deteminant 1, regard A ∈ S^{2}E^{∗} as a selfadjoint operator on E using this identification, and define dA to be the volume element on S^{2}E^{∗} corresponding to the norm ||A||^{2} = Tr(A^{2}). (It is clear that this definition does not depend on X).

Fix a complex number D with ReD > N − 1.

Lemma 1. For any B ∈ S^{2}E such that B > 0

∫_{S_{+}^{2}E^{∗}}(det A)^{(D−N−1)/2}e^{−tr(AB)}dA = Γ_{N}(D/2)(det B)^{−D/2},

Γ_{N}(D/2) = (2π)^{N(N−1)/4} ∏_{l=0}^{N−1} Γ(\frac{D + l}{2} ). (1)

Proof. Denote this integral by f(B). It is easy to see that for any S ∈ GL(E), f(S^{∗}BS) = f(B)| det S|^{−D}. From this it follows that f(B) = C(D)(det B)^{−D/2}.

It remains to compute C(D). Let us pick B such that det B = 1, and identify E^{∗} with E using the isomorphism B : E^{∗} → E. Then E becomes a Euclidean space with the inner product B^{−1}. Thus the constant C(D) has the form

∫_{A:E→E,A^{∗}=A,A≥0}(det A)^{(D−N−1)/2}e^{−trA}dA

Typeset by AMS-TEX

Since this integral is invariant under conjugation by SO(E), we can reduce it to the quotient space S_{+}^{2}E^{∗}/SO(E) = [0, ∞)^{N}/S_{N}. It is known from the theory of symmetric spaces that the pushforward of dA from S_{+}^{2}E^{∗} to S_{+}^{2}E^{∗}/SO(E) has the form b_{N} ∏_{1≤i<j≤N} |s_{i} − s_{j}|ds_{1}...ds_{N}, where s_{i} are the eigenvalues of A ∈ S_{+}^{2}E^{∗}/SO(E), and

b_{N} = \frac{2^{N(N−1)/4}π^{N(N+1)/4}}{∏_{l=1}^{N} Γ(l/2)} ,

Thus, C(D) has the form

(2)  b_{N}N!^{−1} ∫_{[0,∞)^{N}} ∏_{i} t_{i}^{(D−N−1)/2} ∏_{i} e^{−t_{i}} ∏_{1≤i<j≤N} |t_{i} − t_{j}|dt_{1}...dt_{N}.

To compute this integral, we use (a special case of) Selberg’s integral formula:

∫_{0≤t_{i}≤1} ∏_{i} t_{i}^{a} ∏_{i}(1 − t_{i})^{b} ∏_{1≤i<j≤N} |t_{i} − t_{j}|dt_{1}...dt_{N} =

∏_{l=1}^{N} \frac{Γ(a + \frac{l+1}{2})Γ(b + \frac{l+1}{2})Γ(\frac{l}{2} + 1)}{Γ(a + b + \frac{2N−l+3}{2})Γ(\frac{3}{2})}. (3)

Let us set s_{i} = bt_{i} in Selberg’s integral, and send b to +∞. In the limit, using Stirling’s formula, and setting a = (D − N − 1)/2, we get

C(D) = b_{N}N!^{−1} ∏_{l=1}^{N} \frac{Γ(\frac{D+l−N}{2} )Γ(\frac{l}{2} + 1)}{Γ(\frac{3}{2})} =

b_{N}π^{−N/2} ∏_{l=1}^{N} Γ(\frac{D + l − N}{2} )Γ(\frac{l}{2}) = Γ_{N}(D/2).. (4)

The Lemma is proved.

Now consider the function ρ_{D}^{E} on S^{2}E^{∗} which is supported at S_{+}^{2}E^{∗} and is given by the formula

(5)  ρ_{D}^{E}(A) = π^{ND/2}Γ_{N}(D/2)^{−1}(det A)^{(D−N−1)/2}.

By Lemma 1, we have

(6)  ∫_{S_{+}^{2}E^{∗}} ρ_{D}^{E}(A)e^{−tr(AB)}dA = π^{ND/2}(det B)^{−D/2},

ReD > N − 1.

If ReD > N − 1, the function ρ_{D}^{E}(A) is a locally L^{1}-function, so it defines a distribution on S^{2}E^{∗}. Let us compute the Fourier transform of ρ_{D}^{E}(A), i.e.

ρ̂_{D}^{E}(B) = ∫_{S^{2}E^{∗}} ρ_{D}^{E}(A)e^{−itr(AB)}dA,

B ∈ S^{2}E (the integral is understood in the sense of distributions).

Lemma 2.

(7)  ρ̂_{D}^{E}(B) = lim_{J→+0} π^{ND/2} det(iB + J)^{−D/2}.

where J ∈ S^{2}E^{∗}, J > 0.

Remark. Here we use the branch of the function z^{−D/2} which is equal to e^{−D ln z/2} for real positive z.

Proof. By analytic continuation, formula (6) holds for arbitrary B ∈ (S^{2}E^{∗})_{ℂ} with ReB > 0. In particular, we can replace B with iB + J, where B ∈ S^{2}E^{∗}. In the limit when J → +0, we obtain formula (7).

Let μ : S^{2}E → Hom(E^{∗}, E) be the tautological embedding. Let Δ be the polynomial of degree N on S^{2}E, defined by the equation Δ(B) = det(μ(B)). We regard Δ as a differential operator of order N on S^{2}E^{∗} with constant coefficients.

Corollary 3. If ReD > N + 1, then

(8)  Δρ_{D}^{E}(A) = π^{N}ρ_{D−2}^{E}(A).

Proof. The proof is obvious after passing to Fourier transforms.

Let S(S^{2}E^{∗}) be the space of complex-valued Schwarz functions on S^{2}E^{∗}. For any f ∈ S(S^{2}E^{∗}) and ReD > N − 1, define

(9)  I_{D}^{E}(f) = ∫_{S^{2}E^{∗}} ρ_{D}^{E}(A)f(A)dA.

Corollary 4. I_{D}^{E}(f) extends to an entire function of D.

Proof. For ReD > N − 1 − 2K, K ∈ ℤ_{+}, we can define I_{D}^{E}(f) by the formula

(10)  I_{D}^{E}(f) = (−π)^{−KN} ∫_{S^{2}E^{∗}} ρ_{D+2K}^{E}(A)Δ^{K}f(A)dA.

According to Corollary 3, if ReD >> 0, formula (10) defines the same function as (9). Thus, (10) defines an analytic continuation of I_{D}^{E}.

Proposition 5. Let D ≥ 0 be a nonnegative integer, and V be a vector space of dimension D, with a positive definite symmetric bilinear form β. Then for any f ∈ S(S^{2}E^{∗}) one has

(11)  I_{D}^{E}(f) = ∫_{Hom(E,V)} f(x^{∗}(β))dx,

where x^{∗}(β) denotes the inverse image of β under x (a nonnegative symmetric bilinear form on E).

Proof. It is enough to check that (11) holds for functions e^{−tr(AB)}1_{S_{+}^{2}E^{∗}}(A) where 1_{Y} is the characteristic function of the set Y, and Re(B) > 0. This follows from the fact that the function f(A) admits a Fourier representation.

In the case

(12)  f(A) = e^{−tr(AB)}1_{S_{+}^{2}E^{∗}}(A),

evaluating the Gaussian integral on the right hand side of (11), we get π^{ND/2}(det B)^{−D/2}. On the left hand side of (11), by Lemma 1, we get the same. The proposition is proved.

Remark. From Proposition 5 it is clear why formula (9) works only for ReD > N − 1. Indeed, for integer D between 0 and N − 1 the distribution ρ_{D}(A)dA is concentrated on forms of rank ≤ D, so it is a purely singular distribution. Thus, the measure ρ_{D}(A)dA is not absolutely continuous with respect to dA and hence cannot be represented as a function times dA. In fact, it turns out that this distribution is singular (not locally L^{1}) for all D with ReD ≤ N − 1.

Proposition 5 motivates the following definition.

Definition. I_{D}^{E}(f) is called the D-dimenional integral of f.

Another way of computing the D-dimensional integral is using Fourier transforms. By Plancherel formula, it follows from (9) that

(13)  I_{D}^{E}(f) = (2π)^{−N(N+1)/2} ∫_{S^{2}E^{∗}} f̂(B)\overline{ρ̂E ; D̄(B)}dB,

where f̂is the Fourier transform of f. (It is clear that \overline{ρ̂E ; D̄(B)} = lim_{J→+0} π^{ND/2} det(−iB+ J)^{−D/2}).

For ReD < 2, this formula can be written as

(14) I_{D}^{E}(f) = 2^{−N(N+1)/2}π^{N(D−N−1)/2} ∫_{S^{2}E^{∗}} f̂(B)e^{\frac{πiD}{4} σ(B)}| det(B)|^{−D/2}dB,

where σ(B) is the signature of B (the number of pluses minus the number of minuses). If ReD ≥ 2, the distribution \overline{ρ̂E ; D̄(B)} is singular, so in order to define I_{D}^{E}(f) by a direct integration formula, we need to use the operator Δ. Namely, from Lemma 1 it follows that

(15)  Δρ̂_{D}^{E}(B) = (−1)^{N}p_{N}(D)ρ̂_{D+2}^{E}(B), p_{N}(D) = π^{−N} ∏_{l=1}^{N} \frac{D + l − N}{2}

So if ReD < 2 + 2K, we can define I_{D}^{E} by the formula

I_{D}^{E}(f) = 2^{−N(N+1)/2}π^{N(D−N−1)/2} ∏_{J=1}^{K} p_{N}(D − 2J)^{−1}×

∫_{S^{2}E^{∗}} Δ^{K} f̂(B)e^{\frac{πi(D−2K)}{4} σ(B)}| det(B)|^{(−D+2K)/2}dB, (16)

Example. If N = 1, the D-dimensional integral of f is (up to normalization) just the Mellin transform of f:

I_{D}^{E}(f) = π^{D/2}Γ(D/2)^{−1} ∫_{0}^{∞} x^{\frac{D}{2} −1}f(x)dx.

Let S(S_{+}^{2}E^{∗}) be the space of smooth functions f on S_{+}^{2}E^{∗}, such that all the derivatives of f are rapidly decaying. It is easy to see that the restriction map S(S^{2}E^{∗}) → S(S_{+}^{2}E^{∗}) is surjective. So, I_{D}^{E} defines a linear functional S(S_{+}^{2}E^{∗}) → ℂ.

2. D-dimensional integral with parameters.

Now we will extend our construction to a more general situation. Let F ⊂ E be a subspace of dimension M with a volume element dv_{F}. Any symmetric bilinear form A on E defines a form A_{F} on F (the restriction of E).

Also, for any symmetric bilinear form B on E^{∗} such that B_{F^{⊥}} is nondegenerate we can define a form B^{F^{∗}} on F^{∗} as follows. Let (F^{⊥})^{⊥_{B}} be the B-orthogonal complement of F^{⊥}. As B_{F^{⊥}} is nondegenerate, this space is naturally identified with F^{∗} = E^{∗}/F^{⊥}. By the definition, the form B^{F^{∗}} is the restriction of B to (F^{⊥})^{⊥_{B}} = F^{∗}.

Let us find an explicit expression of B^{F^{∗}}. Choose a positive definite inner product on E and thus identify E^{∗} with E. Then we have a decomposition E = F ⊕ F^{⊥}, and B becomes a selfadjoint operator E → E with a decomposition

B = ( \begin{matrix}B_{11} & B_{12} \\ B_{21} & B_{22} \end{matrix} ),

where B_{ij}^{∗} = B_{ji}. Now the operator corresponding to B_{F^{⊥}} is B_{22}, and the operator corresponding to B^{F^{∗}} is B_{11} − B_{12}B_{22}^{−1}B_{21}. From this formula it is obvious that for this operator to be defined, it is sufficient for B_{22} to be invertible.

The following Lemma is a generalization of Lemma 1.

Lemma 6. For ReD > N − 1 and any positive B ∈ S^{2}E, C ∈ S^{2}F^{∗}

∫_{S_{+}^{2}E^{∗}} δ(A_{F} − C)(det A)^{(D−N−1)/2}e^{−tr(AB)}dA =

(17) \frac{Γ_{N}(D/2)}{Γ_{M}(D/2)}e^{−tr(CB^{F^{∗}})}(det C)^{(D−M−1)/2}(det B_{F^{⊥}})^{−D/2},

Remark. The space S^{2}F^{∗} inherits a volume element from F, so the deltafunction δ(A_{F} − C) is well defined.

Proof. Denote the left hand side of (17) by I(B, C). Let S ∈ GL(E) be an automorphism which preserves F. Make a change of variable A → S^{∗}AS. Then A_{F} → S_{F}^{∗} A_{F} S_{F}, where S_{F} is the restriction of S to F. Thus we get

(18)  I(B, C) = | det S|^{N−M}I(SBS^{∗}, (S_{F}^{∗})^{−1}CS_{F}^{−1}).

Now choose B with det B_{F^{⊥}} = 1, and identify E^{∗} with E using the map B : E → E^{∗}. Then E becomes a Euclidean space with the inner product B^{−1}. Now we can identify F^{∗} with F using the map B^{F^{∗}} : F^{∗} → F, thus F is also a Euclidean space.

After these identifications, integral (17) takes the form

(19)  ∫_{A∈S_{+}^{2}E^{∗}} δ(A_{F} − C)(det A)^{(D−N−1)/2}e^{−trA}dA.

Using the decomposition E = F ⊕ F^{⊥}, we can write A in the form

(20)  A = ( \begin{matrix}A_{11} & A_{12} \\ A_{21} & A_{22} \end{matrix} ),

where A_{ij}^{∗} = A_{ji}. This allows us to rewrite (19) as

(21)  ∫_{A∈S_{+}^{2}E^{∗}:A_{11}=C}(det A)^{(D−N−1)/2}e^{−trA}dA.

To compute this integral, let us use the identity det A = det A_{11} det(A_{22}−A_{21}A_{11}^{−1}A_{12}). Since we are integrating over such A that A_{11} = C, we get det A = det C det(A_{22} − A_{21}C^{−1}A_{12}). After the change of variable X = A_{22} − A_{21}C^{−1}A_{12}, integral (21) takes the form (22)

(det C)^{(D−N−1)/2}e^{−trC} ∫_{X≥0}(det X)^{(D−N−1)/2}e^{−trX}dX ∫_{Hom(F^{⊥},F)} e^{−trA_{12}^{∗}C^{−1}A_{12}}dA_{12}

The second integral in (22) is Gaussian and equals

(23)  (2π)^{M(N−M)/2}(det C)^{(M−N)/2}.

The first integral in (22) was computed in Lemma 1 and equals Γ_{N−M}((D −M)/2). Using the identity

(24)  \frac{Γ_{N}(D/2)}{Γ_{M}(D/2)} = (2π)^{M(N−M)/2}Γ_{N−M}((D − M)/2).

we obtain the statement of the Lemma.

Now consider the distribution on S^{2}E^{∗} (depending on C as a parameter)

ρ_{D}^{EF} (A, C) = δ(A_{F} − C)\frac{ρ_{D}^{E}(A)}{ρ_{D}^{F}(C)} =

(25) π^{(N−M)D/2} \frac{Γ_{M}(D/2)}{Γ_{N}(D/2)}δ(A_{F} − C)(det C)^{−(D−M−1)/2}(det A)^{(D−N−1)/2},

for ReD > N − 1.

For f ∈ S(S_{+}^{2}E^{∗}), define a function I_{D}^{EF} (f) on C > 0, by setting

(26)  I_{D}^{EF} (f)(C) = ∫_{S_{+}^{2}E^{∗}} ρ_{D}^{EF} (A, C)f(A)dA.

Since (det C)^{−1} ≤ (det A)^{−1}(trA)^{N−M}, the function I_{D}^{EF} (f) extends continuously to C ≥ 0, i.e. defines a continuous, rapidly decaying function on S_{+}^{2}F^{∗}.

Remark. Although the power of (det C)^{−1} in (25) is more than that of det A, for small det C the intersection of the region of integration in (26) with any fixed ball has area ∼ (det C)^{(N−M)/2} (because of presence of the delta-function), which helps compensate the negative exponent of det C.

Using Lemma 6, it is easy to compute the Fourier transform ρ̂_{D}^{EF} (B, C) of ρ_{D}^{EF} (A, C) with respect to A.

Lemma 7.

(27)  ρ̂_{D}^{EF} (B, C) = lim_{J→+0} π^{(N−M)D/2}e^{−itr(CB^{F^{∗}})} det(iB_{F^{⊥}} + J)^{−D/2}.

Proof. Analogous to the proof of Lemma 2.

Let Δ_{2} be the operator Δ acting on S(S^{2}F^{⊥}). Corollary 8. If ReD > N + 1, then

(28)  Δ_{2}ρ_{D}^{EF} (A, C) = π^{N}ρ_{D−2}^{EF}(A, C).

(here Δ_{2} acts on A). Proof. The proof is immediate after passing to Fourier transforms. Corollary 9. For any f ∈ S(S_{+}^{2}E^{∗}), I_{D}^{EF} (f) is a Schwarz function of C, and for any C ≥ 0 the function I_{D}^{EF} (f)(C), regarded as a function of D, analytically continues to an entire function. Proof. Similar to the Proof of Corollary 4. Proposition 10. Let D ≥ 0 be a nonnegative integer, and V be a vector space of dimension D with a positive definite symmetric bilinear form β. Then for any f ∈ S(S_{+}^{2}E^{∗}) and y : F → V one has

(29)  I_{D}^{EF} (f)(y^{∗}(β)) = ∫_{x∈Hom(E,V):x|_{F} =y} f(x^{∗}(β))dx.

Proof. As in Proposition 5, it is enough to consider f(A) = e^{−tr(AB)}. In this case, on the r.h.s. of (19) is a Gaussian integral, whose value is

π^{(N−M)D/2}e^{−tr(y^{∗}(β)B^{F^{∗}})}(det B_{F^{⊥}})^{−D/2}.

Thus, the Proposition follows from Lemma 6. Definition. I_{D}^{EF} is called the D-dimensional integral with external parameters.

As before, the D-dimensional integral can be computed using Fourier transform. Let f ∈ S(S^{2}E^{∗}). By Plancherel formula,

(30)  I_{D}^{EF} (f)(C) = (2π)^{−N(N+1)/2} ∫_{S^{2}E^{∗}} f̂(B)\overline{ρ̂EF ; D̄ (B, C)}dB,

where f̂ is the Fourier transform of f. Here

\overline{ρ̂EF ; D̄ (B, C)} = lim_{J→+0} π^{(N−M)D/2}e^{itr(CB^{F^{∗}})} det(−iB_{F^{⊥}} + J)^{−D/2}.

For ReD < 2, this formula can be written as

I_{D}^{EF} (f)(C) =

2^{−N(N+1)/2}π^{((N−M)D−N(N+1))/2} ∫_{S^{2}E^{∗}} f̂(B)e^{\frac{πiD}{4} σ(B_{F^{⊥}})}e^{itr(CB^{F^{∗}})}| det(B_{F^{⊥}})|^{−D/2}dB. (31)

If ReD ≥ 2, the distribution \overline{ρ̂EF ; D̄ (B, C)} is singular, so in order to define I_{D}^{EF} (f) by a direct integration formula, we need to use the operator Δ. Namely, from Lemma 6 it follows that

(det C)^{−1}Δρ̂_{D}^{EF} (B, C) = (−1)^{N−M}p_{NM}(D)ρ_{D+2}^{EF}(B),

p_{NM}(D) = π^{M−N} ∏_{l=M+1}^{N} \frac{D + l − N}{2} (32)

So if ReD < 2 + 2K, we can define I_{D}^{EF} by the formula

I_{D}^{EF} (f)(C) = 2^{−N(N+1)/2}π^{((N−M)D−N(N+1))/2}(det C)^{−K} ∏_{J=1}^{K} p_{N}(D − 2J)^{−1}×

∫_{S^{2}E^{∗}} Δ^{K} f̂(B)e^{itr(CB^{F^{∗}})}e^{\frac{πi(D−2K)}{4} σ(B)}| det(B)|^{(−D+2K)/2}dB.

Thus we have defined an operator I_{D}^{EF} : S(S_{+}^{2}E^{∗}) → S(S_{+}^{2}F^{∗}). It is clear that I_{D}^{F G}I_{D}^{EF} = I_{D}^{EG} if E ⊃ F ⊃ G, and I_{D}^{E0} = I_{D}^{E}.

Now we can define the D-dimensional integral of functions with polynomial growth. Let f be a smooth function on S^{2}E^{∗} such that there exists a ∈ ℝ for which f and all its derivatives satisfy the condition ϕ(A) = O(||A||^{a}), ||A|| → ∞ (for any norm on S^{2}E^{∗}). In this case, the integral I_{D}^{EF} (f) is defined for ReD << 0. Indeed, the condition on f implies that the Fourier transform f̂ is a distribution defined on functions with sufficiently many continuous derivatives having polynomial growth at infinity. Thus, we can use formula (30) to define I_{D}^{EF} (f) for ReD << 0.

Remark. The operator I_{D}^{EF} has rather strange properties. For example, if P is any polynomial, then I_{D}^{E}(P (A)) = 0. Indeed, the Fourier transform of P is a distribution supported at 0, so for sufficiently negative D formula (14) gives 0. Thus, I_{D}^{E} is not continuous in any topology in which polynomials are dense.

Example. Let N = 1, M = 0, and f be a rational function, regular at x ≥ 0. Write f as a sum of simple fractions: f(x) = f_{0}(x) + ∑_{i} α_{i}(x + a_{i})^{−n_{i}}, where f_{0} is a polynomial, a_{i} ∈ ℂ \ ℝ_{−}, and n_{i} are positive integers. Then

I_{D}^{E}(f) = π^{D/2} ∑_{i} α_{i} \frac{Γ(n_{i} − D/2)}{Γ(n_{i})}a_{i}^{n_{i}− \frac{D}{2}} .

Thus we see that if f is satisfies the above condition of polynomial growth then I_{D}^{EF} (f)(C) is holomorphic in D for ReD << 0. In general, we cannot expect that this function will analytically continue to the whole complex plane. However, for functions of a certain algebraic nature, it extends meromorphically to the whole complex plane.

3. D-dimensional integral of functions arising from Feynman diagrams.

Definition. A function f on S_{+}^{2}E^{∗} is called a function of Feynman type if it has the form

f(A) = \frac{P (A)}{∏_{j=1}^{l}(tr(AB_{j}) + m_{j}^{2})},

where P is a polynomial, B_{j} ≥ 0, such that (∑ t_{i}B_{i})_{F^{⊥}} is nondegenerate for t_{i} > 0, and m_{j} ≠ 0 are real numbers.

Proposition 11. If f is of Feynman type then I_{D}^{EF} (f) extends meromorphically to the whole complex plane.

Proof. We will prove the statement for P = 1. For general P, the proof is similar to the case P = 1.

Using the formula b^{−1} = ∫_{0}^{∞} e^{−tb}dt, we can rewrite I_{D}^{EF} (f) in the form

(33)  I_{D}^{EF} (f) = ∫_{t_{1},...,t_{l}∈[0,∞)} e^{− ∑ t_{j}m_{j}^{2}}I_{D}^{EF} (e^{−tr(A(∑ t_{j}B_{j}))})dt.

Set B(t) = ∑ t_{j}B_{j}. Using Lemma 6, we can rewrite this in the form (34)

I_{D}^{EF} (f)(C) = π^{(N−M)/2} ∫_{t_{1},...,t_{l}∈[0,∞)} e^{− ∑ t_{j}m_{j}^{2}−tr(CB^{F^{∗}} (t))}(det B_{F^{⊥}}(t))^{−D/2}dt.

Now we will use the following theorem of Bernstein, which is proved with the help of the theory of D-modules.

Bernstein’s theorem. Let Q(t) be any polynomial in l variables. Then there exists a polynomial differential operator L(D) in l variables, with coefficients depending polynomially on D, and a polynomial q in D, such that L(D)Q^{−D/2} = q(D)Q^{−1−D/2}.

This theorem implies the following statement, which is a special case of a result of Bernstein.

Corollary. Assume that Q takes positive values for t_{j} > 0, and let g be a rapidly decaying smooth function defined for t_{i} ≥ 0, such that its derivatives are also rapidly decaying. Then the integral

(35)  I(D, g) = ∫_{t_{i}∈[0,∞)} g(t)Q(t)^{−D/2}dt,

convergent for ReD << 0, extends meromorphically to the whole complex plane.

The corollary is proved by induction in l (the number of t_{j}). The base of induction is trivial. Now, using Bernstein’s theorem and integration by parts, we can write

I(D + 2, g) = q(D)^{−1} ∫ g(t)L(D)(Q(t)^{−D/2})dt =

q(D)^{−1} ∫ (L(D)^{∗}g(t))Q(t)^{−D/2}dt + C(D) = q(D)^{−1}I(D, L(D)^{∗}g) + C(D), (36)

where C(D) is the integral over the boundary, which is meromorphic by the assumption of induction.

Now we apply the corollary to our situation, and obtain the proof of the Proposition for P = 1.

Remark 1. Using Hironaka’s theorem about the existence of a smooth resolution of singularities for a compact projective variety, it is possible to prove that I_{D}^{EF} is meromorphic in D in a much more general situation than in Proposition 11. For example, as was explained to me by M.Kontsevich, one can prove that if f is a rational function without singularities in S_{+}^{2}E^{∗}, then I_{D}^{EF} (f) is meromorphic.

It is also possible to show that for rational f as above the poles of I_{D}^{EF} (f) are all rational, and belong to a finite number of arithmetic progressions.

4. Dimensional regularization of Feynman integrals.

Proposition 11 allows to define the procedure of dimensional regularization of Feynman integrals in quantum field theory.

Suppose that we have a quantum field theory given by some Lagrangian, and we want to compute its correlation functions. According to Feynman rules, such a correlation function equals to the sum of amplitudes of all Feynman diagrams (graphs), which are given by some finite-dimensional integrals.

We will work in the Euclidean setting and in momentum space. For simplicity we will assume that our theory contains only scalar massive fields. Then the amplitude of any Feynman graph Γ is given by the formula

(37)  ∫_{x∈Hom(E,V):x|_{F} =y} g(x)dx, y ∈ Hom(F, V),

where V is the momentum space (a Euclidean space of dimension d with inner product β), E = H_{c}^{1}(Γ \ ∂Γ, ℝ) (the cohomology with compact supports), F is the image of the map H^{0}(∂Γ, ℝ) → H_{c}^{1}(Γ \ ∂Γ, ℝ) in the long exact sequence of cohomology, and g is a function on Hom(E, V) of the form g(z) = f(z^{∗}(β)), where f is a certain function of Feynman type attached to Γ.

Remark. It is clear from the long exact sequence that E/F = H^{1}(Γ, ℝ).

Since cohomology groups are defined over ℤ, they have natural volume elements. This defines the volume element dx in (37).

The problem is that integral (37) is often divergent, since the function g does not decay rapidly enough.

Dimensional regularization is a method to give meaning to the integral (37). This is done as follows. Recall that since f is of Feynman type, I_{D}^{EF} (f) is meromorphic in D.

Definition. Dimensional regularization of integral (37) is the regular part of I_{D}^{EF} (f) at D = d, i.e. \frac{1}{2πi} ∮_{|D−d|=ε} I_{D}^{EF} (f)d ln(D − d).

Remark. In general, we have to consider functions f which (even before the regularization) depend meromorphically on D. This is caused by the fact that our graph Γ may contain divergent proper subgraphs, which have to have been regularized before we start working with Γ. Since regularization is performed by adding D-dependent counterterms, we will have to work with functions which depend meromorphically on D.

5. D-dimensional Stokes formula.

Let F ⊂ E be vector spaces with volume elements, dim E = N, dim F = M, and D be any complex number. Define the space Ω_{EF}^{top} of (Schwarz) top degree forms to be the space S(S_{+}^{2}E^{∗}), and the space Ω_{EF}^{top−1} of forms of degree top − 1 to be the space of Schwarz functions ω : S_{+}^{2}E^{∗} → Hom(E/F, E). That is, Ω_{EF}^{top} = Hom(E/F, E) ⊗ S(S_{+}^{2}E^{∗}).

If D is an integer, and V is a Euclidean space of dimension D with inner product β, then we have a linear map θ : Hom(E, V) → Hom(F, V) given by restriction, and any f ∈ Ω_{EF}^{top} indeed defines a top differential form f(x^{∗}(β))dx on each fiber of θ (dx is the natural volume form on the fiber).

Also, any ω ∈ Ω_{EF}^{top−1} defines a top − 1-form on the fiber. Indeed, let z : (E/F)^{∗} → V be a linear map. We regard z as a constant 1-form on the fibers of θ. Then the top − 1-form corresponding to Ω is defined by the formula (ω ∧ z)(x) = tr(ωβ(x, z))dx, where β(x, z) ∈ Hom(E ⊗ (E/F)^{∗}, ℝ) is regarded as an operator E → E/F.

Recall that the group GL(E) acts on S_{+}^{2}E^{∗} by A → S^{∗}AS, S ∈ GL(E). Thus, any element X of the Lie algebra End(E) of GL(E) defines a vector field on S_{+}^{2}E^{∗}: L_{X}f(A) = \frac{d}{dt}|_{t=0}f(e^{X^{∗}t}Ae^{Xt}).

For any ω = ∑ X_{i} ⊗ f_{i} ∈ Ω_{EF}^{top−1}, X_{i} ∈ Hom(E/F, E), f_{i} ∈ S(S_{+}^{2}E^{∗}), define the form dω ∈ Ω_{EF}^{top} by the formula

(38)  dω = ∑_{i}(L_{X_{i}}f_{i} + Dtr(X_{i})f_{i}),

where X_{i} are regarded as elements of End(E). Thus, we have defined a linear map d : Ω_{EF}^{top−1} → Ω_{EF}^{top} . It is easy to see that if D is an integer, d coincides with the usual differential.

Proposition 12. (Stokes formula.) I_{D}^{EF} (dω) = 0 for any ω ∈ Ω_{EF}^{top−1}.

Proof. It is enough to prove the formula for ReD >> 0; for other values of D the formula follows by analytic continuation.

For ReD >> 0, the formula follows by integration by parts from the definition (26) of I_{D}^{EF}, as L_{X}((det A)^{(D−N−1)/2}dA) = Dtr(X)(det A)^{(D−N−1)/2}dA.