Witten's proof of Wick Formula of QFT

Question

Let $\mathcal{S}$ be a finite dimensional real vector space with a positive definite summetric bilinear form $B$. Let $dv$ be a Lebesgue measure on $\mathcal{S}$ such that $$\int_{\mathcal{S}}e^{-B(v,v)}dv =1\tag{1.10}$$ (convenient normalization).

We want to prove the following. If $$\langle f_1 \dots f_N \rangle_0 := \int_{S} f_1(v)\dots f_N(v) e^{-B(v,v)/2} dv\tag{1.12}$$ where $f_1, \dots, f_N \in S^*$, then as long as $N = 2K$ is even (obviously the integral vanishes for odd $N$), $$\langle f_1 \dots f_N \rangle_0 = \sum_{s \in S_{2K}/ \sim} B^{-1}(f_{s(1)},f_{s_(2)}) \dots B^{-1}(f_{s(2K-1)},f_{s(2K)})\tag{1.13}$$ where $B^{-1}$ is the unique inverse form of $B$, $S_{2K}$ is the symmetric group, and $s_1 \sim s_2$, $s_1,s_2 \in S_{2K}$ if they define the same term in the above.

In the lecture, Witten gives the proof as

Since both sides of [the claimed identity] are symmetric polylinear functions, the proposition reduces to the case $f_1 = \dots = f_{2N} = f$, which is straightforward.

Question: I understand the gist of this argument, but not sure why symmetric polylinearity reduces it to that special case (or then what the straightforward calculation exactly is). Can someone walk through his explanation?

Reference: "Quantum Fields and Strings: A course for mathematicians", vol.1. "Perturbative quantum field theory" lecture 1: Renormalization of Feynman Diagrams (pg 425).

Answer 1

Let us consider the following problem.

Problem: Given a symmetric $N$-linear form $A: V^{\otimes N}\mapsto \mathbb{R}$, we can define a degree-$N$ homogeneous function $Q: V\mapsto \mathbb{R} $ through $Q(x)\equiv A(x,\cdots,x)$. Can we conversely express $A$ in terms of $Q$?

If this is done, $Q = Q'$ immediately implies $A=A'$, which then justifies Witten's method.

A clever solution to this problem invokes partial derivatives, $$A(x_1,\cdots,x_N) = \frac{1}{N!}\frac{\partial}{\partial\lambda_1}\cdots\frac{\partial}{\partial\lambda_N}Q\bigl(\lambda_1 x_1 + \cdots + \lambda_N x_N\bigr)\,\Bigl|_{\lambda_1=\cdots=\lambda_N=0}\,.$$ We can prove this result just by expanding the right-hand side. Upon this, one can also work out other combinatory expressions.

You can refer to this Wikipedia page and the references thereof for more information.

Answer 2

This is a consequence of multilinearity and the inclusion-exclusion principle.

Basic idea

Multilinear forms are completely determined by their contractions: $$T(f_1, \dots, f_N) = S(f_1, \dots, f_N) ~~\text{for all}~~f_1, \dots, f_N \quad \implies \quad T = S. \tag{1}$$ This restates the definition of a function; two functions are the same if their outputs agree for all possible inputs.

It is a surprising result that symmetric multilinear forms are completely determined by their symmetric contractions. If $T$ and $S$ are symmetric, then $$T(g, \dots, g) = S(g, \dots, g) ~~\text{for all}~~g \quad \implies \quad T = S \tag{2}$$ Note that whereas (1) requires $T$ and $S$ to agree for all possible combinations of different inputs, (2) only requires $T$ and $S$ to agree for inputs that are all the same. One is allowed to put in combinations of different inputs into a symmetric form, but those non-symmetric contractions do not provide any additional information about the form beyond the information one could gather from symmetric contractions.

The key idea for proving (2) is to start out with a totally nonsymmetric contraction, then by using symmetry and multilinearity, express it as a weighted sum of symmetric contractions: $$\underbrace{T(f_1, f_2, \dots, f_N)}_\text{non-symmetric contraction} = \sum_i c_i \underbrace{T(g_i, g_i, \dots, g_i)}_\text{symmetric contraction}. \tag{3}$$ Finding these scalars $c_i$ and vectors $g_i$ may be done using a combinatorial argument based on the inclusion-exclusion principle. Below we will illustrate how to determine the $c_i$ and $g_i$ in the special case $N=3$.

Once (3) is shown, the desired result follows since \begin{align*} T(f_1, f_2, \dots, f_N) &= \sum_i c_i T(g_i, g_i, \dots, g_i)\\ &= \sum_i c_i S(g_i, g_i, \dots, g_i) = S(f_1, f_2, \dots, f_N) \end{align*} which implies $T=S$, per (1). Within the first and second lines we use (3) under the assumption that $T$ and $S$ are symmetric, and going from the first line to the second line we use the assumption that $T$ and $S$ agree on symmetric contractions.

Inclusion/exclusion argument for $N=3$

Now we show that (3) holds when $T$ is symmetric, in the special case where $N=3$. This is a large enough $N$ to illustrate the general principle, and small enough that we do not get bogged down in excessively complicated notation.

For convenience, let us dispense with the $T$'s and parantheses and commas, and simply write the action of a form as a formal monomial in its inputs: $$T(a,b,c) = abc$$ $$T(a,a,a) = a a a = a^3.$$ Multilinearity implies that the distributive law holds in this notation, for example, $$(a+a')bc = T(a + a', b, c) = T(a,b,c) + T(a',b,c) = abc + a'bc.$$ Symmetry of $A$ implies that the commutative law holds, for example, $$abc = T(a,b,c) = T(b,a,c) = bac.$$ Hence, without loss of generality we may manipulate these expressions as if they were polynomials. After performing valid polynomial manipulations, whatever equalities we get will also hold for the corresponding multilinear contractions of $T$. However, for combinatorial/counting purposes it is useful keep track of the order of symbols, so for now let us pretend that commutativity does not hold.

We observe the following interesting results: \begin{align*} (a+b)^3 =& aaa + aab + aba + abb + baa + bab + bba + bbb\\ =& \text{ all monomials of length }3\text{ using the symbols }a,b. \end{align*} and similar for other $(a+c)^3$ and $(b+c)^3$, and \begin{align*} (a+b+c)^3 =& aaa + aab + aac + aba + abb + abc + aca + acb + acc\\ &+baa + bab + bac + bba + bbb + bbc + bca + bcb + bcc\\ &+caa + cab + cac + cba + cbb + cbc + cca + ccb + ccc\\ =& \text{ all monomials of length }3\text{ using the symbols }a,b,c. \end{align*} Using the inclusion/exculsion principle, then using symmetry/commutativity, we have \begin{align*} &(a+b+c)^3 - (a+b)^3 - (a+c)^3 - (b+c)^3 + a^3 + b^3 + c^3 \\ &= abc + acb + bac + bca + cab + cba \\ &= 6~abc \end{align*} Notice that subtracting $(a+b)^3$ from $(a+b+c)^3$ removed all terms that did not contain $c$, subtracting $(a+c)^3$ removed all terms that did not contain $b$, and subtracting $(b+c)^3$ removed all terms that did not contain $a$. However, subtracting all three of these terms doubly-subtracted the terms $a^3$, $b^3$, and $c^3$, which is why we had to add those terms back again. This is illustrated in the image below:

To finish up, we solve for $abc$ to get $$\underbrace{abc}_{T(a,b,c)} = \underbrace{\frac{1}{6}\left((a+b+c)^3 - (a+b)^3 - (a+c)^3 - (b+c)^3 + a^3 + b^3 + c^3\right)}_{\sum_i c_i T(g_i, g_i, g_i)}.$$ The left side of this equation is a general non-symmetric contraction, and the right hand side of this equation is a weighted sum of symmetric contractions, as desired.

Larger $N$

For larger $N$, one simply repeats this process of subtracting and adding all possible combinations of symmetric lower order terms to isolate the monomials that contain exactly one of each symbol. For $N=4$, we have \begin{align} 4!~abcd =& (a+b+c+d)^4 \\ &- (a+b+c)^4 - (a+b+d)^4 - (a+c+d)^4 - (b+c+d)^4\\ &+ (a+b)^4 + (a+c)^4 + (a+d)^4 + (b+c)^4 + (b+d)^4 + (c+d)^4\\ &-a^4 - b^4 - c^4 - d^4. \end{align}

For $N=5$, we have \begin{align} 5!~abcde =& (a+b+c+d+e)^5 \\ &- (a+b+c+d)^5 - (a+b+c+e)^5 - \dots \\ &+ (a+b+c)^5 + (a+b+d)^5 + \dots \\ &-(a+b)^5 - (a+c)^5 - \dots \\ &+ a^5 + b^5 + c^5 + d^5 + e^5. \end{align} and so on for larger $N$.