# Math 262A: Lecture 13

Continuing with Lecture 12, we are analyzing the Taylor series

$T_I(\hbar) = \sum\limits_{k=0}^\infty a_k\hbar^k$

of the smooth function

$I(\hbar) = \frac{1}{\sqrt{2\pi\hbar}}\int\limits_{\mathbb{R}} e^{-\frac{1}{\hbar}S(x)} \mathrm{d}x,$

where $S$ is a smooth function on the line such that the above integral converges, and whose Taylor series is of the form

$T_S(x) = \frac{x^2}{2} + \sum\limits_{d=3}^\infty p_d \frac{x^d}{d}.$

I should say here that I’m trying out a new notation: if $F$ is a function of the real variable $y$ which is smooth in a neighborhood of $y=0,$ then I write

$T_F(y) = \sum\limits_{k=0}^\infty F^{(k)}(0)\frac{y^k}{k!},$

this being a formal power series in a single indeterminate which by abuse of notation I also call $y.$ So it does not make sense to write

$F(y) = T_F(y),$

since the LHS is a number and the RHS is a formal power series. However, Taylor’s theorem with remainder tells you that if you take a finite number of terms of the formal power series $T_F$ and trade the formal variable $y$ for a real number $y,$ you get a good approximation to the real number $F(y).$

Back to the problem at hand, what we are trying to do is to express the formal power series $T_I$ in terms of the formal power series $T_S.$ What we currently know is that

$T_I(\hbar) = 1 + \sum\limits_{d=1}^\infty \frac{1}{d!}\sum\limits_{\vdash d} t_d|C_\alpha|p_\alpha\hbar^{\frac{d}{2}-\ell(\alpha)},$

where the internal (finite) sum is over Young diagrams with $d$, with $t_d$ the number of fixed point free involutions in $\mathrm{S}(d)$ and $C_\alpha$ the conjugacy class in $\mathrm{S}(d)$ of permutations of cycle type $\alpha,$ and

$p_\alpha=\prod\limits_{i=1}^{\ell(\alpha)} p_{\alpha_i}$

subject to $p_1=p_2=0.$ So, in completely algebraic language, one could say that we are studying the “Feynman transform”

$\mathcal{F} \colon x^3\mathbb{R}[p_1,p_2,p_3,\dots][[x]] \to \mathbb{R}[p_1,p_2,p_3,\dots][[\hbar]]$

defined by

$\mathcal{F}\left( \sum\limits_{d=3}^\infty p_d \frac{x^d}{d}\right):=\sum\limits_{d=1}^\infty \frac{1}{d!} \sum\limits_{\alpha \vdash d} t_d|C_\alpha|p_\alpha \hbar^{\frac{d}{2}-\ell(\alpha)}.$

The domain of this map is the principal ideal generated by $x^3$ in the algebra of formal power series in $x$ with coefficients being polynomials in the $p_1,p_2,p_3,\dots,$ and the codomain is formal power series in $\hbar$ with coefficients in the same polynomial ring. The problem then is to simplify the output of the Feynman map as a power series in $\hbar,$ i.e. to determine the coefficient of $\hbar^k$ for each $k \in \mathbb{N}.$

What we want to do in this lecture is give a diagrammatic solution to the above algebraic problem. Note that the solution to this problem has analytic value via Laplace’s method, but the solution itself does not involve any analysis and is purely combinatorial.

The first step is to think of the number $|C_\alpha|t_d$ as the cardinality of $C_\alpha \times \mathrm{Inv}(d),$ where again $C_\alpha$ is the conjugacy class of permutations in the symmetric group $\mathrm{S}(d),$ and $\mathrm{Inv}(d)$ is the set of fixed point free involutions in $\mathrm{S}(d).$ Note that $\mathrm{Inv}(d)$ is nonempty if and only if $d$ is even, in which case

$\mathrm{Inv}(d) = C_{(2^{\frac{d}{2}})},$

where $(2^{\frac{d}{2}})$ is the Young diagram with $\frac{d}{2}$ rows of length $2.$ Every pair $(\sigma,\tau) \in C_\alpha \times \mathrm{Inv}(d)$ is a combinatorial map. Associated to this map are two topological maps, which are dual to one another. To construct these topological maps, we use the following construction. First, for each cycle of $\sigma$ we draw a polygon whose number of sides is equal to the length of $\sigma,$ and whose edges are labeled counterclockwise in the cyclic order prescribed by $\sigma.$ At the same time, we place into this polygon the corresponding dual object, namely a vertex in the interior of the polygon together with line segments, called “half edges” or “darts,” which emanate from the vertex, one for each edge of the polygon and labeled in the same way. Note that we need only consider the case where $d$ is even, since $t_d=0$ otherwise, and moreover where all rows of $\alpha$ have length at least $3,$ since $p_1=p_2=0,$ meaning that all polygons in this construction have at least three sides, or dually that all stars have degree at least $3.$ Now we glue the sides of the polygons together according to the pairing permutation $\tau,$ and at the same time pair half edges of stars in the same manner, i.e. using $\tau.$ This produces a pair of dual topological maps $\Gamma,\Gamma'$ on a compact oriented surface $S.$ Our convention is that we view $\Gamma$ as the topological map obtained by glueing stars, and $\Gamma'$ as the topological map obtained by glueing polygons. The figure below illustrates this construction for the combinatorial map

$\left( (1\ 2\ 3)(4\ 5\ 6), (1\ 2)(3\ 4)(5\ 6) \right).$

Now let us consider the Feynman transform

$\mathcal{F}\left( \sum\limits_{d=3}^\infty p_d \frac{x^d}{d}\right):=\sum\limits_{d=1}^\infty \frac{1}{d!} \sum\limits_{\alpha \vdash d} t_d|C_\alpha|p_\alpha \hbar^{\frac{d}{2}-\ell(\alpha)}$

from the point of view of topological rather than combinatorial maps. First, the powers of $\hbar$ occurring in the transformed power series have a clear interpretation: we have

$\hbar^{\frac{d}{2}-\ell(\alpha)} = \hbar^{e(\Gamma)-v(\Gamma)},$

where $v(\Gamma),e(\Gamma)$ are the number of vertices and edges in $\Gamma.$ Observe that actually the difference $e(\Gamma)-v(\Gamma)$ depends only on the underlying graph, not actually on its embedding, so that this is a purely combinatorial parameter. Indeed, it is closely related to the so-called circuit rank of a graph, which is the number of independent cycles in the graph, or equivalently the minimal number of edges which must be deleted in order to obtain an acyclic graph, i.e. a forest. The precise relationship between these parameters is

$e(\Gamma)-v(\Gamma)=r(\Gamma)-c(\Gamma)$

with $r(\Gamma)$ the circuit rank and $c(\Gamma)$ the number of connected components. This is pretty obvious if you think about it: consider the extremal case where we have one connected component consisting of a tree. In this case the difference $r(\Gamma)-c(\Gamma)$ is obviously equal to zero, and if we start adding edges the circuit rank goes up by one every time we add an edge, and moreover the construction is additive in the number of components (graph theory people – did I get that right?).

So, at the coarsest level, the output of the Feynman transform is apparently a generating function for (possibly disconnected) graphs sorted by circuit rank, where we are restricting to graphs in which every vertex has degree at least $3.$ Now this is not quite the case, since many combinatorial maps will give rise to the same topological map. Indeed, if we take the combinatorial map $(\sigma,\tau)$ and conjugate it by an arbitrary permutation $\pi \in \mathrm{S}(d),$ then we get another combinatorial map $(\pi\sigma\pi^{-1},\pi\tau\pi^{-1})$ which obviously yields the same topological map when the construction described above is performed, since we have simply relabeled everything by conjugating. Thus we expect that the correspondence between combinatorial and topological maps should be $d!$-to-one, which would be excellent news, since then the factor

$\frac{1}{d!} |C_\alpha|t_d$

would be equal to one. However, this is not precisely true, since there may be permutations $\pi$ whose conjugation action on the combinatorial map $(\sigma,\tau)$ has no effect, leaving this combinatorial map unaltered; this happens precisely when $\pi$ commutes with both $\pi$ and $\sigma$, or in other words belongs to the centralizer of the subgroup $\langle \sigma,\tau \rangle$ of the symmetric group $\mathrm{S}(d)$. Which they generate. It is thus reasonable to define the automorphism group of a combinatorial map to be this centralizer. We then see that the Feynman transform is indeed a generating function for graphs arranged by circuit rank, but in which each graph is counted with a weight equal to the reciprocal of the order of its automorphism group:

$\mathcal{F}\left( \sum\limits_{d=3}^\infty p_d \frac{x^d}{d}\right):=\sum\limits_\Gamma \frac{\hbar^{r(\Gamma)-1}}{|\mathrm{Aut} \Gamma|}p_\Gamma,$

the sum being over all topological maps of minimum vertex degree $3,$ with

$p_\Gamma = \prod_{v \in \Gamma} p_{\deg(v)},$

where the product is over all vertices of $\Gamma.$ This product is called the “Feynman amplitude” of $\Gamma,$ and it is the only part of the above construction which actually depends on the generating function $S$ we started with — everything else is universal, being exactly the same irrespective of what $S$ is. And that’s the general idea of the Feynman diagram expansion in zero-dimensional scalar-valued quantum field theory. It’s pretty useful, because for example we can say that the leading contribution in the asymptotic expansion of the integral we’re trying to approximate in the $\hbar \to 0$ limit comes from graphs which have minimum vertex degree $3,$ and two independent cycles, and there are only three of those:

We’ll clean this up a bit next week, and then go on to the matrix-valued case, where the expansion depends not just on the underlying graph of a topological map, but also on the topology of the surface into which it is embedded.