Double lines and color counting

The first correlator kept every color index in a quadratic trace. For a longer trace, how can we predict the large-$N$ power before evaluating every sum? We will turn each matrix propagator into a ribbon, count its closed index loops, and check the result against exact finite-$N$ formulas.

Our theory is four-dimensional $\mathcal N=4$ $SU(N)$ Yang–Mills in the conformal vacuum, with theta angle zero. It has sixteen real Lorentzian Poincaré supercharges; four are manifest in an $\mathcal N=1$ presentation. Here $N$ counts fundamental colors; the mathematical gauge rank is $N-1$. The explicit correlators are free, separated-point Euclidean calculations, not interacting predictions deduced from color counting. The other preparation is the meaning of a fixed-‘t Hooft-coupling limit.

A propagator carries two index lines

Retain the ordinary trace, $\operatorname{Tr}(t^at^b)=\delta^{ab}/2$, canonical real scalar components, and $Z=(X^1+iX^2)/\sqrt2$. Write

$$G(x-y)=\frac1{4\pi^2|x-y|^2},\qquad v=\frac G2.$$

The prerequisite derived

$$\langle Z_i{}^j(x)\bar Z_k{}^l(y)\rangle_{SU(N),0} =v\left(\delta_i^l\delta_k^j-\frac1N\delta_i^j\delta_k^l\right). \tag{N08.1}$$

Minahan’s (5.4) instead has coefficient $G$ and omits tracelessness in its large-$N$ treatment. Its canonical matrix is $\sqrt2\,Z$ in our notation; the prerequisite established that normalization map. We retain the factor $1/2$ throughout.

For a moment include the independent identity component and work with a $U(N)$ Gaussian bookkeeping model. Its propagator is just $v\delta_i^l\delta_k^j$. One delta identifies $i$ with $l$; the other identifies $j$ with $k$. Drawing the two identifications separately produces a double line. Every closed chain of such identifications leaves one freely summed color, hence a factor $N$.

One propagator is one ribbon with two oppositely oriented color lines. These are index identifications, not paths in spacetime. At a trace, the matrix order specifies how ribbon ends join.

The order at a trace is cyclic, not arbitrary. A crossing in one sketch can sometimes be removed by redrawing; the invariant test keeps the cyclic order at every vertex and counts the resulting faces. We will do that algebraically, so the answer does not depend on drawing skill. This implements the color-loop argument in Minahan (5.6)–(5.7).

Compare two cubic contractions

Define $\mathcal O_3=\operatorname{Tr}Z^3$ and its conjugate. In the free theory, contractions within either holomorphic insertion vanish, so normal ordering introduces no additional term. Write the two traces as

$$Z_{i_1}{}^{i_2}Z_{i_2}{}^{i_3}Z_{i_3}{}^{i_1}, \qquad \bar Z_{j_1}{}^{j_2}\bar Z_{j_2}{}^{j_3}\bar Z_{j_3}{}^{j_1}.$$

A permutation $\pi$ says that the $a$th $Z$ contracts with the $\pi(a)$th $\bar Z$. Its delta constraints are

$$i_a=j_{\pi(a)+1},\qquad j_{\pi(a)}=i_{a+1}, \tag{N08.2}$$

with subscripts interpreted modulo three.

First take $\pi=(1,3,2)$, meaning the ordered list of its images. The three pairs impose

$$(i_1=j_2,\ j_1=i_2),\qquad (i_2=j_1,\ j_3=i_3),\qquad (i_3=j_3,\ j_2=i_1).$$

There are three independent sums, over $i_1,i_2,i_3$. This contraction contributes $v^3N^3$.

Now take $\pi=(1,2,3)$. Equations (N08.2) force $i_1=i_2=i_3=j_1=j_2=j_3$. Only one color sum remains, and the contribution is $v^3N$. The two contractions have identical spacetime dependence and the same number of propagators, but differ by $N^{-2}$.

All six pairings are exhausted by the following table; the lists are permutations in image notation, not cycle notation.

Pairing lists $\pi$	Number	Free index loops $F$	Contribution per pairing
$(1,3,2),(2,1,3),(3,2,1)$	3	3	$v^3N^3$
$(1,2,3),(2,3,1),(3,1,2)$	3	1	$v^3N$

Thus the exact answer in the stated $U(N)$ model is

$$\boxed{\langle\mathcal O_3(x)\overline{\mathcal O}_3(y)\rangle_{U(N),0} =3v^3(N^3+N).} \tag{N08.3}$$

There is no division by $3!$: it was not included in the definition of the operator. The three leading pairings preserve opposite cyclic order around the two trace vertices. The word “opposite” comes from orienting the boundaries consistently; it does not permit reversing matrix order inside an arbitrary trace.

Why the loss of two faces is a handle

Thicken each propagator to an oriented ribbon and each trace to a disk. Cap every closed color line with a face. This constructs a closed oriented surface. For a connected graph with $V$ disks, $E$ ribbons and $F$ faces, Euler’s relation is

$$\chi=V-E+F=2-2h, \tag{N08.4}$$

where the nonnegative integer $h$ is the number of handles, or genus. For our cubic example $V=2$ and $E=3$. Three faces give $\chi=2$, a sphere: the planar contraction. One face gives $\chi=0$, a torus: a genus-one contraction.

This genus is not a perturbative loop order. Both diagrams above come from free Wick contractions, with no interaction vertex. It is the topology of their color ribbons that differs.

Now connect the topology to the coupling. Put $g=g_{\rm YM}>0$ and introduce rescaled fields $\mathsf Z=gZ$, with all other adjoint fields rescaled consistently. The action has an overall $1/g^2=N/\lambda$, where $\lambda=g^2N$. A rescaled propagator supplies $\lambda/N$ and an ordinary single-trace interaction vertex supplies $N/\lambda$, up to $N$-independent coefficients. In particular,

$$\langle\mathsf Z_i{}^j\overline{\mathsf Z}_k{}^l\rangle_{U(N),0} =\frac{\lambda G}{2N}\delta_i^l\delta_k^j.$$

Suppose a connected ribbon graph contains $V_{\rm int}$ interaction vertices and $b$ inserted ordinary traces, with no extra $N$ multiplying the insertions. Cap those traces as disks in (N08.4), so $V=V_{\rm int}+b$. Its color-coupling factor is

$$N^F\left(\frac{\lambda}{N}\right)^E \left(\frac{N}{\lambda}\right)^{V_{\rm int}} =\lambda^{E-V_{\rm int}}N^{2-2h-b}. \tag{N08.5}$$

This derives the topological power rather than assigning it by diagram appearance. The scalar, gauge, fermion and ghost numerators can alter coefficients or make contributions cancel; they do not change this index count for a specified ribbon term. Minahan’s discussion following (5.12) identifies the same fixed combination $g^2N$.

For two cubic traces in rescaled fields, (N08.3) becomes

$$\langle\operatorname{Tr}\mathsf Z^3 \operatorname{Tr}\overline{\mathsf Z}^{\,3}\rangle_{U(N),0} =3\left(\frac{\lambda G}{2}\right)^3 \left(1+\frac1{N^2}\right). \tag{N08.6}$$

The unnormalized rescaled two-point function is order one. Replacing each trace by $\operatorname{Tr}/N$ divides it by another $N^2$. Returning instead to canonical fields multiplies a correlator with total inserted length $J_{\rm tot}$ by $g^{-J_{\rm tot}}$; its fixed-$\lambda$ power is $N^{J_{\rm tot}/2+2-2h-b}$. These are different observable normalizations, not contradictory genus rules.

Restore the actual traceless theory

The pure double-line propagator has so far included a free Abelian center. Restore $SU(N)$ exactly by writing $Z=X+cI_N$, where $\operatorname{Tr}X=0$ and the independent complex Gaussian center has $\langle c\bar c\rangle=v/N$. Then

$$\operatorname{Tr}Z^3 =\operatorname{Tr}X^3+3c\operatorname{Tr}X^2+Nc^3.$$

The three terms are mutually orthogonal in their free two-point function: they have different holomorphic center degrees. The prerequisite gives $\langle\operatorname{Tr}X^2\operatorname{Tr}\bar X^2\rangle =2(N^2-1)v^2$. Using $\langle c^3\bar c^3\rangle=3!(v/N)^3$ therefore gives

$$\begin{aligned} \langle\operatorname{Tr}X^3\operatorname{Tr}\bar X^3\rangle &=3v^3(N^3+N) -\frac{18(N^2-1)}N v^3-\frac6N v^3\\ &=\boxed{\frac{3(N^2-1)(N^2-4)}N\,v^3}. \end{aligned} \tag{N08.7}$$

This finite-$N$ result is positive for $N\geq3$ and vanishes for $N=2$. Relative to the leading $3N^3v^3$, it is $1-5/N^2+4/N^4$. In particular the negative subleading coefficient is not a negative number of genus-one surfaces. It includes the trace-removal terms in (N08.1). The convenient $U(N)$ topology and the exact $SU(N)$ answer must be distinguished.

All statements here concern local field contractions. No global equivalence between $U(N)$ and a direct product $SU(N)\times U(1)$ has been assumed.

Practice

1. Length four and the range of an approximation

In the same $U(N)$ Gaussian model, compute the free norm of $\operatorname{Tr}Z^4$ by counting the faces of all $24$ pairings. Give the error relative to its planar term and the smallest integer $N$ for which that error is at most one percent. Does this establish a bound for operators whose length grows with $N$?

Hint 1

Let $c=(1234)$ denote advancing one slot. Eliminate the $j$ indices in (N08.2). The free $i$ indices are the cycles of $c\pi^{-1}c\pi$.

Hint 2

Four pairings reverse cyclic order and have four faces. Check the remaining pairings for two faces, retaining the factor $v^4$.

Solution

The relation is $i_a=i_{c\pi^{-1}c\pi(a)}$, so each cycle supplies one independent color sum. The four cyclic-order-reversing pairings have $F=4$; the other twenty have $F=2$. Euler’s formula with $V=2,E=4$ assigns genus zero and one respectively. Hence

$$\langle\operatorname{Tr}Z^4\operatorname{Tr}\bar Z^4\rangle_{U(N),0} =v^4(4N^4+20N^2).$$

The correction divided by the leading term is $5/N^2$, so $N\geq23$ meets the stated one-percent criterion. The error divided by the exact answer would instead be $5/(N^2+5)$; the denominator of an error estimate must be specified.

This result holds at fixed length four. The number of pairings grows with length: even the coefficient multiplying $N^{-2}$ need not remain bounded in a family of growing operators. A fixed-length genus expansion does not prove a uniform approximation in that different limit. Minahan, qualification following (5.7).

2. One trace versus two traces

Still in the $U(N)$ bookkeeping model, let $P=\operatorname{Tr}Z^4$ and $Q=(\operatorname{Tr}Z^2)^2$. Compute their two-point Gram matrix, meaning the matrix of pairwise free norms and overlaps. Show that the unit-normalized overlap tends to zero as $1/N$. Is this by itself an anomalous-dimension mixing calculation?

Hint 1

The slots of $P$ advance by $\alpha=(1234)$, while those of $Q$ advance by $\beta=(12)(34)$. Count cycles of $\alpha\pi^{-1}\beta\pi$. For the norm of $Q$, use $\beta$ at both ends.

Hint 2

For the mixed overlap, sixteen permutations have three faces and eight have one. For the norm of $Q$, eight have four faces and sixteen have two.

Solution

The complete Gram matrix in the ordered basis $(P,Q)$ is

$$\mathcal G=v^4 \begin{pmatrix} 4N^4+20N^2&16N^3+8N\\ 16N^3+8N&8N^4+16N^2 \end{pmatrix}.$$

The mixed graph has three trace disks. Every trace is connected to the one length-four trace, so (N08.5) gives $N^{-1}$ for its planar rescaled-field overlap. Its canonical-field version has $J_{\rm tot}=8$ and therefore $N^3$, as the answer shows. Some norm-$Q$ contractions are disconnected; applying the connected formula to their entire union would be incorrect.

Dividing the mixed entry by the square root of the two diagonal entries gives

$$\rho_N= \frac{\sqrt2(2N^2+1)} {N\sqrt{(N^2+5)(N^2+2)}} =\frac{2\sqrt2}{N}+O(N^{-3}).$$

This is a free inner-product overlap. It shows that trace number is not an exact orthogonality label at finite $N$. No interaction insertion, logarithmic divergence or renormalization matrix has been calculated, so it does not establish an anomalous dimension. At $N=1$, $P=Q$ and $\rho_1=1$, a useful check of the exact expression.

3. A finite-rank identity defeats an asymptotic normalization

For a traceless complex two-by-two matrix $X$, prove that $\operatorname{Tr}X^3=0$ and $\operatorname{Tr}X^4=(\operatorname{Tr}X^2)^2/2$. Then, for $SU(N)$ with $N\geq3$, normalize $\operatorname{Tr}X^3$ to have two-point coefficient one multiplying $|x-y|^{-6}$. Explain why that normalization cannot be continued to $N=2$.

Hint 1

Use the two-by-two characteristic equation $X^2-(\operatorname{Tr}X)X+\det(X)I=0$. Tracelessness also gives $\det X=-\operatorname{Tr}X^2/2$.

Hint 2

In (N08.7), insert $v=1/(8\pi^2|x-y|^2)$ and divide the operator by the square root of its distance-independent coefficient.

Solution

Cayley–Hamilton gives $X^2=(\operatorname{Tr}X^2)I/2$. Multiplying by $X$ and tracing makes the cubic trace vanish. Squaring and tracing gives $\operatorname{Tr}X^4=2(\operatorname{Tr}X^2)^2/4$. These are operator polynomial identities for the free commuting scalar entries; they require no restriction to diagonal field configurations.

For $N\geq3$, define

$$\widehat{\mathcal O}_3= (8\pi^2)^{3/2} \sqrt{\frac{N}{3(N^2-1)(N^2-4)}}\, \operatorname{Tr}X^3.$$

Equation (N08.7) then gives $\langle\widehat{\mathcal O}_3(x) \widehat{\overline{\mathcal O}}_3(y)\rangle_0=|x-y|^{-6}$. At $N=2$ there is a zero operator, not an operator with a very small positive norm. The divergent prefactor cannot normalize it. Finite-rank trace relations must be imposed before selecting an independent operator basis.

Color topology now predicts powers of $N$ with the normalization and gauge-group correction stated. It does not determine the functions of $\lambda$ multiplying those powers, prove convergence of their perturbation series, or establish a strong-coupling answer. Continue with Why composite operators mix to calculate what interaction insertions do to operators with the same quantum numbers.

A first AdS5 times S5 dictionary compares this color expansion with gravitational interactions. The coupling normalization and curvature scale explain why large rank alone does not guarantee a smooth string geometry.