Compute a first gauge-invariant correlator

Protected and unprotected operators identified $\mathcal O_2=\operatorname{Tr}Z^2$ as a scalar of dimension two. That fixes the distance dependence of its conformal two-point function, but it does not tell us the coefficient for a particular normalization of $Z$. We will calculate that coefficient, keeping the finite number of colors throughout.

The setting is four-dimensional $\mathcal N=4$ $SU(N)$ Yang–Mills, $N\geq2$, in its conformal vacuum, with all scalar expectation values zero. The underlying Lorentzian theory has sixteen real Poincaré supercharges; an $\mathcal N=1$ presentation makes four manifest. We compute the zero-coupling, separated-point Euclidean correlator. No large-$N$ limit or nonrenormalization theorem enters the calculation.

Start with the component propagator

Use Hermitian fundamental generators $t^a$, $a=1,\ldots,N^2-1$, and the ordinary matrix trace:

$$\operatorname{Tr}(t^at^b)=\frac12\delta^{ab}, \qquad X^I=X^{Ia}t^a,\qquad Z=\frac{X^1+iX^2}{\sqrt2}=Z^at^a. \tag{N07.1}$$

The six real scalars have canonical free Euclidean action $S_0=\frac12\int d^4x\,\sum_{I,a}(\partial_\mu X^{Ia})^2$. The complex conjugate on this integration contour is $\bar Z=Z^\dagger$. Let $r=|x-y|>0$.

For one real component, the inverse kinetic operator satisfies $-\partial^2G(x)=\delta^{(4)}(x)$. Rotational invariance and the four-dimensional radial Laplacian give $G''+3G'/r=0$ away from the origin. The solution decaying at infinity is $G=c/r^2$. Integrating the Green-function equation over a small ball fixes its coefficient:

$$1=-\int_{S^3_r}dS\,\partial_rG =-(2\pi^2r^3)\frac{-2c}{r^3}=4\pi^2c, \qquad G(r)=\frac1{4\pi^2r^2}. \tag{N07.2}$$

The two independent real fields in $Z$ now give

$$\langle Z^a(x)\bar Z^b(y)\rangle=\delta^{ab}G(r), \qquad \langle Z^a(x)Z^b(y)\rangle=0. \tag{N07.3}$$

In the second expression the $X^1X^1$ and $i^2X^2X^2$ contractions cancel. This cancellation will remove self-contractions within $\operatorname{Tr}Z^2$ in the free theory. It does not remove the short-distance singularity as the two operator insertions approach each other.

Convert component indices into matrix indices

Write $Z_i{}^j$ for the matrix entries. We need a sum of two generators. To derive it, expand an arbitrary complex $N$-by-$N$ matrix $M$ into its trace and traceless parts:

$$M=\frac{\operatorname{Tr}M}{N}I_N +2\sum_a\operatorname{Tr}(t^aM)t^a.$$

Comparing coefficients of each entry of $M$ gives the completeness identity

$$\sum_a(t^a)_i{}^j(t^a)_k{}^l =\frac12\left(\delta_i^l\delta_k^j -\frac1N\delta_i^j\delta_k^l\right). \tag{N07.4}$$

The first term exchanges the two matrix indices. The second removes the identity matrix, which is not an $SU(N)$ generator. Thus

$$\langle Z_i{}^j(x)\bar Z_k{}^l(y)\rangle =\frac{G(r)}2 \left(\delta_i^l\delta_k^j -\frac1N\delta_i^j\delta_k^l\right). \tag{N07.5}$$

As a quick check, setting $j=i$ and summing gives zero. The propagator cannot produce a fluctuating trace in a traceless field.

Perform every contraction of the quadratic trace

Define the normal-ordered composite $\mathcal O_2=:\!\operatorname{Tr}Z^2\!:$ and its conjugate $\overline{\mathcal O}_2=:\!\operatorname{Tr}\bar Z^2\!:$. Normal ordering means subtracting contractions within one insertion, using the free vacuum. Here these subtractions vanish by (N07.3). Under a gauge transformation $Z(x)\mapsto U(x)Z(x)U(x)^{-1}$, each trace is separately invariant. No Wilson line between $x$ and $y$ is required.

In matrix indices, the expression to contract is

$$\sum_{i,j,k,l} \left\langle Z_i{}^j(x)Z_j{}^i(x) \bar Z_k{}^l(y)\bar Z_l{}^k(y) \right\rangle.$$

There are exactly two nonzero Wick pairings: the first $Z$ can contract with either $\bar Z$, after which the second pairing is fixed. Denote the propagator (N07.5) by $C_i{}^j{}_k{}^l$. Wick’s theorem gives

$$\langle\mathcal O_2(x)\overline{\mathcal O}_2(y)\rangle =\sum_{i,j,k,l} \left(C_i{}^j{}_k{}^l C_j{}^i{}_l{}^k +C_i{}^j{}_l{}^k C_j{}^i{}_k{}^l\right). \tag{N07.6}$$

There is no extra factor $1/2!$: the operator was defined without one. The two summands are equal by relabeling $k$ and $l$, but let us evaluate one before multiplying by two. Removing its common $G^2/4$, its four terms are

$$\begin{aligned} &\sum_{i,j,k,l} \left(\delta_i^l\delta_k^j-\frac{\delta_i^j\delta_k^l}{N}\right) \left(\delta_j^k\delta_l^i-\frac{\delta_j^i\delta_l^k}{N}\right)\\ &\hspace{1cm}=N^2-\frac{N}{N}-\frac{N}{N} +\frac{N^2}{N^2}=N^2-1. \end{aligned} \tag{N07.7}$$

For the first product, $l=i$ and $k=j$ leave two independent sums. Each cross term forces all four indices equal, leaving one sum. For the last product, $i=j$ and $k=l$ again leave two sums. In particular, the trace subtraction must be used in both propagators.

The complete answer is

$$\boxed{\displaystyle \langle\mathcal O_2(x)\overline{\mathcal O}_2(y)\rangle_0 =\frac{N^2-1}{2}\,G(r)^2 =\frac{N^2-1}{32\pi^4r^4}.} \tag{N07.8}$$

The subscript $0$ specifies zero coupling. The positive coefficient counts the $N^2-1$ adjoint components, with the operator and Wick normalizations included. The power $r^{-4}$ agrees with dimension two. The correlator $\langle\mathcal O_2\mathcal O_2\rangle_0$ instead vanishes: every possible propagator would join two holomorphic fields.

Work the smallest non-Abelian example

For $SU(2)$, use $t^a=\sigma^a/2$, where $\sigma^a$ are the Pauli matrices. An arbitrary complex adjoint field is

$$Z=\frac12 \begin{pmatrix} z_3&z_1-iz_2\\ z_1+iz_2&-z_3 \end{pmatrix}, \qquad \operatorname{Tr}Z^2=\frac12(z_1^2+z_2^2+z_3^2). \tag{N07.9}$$

Using $\langle z_a(x)\bar z_b(y)\rangle=\delta_{ab}G$, each square has two cross contractions. Therefore $\langle\mathcal O_2\overline{\mathcal O}_2\rangle_0 =\frac14(3)(2G^2)=3G^2/2$, as (N07.8) requires. The matrices being integrated over need not be diagonal or mutually commuting. Restricting them to a classical Coulomb vacuum would change the problem.

A direct Gaussian integral provides another check. Replace the fields at the two points by complex Gaussian component arrays $z_a$ and $w_a$ with unit self-covariance and $\langle z_a\bar w_b\rangle=c\,\delta_{ab}$, $|c|<1$. Construct the two matrices in (N07.9) and integrate their trace product. For $c=0.3$, the result is $0.135$; for $c=0.6i$, it is $-0.54$. Both equal $3c^2/2$. A complex off-diagonal covariance can give a complex or negative cross correlator; the covariance matrix itself remains positive. This finite Gaussian test checks the Wick coefficient, without representing an interacting path integral.

Normalize the operator and compare the Abelian center

An operator with unit separated two-point coefficient is

$$\widehat{\mathcal O}_2 =4\pi^2\sqrt{\frac{2}{N^2-1}}\,\mathcal O_2, \qquad \langle\widehat{\mathcal O}_2(x) \widehat{\overline{\mathcal O}}_2(y)\rangle_0 =\frac1{r^4}. \tag{N07.10}$$

This is an operator normalization, not a change in the gauge coupling. It also differs from replacing the ordinary trace by $\operatorname{tr}_N=\operatorname{Tr}/N$: that replacement divides the two-point coefficient by $N^2$ and leaves its other factors.

For comparison, a $U(N)$ adjoint has the additional generator $t^0=I_N/\sqrt{2N}$. Write $Z=Z_{\mathrm{su}}+z_0I_N/\sqrt{2N}$. In the free Gaussian theory, $z_0$ and the traceless components are independent, so

$$\operatorname{Tr}Z^2 =\operatorname{Tr}Z_{\mathrm{su}}^2+\frac12z_0^2, \qquad \langle\operatorname{Tr}Z^2(x)\operatorname{Tr}\bar Z^2(y)\rangle_{U(N),0} =\frac{N^2}{2}G(r)^2. \tag{N07.11}$$

The difference is $G^2/2$, exactly the center’s contribution. Equivalently, including $t^0$ cancels the subtraction in (N07.5). This is a decomposition of local fields and their free contractions; it does not assert that the global group $U(N)$ is the direct product $SU(N)\times U(1)$.

To compare with a common source, Minahan’s (5.3)–(5.5), including footnote 7, uses a matrix propagator $G\,\delta_i^l\delta_k^j$ and omits the traceless subtraction in its large-$N$ calculation. At the level of canonical fields the translation is $Z_{\mathrm{MIN}}=\sqrt2\,Z$. For length two its normalized expression therefore becomes

$$\Psi_{2,\mathrm{MIN}} =\frac{4\pi^2}{\sqrt2N}\operatorname{Tr}Z_{\mathrm{MIN}}^2 =\frac{4\pi^2\sqrt2}{N}\operatorname{Tr}Z^2.$$

Its exact $SU(N)$ coefficient is $1-1/N^2$, whereas its coefficient in the stated leading large-$N$ approximation is one. There is no disagreement with (N07.10): exact finite-$N$ normalization replaces $N$ by $\sqrt{N^2-1}$. At this length the difference just calculated comes from tracelessness; a general analysis of contraction topology is a further question.

Exercises

1. Change the flavor multiplicity

Let $W=(X^3+iX^4)/\sqrt2$ and $\mathcal P=:\!\operatorname{Tr}(ZW)\!:$ in the same $SU(N)$ theory. Compute $\langle\mathcal P(x)\overline{\mathcal P}(y)\rangle_0$, where $\overline{\mathcal P}=\operatorname{Tr}(\bar W\bar Z)$. Find the unit-normalized operator. Explain why cyclicity of the trace does not restore the factor of two in (N07.8).

Hint 1

The component covariances $\langle Z\bar W\rangle$ and $\langle W\bar Z\rangle$ vanish. Count the allowed pairings before summing color indices.

Hint 2

The trace gives $\mathcal P=\frac12\sum_a Z^aW^a$. There is only one way to match the two different flavors with their conjugates.

Solution

Wick’s theorem gives

$$\langle\mathcal P(x)\overline{\mathcal P}(y)\rangle_0 =\frac14\sum_{a,b} \langle Z^a\bar Z^b\rangle\langle W^a\bar W^b\rangle =\frac{N^2-1}{4}G(r)^2.$$

In matrix indices this is just one of the color sums in (N07.7). The second Wick pairing would require propagators between different flavors, so it contributes zero. Consequently $\widehat{\mathcal P}=8\pi^2\mathcal P/\sqrt{N^2-1}$ has correlator $r^{-4}$. Cyclicity says $\operatorname{Tr}ZW=\operatorname{Tr}WZ$; it does not turn the vanishing cross-flavor propagator into a nonzero one. This is the smallest example of the flavor-dependent multiplicities in Minahan’s (5.8)–(5.9).

2. Remove the center inside a matrix integral

Start with a $U(N)$ Gaussian matrix, not an $SU(N)$ one. Define $\mathcal U=\operatorname{Tr}Z^2$ and $\mathcal T=(\operatorname{Tr}Z)^2$. Find their two-point coefficient matrix in units of $G^2$. Show that $\mathcal S=\mathcal U-\mathcal T/N$ has the $SU(N)$ answer and is orthogonal to $\mathcal T$.

Hint 1

Use $Z=Z_{\mathrm{su}}+z_0I_N/\sqrt{2N}$. Then $\operatorname{Tr}Z=\sqrt{N/2}\,z_0$.

Hint 2

Only the center in $\mathcal U$ can contract with $\mathcal T$. Its elementary square has correlator $2G^2$.

Solution

The decomposition gives $\mathcal U=\mathcal O_{2,\mathrm{su}}+z_0^2/2$ and $\mathcal T=Nz_0^2/2$. Hence

$$\frac1{G^2} \begin{pmatrix} \langle\mathcal U\overline{\mathcal U}\rangle& \langle\mathcal U\overline{\mathcal T}\rangle\\ \langle\mathcal T\overline{\mathcal U}\rangle& \langle\mathcal T\overline{\mathcal T}\rangle \end{pmatrix} =\frac12 \begin{pmatrix} N^2&N\\ N&N^2 \end{pmatrix}.$$

Subtracting the center yields

$$\langle\mathcal S\overline{\mathcal S}\rangle =\left(\frac{N^2}{2}-1+\frac12\right)G^2 =\frac{N^2-1}{2}G^2,\qquad \langle\mathcal S\overline{\mathcal T}\rangle =\left(\frac N2-\frac N2\right)G^2=0.$$

Indeed $\mathcal S=\operatorname{Tr}Z_{\mathrm{su}}^2$ as an operator, not merely in this two-point function. For $N=1$ it vanishes, while the $U(1)$ center remains. This last check does not introduce a nontrivial $SU(1)$ gauge theory.

3. Diagnose a coincident-point contraction

Change the operator to $\mathcal B_{\mathrm{bare}}=\operatorname{Tr}(Z\bar Z)$ in $SU(N)$. Use a common free-field regulator with $\langle Z^a(x)\bar Z^b(x)\rangle =\delta^{ab}C_\epsilon$ and separated propagator $\delta^{ab}G_\epsilon(r)$ tending to (N07.2). Find its one-point function and full two-point function. Construct $\mathcal B=:\!\operatorname{Tr}(Z\bar Z)\!:$ and take the separated-point limit. Does the resulting $r^{-4}$ prove protection?

Hint 1

Now one pairing contracts the two fields within each insertion. Separate this disconnected contribution from contractions joining the insertions.

Hint 2

Write $\mathcal B_{\mathrm{bare}} =\frac12\sum_a Z^a\bar Z^a$. Recall the singlet $\mathcal K$ and traceless tensor $\mathcal O^{IJ}$ from the prerequisite.

Solution

Put $D=N^2-1$ for the number of adjoint components. A contraction within one insertion gives $\langle\mathcal B_{\mathrm{bare}}\rangle=DC_\epsilon/2$. For two insertions Wick’s theorem gives

$$\langle\mathcal B_{\mathrm{bare}}(x) \mathcal B_{\mathrm{bare}}(y)\rangle =\frac{D^2}{4}C_\epsilon^2 +\frac D4G_\epsilon(r)^2.$$

The first term is the product of the one-point functions. There is only one nonzero cross pairing: $Z(x)$ with $\bar Z(y)$ and $\bar Z(x)$ with $Z(y)$. Define $\mathcal B=\mathcal B_{\mathrm{bare}}-DC_\epsilon/2$. For $r>0$, removing the regulator gives $\langle\mathcal B(x)\mathcal B(y)\rangle_0 =DG(r)^2/4$. Normal ordering removes the self-contraction, not the singularity of this answer at $r=0$.

Finally, $\mathcal B=(\mathcal O^{11}+\mathcal O^{22})/2+\mathcal K/6$ at the level of these free representatives. It has a nonzero Konishi component. Its free dimension two therefore does not imply that it remains a single dimension-two primary at nonzero coupling. The contraction calculation and the protection argument answer different questions.

You can now normalize a short gauge-invariant scalar operator without guessing a factor from its length. The next question is how to organize the much larger set of contractions for longer traces, and which color powers survive when the number of colors becomes large.

Double lines and color counting answers that question by comparing two cubic-trace contractions, deriving the genus power, and restoring the exact traceless correction.

To turn four-point consistency into a quantitative restriction, enter the superconformal bootstrap. It introduces crossing, descendants and a protected analytic bound from this four-dimensional starting point.