Protected operators and unprotected operators

Two operators built from two scalars can have the same engineering dimension and different quantum dimensions. We will identify the distinction before computing a Feynman diagram. Then a short perturbative check will show that the distinction has an observable consequence.

Start from the scalar vacuum space and what BPS and protected mean. Here we return to the conformal vacuum, with all scalar expectation values zero. The theory is four-dimensional $\mathcal N=4$ $SU(N)$ Yang–Mills, $N\geq2$, with sixteen real Poincaré supercharges and sixteen conformal partners. An $\mathcal N=1$ description makes four Poincaré charges manifest.

The six Hermitian adjoint scalars are $X^I$, $I=1,\ldots,6$, with canonical kinetic term $\sum_I\operatorname{Tr}(\mathscr D X^I)^2$, $\mathscr D=\partial-ig[A,\,\cdot\,]$, and ordinary fundamental trace $\operatorname{Tr}(t^at^b)=\delta^{ab}/2$. Local operators must be gauge invariant. Products at one point below mean renormalized composites; at zero coupling they can be defined by normal ordering. The expressions specify their weak-coupling representatives, not a regulator-independent prescription for multiplying arbitrary distributions.

First separate the R-symmetry representations

Define the symmetric quadratic array and its trace:

$$B^{IJ}=\operatorname{Tr}(X^IX^J)=B^{JI}, \qquad \mathcal K=\sum_{I=1}^6B^{II}. \tag{N06.1}$$

Symmetry follows from the color trace, not from commuting color matrices. Under $SO(6)_R$, $B$ is a symmetric rank-two tensor. It has $6\cdot7/2=21$ components. Removing its one trace gives

$$\mathcal O^{IJ} =B^{IJ}-\frac{\delta^{IJ}}6\mathcal K, \qquad \sum_I\mathcal O^{II}=0. \tag{N06.2}$$

Thus the quadratic operators separate into a twenty-dimensional traceless tensor and a singlet. In $SU(4)_R$ notation these representations are called $\mathbf{20}'=[0,2,0]$ and $\mathbf1=[0,0,0]$. The three integers are Dynkin labels identifying an irreducible representation; the prime distinguishes this twenty-dimensional representation from other representations of the same dimension. No representation table is needed for the subtraction in (N06.2).

Now choose

$$Z=\frac{X^1+iX^2}{\sqrt2},\qquad \mathcal O_2=\operatorname{Tr}Z^2 =\frac12\bigl(\mathcal O^{11}-\mathcal O^{22} +2i\mathcal O^{12}\bigr). \tag{N06.3}$$

The trace subtraction cancels. Equivalently, write $Z=Y_IX^I$ with $Y=(1,i,0,0,0,0)/\sqrt2$. Its complex bilinear square is $Y_IY_I=0$, although its positive norm is $\overline Y_IY_I=1$. Therefore $Y_IY_J$ is traceless. These two contractions must not be confused.

For example, in the $SU(2)$ color algebra choose $Z=z\,\mathrm{diag}(1,-1)/2$. Then $\operatorname{Tr}Z^2=z^2/2$, so this is a nonzero operator despite $\operatorname{Tr}Z=0$. Gauge rank and R-representation dimension have different meanings: the color rank here is one, and the protected quadratic tensor still has twenty components.

Turn chirality into a dimension equation

Radial quantization associates a local primary with a state on $S^3$. The dilation generator has eigenvalue $\Delta$, and the radial adjoint of a Poincaré supercharge is a conformal supercharge. A superconformal primary is annihilated by all conformal supercharges. These are the positive-norm assumptions from the prerequisite; they are not the massive-particle algebra on the Coulomb branch. The conformal adjoints are reviewed in Simmons-Duffin, (108)–(116).

Choose the manifest $\mathcal N=1$ subalgebra with $R_1(\theta)=1$. Its supercharges have charges $-1$ and $+1$, and the three adjoint chiral scalars have $r=2/3$. On a scalar primary, the spin terms in the radial anticommutator have zero expectation. Normalize the two spinor components by a common positive factor so that their first-descendant Gram matrices are

$$G_{\bar q}=\left(\Delta-\frac32r\right)I_2, \qquad G_q=\left(\Delta+\frac32r\right)I_2. \tag{N06.4}$$

For example, $\|\bar q_{\dot\alpha}|\mathcal O\rangle\|^2 =\langle\mathcal O|\{\bar q_{\dot\alpha}^{\dagger}, \bar q_{\dot\alpha}\}|\mathcal O\rangle$ for a unit-normalized scalar primary, since the conformal charge already kills the state. The relative factor $3/2$ belongs to the four-dimensional superconformal algebra. It is not the three-dimensional coefficient. These scalar relations agree with CDI, tables 10–11; first-level positivity alone is not a full multiplet classification.

A scalar chiral primary obeys $\bar q_{\dot\alpha}|\mathcal O\rangle=0$ for both components. Positivity then gives

$$\Delta=\frac32r. \tag{N06.5}$$

The quadratic $\mathcal O_2$ is chiral, with $r=4/3$, and hence has $\Delta=2$ whenever its chiral primary survives. The R generator here is the fixed superconformal generator inside $SU(4)_R$, not an arbitrary trial flavor mixture. An elementary gauge-dependent $Z$ is not itself a gauge-invariant local state; applying (N06.5) to $\mathcal O_2$ avoids that problem.

Why this operator is half-BPS

The full theory supplies more annihilating charges than the manifest chiral condition. Recall the antisymmetric scalar tensor $\mathsf X^{AB}$, with $Z=\mathsf X^{14}$, and undotted charges $Q_{\alpha C}$ in $\overline{\mathbf4}$. The scalar transformations have the index structure

$$[Q_{\alpha C},\mathsf X^{AB}] \ \propto\ \delta_C^A\chi_\alpha^B-\delta_C^B\chi_\alpha^A, \qquad [\bar Q_{\dot\alpha}^{\,C},\mathsf X^{AB}] \ \propto\ \epsilon^{ABCD}\bar\chi_{\dot\alpha D}. \tag{N06.6}$$

Only the zero entries are used here; the nonzero common transformation coefficients are suppressed. Setting $A=1,B=4$ gives

$$Q_{\alpha2}\mathcal O_2=Q_{\alpha3}\mathcal O_2 =\bar Q_{\dot\alpha}^{\,1}\mathcal O_2 =\bar Q_{\dot\alpha}^{\,4}\mathcal O_2=0. \tag{N06.7}$$

The product rule preserves these zeros inside the trace. There are four undotted and four dotted components: half of the sixteen components of the complexified Poincaré algebra. This is the conventional half-BPS classification. The theory still has sixteen real Lorentzian Poincaré charges; radial adjoints of the displayed components are conformal charges, not another sixteen Poincaré charges.

One can check the count without relying on names. Represent a scalar index pair by $c_A^\dagger c_B^\dagger|0\rangle$, with four genuine fermionic oscillators satisfying $\{c_A,c_B^\dagger\}=\delta_{AB}$. A free supercharge removes an occupied internal index or adds an unoccupied one, while creating a bosonic spinor oscillator:

$$Q_{\alpha A}\sim a_\alpha^\dagger c_A,\qquad \bar Q_{\dot\alpha}^{\,A}\sim b_{\dot\alpha}^\dagger c_A^\dagger. \tag{N06.8}$$

On the $14$ state, removal vanishes for $A=2,3$ and addition vanishes for $A=1,4$. Acting on both scalar factors gives (N06.7) again, including the fermionic signs. This is a free representation check based on Minahan, (4.7)–(4.10), not a proof by free oscillators of an interacting nonrenormalization theorem.

That last step needs the full representation theory. In a unitary $\mathcal N=4$ SCFT, scalar half-BPS primaries in $[0,p,0]$ have $\Delta=p$. These multiplets cannot occur as the constituents into which a long multiplet splits at its unitarity threshold: they are absolutely protected under continuous deformations preserving this algebra. This is the additional input supplied by CDI, (2.53)–(2.54). It applies at finite $N$ to every nonzero independent operator of this type. Finite-color trace identities can reduce the independent operators; they do not shift a surviving half-BPS dimension.

For $p=2$, this is the stress-tensor multiplet. Every component of its $\mathbf{20}'$ primary has dimension two because R rotations commute with dilations. A neutral component need not be chiral under our fixed $\mathcal N=1$ choice, or be annihilated by the same eight components as $\mathcal O_2$. Belonging to the protected multiplet is the relevant statement.

Konishi can change its dimension

The singlet $\mathcal K$ in (N06.1) is the weak-coupling representative of the Konishi primary. In complex scalar notation,

$$\mathcal K=2\sum_{i=1}^3\operatorname{Tr}(\phi_i^\dagger\phi_i), \qquad \phi_i=\frac{X^{2i-1}+iX^{2i}}{\sqrt2}. \tag{N06.9}$$

It is R-neutral and is not chiral: varying $\phi_i^\dagger$ produces a nonzero fermion-scalar composite. At zero coupling $\Delta_0=2$, so both first-level matrices in (N06.4) are $2I_2$, not zero. Free Konishi nevertheless has null descendants at higher level. It is semishort at the scalar-singlet long-multiplet threshold $\Delta=2$.

The full unitarity conditions allow a singlet long primary with $\Delta>2$. The recombination rule allows the free semishort multiplet to join additional free-point multiplets when the interaction is turned on. This differs from the absolute protection of $[0,2,0]$. The boundary and recombination statements follow from CDI, tables 19–20 and (2.53). Thus free shortening alone cannot prove that Konishi stays at dimension two.

A perturbative diagnostic confirms that it moves. Let $\lambda=g^2N$ and take the planar, weak-coupling limit. We use the sourced one-loop result, specialized to symmetric quadratic polarizations $T_{IJ}$:

$$\mathcal O_T=T_{IJ}B^{IJ},\qquad (\Gamma^{(1)}T)_{IJ} =\frac{\lambda}{8\pi^2}\delta_{IJ}\sum_KT_{KK}. \tag{N06.10}$$

Here $\Gamma^{(1)}$ is the order-$\lambda$ change of the dilation operator, not a classical R generator. Equation (N06.10) follows by restricting Minahan, (5.24)–(5.26) to two fields: exchange acts trivially, and the two cyclic neighbor terms each contribute half the flavor trace. Its diagrammatic derivation is additional dynamical input; we are applying it, not deriving a loop integral from shortening.

The source’s canonical matrix scalar is $\sqrt2$ times ours: this matches its propagator (5.4) and quartic vertex (5.11) to our trace convention with the same $g$. We do not use its large-$N$ matrix propagator as an exact finite-$N$ correlator.

For a traceless $T$, (N06.10) is zero. For the singlet $T_{IJ}=\delta_{IJ}$, its trace is six, giving

$$\Delta_{\mathcal O^{IJ}}=2,\qquad \Delta_{\mathcal K} =2+\frac{3\lambda}{4\pi^2}+O(\lambda^2) \quad\text{in this perturbative limit}. \tag{N06.11}$$

The first equality has the independent exact protection argument. The positive one-loop coefficient in the second is evidence of a nonconstant dimension, not its exact all-coupling value.

For a useful consequence, $B^{11}=\mathcal O^{11}+\mathcal K/6$ combines the protected and unprotected sectors. At nonzero coupling it is not one scaling eigenoperator. A shared engineering dimension does not justify assigning the same exact dimension to each term.

Know what has actually been protected

In the conformal vacuum, a scalar scaling eigenoperator has a separated two-point function proportional to $|x|^{-2\Delta}$. Protecting $\Delta$ fixes this exponent. It does not by itself fix the coefficient, a higher-point function or a line expectation value. Multiplying an operator by a coupling-dependent nonzero constant already changes its coefficient without changing its representation. Further nonrenormalization statements require their own normalization and Ward identities.

The next concrete question is therefore a normalization calculation: what coefficient follows from the explicitly trace-normalized fields at zero coupling? Keep that question separate from the shortening argument established here.

Exercises

1. Find the protected part of a neutral operator

Consider $\mathcal A=\operatorname{Tr}[(X^1)^2+(X^2)^2]$. Separate it into a traceless tensor component and a multiple of $\mathcal K$. Compare with $\mathcal B=\operatorname{Tr}[(X^1)^2+(X^2)^2-(X^3)^2-(X^4)^2]$. Both have zero charge under rotations in each of the three scalar planes. Does this make both unprotected?

Hint 1

The coefficient matrices are diagonal. Their traces are two and zero.

Hint 2

Use $T=T_0+(\operatorname{tr}T)I_6/6$. A zero Cartan weight does not specify an entire R representation.

Solution

The decomposition is

$$\mathcal A =\mathcal O^{11}+\mathcal O^{22}+\frac13\mathcal K, \qquad \mathcal B =\mathcal O^{11}+\mathcal O^{22} -\mathcal O^{33}-\mathcal O^{44}.$$

The first operator combines the protected tensor with the singlet. It is not a single scaling eigenoperator once their dimensions differ. The second lies entirely in the protected $\mathbf{20}'$, so its dimension is two. Under the one-loop diagnostic, $\Gamma^{(1)}\mathcal A=\lambda\mathcal K/(4\pi^2)$ and $\Gamma^{(1)}\mathcal B=0$. These agree with applying $\gamma_{\mathcal K}^{(1)}$ to the singlet coefficient $1/3$.

Neutrality under the three commuting rotations is weaker than being an R singlet: a nontrivial representation can contain zero-weight states. Nor is $\mathcal B$ required to obey the same fixed-subalgebra chirality condition as $\operatorname{Tr}Z^2$.

2. Change the null polarization

Let $Z_\rho=(X^1+i\rho X^2)/\sqrt2$ with real $\rho$. Decompose $\operatorname{Tr}Z_\rho^2$ into its traceless tensor and singlet parts. For which $\rho$ is it purely protected? Compute its image under (N06.10), and explain why a nonzero image need not be proportional to the original operator.

Hint 1

The relevant trace is the complex bilinear $Y_\rho\cdot Y_\rho$, not the positive norm $\overline Y_\rho\cdot Y_\rho$.

Hint 2

$Y_\rho\cdot Y_\rho=(1-\rho^2)/2$. Contract (N06.2) with $Y_{\rho I}Y_{\rho J}$.

Solution

Direct expansion gives

$$\operatorname{Tr}Z_\rho^2 =\frac12\left(\mathcal O^{11}-\rho^2\mathcal O^{22} +2i\rho\mathcal O^{12}\right) +\frac{1-\rho^2}{12}\mathcal K.$$

The singlet coefficient vanishes only at $\rho=\pm1$. These give the chosen chiral highest-weight representative and its conjugate, both protected. In contrast the positive norm $(1+\rho^2)/2$ never vanishes, so it could not detect tracelessness.

For general $\rho$,

$$\Gamma^{(1)}\operatorname{Tr}Z_\rho^2 =\frac{\lambda(1-\rho^2)}{16\pi^2}\mathcal K.$$

This also equals the singlet coefficient times $3\lambda/(4\pi^2)$, an independent factor check. Unless the protected part is absent, the original operator is a sum of two scaling sectors, so a single anomalous dimension is not defined for that sum.

3. Finite color rank is not a shift of a protected dimension

For a traceless complex two-by-two color matrix $Z$, derive the identities for $\operatorname{Tr}Z^3$ and $\operatorname{Tr}Z^4$ in terms of $\operatorname{Tr}Z^2$. Interpret them for half-BPS local operators. Contrast this with a proposed interacting Konishi primary of dimension $2+\epsilon$, $\epsilon>0$: which assertion is excluded by representation theory, and which is allowed?

Hint 1

The two-by-two characteristic equation gives $Z^2-\tfrac12\operatorname{Tr}(Z^2)I_2=0$. It applies without diagonalizing $Z$.

Hint 2

A trace identity removes an independent operator. Konishi instead changes the dimension of a nonzero operator in an allowed long representation.

Solution

Multiplying the characteristic equation by $Z$ and taking the trace gives $\operatorname{Tr}Z^3=\tfrac12\operatorname{Tr}(Z^2)\operatorname{Tr}Z=0$. Squaring it and tracing gives

$$\operatorname{Tr}Z^4 =\frac12\bigl(\operatorname{Tr}Z^2\bigr)^2.$$

At the free point these are exact polynomial identities. In the half-BPS operator algebra they express the corresponding finite-rank relations, with coincident products defined in a compatible renormalized basis. The cubic single trace supplies no nonzero primary at $N=2$. The quartic is not independent of the product of two quadratic operators; the surviving half-BPS operator has dimension four. Arbitrary normalizations can change the written coefficient relating representatives, not the existence of the relation or the protected dimension.

A nonzero half-BPS $[0,4,0]$ primary with dimension $4+\epsilon$ is excluded while the full unitary $\mathcal N=4$ algebra is maintained. A scalar-singlet Konishi primary with dimension $2+\epsilon$ is allowed and is realized perturbatively. The latter statement neither requires an independent cubic trace nor follows from a color trace identity.

Next, compute a first gauge-invariant correlator. The free Wick calculation fixes the normalization of this protected quadratic operator and separates the finite-color answer from its large-N approximation.