Enter the superconformal bootstrap

An exact partition function gives a number. A conformal four-point function contains infinitely many dimensions and interaction coefficients. The bootstrap asks which choices of those data could belong to a consistent theory. Its first useful result need not require a large optimization program: a single protected coefficient can already rule out a proposed theory.

Prepare with operators and the OPE, shortening and protected quantities, and the reproduction-to-research method. Either the N=4 correlation-function entrance or the ABJM supersymmetry entrance is sufficient; the four-dimensional setup needed below is introduced here. The calculation is an established analytic reproduction, followed by a controlled change of assumption and a numerical extraction test.

Crossing before supersymmetry

Let $\phi$ be an identical real scalar primary with dimension $\Delta_\phi$ and two-point function $\langle\phi(x)\phi(0)\rangle=|x|^{-2\Delta_\phi}$. At separated Euclidean points, conformal symmetry leaves two invariant ratios,

$$\begin{gathered} u=\frac{x_{12}^2x_{34}^2}{x_{13}^2x_{24}^2},\qquad v=\frac{x_{14}^2x_{23}^2}{x_{13}^2x_{24}^2},\\ \langle\phi_1\phi_2\phi_3\phi_4\rangle =\frac{g(u,v)}{(x_{12}^2x_{34}^2)^{\Delta_\phi}}. \end{gathered} \tag{D11.1}$$

Here $x_{ij}=x_i-x_j$. A conformal transformation can place three points at $0$, $1$ and infinity and the fourth at $z$ in a plane. Then $u=z\bar z$ and $v=(1-z)(1-\bar z)$; in Euclidean configurations $\bar z$ is the complex conjugate. Analytic continuation later permits independent variables, but we will not need a Lorentzian limit.

Interchanging points 1 and 3 leaves the correlator unchanged and interchanges $u,v$. The ratio of the two prefactors is $(u/v)^{\Delta_\phi}$, so

$$v^{\Delta_\phi}g(u,v)=u^{\Delta_\phi}g(v,u). \tag{D11.2}$$

This is crossing symmetry. It compares two expansions of the same observable, rather than comparing different theories. Pairing $(12)$ and $(34)$ and inserting the OPE gives

$$g(u,v)=1+\sum_{\mathcal O\ne1}p_{\mathcal O} g_{\Delta,\ell}(u,v),\qquad p_{\mathcal O}=|\lambda_{\phi\phi\mathcal O}|^2\geq0. \tag{D11.3}$$

Choose an orthonormal basis of physical states in radial quantization. The contribution of one primary and all its descendants is its conformal block $g_{\Delta,\ell}$. Positivity follows because the coefficient is the norm of an OPE amplitude in that basis. With a nonorthogonal descendant basis, the projector contains its inverse Gram matrix; omitting it changes the block. Identical-scalar Bose symmetry keeps even spins. The scalar block is not simply $u^{\Delta/2}$: that is only its leading OPE term.

As a check, three Gaussian Wick pairings give $g=1+u^{\Delta_\phi}+(u/v)^{\Delta_\phi}$. Substitution satisfies D11.2 term by term after permutation. It is a useful crossing test, although a generalized-free correlator alone need not supply the local stress tensor of the desired interacting theory.

To see the logic of numerical exclusion, define $F_{\mathcal O}=v^{\Delta_\phi}g_{\Delta,\ell}(u,v)-u^{\Delta_\phi}g_{\Delta,\ell}(v,u)$, including $F_1=v^{\Delta_\phi}-u^{\Delta_\phi}$. Crossing says $F_1+\sum p_{\mathcal O}F_{\mathcal O}=0$. A linear operation $\alpha$ with $\alpha(F_1)>0$ and $\alpha(F_{\mathcal O})\geq0$ on every allowed nonidentity operator gives a contradiction. Derivatives at $z=\bar z=1/2$ are common choices. Checking a finite sample of dimensions is not a proof of the required inequality on an unbounded continuum.

Which theory and which protected sector?

The benchmark concerns a unitary local four-dimensional N=4 SCFT, with 16 real Poincaré supercharges and their conformal partners. We take a single stress-tensor multiplet and assume the absence of conserved higher-spin-current multiplets. We are on flat space in its conformal vacuum, with no curved-background sources or preserved-subalgebra reduction of the physical theory. The construction below selects an N=2 subalgebra with eight Poincaré charges to organize protected operators; it does not change the full supersymmetry.

Its stress tensor belongs to a multiplet whose lowest operator $\mathcal O^{IJ}$ has dimension two and is a symmetric traceless tensor of $SO(6)_R\simeq SU(4)_R$. This representation has $6\cdot7/2-1=20$ components and is called $20'$. In SYM it is proportional to $\operatorname{Tr}(\Phi^I\Phi^J)-\delta^{IJ}\operatorname{Tr}(\Phi^K\Phi^K)/6$, with normalization fixed by its two-point function. The anomaly coefficient is $c=a$; for an SYM gauge algebra $\mathfrak g$, $c=\dim\mathfrak g/4$. Thus $SU(N)$ has $c=(N^2-1)/4$ and gauge rank $N-1$. Adding the free $U(1)$ center changes the theory and adds $1/4$ to $c$.

The following protected-sector construction is a sourced representation-theory input, not a new proof of the full N=4 Ward identities. Choose a nilpotent combination of superconformal generators, retain operators killed by it, and identify operators differing by an exact term. With an appropriate R polarization on a plane, antiholomorphic translations become exact. Correlators of the resulting classes depend only on the holomorphic coordinate. This is an auxiliary chiral algebra. Its two-dimensional state metric need not be positive; physical positivity will be imposed on the original four-dimensional multiplets.

Under the selected N=2 algebra, part of $\mathcal O^{IJ}$ is a triplet of an $SU(2)_F$ flavor symmetry commuting with that subalgebra. It becomes a holomorphic current $j^a(z)$, $a=1,2,3$, with

$$\begin{gathered} j^a(z)j^b(0)\sim\frac{k_{2d}\delta^{ab}}{z^2} +\frac{i f^{ab}{}_d j^d(0)}{z},\\ f^{abc}=\sqrt2\epsilon^{abc},\qquad k_{2d}=-2c. \end{gathered} \tag{D11.4}$$

The negative level is expected. It is neither a three-dimensional CS level nor a negative physical stress-tensor norm. The relation $k_{2d}=-2c$ and the map from four-dimensional multiplets are the specific N=4 inputs verified in BRV2. They cannot be imported unchanged into three-dimensional ABJM. ABJM has a different superconformal algebra and different block/shortening data, even in its maximally supersymmetric cases.

A protected crossing calculation

Divide the current four-point function by $k_{2d}^2$ and remove its position prefactor. Write the resulting tensor as

$$R^{abcd}(z)=A(z)\delta^{ab}\delta^{cd} +B(z)\delta^{ac}\delta^{bd}+C(z)\delta^{ad}\delta^{bc}. \tag{D11.5}$$

The double-pole terms in D11.4 give its three disconnected pairings. The simple poles give products of structure constants. For example, contracting their Levi-Civita tensors gives $f^{ace}f^{bde}=2(\delta^{ab}\delta^{cd}-\delta^{ad}\delta^{bc})$. Summing these pole contributions yields

$$\begin{aligned} A(z)&=1+\frac{2z^2}{k_{2d}(1-z)},\\ B(z)&=z^2-\frac{2z}{k_{2d}(1-z)},\\ C(z)&=\frac{z^2}{(1-z)^2}+\frac{2z}{k_{2d}}. \end{aligned} \tag{D11.6}$$

There is no additional regular term: as a function of an insertion position, a current correlator is meromorphic, its poles are fixed by the OPE, and its falloff at infinity fixes the remaining polynomial ambiguity. These expressions reproduce the normalized correlator in BRV2 (2.13), with the unrescaled OPE level kept in the connected terms.

Exchanging insertions 1 and 3 sends the first tensor structure to the third and leaves the second fixed. Since the holomorphic currents have weight one, the prefactor gives

$$A(z)=\frac{z^2}{(1-z)^2}C(1-z),\qquad B(z)=\frac{z^2}{(1-z)^2}B(1-z). \tag{D11.7}$$

For the first relation, the right-hand side is $1+2z^2/[k_{2d}(1-z)]$, exactly $A(z)$. Substituting into the second gives $z^2-2z/[k_{2d}(1-z)]$. We have derived and checked an actual protected crossing relation, including its tensor permutation. Requiring the singlet function alone to be invariant under $z\mapsto1-z$ would be incorrect because crossing mixes the flavor channels.

Project onto the singlet in the $(12)$ channel using $\delta^{ab}\delta^{cd}/3$. Summing the indices gives $3A+B+C$, so

$$\widehat h_1(z)=3+z^2+\frac{z^2}{(1-z)^2} -\frac{2z^2}{c(1-z)} =3+\left(2-\frac2c\right)(z^2+z^3)+O(z^4). \tag{D11.8}$$

The factor three in the identity term is a projector normalization, not an extra vacuum. Notice that D11.7 holds for any nonzero $c$. Crossing of this protected rational function by itself has not yet enforced physical four-dimensional positivity.

Build a block and extract one physical coefficient

For holomorphic global conformal symmetry, a primary $|h\rangle$ satisfies $L_1|h\rangle=0$ and $L_0|h\rangle=h|h\rangle$. Normalize it by $\langle h|h\rangle=1$. The algebra $[L_1,L_{-1}]=2L_0$ implies recursively

$$\langle h|L_1^nL_{-1}^n|h\rangle=n!(2h)_n, \qquad K_h(z)=z^h\sum_{n=0}^{\infty} \frac{(h)_n^2}{(2h)_n n!}z^n. \tag{D11.9}$$

Here $(h)_n=h(h+1)\cdots(h+n-1)$, with $(h)_0=1$. The two identical-external-operator three-point matrix elements each supply $(h)_n$; division by the descendant norm produces the coefficient. Thus $K_h=z^h\,{}_2F_1(h,h;2h;z)$. This is a global block including all holomorphic derivatives, not a full Virasoro block.

One check uses the Casimir equation $[z^2(1-z)\partial_z^2-z^2\partial_z]K_h=h(h-1)K_h$. Substitution of the series gives $a_{n+1}/a_n=(h+n)^2/[(2h+n)(n+1)]$, exactly D11.9. The OPE boundary condition chooses the $z^h$ solution instead of its other Casimir solution. For $h=2$, define the source’s rescaled block $h_2=-K_2/2$; then $h_2=-z^2(1+z+9z^2/10+\cdots)/2$. Its minus sign does not mean a physical squared OPE coefficient is negative.

Only three nonidentity multiplet types contribute to the $h_2$ term in the protected singlet. The following entries are the precise short-multiplet decomposition input from BRV2 Table 2, with physical coefficients in its Table 3:

Four-dimensional multiplet	Physical squared OPE coefficient	Contribution through the $h_2$ block
Stress-tensor multiplet, scalar $[0,2,0]$	$2/c$	$(1/2)h_2$ per unit coefficient
Quarter-BPS scalar $[2,0,2]$, dimension four	$b$	$-6h_2$ per unit coefficient
Semishort singlet with higher-spin currents	$\alpha_0$	$(3/2)h_2$ per unit coefficient

The square brackets are $SU(4)_R$ Dynkin labels, not spins or gauge charges. Other allowed terms start at higher powers or vanish in this protected projection. For degenerate operators, $b$ and $\alpha_0$ denote sums of their squared coefficients and remain nonnegative.

From D11.8 and $h_2=-z^2/2+\cdots$, the coefficient of $h_2$ is $-4+4/c$. Matching the table therefore gives

$$-4+\frac4c=\frac1c-6b+\frac32\alpha_0, \qquad b=\frac23-\frac1{2c}+\frac{\alpha_0}{4}. \tag{D11.10}$$

This is where four-dimensional unitarity enters. Our no-higher-spin hypothesis sets $\alpha_0=0$. Since $c>0$ and $b\geq0$, D11.10 implies

$$c\geq\frac34. \tag{D11.11}$$

The SYM check is immediate but informative: $SU(2)$ has $c=3/4$, giving $b=0$; $SU(3)$ has $c=2$, giving $b=5/12$. Saturation means that this multiplet has zero OPE weight in the selected product. It does not determine every unprotected dimension, fix the Yang–Mills coupling or prove a complete classification of SCFTs.

Change the hypothesis, then test a numerical extraction

Suppose the assumed absence of higher-spin multiplets is replaced by an independently justified upper bound $0\leq\alpha_0\leq\varepsilon$. D11.10 then implies only $c\geq6/(8+3\varepsilon)$. At $\varepsilon=1/3$ this weakens to $c\geq2/3$. The parameter $\varepsilon$ bounds an OPE weight of an exactly semishort multiplet; it is not a small anomalous dimension or a continuously varied gauge coupling. The result is necessary, not sufficient to construct a theory.

If no upper bound on $\alpha_0$ is imposed, $c<3/4$ instead requires $\alpha_0\geq2/c-8/3$. At $c=1/4$ the required minimum is $16/3$. This explains why the interacting-theory argument cannot exclude a free Abelian vector multiplet. At the free limit, short multiplets can combine into a long multiplet as the coupling is turned on. Their protected contributions cancel in the combination indicated by D11.10; one must not count the threshold long block and its short constituents twice.

Now hold $\alpha_0=0$ and $c$ fixed. A tempting numerical shortcut is to sample the rational function at a small real $s>0$ and define

$$b_s=\frac16\left[\frac1c-\frac{\widehat h_1(s)-3}{h_2(s)}\right]. \tag{D11.12}$$

This includes higher blocks. Dividing the series, rather than assuming they vanish, gives

$$b_s=b+\left(\frac25-\frac1{15c}\right)(s^2+s^3)+O(s^4). \tag{D11.13}$$

At $c=3/4$, a positive $b_s$ at finite $s$ is therefore expected even though $b=0$ exactly. Increasing arithmetic precision does not remove this bias.

There is also a simple rigorous bound on the analytic bias. For $3/4\leq c\leq2$ and $0<s<1$, write $F_2={}_2F_1(2,2;4;s)\geq1$ and $D(s)=\widehat h_1(s)-3-(-4+4/c)h_2(s)$. Its coefficients begin at $s^4$ and are

$$d_m=m-1-\frac2c -12\left(1-\frac1c\right)\frac{m-1}{m(m+1)},\qquad m\geq4.$$

They obey $0\leq d_m\leq m$: the expression is affine in $1/c$, so it suffices to check the two endpoints, where these inequalities reduce to positive integer polynomials. Hence

$$0\leq b_s-b=\frac{D(s)}{3s^2F_2(s)} \leq\frac{s^2(4-3s)}{3(1-s)^2}. \tag{D11.14}$$

The last step sums $\sum_{m\geq4}m s^m=s^4(4-3s)/(1-s)^2$. This bound concerns finite-$s$ contamination, not floating-point roundoff. At very small $s$, the subtraction $\widehat h_1(s)-3$ loses digits unless the precision is increased or its factored expression is used.

The following short computation uses the nonsubtractive expression and the independent Euler integral $F_2(s)=6\int_0^1t(1-t)(1-st)^{-2}dt$:

import mpmath as mp
mp.mp.dps = 60
c = mp.mpf(3)/4
for s in map(mp.mpf, ['0.1', '0.05', '0.025']):
    F2 = 6*mp.quad(lambda t: t*(1-t)/(1-s*t)**2, [0, 1])
    hminus3 = s*s*(1 + 1/(1-s)**2 - 2/(c*(1-s)))
    bs = (1/c + 2*hminus3/(s*s*F2))/6
    bound = s*s*(4-3*s)/(3*(1-s)**2)
    print(mp.nstr(s, 5), mp.nstr(bs, 12), mp.nstr(bound, 12))

The first two output columns are approximately 0.1, 0.00346102081132; 0.05, 0.000818950857840; and 0.025, 0.000199444287499. Each estimate lies below its printed bound. The estimate tends to zero with quadratic leading bias. Compare the integral with a separately summed descendant series, then change $s$; those test different errors. A numerical optimizer for unprotected dimensions needs more: the full superblocks, all allowed spins, continuous-dimension positivity, controlled derivative truncation and numerical conditioning. Our small coefficient extraction is not such an optimizer.

Reading sequence and a bounded next project

Literature checked 20 October 2026. CFT supplies the general OPE/projector and crossing preparation. BRV introduces the N=4 bootstrap setting; BRV2 provides the explicit protected decomposition used here. The latter’s normalization is used consistently above; formulas from its earlier short paper should not be combined without translating their block normalization.

A useful next reproduction is the full protected block sum in BRV2, followed by one separating functional with a stated range and tail control. A later direction is CDP’s use of integrated localization constraints to make bounds depend on the complex gauge coupling, with explicit $SU(2)$ and $SU(3)$ examples. Those numerical bounds and assumption-dependent islands are different claims from D11.11. This selected reading path is not a claim that the subject stops at those papers, nor that reproducing them would be novel.

For a first research note, report the coefficient or proposed gap, the complete allowed multiplet set, the functional, and the error or positivity certificate. A failed search for a separating functional means the chosen computation did not exclude the hypothesis; it does not construct an SCFT. An ABJM continuation requires its own three-dimensional stress multiplet and Ward identities before substituting any analogous protected coefficient.

Exercises

1. A crossing test that contains no information

For D11.2 define $f(z)=v^{\Delta_\phi}g(u,v)-u^{\Delta_\phi}g(v,u)$ along the real diagonal $z=\bar z$. Explain why testing only $z=1/2$ cannot exclude any proposed $g$. Which Taylor derivatives there can carry information?

Hint 1

Along the diagonal, $u=z^2$ and $v=(1-z)^2$. Exchange $z$ and $1-z$.

Hint 2

Write $z=1/2+t$ and compare $f(t)$ with $f(-t)$.

Solution

The definition itself gives $f(1-z)=-f(z)$, whether or not $g$ solves crossing. At $z=1/2$ the two terms coincide, so their difference vanishes identically. The same antisymmetry forces all even Taylor derivatives there to vanish. Odd derivatives, beginning with the first, can test crossing. For example, the deliberately incomplete identity-only proposal $g=1$ at $\Delta_\phi=2$ gives $f=(1-z)^4-z^4$, with $f'(1/2)=-1$. It passes the value test and fails the first derivative. A zero at the symmetric point is not a consistency certificate.

2. What does a negative coefficient rule out?

Take $c=1/2$. Compute $b$ if $\alpha_0=0$, then find the minimum allowed $\alpha_0$. Compare this with the weakened assumption $\alpha_0\leq1$. Does the rational current correlator fail crossing at this $c$?

Hint 1

Use D11.10 before making any claim about whether a theory exists.

Hint 2

The protected crossing identities in D11.7 did not use positivity.

Solution

With no higher-spin contribution, $b=2/3-1=-1/3$, contradicting positivity. Restoring $\alpha_0$ gives $b=-1/3+\alpha_0/4$, requiring $\alpha_0\geq4/3$. An upper bound of one is therefore still incompatible. Equivalently, $\varepsilon=1$ requires $c\geq6/11$, whereas $1/2<6/11$.

The rational current correlator continues to satisfy D11.7. The contradiction is with the four-dimensional multiplet decomposition, positivity and the additional upper bound, not with its rational crossing identity alone. Allowing $\alpha_0=4/3$ makes this one coefficient zero but does not establish all other consistency conditions or the existence of a local theory. Four-dimensional positive norms must not be imposed directly on the auxiliary negative-level current algebra.

3. Remove one numerical bias without overclaiming

At $c=3/4$, combine $b_s$ and $b_{s/2}$ to cancel their leading error. Determine the first surviving power and its sign. Prove the coefficient inequalities used in D11.14 over the whole stated range of $c$.

Hint 1

In D11.13 the coefficients of $s^2$ and $s^3$ are equal. Cancellation of one does not cancel the other.

Hint 2

For $d_m$, evaluate the endpoints $c=3/4$ and $c=2$, and use $m\geq4$.

Solution

Write $A=2/5-1/(15c)$. Then $b_{s/2}=b+A s^2/4+A s^3/8+O(s^4)$. The combination $\widetilde b_s=(4b_{s/2}-b_s)/3$ is

$$\widetilde b_s=b-\frac{A}{6}s^3+O(s^4).$$

At $c=3/4$, $A=14/45$, so the leading residual is $-7s^3/135$. A small negative extrapolated estimate is therefore compatible with the exact nonnegative value $b=0$. Its error is cubic, not quartic; increasing arithmetic precision cannot remove it.

At the lower endpoint, $d_m=m-11/3+4(m-1)/[m(m+1)]$. This is positive since $m\geq4$; also $4(m-1)/[m(m+1)]\leq1$, so $d_m<m$. At the upper endpoint, $d_m=m-2-6(m-1)/[m(m+1)]$. The subtracted fraction is at most $6/m\leq3/2$, so $d_m\geq m-7/2>0$, and its upper bound by $m$ is immediate. Affinity in $1/c$ proves both inequalities throughout the interval.

The new extrapolation needs its own error estimate if used as a certificate. The positive bound in D11.14 applies to $b_s-b$, not automatically to a signed linear combination. The analytic value remains the reference against which this numerical exercise is judged.