Deformed spheres and CFT data

What does a fourth derivative of a sphere partition function measure? It contains connected four-point information, but it also remembers how the action depends nonlinearly on the source. Ignoring that dependence can produce a finite, reproducible wrong answer. We will expose the discrepancy in the smallest ABJM theory, then change the sphere shape while keeping the mass source fixed.

Prepare with the reproduction method, ABJM masses and squashing, and current normalizations. This is a reproduction and controlled training extension. It does not determine an unknown interacting four-point function or establish a new duality.

Fix two source directions

Use compact $U(1)_1\times U(1)_{-1}$ ABJM: $N=1$, levels exactly $+1,-1$, both gauge centers retained, and no extra quotient. Its local undeformed conformal theory is the free maximally supersymmetric multiplet, containing four complex scalars and their fermions. There are sixteen real Poincaré supercharges, with sixteen conformal partners at the conformal point. The sphere calculation uses the manifest three-dimensional $\mathcal N=2$ construction; the generic toric background supplies one rigid label direction of each R charge. This is a statement about the constructed subalgebra, not a count of every symmetry at a special undeformed point.

Keep the prerequisite’s toric metric, with $\ell_1=Lb$, $\ell_2=L/b$ and $f^2=\ell_1^2\sin^2\chi+\ell_2^2\cos^2\chi$. Its complete rigid sources are

$$H=\frac{i}{f},\qquad V=C=0,\qquad A^{(R)}=\frac12\left(1-\frac{\ell_1}{f}\right)d\varphi +\frac12\left(1-\frac{\ell_2}{f}\right)d\vartheta . \tag{D13.1}$$

The space is the unquotiented sphere with its usual spin structure. In particular, the round limit uses the positive auxiliary branch $H=i/L$. Take two dimensionless real masses $x,y$, meaning physical sources $x/L,y/L$. In the ordered chiral list $(A_1,A_2,B_2,B_1)$ choose

$$T_1=\frac12\operatorname{diag}(1,1,-1,-1),\qquad T_2=\frac12\operatorname{diag}(1,-1,1,-1),\qquad \operatorname{Tr}_4(T_aT_b)=\delta_{ab}. \tag{D13.2}$$

These are the first two Cartan generators of Binder–Chester–Pufu (3.7), after the displayed field ordering. They are flavor generators, not gauge generators. Each superpotential word has total charge zero. Freeze each background multiplet at $a_a=0$, $\sigma_a=(x,y)_a/L$ and $D_a=-H\sigma_a$, with vanishing gauginos. Thus changing a mass also changes its auxiliary source; it is not a change of only the fermion mass.

The charge eigenvalues of $xT_1+yT_2$ are $p,q,-q,-p$, where $p=(x+y)/2$ and $q=(x-y)/2$. In the integer-charge cover used in the prerequisite, the one-direction source is $M=x/2$. This fixes the factor of two when comparing derivatives.

Use the inherited double-sine prescription and Abel limit of the Cartan integral, with no added integer source counterterms or new dimensionful subtraction scale. Set $F_b=-\log Z_b$ on the branch continuous from zero real mass. The rank-one common-center integration already established in the prerequisite now gives

$$\boxed{Z_b(x,y)=D_b(p)D_b(q).} \tag{D13.3}$$

Indeed the two matter factors before that integration are $D_b(s-t+p)$ and $D_b(t-s-q)$; the normalized delta sets $s-t=0$, and $D_b$ is even. This is Kubo–Nosaka–Pang (4.11) with their ABJM masses $(m_1,m_2,m_3)=(x,y,0)$ and dual variables $(\zeta,m)=(p,-q)$. Their dual theory is a three-dimensional $\mathcal N=4$ gauge theory, not four-dimensional SYM. No large-rank approximation has entered (D13.3).

Reproduce the fourth derivative

At the round point $D_1(z)=1/(2\cosh\pi z)$, so

$$\begin{aligned} F_1(x,y)&=\log4+\log\cosh(\pi p)+\log\cosh(\pi q)\\ &=\log4+\frac{\pi^2}{4}(x^2+y^2) -\frac{\pi^4}{96}(x^4+6x^2y^2+y^4)+O((|x|+|y|)^6). \end{aligned} \tag{D13.4}$$

The expansion follows from $\log\cosh z=z^2/2-z^4/12+O(z^6)$; the factors $1/2$ in the generators must remain inside $p,q$. At $x=y=0$,

$$F_{xx}=\frac{\pi^2}{2},\qquad F_{xy}=0,\qquad \boxed{F_{xxxx}=F_{xxyy}=-\frac{\pi^4}{4}.} \tag{D13.5}$$

The boxed result reproduces BCP (3.43). A fourth derivative with respect to the corresponding dimensional mass has an extra factor $L^4$ at fixed radius. A negative fourth derivative does not contradict positivity of a current norm; that positivity constrains the relevant two-point coefficient.

For a normalized generator (D13.2), define $\tau_F$ by $\langle j_\mu(x)j_\nu(0)\rangle=\tau_F(\delta_{\mu\nu}\partial^2-\partial_\mu\partial_\nu)|x|^{-2}/(16\pi^2)$ at separated points. Here the four free chirals give $\tau_F=\sum_i(T_a)_{ii}^2=1$. In the enhanced theory the sourced Ward normalization is $\tau_F=c_T/16$, with $c_T=1$ for one real scalar. Equivalently, in this two-mass slice $c_T=32\operatorname{Re}F_{xx}/\pi^2=16$; Chester–Kalloor–Sharon (2.8) uses $m_\pm=y\pm x$, accounting for its factor 64. Comparing its stress-tensor tensor structure with the prerequisite gives

$$\tau_{RR}=\frac{c_T}{16}=1,\qquad C_T=\frac{3c_T}{32\pi^2}=\frac{3}{2\pi^2}. \tag{D13.6}$$

The equality of these flavor and R coefficients uses this normalized enhanced multiplet. It is not a rule for an arbitrary flavor generator or a generic $\mathcal N=2$ theory.

Keep the quadratic source term

Write a one-parameter action as $S(x)=S_0+x\mathcal V+x^2\mathcal W/2$. The symbols $\mathcal V=S_{,x}(0)$ and $\mathcal W=S_{,xx}(0)$ are integrated operators; the latter is sometimes called a seagull insertion. Expanding $-\log\langle e^{-x\mathcal V-x^2\mathcal W/2}\rangle_0$ in connected cumulants gives

$$F_{xxxx}=-\langle\mathcal V^4\rangle_c +6\langle\mathcal V^2\mathcal W\rangle_c -3\langle\mathcal W^2\rangle_c. \tag{D13.7}$$

For example, the cubic cumulant has three placements of $\mathcal W$, each supplying $-x^2/2$; multiplication by $4!$ produces the coefficient six after taking $-\log Z$. The formula assumes no cubic or quartic term in this chosen action. A more general source family requires those derivatives too.

Here the conformal scalar operator at $b=1$, in units $L=1$, is $-\nabla^2+3/4+w^2-iw$. The term $w^2$ produces $\mathcal W$; deleting it changes the theory being differentiated. BCP’s opposite sign for the imaginary linear term is translated by $w\mapsto-w$. Our paired masses make the resulting even derivatives identical.

An independent mode calculation makes the size of the error visible. Put $a_n=n^2-1/4$ for $n=1,2,\ldots$. Scalar modes have degeneracy $n^2$; the Dirac eigenvalues are $\pm i(n+1/2)$ with degeneracy $n(n+1)$ for each sign. A pair of chirals of masses $\pm w$ contributes the following fourth-order coefficient, which is convergent even though the undifferentiated determinant needs its prescription:

$$[w^4]F_{\rm pair}= \sum_{n=1}^\infty\left[ n^2\left(\frac1{a_n^2}-\frac{(2a_n+1)^2}{2a_n^4}\right) +\frac{n(n+1)}{(n+1/2)^4}\right]. \tag{D13.8}$$

The first bracket comes from $\log[(a_n+w^2)^2+w^2]$; the second from $-2n(n+1)\log[(n+1/2)^2+w^2]$. After multiplication by $4!$, partial fractions telescope to $-192\sum_{n\geq1}(2n-1)^{-4}=-2\pi^4$. Four chirals of masses $\pm x/2$ supply $2/2^4=1/8$ of this pair response, recovering (D13.5).

If the scalar $w^2$ term is omitted, its bracket becomes $-n^2/(2a_n^4)$. The difference between the complete and linear-source sums is

$$\frac{24}{8}\sum_{n=1}^\infty \left(-\frac{n^2}{a_n^2}-\frac{2n^2}{a_n^3}\right) =-\frac{9\pi^2}{4}. \tag{D13.9}$$

Thus the linear-insertion result is $-\pi^4/4+9\pi^2/4$, not the physical derivative. This reproduces the correction in BCP (3.40)–(3.43). These are source-action insertions, not freely adjustable background counterterms. Nor should the free-theory singlet contribution be carried unchanged into the strongly coupled formula (3.45).

Change the geometry at fixed real mass

Now keep the same charges, $x,y$, scale $L$ and prescription, but vary $b$. The prerequisite’s finite determinant difference uses

$$K_b(t)=\frac{\sinh[(b+b^{-1})t/2]}{\sinh(bt)\sinh(t/b)},\qquad \log\frac{D_b(w)}{D_b(0)}= \int_0^\infty\frac{dt}{t}K_b(t)[\cos(2wt)-1]. \tag{D13.10}$$

Consequently $F_{xx}(b;0,0)=2\int_0^\infty tK_b(t)\,dt$. Exchange of the two circle scales makes its first $b$ derivative vanish at one. Differentiating both denominator factors gives

$$\left.\partial_b^2K_b(t)\right|_1 =\frac{-t\coth t+2t^2\operatorname{csch}^2t}{\sinh t}, \qquad \boxed{F_{bbxx}(1;0,0)=2\pi^2-\frac{\pi^4}{4}.} \tag{D13.11}$$

To evaluate it, use $\int_0^\infty t/\sinh t\,dt=\pi^2/4$ and $\int_0^\infty t^3/\sinh t\,dt=\pi^4/8$, obtained by summing positive exponentials. The identity $(\operatorname{csch}t)''=\operatorname{csch}t+2\operatorname{csch}^3t$ and integration by parts then give $\int t^3\operatorname{csch}^3t\,dt=3\pi^2/4-\pi^4/16$. All boundary terms vanish. On a compact neighborhood of $b=1$ the differentiated kernels have integrable small-$t$ limits and uniform exponential tails, justifying differentiation under the integral.

Equation (D13.11) is also the rank-one specialization of CKS (3.7): $F_{m_+m_+}=\pi^2/4$, $F_{m_+m_+m_+m_+}=-\pi^4/8$, and the mixed $m_+,m_-$ derivative vanishes. It is a definite integrated response to this complete background family, not an arbitrary metric variation or a separated four-point function by itself.

The executed mode sums and real-kernel integrals agree with the formulas at 70-digit working precision; finite mode cutoffs and finite differentiation steps have separate errors. For example, the uncorrected pair-mode cutoff $n\leq100$ still misses the fourth response by about $0.001204$. More working digits do not remove that cutoff error.

A dated reading route and a bounded next calculation

The literature check for this dossier is dated 20 October 2026. BCP explains how mass derivatives constrain integrated stress-multiplet correlators, including source terms. CKS derives relations between squashing and mass derivatives using a specified complex mass path. KNP studies levels $\pm1$, with the dual adjoint mass tuned to $m_b=i(b^2-3)/(4b)$, and obtains the perturbative large-rank Airy expansion on its stated locus. Its derivation for odd $b^2\geq3$ and numerical continuation to other $b>1$ have different evidential status; neither is a generic exact answer at arbitrary level and real mass. The tuned $b=1$ endpoint instead has $m_b=-i/2$ and a divergent determinant, as its footnote 4 explains. Its dual variables and its three-dimensional matter theory must remain in the comparison.

A concrete next assignment is to extract the same fourth derivative at a chosen higher rank from a convergent regulated integral, comparing changes of step size, cutoff and contact prescription separately. The rank-one results above are calibration values, not predictions for that rank. An apparent discrepancy must survive those checks and the source dictionary before it can be interpreted as new CFT information. This selected reading route makes no claim that such a variation is novel or that the literature ends with these papers.

Exercises

1. Rotate a normalized mass generator

Set $x=t\cos\alpha$, $y=t\sin\alpha$ at $b=1$. Show that the generator has unit squared trace. Find the second and fourth $t$ derivatives at zero, and explain why equal two-point normalization does not force equal fourth response.

Hint 1

Use the orthogonality of the two generators, then substitute into the polynomial in (D13.4).

Hint 2

The fourth invariant contains $6\cos^2\alpha\sin^2\alpha$, whereas the square of the quadratic invariant contains coefficient two.

Solution

For $T=\cos\alpha\,T_1+\sin\alpha\,T_2$, orthogonality gives $\operatorname{Tr}T^2=1$. Thus

$$F_{tt}(0)=\frac{\pi^2}{2},\qquad F_{tttt}(0)=-\frac{\pi^4}{4} \bigl[1+\sin^2(2\alpha)\bigr].$$

Indeed $\operatorname{Tr}T^4=[1+\sin^2(2\alpha)]/4$ is not fixed by $\operatorname{Tr}T^2$. At $\alpha=0$ the fourth response is $-\pi^4/4$; at $\alpha=\pi/4$ it is $-\pi^4/2$. The latter direction has eigenvalues $1/\sqrt2,0,0,-1/\sqrt2$. These different eigenvalue sets are not conjugate flavor generators. The comparison is a local source-direction calculation; a generic irrational rotation is not being declared a new compact unit-period subgroup.

2. Diagnose a finite differentiation step

For $f(x)=F_1(x,0)$, define

$$\mathscr D_h= \frac{f(2h)-4f(h)+6f(0)-4f(-h)+f(-2h)}{h^4}.$$

Find its leading bias, construct a fourth-order improved estimate, and bound amplification of an absolute input uncertainty $\delta$ in each value of $f$. Explain whether taking $h$ as small as possible is always sensible.

Hint 1

Expand through $x^6$. The stencil annihilates the constant and quadratic terms, but not the sixth-order term.

Hint 2

Combine the estimates at $h$ and $h/2$ to cancel a term proportional to $h^2$. For the input bound, add absolute stencil coefficients.

Solution

Taylor expansion gives $\mathscr D_h=f^{(4)}(0)+h^2f^{(6)}(0)/6+O(h^4)$. Since $f=\log4+2\log\cosh(\pi x/2)$, $f^{(6)}(0)=\pi^6/2$, and

$$\mathscr D_h=-\frac{\pi^4}{4}+\frac{\pi^6}{12}h^2+O(h^4), \qquad \mathscr R_h=\frac{4\mathscr D_{h/2}-\mathscr D_h}{3} =f^{(4)}(0)+O(h^4).$$

For $h=0.08,0.04,0.02$, the direct biases are approximately $0.502627,0.127543,0.0320060$; the improved biases are $0.00251550,0.000160266,0.0000100653$. They exhibit the predicted orders, not arbitrary precision. Input uncertainty contributes at most $16\delta/h^4$ to the direct estimate. Decreasing $h$ suppresses truncation bias while amplifying numerical or integral error. One must vary both the step and the input accuracy, retaining the source prescription throughout.

3. Follow a tuned complex path

Use KNP’s rank-one dual variables, for which $\mathcal F(b;\zeta,m)=-\log[D_b(\zeta)D_b(m)]$. Set $\zeta=0$ and $m=m_b=i(b^2-3)/(4b)$. At $b_0=\sqrt3$ the mass vanishes. Given $D_{\sqrt3}(m)=[1+2\cosh(2\pi m/\sqrt3)]^{-1}$, compare $d^2\mathcal F(b;0,m_b)/db^2$ with the fixed-mass partial derivative $\mathcal F_{bb}$ at that point.

Hint 1

Differentiate the composite function twice before setting $m=0$. Evenness in $m$ removes two terms, but not the term containing $(m_b')^2$.

Hint 2

At $b_0$, $m_b'=i/2$. Compute $\mathcal F_{mm}$ from the supplied elementary determinant, retaining $i^2=-1$.

Solution

The chain rule gives

$$\frac{d^2\mathcal F}{db^2} =\mathcal F_{bb}+2m_b'\mathcal F_{bm} +(m_b')^2\mathcal F_{mm}+m_b''\mathcal F_m.$$

At zero mass, $\mathcal F_m=\mathcal F_{bm}=0$ by evenness. The elementary formula gives $\mathcal F_{mm}=8\pi^2/9$, so

$$\left.\frac{d^2\mathcal F(b;0,m_b)}{db^2}\right|_{b_0} =\left.\mathcal F_{bb}\right|_{b_0,m=0}-\frac{2\pi^2}{9}.$$

The negative addition comes from the imaginary tangent to the mass path. Near this point the determinant has no zero or pole, so the local analytic continuation is well defined; this does not authorize an arbitrary global contour continuation. The two derivatives probe different source families even though they pass through the same background. Neither should be identified with the round-point response (D13.11).

Throughout, allowed quadratic flavor Chern–Simons contact changes shift the complex mass Hessian but not these fourth mass derivatives. For example, the integer-charge cover gives $\Delta F=i\pi\kappa(x/2)^2$ with $\kappa\in\mathbb Z$. Its sign is fixed by the positive-$H$ background, and its fourth derivative vanishes. It is independent of $b$, so it also leaves $F_{bbxx}$ unchanged. This statement presumes the declared conformal prescription; introducing arbitrary new dimensionful source counterterms would change the observable under comparison.