Superconformal currents and two-point functions

A conserved current has a fixed scaling dimension, but that does not fix the size of its fluctuations. A two-point coefficient measures those fluctuations once the current’s charge normalization is fixed. Supersymmetry can relate that coefficient to another one; it does not normally supply its numerical value.

Use source derivatives and contact terms, shortening and protected quantities, and stress-tensor improvements as preparation. We will derive a current correlator, turn it into a response to a smooth source, and calculate its normalization for a free chiral multiplet. The paper’s squashed-sphere response will then have a precisely identified quantity to measure.

Fix the theory, vacuum and current

Work in a unitary three-dimensional $\mathcal N=2$ superconformal field theory, in its conformally invariant vacuum on flat Euclidean $\mathbb R^3$. Its Lorentzian parent has four real Poincaré and four real conformal supercharges. A particular curved source configuration need not preserve that full algebra. No gauge rank is fixed here; if the theory has a gauge description, the currents below are gauge-invariant local operators.

Let $j_\mu^{(R)}$ be the superconformal R-current: its charge is the R-generator in the superconformal algebra, with the two Poincaré families carrying charges $-1,+1$. It belongs to the same superconformal multiplet as the stress tensor $T_{\mu\nu}$ and the supersymmetry currents. Choose the conformal stress tensor, including the improvement specified by this multiplet. At separated points,

$$\partial^\mu j_\mu^{(R)}=0,\qquad \partial^\mu T_{\mu\nu}=0,\qquad T^\mu{}_\mu=0, \qquad \Delta_j=2,\quad\Delta_T=3.$$

These equations have contact terms when inserted into correlation functions with charged operators. We first keep $x\ne0$, so no such distribution is being discarded without a prescription.

A conserved flavor current can be added to a trial R-current. That changes the operator whose correlator is measured. Conservation alone does not identify the R-generator of the superconformal algebra, just as conservation alone did not choose the conformal stress-tensor improvement.

Derive the current’s tensor structure

Write $r_x=|x|$ for distance, reserving $r$ later for a field’s R-charge. Translation and rotation invariance, together with dimension two, give the parity-even ansatz

$$G_{\mu\nu}(x)=\langle j_\mu^{(R)}(x)j_\nu^{(R)}(0)\rangle =a\frac{\delta_{\mu\nu}}{r_x^4} +b\frac{x_\mu x_\nu}{r_x^6}.$$

For $x\ne0$, direct differentiation in three dimensions yields $\partial_\mu(\delta_{\mu\nu}r_x^{-4})=-4x_\nu/r_x^6$ and $\partial_\mu(x_\mu x_\nu r_x^{-6})=-2x_\nu/r_x^6$. Conservation therefore requires $b=-2a$. Define

$$I_{\mu\nu}(x)=\delta_{\mu\nu}-2\frac{x_\mu x_\nu}{r_x^2}, \qquad G_{\mu\nu}(x)=C_J\frac{I_{\mu\nu}(x)}{r_x^4}. \tag{O07.1}$$

$I$ is the orthogonal reflection matrix appearing in conformal inversion. Full conformal covariance gives this same structure and forbids an independent parity-odd two-point structure at separated points. This does not require the theory itself to be parity invariant: a parity-odd contact term can remain.

The coefficient $C_J$ is positive for a nonzero Hermitian current by the positive radial-quantization norm. This is not a claim that every coordinate component of $G$ is positive: along $x=(r_x,0,0)$, $G_{11}=-C_J/r_x^4$. The reflected vector index enters the norm.

To match the paper, introduce the differential operator $\mathcal D_{\mu\nu}=\delta_{\mu\nu}\partial^2-\partial_\mu\partial_\nu$. It is identically transverse. Since $\partial^2r_x^{-2}=2r_x^{-4}$ and $\partial_\mu\partial_\nu r_x^{-2} =-2\delta_{\mu\nu}r_x^{-4}+8x_\mu x_\nu r_x^{-6}$, we find

$$\boxed{ G_{\mu\nu}(x) =\frac{\tau_{RR}}{16\pi^2}\mathcal D_{\mu\nu}\frac1{r_x^2} =\frac{\tau_{RR}}{4\pi^2}\frac{I_{\mu\nu}(x)}{r_x^4}, \qquad C_J=\frac{\tau_{RR}}{4\pi^2}.} \tag{O07.2}$$

This defines the dimensionless $\tau_{RR}$ of P v3, (8.3). A different printed prefactor defines a different numerical coefficient. The position-space power is fixed by dimension; the factor $16\pi^2$ is a normalization choice.

Turn that coefficient into a source response

Let $a_\mu$ source this current with linear action $+\int a_\mu j^{(R)\mu}$, and set $\mathcal W[a]=-\log Z[a]$. In the paper’s background dictionary this is the combination $A_\mu-3V_\mu/2$, not the matter derivative’s $A_\mu-V_\mu/2$. One can probe the flat correlator by varying $A=a$ with the other sources zero. This arbitrary probe is not claimed to preserve supersymmetry.

The preceding lesson gives the exact response identity

$$\mathcal K_{\mu\nu}(x,y) =\frac{\delta^2\mathcal W}{\delta a_\mu(x)\delta a_\nu(y)} =\left\langle S_{,a_\mu a_\nu}\right\rangle -\langle j_\mu^{(R)}(x)j_\nu^{(R)}(y)\rangle_c. \tag{O07.3}$$

The explicit second derivative is local. For the nonlocal part we consequently need minus the Fourier transform of (O07.2). Use $\widehat f(p)=\int d^3x\,e^{ip\cdot x}f(x)$ and inverse measure $d^3p/(2\pi)^3$. The transform of $1/r_x^2$ is obtained without treating its singularity as an ordinary integrable function at infinity:

$$\begin{aligned} \frac1{r_x^2}&=\int_0^\infty du\,e^{-u r_x^2},\\ \widehat{r_x^{-2}}(p) &=\pi^{3/2}\int_0^\infty du\,u^{-3/2}e^{-p^2/(4u)} =\frac{2\pi^2}{|p|},\qquad p\ne0. \end{aligned} \tag{O07.4}$$

These formulas specify the tempered-distribution transform; a damping regulator followed by its limit gives the same result. Differentiation becomes multiplication by $-ip_\mu$. Define the transverse projector $P^T_{\mu\nu}(p)=\delta_{\mu\nu}-p_\mu p_\nu/p^2$. Then

$$\widehat G_{\mu\nu}^{\rm nonlocal}(p) =-\frac{\tau_{RR}}8|p|P^T_{\mu\nu},\qquad \boxed{\widehat{\mathcal K}_{\mu\nu}^{\rm nonlocal}(p) =\frac{\tau_{RR}}8|p|P^T_{\mu\nu}.} \tag{O07.5}$$

The minus sign in the first formula is compatible with $C_J>0$. Fourier transforming the singular correlator requires its local completion; it is not the variance of an unsubtracted pointwise random variable. Positivity of the conformal current norm and the sign convention for the source Hessian are distinct statements.

For a real smooth decaying source, its quadratic nonlocal response is therefore

$$\operatorname{Re}\mathcal W^{(2)}_{\rm nonlocal}[a] =\frac{\tau_{RR}}{16}\int\frac{d^3p}{(2\pi)^3} |p|\,\widehat a_\mu(-p)P_T^{\mu\nu}(p)\widehat a_\nu(p). \tag{O07.6}$$

The factor $1/2$ is the Taylor-expansion factor. A pure gradient source drops out because $P_Tp=0$. The source has dimension one, $\widehat a$ dimension $-2$, and the integral is dimensionless. A local counterterm adds a polynomial momentum kernel; it cannot change the nonanalytic $|p|$ coefficient.

In particular the Chern–Simons completion contributes $\mathcal C_{\mu\nu}=+i\kappa\epsilon_{\mu\nu\rho}\partial^\rho\delta^{(3)}(x)/(2\pi)$ to $\mathcal C=-\mathcal K$, with the sign fixed in the preceding lesson. Its coefficient requires the actual R-source global data; do not import a unit flavor group’s integer lattice without checking those data. A real local counterterm can affect the real response too if the prescription allows a new dimensionful scale. Exercise 3 makes that limitation explicit.

What supersymmetry adds to the stress tensor

Conservation fixes tensor shapes but would leave $C_J$ and the stress-tensor coefficient independent in a generic conformal theory. The additional superconformal Ward identity relates the norms of the currents within their common multiplet. In the stated R-charge and stress-tensor normalization, its component form is P, (8.3):

$$\langle T_{\mu\nu}(x)T_{\rho\sigma}(0)\rangle =\frac{\tau_{RR}}{64\pi^2} \left(\mathcal D_{\mu\rho}\mathcal D_{\nu\sigma} +\mathcal D_{\mu\sigma}\mathcal D_{\nu\rho} -\mathcal D_{\mu\nu}\mathcal D_{\rho\sigma}\right)\frac1{r_x^2}. \tag{O07.7}$$

We use this supersymmetric identity as the input relating the two operators; the current-conservation calculation alone did not prove it. Its conservation follows immediately from $\partial^\mu\mathcal D_{\mu\nu}=0$. Its trace vanishes because in three dimensions $\mathcal D_{\mu\mu}=2\partial^2$ and $\mathcal D_{\mu\rho}\mathcal D_{\mu\sigma}=\partial^2\mathcal D_{\rho\sigma}$.

To translate to another common convention, define $C_T$ by

$$\langle T_{\mu\nu}(x)T_{\rho\sigma}(0)\rangle =\frac{C_T}{r_x^6} \left[\frac12(I_{\mu\rho}I_{\nu\sigma}+I_{\mu\sigma}I_{\nu\rho}) -\frac13\delta_{\mu\nu}\delta_{\rho\sigma}\right]. \tag{O07.8}$$

For $x=(r_x,0,0)$ the $23,23$ component on the right is $C_T/(2r_x^6)$. On the left of (O07.7) its differential operator reduces to $\mathcal D_{22}\mathcal D_{33}r_x^{-2}=48/r_x^6$. Thus the explicit dictionary is

$$\boxed{C_T=\frac{3\tau_{RR}}{2\pi^2}=6C_J.} \tag{O07.9}$$

Both tensor expressions have dimension six. The factor six is specific to these definitions and the superconformal R-current, not a relation for an arbitrary flavor current or an arbitrary improved stress tensor.

Worked example: a free chiral multiplet

Take a massless gauge-neutral chiral multiplet, with scalar $\phi$ and two-component fermion $\psi$. Its physical superconformal scalar charge is $r=1/2$; the fermion charge is $r-1=-1/2$. The auxiliary has no propagating separated-point contribution. Use the independent Euclidean row $\chi$ so that the fermion action is $\chi\gamma^\mu\partial_\mu\psi$. In the earlier component convention $\chi=-i\widetilde\psi^TC$. The gamma matrices obey $\{\gamma_\mu,\gamma_\nu\}=2\delta_{\mu\nu}$ and $\operatorname{tr}1=2$.

For a scalar of charge $q_s$ and a fermion of charge $q_f$, the plus-source currents and free propagators are

$$\begin{aligned} j_\mu^s&=iq_s(\widetilde\phi\,\partial_\mu\phi -\partial_\mu\widetilde\phi\,\phi), &G_s(x)&=\frac1{4\pi r_x},\\ j_\mu^f&=-iq_f\chi\gamma_\mu\psi, &S(x)&=\langle\psi(x)\chi(0)\rangle =-\gamma^\mu\partial_\mu G_s(x) =\frac{\gamma\cdot x}{4\pi r_x^3}. \end{aligned} \tag{O07.10}$$

The current agrees with P, (7.13) after the stated row translation. The sign in $S$ follows from $-\partial^2G_s=\delta^{(3)}$.

The two scalar cross-contractions give $\langle j_\mu^s j_\nu^s\rangle =2q_s^2(\partial_\mu G_s\partial_\nu G_s-G_s\partial_\mu\partial_\nu G_s)$. Inserting $G_s$ gives $q_s^2I_{\mu\nu}/(8\pi^2r_x^4)$. For the fermion, the closed Grassmann contraction and the two factors $-i$ give $q_f^2\operatorname{tr}[\gamma_\mu S(x)\gamma_\nu S(-x)]$. Since $S(-x)=-S(x)$ and $\operatorname{tr}(\gamma_\mu\gamma\cdot x\gamma_\nu\gamma\cdot x) =4x_\mu x_\nu-2\delta_{\mu\nu}r_x^2$, its result is the same tensor with coefficient $q_f^2$.

The scalar and fermion sectors have no connected mixed contraction, so

$$\tau_{RR}^{\rm free} =\frac{(1/2)^2+(-1/2)^2}{2}=\frac14, \qquad C_T^{\rm free}=\frac3{8\pi^2}. \tag{O07.11}$$

This reproduces the value stated after the paper’s (8.3) using flat-space fields. It is separate from evaluating any sphere determinant. Giving the fermion the scalar’s charge happens to leave its squared contribution unchanged at $r=1/2$, so that numerical answer alone would not detect the error. Mixed current correlations do detect it; exercise 2 uses them.

Identify the sphere quantity before differentiating it

For the specific supersymmetric squashed-sphere family of the paper, with its superconformal R-multiplet and prescribed source counterterms, the result to be derived geometrically is

$$\left.\frac{d^2}{db^2}\operatorname{Re}F(b)\right|_{b=1} =\frac{\pi^2}{2}\tau_{RR},\qquad F(b)=-\log Z_{S_b^3}. \tag{O07.12}$$

Here $b$ is that family’s dimensionless squashing parameter, and $b=1$ is the round point. Equation (O07.12) is P, (8.2), not a formula for every deformation called “squashing.” Its derivation must specify which metric and auxiliary sources vary, the integrated current kernel and its regulator. The paper performs those steps in (8.4)–(8.10); they are the next sphere calculation, not consequences of changing a metric name in (O07.6).

For that background family the allowed contact contributions relevant to this response are imaginary in the conformal supersymmetric prescription, as explained after (8.10). The real part then isolates the stated coefficient. This does not erase the contact data from the full complex partition function, and it does not make arbitrary real finite counterterms harmless in another prescription.

The distinction to carry forward is now numerical as well as conceptual: $\tau_{RR}$ is the coefficient of a specified physical current, $C_T$ uses the tensor definition (O07.8), and a sphere derivative measures them only after its source path and scheme have been fixed.

Exercises

1. Measure the nonlocal coefficient with a smooth probe

In flat space take $L>0$ and a real source amplitude $a_0$ of dimension one: $a(x)=(a_0/L)(-x^2,x^1,0)e^{-|x|^2/(2L^2)}$. Show that it is transverse and compute (O07.6). Explain what happens if a smooth gradient $\partial_\mu\alpha$ is added. Use $x^1,x^2$ here as coordinate labels.

Hint 1

The radial Gaussian has transform $(2\pi)^{3/2}L^3e^{-L^2p^2/2}$. With our Fourier sign, multiplying by $x^j$ is $-i\partial_{p_j}$.

Hint 2

The squared transverse factor is $p_1^2+p_2^2$. Its angular integral is $8\pi p^2/3$, and $\int_0^\infty p^5e^{-L^2p^2}dp=L^{-6}$.

Solution

The two derivatives of the Gaussian cancel in $\partial\cdot a$; derivatives of the explicit coordinate prefactors contribute zero. The transform is $\widehat a=i a_0(2\pi)^{3/2}L^4(-p_2,p_1,0)e^{-L^2p^2/2}$, which is perpendicular to $p$. Hence

$$\operatorname{Re}\mathcal W^{(2)}_{\rm nonlocal} =\frac{\tau_{RR}a_0^2L^8}{16}\frac{8\pi}{3} \int_0^\infty dp\,p^5e^{-L^2p^2} =\frac{\pi\tau_{RR}}6(a_0L)^2.$$

This is dimensionless and nonnegative. At fixed dimensionless amplitude $a_0L$ it is independent of the width, as expected at a fixed point for this scaled source family. A gradient has transform $-ip_\mu\widehat\alpha$ and is annihilated by $P_T$, including in cross terms. This statement concerns the nonlocal quadratic response; allowed local terms require their separate treatment.

2. A trial current is not a new superconformal generator

Take two decoupled free chirals and a flavor generator $F$ which gives them charges $+1,-2$. Each multiplet’s scalar and fermion have the same flavor charge. Compute the two-point coefficient matrix $\tau_{AB}$ for the physical R-current and this flavor current, using the normalization (O07.2). Then find the coefficient of $R_t=R+tF$. Does it become the physical $\tau_{RR}$ for every $t$?

Hint 1

For two currents $A,B$, one free multiplet contributes $(q_s^Aq_s^B+q_f^Aq_f^B)/2$. Independent multiplets add.

Hint 2

For physical R, the two component charges are $+1/2,-1/2$. Their contributions to a mixed R-flavor coefficient cancel, while their squares do not.

Solution

The coefficients are

$$\begin{pmatrix}\tau_{RR}&\tau_{RF}\\\tau_{FR}&\tau_{FF}\end{pmatrix} =\begin{pmatrix}1/2&0\\0&5\end{pmatrix}, \qquad \tau_{R_tR_t}=\frac12+5t^2.$$

Each physical R contribution is $1/4$. The flavor entries are $1^2$ and $(-2)^2$. For either multiplet, $[(1/2)q_F+(-1/2)q_F]/2=0$. This would fail if the fermion were incorrectly assigned scalar R-charge $+1/2$.

$R_t$ is a conserved trial symmetry, but the fixed SCFT’s superconformal generator remains $R$. Its scalar dimensions stay $1/2$; they do not become $1/2+t$ and $1/2-2t$. The physical stress coefficient is $C_T=3/(4\pi^2)$ from (O07.9) with $\tau_{RR}=1/2$, not from substituting the trial-current coefficient. This example establishes the distinction explicitly; it is not a proposed minimization theorem for arbitrary interacting theories.

3. Taking the real part is not a universal subtraction

Allow a broader source prescription with the real local term $\Delta\mathcal W=c(4M)^{-1}\int d^3x\,f_{\mu\nu}f^{\mu\nu}$, where $f=da$, $c$ is dimensionless and $M>0$ is a fixed mass scale. Find its momentum Hessian and its value for the probe in exercise 1. Can it change $\tau_{RR}$? How would changing $L$ at fixed $a_0L$ distinguish the two responses?

Hint 1

Integration by parts gives $\int f^2=2\int d^3p/(2\pi)^3\,p^2\widehat a(-p)P_T\widehat a(p)$. Remember the $1/2$ in the Hessian expansion.

Hint 2

Relative to exercise 1 there is one additional factor $p$ in the radial integral: $\int_0^\infty dp\,p^6e^{-L^2p^2}=15\sqrt\pi/(16L^7)$.

Solution

The added kernel is local and transverse:

$$\Delta\widehat{\mathcal K}_{\mu\nu} =\frac cM(p^2\delta_{\mu\nu}-p_\mu p_\nu),\qquad \Delta\mathcal W=\frac{5c\pi^{3/2}}{4M}a_0^2L.$$

The polynomial kernel cannot alter the nonanalytic $|p|$ coefficient in (O07.5), so it does not change the defined $\tau_{RR}$. It does change a real integrated response at a fixed width. Setting $u=a_0L$ gives the combined result $\pi\tau_{RR}u^2/6+5c\pi^{3/2}u^2/(4ML)$. The first term is constant in $L$ and the second scales as $1/L$.

This term introduces a scale into the source prescription and is not a conformally covariant counterterm for the scale-free response. Allowing it does not deform the zero-source SCFT, but it changes contact conventions. The special real-part protection of the specified supersymmetric sphere prescription cannot be inferred merely from the word “real.”

What squashing the sphere measures uses this current normalization in a complete supersymmetric background and derives the prescribed round-point response.

Paper guides

Connect this lesson to the original papers and their notation.

• Paper guide: Section 8, physical observables