Put N=4 SYM on the four-sphere

The circular Wilson loop has a supersymmetry on flat space. To use that symmetry in a sphere path integral, we need more than a curved metric: we need its spinor, the curved action it preserves, and a charge whose square is a specified bosonic symmetry.

This lesson builds those ingredients. It uses the Wilson-loop spinor condition, spin connections, fixed background sources, curvature improvements, and the general field-theory localization argument. We do not yet find the localization locus or evaluate a determinant.

The theory is massless four-dimensional $SU(N)$ N=4 SYM, $N\geq2$. Its flat-space parent has sixteen real Poincaré supercharges and sixteen conformal supercharges. We organize it as an N=2 vector multiplet and an adjoint hypermultiplet, then select one odd charge for localization. Neither this organization nor the choice of charge changes the gauge rank or removes the extra supersymmetries of the massless parent. Euclidean spinors are complexified; counting their complex coordinates is not an additional doubling of physical supercharges.

Start with a metric and a spin connection

Let $\ell>0$ be the sphere radius. Stereographic coordinates $x^\mu$, $\mu=1,\ldots,4$, cover the north pole and omit the south pole. Put

$$f=1+\frac{x^2}{4\ell^2},\qquad ds^2=f^{-2}\delta_{\mu\nu}dx^\mu dx^\nu,\qquad e^a=f^{-1}dx^a,\qquad |x|=2\ell\tan\frac{\theta}{2}. \tag{N12.1}$$

Here $\theta$ is the polar angle from the north pole. In particular, the equator is the coordinate sphere $|x|=2\ell$, not $|x|=\ell$. The scalar curvature in our ordinary positive-sphere convention is $R_+=12/\ell^2$. This is the four-dimensional convention of Pestun, not the negative-curvature convention used for the target paper’s three-dimensional spheres.

Write $\Omega=-\log f$. Cartan’s equation gives $\omega_{\mu ab}=\delta_{a\mu}\partial_b\Omega-\delta_{b\mu}\partial_a\Omega$. Use full complex Euclidean Clifford matrices with $\{\Gamma_a,\Gamma_b\}=2\delta_{ab}$ and $\Gamma_{ab}=[\Gamma_a,\Gamma_b]/2$. The spinor derivative is therefore

$$\nabla_\mu =\partial_\mu+\frac14\omega_{\mu ab}\Gamma_{ab} =\partial_\mu-\frac{\Gamma_\mu\not{x}-x_\mu} {4\ell^2f}, \qquad \not{x}=x^a\Gamma_a. \tag{N12.2}$$

The last $\Gamma_\mu$ has a flat index identified with the coordinate label; the curved Clifford matrix is $f^{-1}\Gamma_\mu$. This factor matters in the next equations.

Solve the conformal spinor equations

A conformal Killing spinor is accompanied by a second spinor $\widetilde\epsilon$. On this sphere their equations and solutions are

$$\begin{aligned} \nabla_\mu\epsilon&=f^{-1}\Gamma_\mu\widetilde\epsilon, & \nabla_\mu\widetilde\epsilon&= -\frac{f^{-1}}{4\ell^2}\Gamma_\mu\epsilon,\\ \epsilon(x)&=f^{-1/2}(\epsilon_s+\not{x}\,\epsilon_c), & \widetilde\epsilon(x)&= f^{-1/2}\left(\epsilon_c-\frac{\not{x}}{4\ell^2}\epsilon_s\right). \end{aligned} \tag{N12.3}$$

The constant parameters $\epsilon_s$ and $\epsilon_c$ have opposite ten-dimensional chirality in the full Clifford notation. The ten-dimensional labels package the four-dimensional fields; no extra spacetime is being integrated over. We normalize $\epsilon$ to be dimensionless below, so $\epsilon_c$ and $\widetilde\epsilon$ have dimension inverse length.

Let us verify the first equation, rather than infer it from the flat answer. Substitution of (N12.2) gives

$$\begin{aligned} \nabla_\mu\epsilon &=f^{-1/2}\left[ \Gamma_\mu\epsilon_c- \frac{\Gamma_\mu\not{x}}{4\ell^2f} (\epsilon_s+\not{x}\epsilon_c)\right]\\ &=f^{-3/2}\Gamma_\mu \left[\left(f-\frac{x^2}{4\ell^2}\right)\epsilon_c -\frac{\not{x}}{4\ell^2}\epsilon_s\right] =f^{-1}\Gamma_\mu\widetilde\epsilon. \end{aligned}$$

We used $\not{x}^2=x^2$. Replacing the pair $(\epsilon_s,\epsilon_c)$ by $(\epsilon_c,-\epsilon_s/(4\ell^2))$ proves the second equation by the same calculation. Commuting the two derivatives then gives $[\nabla_\mu,\nabla_\nu]\epsilon =f^{-2}\Gamma_{\mu\nu}\epsilon/(2\ell^2)$, the spin curvature of a positive-curvature sphere. At fixed $x$ and $\ell\to\infty$, (N12.3) returns $\epsilon_s+\not{x}\epsilon_c$.

These are sections, not sixteen globally constant numbers in one frame. To see regularity at the missing pole explicitly, use the orientation-preserving southern coordinates $y=4\ell^2 Jx/x^2$, where $J=\operatorname{diag}(-1,1,1,1)$. The orthonormal frames differ by the proper rotation $J(1-2nn^T)$, $n=x/|x|$. Its spin transition can be chosen as $\epsilon_{\rm south}=\Gamma_1\not{n}\,\epsilon_{\rm north}$. It converts (N12.3) to the same formula in $y$, with constants

$$\epsilon_s^{\rm south}=2\ell\Gamma_1\epsilon_c,\qquad \epsilon_c^{\rm south}=-\frac{\Gamma_1\epsilon_s}{2\ell}.$$

Both southern spinors are smooth at $y=0$. Direction dependence of northern components as $x\to\infty$ was a frame effect. The solution and its source conventions are Pestun (2.12)–(2.13), Appendix B.

Choose the charge that preserves the equatorial loop

We now specify a particular solution. For explicit bilinears it is useful to use Pestun’s real sixteen-component chiral blocks $G^M,\widetilde G^M$:

$$\Gamma_{\!P}^{M}= \begin{pmatrix}0&\widetilde G^M\\G^M&0\end{pmatrix}, \qquad G^0=1,\quad \widetilde G^0=-1,\qquad G^m=\widetilde G^m=(G^m)^T,\quad m=1,\ldots,9. \tag{N12.4}$$

The full matrices $\Gamma_{\!P}^M$ have signature $(9,1)$; spacetime indices on the sphere remain $1,\ldots,4$ and Euclidean. Products between chiral spaces alternate the two blocks; in particular $G^{MN}=\widetilde G^{[M}G^{N]}$, with antisymmetrization including $1/2$. Lowering the source time index changes its sign. The first four blocks satisfy the ordinary positive Clifford algebra, so the verification of (N12.3) applies unchanged. A transpose in the bilinears below is the source’s symmetric spinor pairing, not an imposed Euclidean complex-conjugation condition.

In Pestun’s Appendix A basis, choose

$$\epsilon_s=(0_8,1,0_7)^T,\qquad G^{5678}\epsilon_s=\epsilon_s,\qquad G^{1234}\epsilon_s=-\epsilon_s,\qquad \epsilon_c=\frac{G^0\widetilde G^1G^2}{2\ell}\epsilon_s =\frac{G^1G^2}{2\ell}\epsilon_s. \tag{N12.5}$$

The first projection selects the manifest N=2 subalgebra, retaining eight Poincaré and eight conformal labels before further choices. The second fixes the four-dimensional chirality at the north pole. The selected charge is $Q=Q_\epsilon$ with a fixed commuting label $\epsilon$. It is one odd operation on fields, lying in the rigid sphere subalgebra described by Pestun. Multiplying by an external odd constant produces an ordinary even supersymmetry variation.

Here is a direct link to the previous Wilson-loop convention. There the internal matrix was $S=n^I\Gamma_{4+I}$, $S^2=1$, and a coordinate-radius-$R$ circle required $\epsilon_c=i\Gamma_{12}S\epsilon_s/R$. Set $\Gamma_{\!P}^0=iS$ in the full Clifford presentation. Since $\Gamma_{\!P,E}^0=i\Gamma_{\!P}^0$, this means $S=-\Gamma_{\!P,E}^0$, the same scalar-direction reflection used previously. Equation (N12.5) is precisely the circle condition at $R=2\ell$. The factor $f^{-1/2}$ in the sphere spinor and the factor $f^{-1}$ multiplying both tangent and speed do not disturb its local cancellation.

The chosen solution obeys $\epsilon^T\epsilon=1$ and $\epsilon^T\widetilde\epsilon=0$. Its bosonic bilinear $v^M=\epsilon^TG^M\epsilon$, with a frame-to-coordinate conversion for spacetime components, is

$$\begin{aligned} v&=\frac1\ell (-x^2\partial_1+x^1\partial_2-x^4\partial_3+x^3\partial_4),\\ |v|_g&=\sin\theta,\qquad v^0=1,\qquad v^9=-\cos\theta,\qquad v^{5,6,7,8}=0. \end{aligned} \tag{N12.6}$$

For example, these results follow by expanding $f^{-1}(\epsilon_s+\not{x}\epsilon_c)^TG^M (\epsilon_s+\not{x}\epsilon_c)$: the spatial terms are $(-x^2,x^1,-x^4,x^3)/(\ell f)$ in the orthonormal frame; the ninth component is $-(1-x^2/(4\ell^2))/f$. The equal rotations have period $2\pi\ell$. They vanish at both poles, although the spinor itself never vanishes. These statements agree with Pestun’s selected charge and (3.5).

Identify its square before using localization

For this paragraph use source fields $A_\mu^P,\phi_A^P$ and $D_\mu^P=\nabla_\mu+[A_\mu^P,\cdot]$. The scalar labels are $A=0,5,\ldots,9$. The compact gauge algebra is anti-Hermitian before the scalar continuation; the convergent contour has $\phi_0^P=i\phi_0^{P,E}$ with $\phi_0^{P,E}$ anti-Hermitian. Define

$$\Lambda=v^\mu A_\mu^P+\phi_0^P-\cos\theta\,\phi_9^P.$$

The component algebra Pestun (2.7) gives

$$\begin{aligned} Q^2A_\mu^P&=-\mathcal L_v A_\mu^P+D_\mu^P\Lambda,\\ Q^2\phi_A^P&=-v^\mu\partial_\mu\phi_A^P -[\Lambda,\phi_A^P]+\mathcal J_{AB}\phi_B^P,\\ \mathcal J_{56}&=\ell^{-1},\qquad \mathcal J_{78}=-\ell^{-1},\qquad \mathcal J_{BA}=-\mathcal J_{AB}, \end{aligned} \tag{N12.7}$$

with all other $\mathcal J_{AB}$ zero. We have defined $\mathcal J$ as the coefficient appearing in $Q^2$; it is the negative of the source’s separately named $R$ generator in (2.9).

To check the gauge-field expression, expand $-v^\nu F_{\nu\mu}^P+D_\mu^P(v^A\phi_A^P)$ and use $v^\nu F_{\nu\mu}^P=\mathcal L_vA_\mu^P-D_\mu^P(v\cdot A^P)$. For the scalars, the translation and reduced internal derivative combine into $-v^\mu\partial_\mu\phi_A^P-[\Lambda,\phi_A^P]$. The remaining bilinear $2\widetilde\epsilon^T G_{AB}\epsilon$ has entries $\mathcal J_{56},\mathcal J_{78}$ above. There is no dilation term because $\epsilon^T\widetilde\epsilon=0$. Thus the square consists of a sphere rotation, a field-dependent gauge transformation, and an internal rotation. It is not identically zero.

The unextended fermion algebra also contains its field equation. Localization must not use that equation inside the path integral. For this one charge, introduce seven even adjoint auxiliary fields $K_i$ and commuting spinors $\nu_i$ satisfying

$$\epsilon^TG^M\nu_i=0,\qquad \nu_i^TG^M\nu_j=\delta_{ij}v^M,\qquad \sum_i\nu_i\nu_i^T+\epsilon\epsilon^T =\frac12v_M\widetilde G^M. \tag{N12.8}$$

These are finite algebraic identities. They allow the terms $Q\Psi\supset K_i\nu_i$ and $QK_i=-\nu_i^TG^MD_M^P\Psi$ to cancel the unwanted fermion equation in $Q^2$: its coefficient is $-\sum_i\nu_i\nu_i^T+\tfrac12v_M\widetilde G^M-\epsilon\epsilon^T=0$. The first identity also shows directly that the new term changes neither bosonic equation in (N12.7). The full sourced transformation is $Q\Psi=\tfrac12F_{MN}^PG^{MN}\epsilon -2\phi_A^P\widetilde G^A\widetilde\epsilon+K_i\nu_i$: the curvature term follows from $\tfrac12G^{\mu A}\phi_A^P\nabla_\mu\epsilon$ and $G^{\mu A}\widetilde G_\mu=-4\widetilde G^A$. Pestun proves the auxiliary closure in (2.26)–(2.30) and Appendix C.

One can check existence here without solving another differential equation. Let $w=(-x^2,x^1,-x^4,x^3)$ and $U=f^{-1/2}(1-w^aG^aG^9/(2\ell))$. Then $U^TU=1$, $\epsilon=U\epsilon_s$, and $\nu_i=UG^{i8}\epsilon_s$, $i=1,\ldots,7$, satisfy (N12.8). Their auxiliary frame is allowed to change between coordinate patches. This construction closes the selected operation; it does not furnish a finite auxiliary completion of every N=4 supercharge.

The Euclidean source auxiliary contour is continued along with $\phi_0^P$, as in Pestun Section 3. Off-shell closure alone is not a proof that a regulator and contour have a vanishing $Q$-boundary term. Those are the integration assumptions of the localization argument already stated in the prerequisite. We have now identified the charge and its bosonic square to which that argument will apply.

Derive the curvature term

Return to the Hermitian absorbed fields of the Wilson-loop lesson: $\mathcal A=gA$, $\Phi_I=gX_I$, $\operatorname{Tr}_N(t^at^b)=\delta^{ab}/2$, and $D=\nabla-i[\mathcal A,\cdot]$. The gauge coupling $g$ is dimensionless. Every scalar has engineering dimension one. It is enough first to work with one canonically normalized real scalar.

Let $g_{\mu\nu}=e^{2\Omega}\delta_{\mu\nu}$ and relate its curved field $X$ to a flat field $\varphi$ by $X=e^{-\Omega}\varphi$. Direct expansion gives

$$\begin{aligned} \sqrt g\,(\partial X)^2 &=(\partial\varphi)^2 -\partial_\mu(\varphi^2)\partial_\mu\Omega +\varphi^2(\partial\Omega)^2,\\ R_+&=-6e^{-2\Omega} \left[\partial^2\Omega+(\partial\Omega)^2\right],\\ \sqrt g\left[(\partial X)^2+\frac{R_+}{6}X^2\right] &=(\partial\varphi)^2 -\partial_\mu\!\left(\varphi^2\partial_\mu\Omega\right). \end{aligned} \tag{N12.9}$$

The coefficient $1/6$ cancels both non-boundary terms. The surface term vanishes for a smooth sphere field expressed in these two patches, or for the explicitly decaying flat test fields used below; it cannot be discarded for arbitrary singular boundary data.

The same argument works inside the color trace with gauge-covariant derivatives. The Yang–Mills term is already Weyl invariant in four dimensions. The scalar commutator potential has weight four, and the fermion has weight $3/2$: $\not{\nabla}_g(e^{-3\Omega/2}\psi)= e^{-5\Omega/2}\not{\partial}\psi$. Consequently the kinetic and Yukawa terms need no additional ordinary fermion mass. The physical bosonic action is

$$S_{\rm bos}=\frac1{g^2}\int_{S^4}\!\sqrt g\,d^4x\, \operatorname{Tr}_N\left[ \frac12\mathcal F_{\mu\nu}\mathcal F^{\mu\nu} +D_\mu\Phi_I D^\mu\Phi_I -\frac12[\Phi_I,\Phi_J][\Phi_I,\Phi_J] +\frac{2}{\ell^2}\Phi_I\Phi_I \right]. \tag{N12.10}$$

The sums over $I,J=1,\ldots,6$ are ordered sums; commutators of Hermitian matrices are anti-Hermitian, so the potential has the positive sign in the energy. This is the continued Hermitian form of Pestun (2.2). For comparison of actions, $A^P=-i\mathcal A$, the five spatial source scalars are $-i\Phi_I$, and $\phi_0^{P,E}=-i\Phi_6$, with $\phi_0^P=i\phi_0^{P,E}$. The positive source pairing $-\operatorname{Tr}_N$ then becomes the displayed Hermitian trace.

In canonical components the scalar quadratic action is $\tfrac12\int\sqrt g[(DX_I^a)^2+2(X_I^a)^2/\ell^2]$. Its operator is $-\nabla^2+2/\ell^2$. This is a curvature-induced gap, not a freely adjustable hypermultiplet mass; it vanishes in the flat limit. A change of the improvement coefficient changes the background theory and the spinor symmetry, even if the flat kinetic action is unchanged. The classical Weyl calculation also does not assert equality of quantum partition functions without their Weyl anomaly.

A propagator checks the background

Consider one free canonical scalar component, with gauge interactions switched off. The conformal map predicts its sphere Green function from $G_{\mathbb R^4}(x,y)=1/(4\pi^2|x-y|^2)$:

$$G_{S^4}(x,y)=\frac{f(x)f(y)}{4\pi^2|x-y|^2} =\frac1{8\pi^2\ell^2(1-\cos\gamma)}, \tag{N12.11}$$

where $\gamma$ is the angular separation of the two points. To verify it, use the operator identity following from (N12.9), $(-\nabla^2+2/\ell^2)(f\varphi)=f^3(-\partial^2\varphi)$. Away from $y$, (N12.11) solves the homogeneous equation. Close to $y$, it has the required singularity $1/(4\pi^2d^2)$ in proper distance $d$.

The normalization has a useful global check. Place $y$ at the north pole and integrate the equation over a cap of polar radius $\theta$. Its outward derivative flux and its mass-term integral are respectively $\cos^4(\theta/2)$ and $1-\cos^4(\theta/2)$; their sum is one. At the opposite pole the propagator is finite. Omitting the curvature term would lose this check and introduce a constant zero mode, so the massless flat propagator would not define an inverse on the compact sphere.

Exercises

1. Change the improvement coefficient

Replace the canonical quadratic action by $\tfrac12\int\sqrt g[(\partial X)^2+\xi R_+X^2]$. Find the constant-mode eigenvalue and the eigenvalue on a coordinate function $Y^A$ restricted from the embedding sphere $\sum_{A=1}^5(Y^A)^2=\ell^2$. When is the quadratic form strictly positive? Does positivity alone recover the chosen supersymmetry?

First hint

A constant has zero Laplacian. Use the polar coordinate $Y^5=\ell\cos\theta$ to compute the other eigenvalue directly.

Second hint

For a function of $\theta$ only, $\nabla^2h=\ell^{-2}(\sin^{-3}\theta)\partial_\theta(\sin^3\theta\,\partial_\theta h)$. The Laplacian quadratic form is nonnegative.

Solution

The constant eigenvalue is $12\xi/\ell^2$. Direct differentiation gives $-\nabla^2Y^5=4Y^5/\ell^2$; rotations give the same answer for the other four coordinates. Their eigenvalue is $(4+12\xi)/\ell^2$.

Since $-\nabla^2$ is nonnegative on a compact sphere, $\xi>0$ makes the whole quadratic form strictly positive. At $\xi=0$ the constants are zero modes; at $\xi<0$ they are negative modes. The conformal value $\xi=1/6$ gives respectively $2/\ell^2$ and $6/\ell^2$.

Many positive values of $\xi$ exist, but only $1/6$ gives the Weyl cancellation (N12.9) and the corresponding massless N=4 background action. Positivity is an integration property; it does not prove a supersymmetry Ward identity or $Q$-exactness.

2. Move the circle without changing the charge

Keep the charge (N12.5), the scalar direction, and the north-patch frame fixed. Replace the equatorial circle by $x=(R\cos t,R\sin t,0,0)$, $R>0$. For which $R$ does this same charge annihilate the loop? Explain why being an orbit of $v$ is insufficient.

First hint

The metric rescales tangent and speed by the same factor. It suffices to test the local spinor condition at $x=(R,0,0,0)$.

Second hint

In the source blocks the unit-tangent kernel at that point is $G^2-G^0$. The minus sign comes from lowering the source time index. Also $G^1\epsilon_c=G^2\epsilon_s/(2\ell)$.

Solution

Ignoring the nonzero common conformal factor, the kernel gives

$$(G^2-1)\left(\epsilon_s+\frac{R}{2\ell}G^2\epsilon_s\right) =\left(1-\frac{R}{2\ell}\right)(G^2-1)\epsilon_s.$$

The last spinor is nonzero: $G^2\epsilon_s$ has the opposite four-dimensional chirality from $\epsilon_s$. Thus this charge requires $R=2\ell$, and the previous circle identity proves cancellation everywhere when that condition holds. This is $\theta=\pi/2$ and proper radius $\ell$.

Every such circle is an orbit of the spacetime vector in (N12.6). That tests invariance under part of $Q^2$, not annihilation by $Q$. A different latitude can be studied with a changed supercharge or scalar coupling, but it does not pass the fixed-input problem posed here.

3. Check the action on a smooth nonconstant profile

For one canonical free scalar choose $X(\theta)=c(1+\alpha\cos\theta)$, with real $c,\alpha$. Compute the conformally coupled action on $S^4$ and compare it with the flat action of $\varphi=X/f$ on the stereographic patch. Explain the radius dependence and whether this profile is a localization saddle.

First hint

The sphere averages are $\langle\cos\theta\rangle=0$ and $\langle\cos^2\theta\rangle=1/5$. The volume is $8\pi^2\ell^4/3$.

Second hint

The gradient squared is $c^2\alpha^2\sin^2\theta/\ell^2$. In the flat patch use $\cos\theta=(1-\rho^2/(4\ell^2))/f$ with $\rho=|x|$.

Solution

The kinetic contribution is $2c^2\alpha^2\operatorname{Vol}(S^4)/(5\ell^2)$. The curvature contribution is $c^2\operatorname{Vol}(S^4)(1+\alpha^2/5)/\ell^2$. Hence

$$S[X]=\frac{8\pi^2\ell^2c^2}{3} \left(1+\frac{3\alpha^2}{5}\right).$$

The constant and first-coordinate pieces are orthogonal eigenfunctions with eigenvalues $2/\ell^2$ and $6/\ell^2$, which gives the same answer. Since $c$ has mass dimension one, $\ell^2c^2$ is dimensionless. If one changes $\ell$ while holding $\ell c$ fixed, the action is unchanged; holding $c$ fixed is a different comparison.

The transformed flat profile is $\varphi(\rho)=c[(1+\alpha)+(1-\alpha)\rho^2/(4\ell^2)]/f^2$. Its ordinary action is $\pi^2\int_0^\infty d\rho\,\rho^3(\varphi')^2$. The boundary term in (N12.9) vanishes at zero and infinity, so that integral equals the answer above; direct radial integration confirms it. The profile is a test of the curved kinetic action. We have imposed no equations obtained by setting fermion variations to zero, so it has not been identified as a localization saddle.

The background data to carry forward are the spinor (N12.3), the chosen constants (N12.5), its nonzero bosonic square (N12.7), and the curvature coupling (N12.10). Next, find the localization locus and its measure by choosing a suitable $Q$-exact deformation and solving its vanishing conditions; the sphere construction alone does not determine that locus or its measure.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading Pestun's four-sphere localization paper