Find the localization locus and its measure

Which field configurations remain when the localization parameter becomes large? The answer is not obtained by minimizing the physical Yang–Mills action. We minimize a chosen supersymmetry variation, retain its collective coordinates, and integrate the other modes around them. The physical action then supplies a weight on the surviving configurations.

Use the four-sphere charge and contour and matrix measures, roots, and eigenvalues. We keep massless four-dimensional $SU(N)$ N=4 SYM, $N\geq2$, on a sphere of radius $\ell$. Its flat parent has sixteen real Poincaré supercharges; the N=2 vector–adjoint-hyper organization and the one odd localization charge $Q$ are choices within that theory. The gauge rank remains $N-1$.

The deformation and its integration contour

The previous lesson constructed a commuting spinor label $\epsilon$, its partner $\widetilde\epsilon$, and seven auxiliary spinors $\nu_I$. They make $Q$ close off shell. In the source’s chiral-block notation,

$$Q\Psi=\frac12F^P_{MN}G^{MN}\epsilon -2\phi_A^P\widetilde G^A\widetilde\epsilon+K_I^P\nu_I.$$

Here $M,N=0,\ldots,9$ package the four-dimensional gauge and scalar fields; they do not introduce extra integration coordinates. The seven auxiliaries complete this selected charge, not every N=4 transformation.

Choose $V=\int\sqrt g\,(\Psi,\overline{Q\Psi})$ and deform $S$ to $S+tQV$, $t>0$. On the chosen contour, $(QV)_{\rm bos}$ is the nonnegative norm of $Q\Psi$. The bar is extended analytically off that contour. Since $Q^2$ includes a complexified gauge transformation, literal tangency to the real field contour is not automatic: the localization Ward identity uses a compatible gauge fixing, regulator, and integration by parts without a field-space boundary contribution. Positivity supplies convergence in normal directions; it does not replace that Ward identity. These are the contour qualifications in Pestun (3.1)–(3.4) and p. 20.

We now translate to Hermitian matrices. Write the gauge field as $\mathcal A$, its curvature as $\mathcal F=d\mathcal A-i\mathcal A\wedge\mathcal A$, and $D=\nabla-i[\mathcal A,\cdot]$. The vector scalars are $a(x),\sigma(x)$; the hypermultiplet contains $q_i(x)$, $i=5,\ldots,8$. All have dimension inverse length. The continued auxiliaries $k_I$ have dimension inverse length squared. Precisely,

$$\begin{gathered} A^P=-i\mathcal A,\quad \phi_0^{P,E}=-ia,\quad \phi_0^P=i\phi_0^{P,E},\\ \phi_9^P=-i\sigma,\quad \phi_i^P=-iq_i,\quad K_I^P=iK_I^{P,E},\quad K_I^{P,E}=-ik_I. \end{gathered} \tag{N13.1}$$

The positive source pairing $-\operatorname{Tr}_N$ becomes the ordinary fundamental trace on Hermitian fields, with $\operatorname{Tr}_N(t^at^b)=\delta^{ab}/2$. Put $C_bX=-i[b,X]$ and $\|X\|^2=\operatorname{Tr}_N X^2$, including any displayed tensor-index sums. In particular, curvature norms below sum over ordered pairs $\mu,\nu$.

Complete the squares before imposing field equations

Let $\theta$ be polar angle from the north pole and $s_2=\sin^2(\theta/2)$, $c_2=\cos^2(\theta/2)$. With orientation $\epsilon_{1234}=+1$, define $\mathcal F^\pm=(\mathcal F\pm *\mathcal F)/2$. The spinor bilinears determine real two-forms $w^\pm$ of the indicated duality and functions $w_I=2\nu_I^T\widetilde\epsilon$. Their normalization is

$$w^\pm_{\mu\nu}w^{\pm\mu\nu}=\ell^{-2},\qquad \sum_{j=5}^7 w_j^2=\ell^{-2},\qquad w_1=\cdots=w_4=0.$$

For example, $w^-_{\mu\nu}=\widetilde\epsilon_L^TG^9G_{\mu\nu}\epsilon_L/s_2$ and $w^+_{\mu\nu}=\widetilde\epsilon_R^TG^9G_{\mu\nu}\epsilon_R/c_2$, where the projections are $(1\pm G^{1234})/2$. The denominators are used away from the poles; the weighted squares have the appropriate limiting meaning.

After integration by parts on the smooth closed sphere, the vector contribution is

$$\begin{aligned} \mathcal L_v={}& s_2\|\mathcal F^-+w^-\sigma\|^2 +c_2\|\mathcal F^++w^+\sigma\|^2\\ &+\|Da\|^2+\|D\sigma\|^2+\|C_a\sigma\|^2 +\sum_{j=5}^7\|k_j+w_j a\|^2. \end{aligned} \tag{N13.2}$$

One can check the essential completion without performing every spinor contraction. Orthogonality of self-dual and anti-self-dual tensors gives

$$\begin{aligned} &s_2\|\mathcal F^-+w^-\sigma\|^2+ c_2\|\mathcal F^++w^+\sigma\|^2\\ &=\frac12\|\mathcal F\|^2 +\frac{\cos\theta}{2}\operatorname{Tr}_N (\mathcal F_{\mu\nu}(*\mathcal F)^{\mu\nu})\\ &\quad+2\operatorname{Tr}_N \bigl[\sigma(s_2w^-+c_2w^+)_{\mu\nu}\mathcal F^{\mu\nu}\bigr] +\ell^{-2}\operatorname{Tr}_N\sigma^2. \end{aligned} \tag{N13.3}$$

These are the curvature, spinor-induced mixing, and scalar terms in the expanded fermion norm. Likewise, the auxiliary and $a$ terms are $\sum k_j^2+2a\sum w_jk_j+a^2/\ell^2=\sum(k_j+w_j a)^2$ inside the trace. This last identity already shows why setting $k_j=0$ prematurely would force the wrong conclusion $a=0$.

For completeness, the hypermultiplet squares are

$$\begin{aligned} \mathcal L_h={}& \sum_i\|C_aq_i\|^2 +\sum_{m,j}\left\|D_mq_j-\sum_i c_{mij}q_i\right\|^2\\ &+\frac12\sum_{i,j}\|C_{q_i}q_j\|^2 +\frac{3}{4\ell^2}\sum_i\|q_i\|^2 +\sum_{I=1}^4\|k_I\|^2. \end{aligned} \tag{N13.4}$$

Here $m=1,2,3,4,9$, with the last reduced derivative $D_9q=C_\sigma q$, and $c_{mij}=\widetilde\epsilon^TG^mG^{ij}\epsilon$. These coefficients are antisymmetric in $i,j$ and satisfy $\sum_{m,j}c_{mij}c_{mkj}=\delta_{ik}/(4\ell^2)$. Expanding the derivative square therefore contributes $q_iq_i/(4\ell^2)$; completing it from the original $q_iq_i/\ell^2$ term leaves the strictly positive $3/(4\ell^2)$ remainder. This is the useful part of the hypermultiplet calculation: no nonzero smooth $q_i$ can make (N13.4) vanish.

Equations (N13.2)–(N13.4) are the Hermitian translation of Pestun (3.13)–(3.16) and the following hypermultiplet formula. Their equality to the original deformation is an integrated identity after the stated integrations by parts. We will not apply that smooth-field argument unaltered to singular configurations.

The Bianchi identity removes an apparent extra branch

Vanishing of (N13.4) gives $q_i=0$ and $k_1=\cdots=k_4=0$. Away from the poles, both curvature-square weights are positive, so (N13.2) requires

$$D\sigma=0,\qquad \mathcal F=-(w^-+w^+)\sigma. \tag{N13.5}$$

The second equation is not an arbitrary permissible choice of curvature. Every connection obeys $D\mathcal F=0$. Consequently $(dw)\sigma=0$, where $w=w^-+w^+$.

Let us compute this obstruction explicitly. In the northern stereographic chart, set $\rho=|x|$, $f=1+\rho^2/(4\ell^2)$ and

$$\Omega=dx^1\wedge dx^2+dx^3\wedge dx^4,\qquad \eta=x^1dx^2-x^2dx^1+x^3dx^4-x^4dx^3.$$

Then $d\eta=2\Omega$. Evaluating the chosen spinor bilinears gives

$$w=-\frac{d\rho\wedge\eta}{\ell\rho f^2}, \qquad dw=\frac{2\,d\rho\wedge\Omega}{\ell\rho f^2}, \qquad |dw|_g^2=\frac{4f^2}{\ell^2\rho^2}>0. \tag{N13.6}$$

The last norm is the ordinary three-form norm, summing distinct increasing triples. To check it, rotate a point to $(\rho,0,0,0)$: only the $134$ component of $d\rho\wedge\Omega$ remains, and raising three indices multiplies its squared coefficient by $f^6$. Differentiating the radial prefactor in $w$ gives no extra term because $d\rho\wedge d\rho=0$.

Thus $\sigma=0$ and $\mathcal F=0$ at every point away from the poles. Smoothness extends the result to them. A flat connection on simply connected $S^4$ is gauge equivalent to zero; in particular, a smooth flat bundle here has no nonzero instanton number. The remaining equation $Da=0$ becomes ordinary constancy. We have obtained

$$\boxed{ \mathcal A=0,\quad a(x)=a,\quad \sigma=q_i=0,\quad k_j=-w_j a\ (j=5,6,7),\quad k_1=\cdots=k_4=0 } \tag{N13.7}$$

up to gauge transformations, with constant traceless Hermitian $a$. This fills in the nonvanishing-$dw$ step in Pestun (3.17)–(3.18). The auxiliaries generally do not vanish. We solved the deformation equations, not the auxiliary equations of the undeformed physical action.

A smooth trial field that fails the test

Could a genuine gauge potential partly cancel $w\sigma$? Choose a fixed Hermitian color generator $T$, constant $\sigma=\sigma_0T$, and the smooth Abelian-in-color connection $\mathcal A=\beta\sigma_0T\,v^\flat$, $\beta\in\mathbb R$. Here $v$ is the sphere rotation generated by $Q^2$, and $v^\flat=\eta/(\ell f^2)$ is its globally smooth metric-dual one-form. Set $a=q_i=k_I=0$.

Compute $\mathcal F=d\mathcal A$ before substituting in the squares. On the radial axis its orthonormal components are $\mathcal F_{12}=2\beta\sigma_0T\cos\theta/\ell$ and $\mathcal F_{34}=2\beta\sigma_0T/\ell$. The resulting density is

$$\mathcal L_v= \frac{\sigma_0^2\operatorname{Tr}_N T^2}{\ell^2} \left[1-8\beta\cos\theta+4\beta^2(1+3\cos^2\theta)\right].$$

Using $\langle\cos\theta\rangle_{S^4}=0$, $\langle\cos^2\theta\rangle_{S^4}=1/5$, and $\operatorname{vol}(S^4)=8\pi^2\ell^4/3$ yields

$$\int_{S^4}\sqrt g\,\mathcal L_v =\frac{8\pi^2\ell^2\sigma_0^2}{3}\operatorname{Tr}_N T^2 \left(1+\frac{32}{5}\beta^2\right)>0 \quad(\sigma_0\ne0). \tag{N13.8}$$

The term linear in $\beta$ can reduce the density in one region but increases it elsewhere. The minimum in this family is at $\beta=0$. This explicit test respects the Bianchi identity; it is a diagnostic family, while (N13.5)–(N13.6) gave the general smooth-field exclusion.

Why the two poles still matter

At the north pole $s_2=0$, so the curvature squares permit $\mathcal F^-\ne0$ with $\mathcal F^+=0$. At the south pole $c_2=0$, the roles reverse. A smooth field cannot have support at just one point, so these are absent from (N13.7). They enter through limits of shrinking instanton configurations and the compactification and regularization of their moduli spaces. One must not substitute a delta-function curvature into $\mathcal F^2$ and multiply distributions.

Near each fixed point, $Q^2$ rotates two tangent planes with frequencies $\epsilon_1=\epsilon_2=1/\ell$. The local instanton integral is evaluated with these rotation weights; this is the meaning of the equivariant, or Omega-background, prescription here. We denote the resulting factors by $Z_N(a)$ and $Z_S(a)$. Their construction is the additional polar input of Pestun Section 5, beyond the smooth-square proof. That section presents the counting for $U(N)$; using it for our specified $SU(N)$ theory requires separating the central sector, rather than silently importing an extra Abelian factor.

The instanton sign convention deserves care. With the site’s Hermitian curvature and orientation, $k_{\rm site}=\int\operatorname{Tr}_N\mathcal F\wedge\mathcal F/(8\pi^2)$ and the Euclidean theta term is $-i\theta_{\rm site}k_{\rm site}$. Pestun instead writes $k_P=\int\operatorname{tr}F^P\wedge F^P/(8\pi^2)$ using a literal fundamental matrix trace. Since $F^P=-i\mathcal F$,

$$k_P=-k_{\rm site},\qquad \theta_P=-\theta_{\rm site}. \tag{N13.9}$$

The second relation matches the actions. The source’s footnote 7 explicitly distinguishes that trace from its positive pairing $(X,Y)=-\operatorname{tr}XY$. Thus its north-pole “instantons” have negative $k_{\rm site}$; the south-pole configurations have positive $k_{\rm site}$. This is a translation, not a change of the equations $\mathcal F^\pm=0$. We keep north/south factor labels; on the usual real contour the two factors pair by conjugation with the correspondingly translated theta parameter.

Normal fluctuations give a ratio of Gaussian integrals

The remaining constant matrix is a collective coordinate, meaning a parameter along the locus. It must be integrated, not inserted as a zero eigenvalue in a Gaussian denominator.

Gauge fixing also matters before expanding. Extend $Q$ by the odd BRST operation implementing infinitesimal gauge transformations and introduce its ghosts. Pestun’s constant ghost sector makes the combined square a rotation plus a constant gauge transformation; it avoids treating residual gauge transformations as physical zero modes. This is the role of the gauge-fixed complex, (4.12), (4.16)–(4.19).

At a finite compatible regulator, write the quadratic action for real bosonic normal coordinates $b$, Grassmann coordinates $\psi$, and ghost pairs $(\bar c,c)$:

$$S_2=\frac{t}{2}b^TK_B(a)b +\frac{t}{2}\psi^TK_F(a)\psi +t\bar c\,M_{\rm FP}(a)c. \tag{N13.10}$$

$K_F$ is antisymmetric. Gaussian integration gives an inverse square root for the real bosons, a Pfaffian for the fermions, and a determinant for the complex ghost pairs. After consistent normalization of paired powers of $t$, the structure is

$$Z_{\rm quad}(a)=C_{\rm reg}\, \frac{\operatorname{Pf}'K_F(a)\, \det' M_{\rm FP}(a)} {\sqrt{\det'K_B(a)}}. \tag{N13.11}$$

A Pfaffian is the Grassmann Gaussian whose square is the determinant; its sign or phase still requires an orientation and continuation prescription. Here $K_F$ excludes the ghosts. If ghosts are included in a combined fermionic operator, do not multiply by their determinant again.

The primes remove the tangent coordinates $a$, gauge directions already fixed, and constant gauge ghost modes handled by the residual quotient. Polar instanton collective coordinates belong to their own factors. A prime does not delete every spacetime-dependent Cartan fluctuation: those are generally normal modes and contribute.

This is the finite Gaussian content of Pestun (4.20)–(4.22). Passing to infinitely many modes requires a supersymmetry-compatible ultraviolet prescription, phases, and allowed local counterterms. Equality of bosonic and fermionic component counts alone neither determines this ratio nor proves it is one. The next calculation evaluates the operators and polar factors in the appropriate N=4 limit.

The residual gauge quotient supplies the Vandermonde

Constant gauge transformations still conjugate $a$. For $SU(2)$ write $a=\operatorname{diag}(u,-u)=2u\,t_3$, $t_i=\sigma_i/2$. For $\lambda=\lambda_1t_1+\lambda_2t_2$, the orbit tangent is $\delta a=i[\lambda,a]$, or

$$\begin{pmatrix}\delta a^1\\\delta a^2\end{pmatrix} = \begin{pmatrix}0&-2u\\2u&0\end{pmatrix} \begin{pmatrix}\lambda_1\\\lambda_2\end{pmatrix}, \qquad \det M_{\rm orbit}=(2u)^2. \tag{N13.12}$$

The Cartan metric is $\operatorname{Tr}_2(da)^2=2\,du^2$. Integrating over the entire real $u$ axis counts the two eigenvalue orderings twice, so divide by $|W|=2$. Alternatively integrate one chamber, $u\geq0$, without that factor. The continuous diagonal stabilizer contributes a convention-dependent group-volume constant. At $u=0$ it enlarges and this orbit coordinate chart degenerates; the vanishing Jacobian is the expected result.

For $SU(N)$ the same two-dimensional calculation applies to each pair of eigenvalues. Define $\Delta(a)=\prod_{\alpha>0}\alpha(a)$, where the positive roots are $\alpha_{ij}(a)=a_i-a_j$, $i<j$. With the induced measure on $\sum_i a_i=0$, the answer has the structure

$$Z_{S^4}= \frac{C}{N!}\int_{\mathfrak h}da\, \Delta(a)^2\,e^{-S_{\rm cl}(a)} Z_{\rm 1-loop}'(a)\,Z_N(a)Z_S(a). \tag{N13.13}$$

$S_{\rm cl}$ is the physical action evaluated on (N13.7). In this equation $Z_{\rm 1-loop}'$ excludes the finite residual orbit determinant $\Delta^2$; a convention that includes it must omit the extra displayed factor. The constant $C$ includes measure conventions and cancels from normalized observables. A $Q$-closed Wilson loop inserts its value on the locus in the numerator.

The integral has $N-1$ real Cartan variables. Replacing $SU(N)$ by $U(N)$ adds the central scalar coordinate and changes normalized observables: it is a different theory. We have derived the integration variables and measure, but have not yet evaluated the physical weight, fluctuation product, or polar factors.

Exercises

1. The auxiliary field carries the zero direction

For one commuting color component define dimensionless variables $A=\ell a$, $B_j=\ell^2k_j$, and $W_j=\ell w_j$. Include all seven auxiliaries, so $\sum_jW_j^2=1$, and consider the quadratic form $\mathcal Q=\sum_j(B_j+W_jA)^2$. Find its kernel and the eigenvalues of its matrix in the Euclidean metric $dA^2+\sum_jdB_j^2$. Explain why imposing $B=0$ before finding the locus is wrong, and why a zero direction here does not imply a divergent physical partition function.

Hint 1

Separate $B$ into its component parallel to $W$ and six perpendicular components.

Hint 2

The remaining two-variable form is $(B_\parallel+A)^2$. Compare the deformation with the physical action, rather than replacing one by the other.

Solution

The six perpendicular components contribute their ordinary squares. In the $(A,B_\parallel)$ plane the matrix is $\left(\begin{smallmatrix}1&1\\1&1\end{smallmatrix}\right)$, with eigenvalues $0,2$. Thus the complete quadratic matrix has spectrum $0,1,1,1,1,1,1,2$ and kernel $(A,B)=(A,-WA)$. These are eigenvalues of the quadratic form matrix; the second-derivative Hessian is twice that matrix.

Setting $B=0$ first changes $\mathcal Q$ to $A^2$ and removes its actual tangent direction. The dimensionless variables also show that the shifted auxiliary has the correct dimension two. In the chosen auxiliary frame the preceding lesson gives $W_7=-1$ and the other $W_j=0$, a direct example of the same kernel. The collective $a$ is subsequently weighted by the undeformed action and any insertions. Its being a zero mode of $QV$ says nothing by itself about a zero mode or divergence of the physical action.

2. Change the rank and check the measure

For $SU(3)$ use $a=\operatorname{diag}(u,v,-u-v)$. Derive its Cartan metric, squared Vandermonde and Weyl factor. For the defined finite-dimensional test weight $e^{-\kappa\operatorname{Tr}_3a^2}$, $\kappa>0$, find the scaling with $\kappa$ of the complete Cartan integral. What scaling would the incorrect flat eigenvalue measure give?

Hint 1

Compute $\operatorname{Tr}(da)^2$ before choosing a volume element. There are three positive roots.

Hint 2

Rescale both $u$ and $v$ by $\kappa^{-1/2}$ and count the degree of the root product.

Solution

The metric is $2du^2+2du\,dv+2dv^2$, with matrix $\left(\begin{smallmatrix}2&1\\1&2\end{smallmatrix}\right)$ and determinant $3$. Therefore the induced volume is $\sqrt3\,du\,dv$. The three differences give

$$\frac{\sqrt3}{6}\,du\,dv\, (u-v)^2(2u+v)^2(u+2v)^2.$$

The denominator $6$ counts all permutations when the whole plane is integrated. Since $\operatorname{Tr}_3a^2=2(u^2+uv+v^2)$, rescaling the two coordinates supplies $\kappa^{-1}$ and the degree-six polynomial supplies $\kappa^{-3}$. The answer scales as $\kappa^{-4}$, agreeing with an eight-dimensional Lie-algebra Gaussian. Omitting the Vandermonde instead gives $\kappa^{-1}$ and loses six gauge-orbit directions. This is a measure check, not an evaluation of the four-dimensional fluctuation operators.

3. Equal counts are not equal determinants

Let $a$ be a collective coordinate. Two real normal coordinates $x,y$ and one Grassmann pair have $S_2=t\lambda(a)(x^2+y^2)/2+t\nu(a)\bar\psi\psi$, with $t,\lambda>0$. Normalize the bosonic measure by $dx\,dy/(2\pi)$ and choose $\int d\bar\psi\,d\psi\,\bar\psi\psi=-1$. Find the normal factor, compare $\nu=\lambda$ with $\nu=2\lambda$, and explain why adding $a$ as a third Gaussian coordinate fails.

Hint 1

The fermionic exponential stops after its linear term. The bosonic integral is a polar-coordinate Gaussian.

Hint 2

The undeformed action may depend on $a$ even though the deformation’s quadratic form has a zero direction along it.

Solution

The bosonic factor is $\int_0^\infty r\,dr\,e^{-t\lambda r^2/2}=1/(t\lambda)$. The Grassmann factor is $\int d\bar\psi\,d\psi\,(1-t\nu\bar\psi\psi)=t\nu$. Their product is $\nu/\lambda$. It is one for matched coefficients and two for the changed coefficient, despite identical field counts. More generally, an unmatched $a$ dependence changes the collective-coordinate integrand.

Putting $a$ into the normal quadratic matrix gives $\operatorname{diag}(\lambda,\lambda,0)$; its inverse determinant is undefined. The correct operation is to integrate $a$ separately with its physical weight and residual gauge measure. A genuine supersymmetric cancellation requires the operator pairing and its compatible regulator, not just this toy model’s number of variables.

The checkpoint is now concrete: explain why only one constant adjoint scalar survives smoothly, why its auxiliary is nonzero, and why its eigenvalues carry $\Delta^2/|W|$. Next, derive why the matrix model becomes Gaussian by evaluating the physical action, normal determinant and polar sectors in the massless N=4 theory.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading Pestun's four-sphere localization paper