A first gauge-theory matrix integral

The free chiral determinant integrated every matter harmonic at a fixed mass. In a gauge theory, part of that mass is a dynamical scalar. Localization leaves us one more integral to do.

Return to one-flavor SQED: three-dimensional $\mathcal N=2$, compact gauge group $U(1)$, two chiral multiplets $\Phi_+,\Phi_-$ of charges $+1,-1$, gauge Chern–Simons level $k=0$, and superpotential $W=0$. The full Lorentzian theory has four real supercharges and gauge rank one. We use its Euclidean complexification on the same round sphere as the determinant lesson, with radius $\ell$, $H=-i/\ell$, $A=V=C=0$ and $R=-6/\ell^2$. Both spinor families are preserved.

Use the charged matter action, FI completion, and Cartan measures as preparation. We will keep every sign and measure factor visible, then evaluate one special case without assuming the infrared R-charge.

Identify the remaining integration variable

The supersymmetric Yang–Mills deformation and its gauge-fixing partners localize the gauge sector to

$$f_{\mu\nu}=0,\qquad \partial_\mu\sigma=0,\qquad D+\sigma H=0. \tag{O04.1}$$

These are zeros of the deformation, not the equations obtained by eliminating $D$ from the original interacting action. In our convention,

$$a_\mu=0,\qquad \sigma=\frac{s}{\ell},\qquad D=\frac{is}{\ell^2},\qquad s\in\mathbb R. \tag{O04.2}$$

The auxiliary fluctuation contour is $D=i\sigma/\ell+ih$ with real $h$: the term $-(D+\sigma H)^2$ becomes $h^2$. The ordinary scalar $\sigma$ stays real. On $S^3$, every smooth flat $U(1)$ connection is gauge equivalent to zero, and every smooth $U(1)$ bundle is topologically trivial: $\pi_1(S^3)=0$ and $H^2(S^3,\mathbb Z)=0$. There is no flux sum in this partition function without inserted defects. This is different from a calculation on $S^2\times S^1$.

The variable $s$ is a constant Lie-algebra scalar, not a gauge holonomy. Compactness of the gauge group does not make its range periodic. It fixes the unit electric charge and the flux lattice.

Willett, (3.11)–(3.22), derives the gauge/ghost reduction. The longitudinal connection, nonconstant scalar and ghost determinants cancel. The constant scalar mode must be omitted from those Gaussian determinants because it is integrated separately. Constant gauge transformations are divided out rather than counted as ghost zero modes. The original KWY calculation, (3.10)–(3.20), treats these same steps.

For $U(1)$ there are no nonzero roots and its Weyl group has order one. Thus there is no Vandermonde, no hyperbolic root product, and no factor $1/2$. Exchanging $\Phi_+$ with $\Phi_-$ is not an extra Weyl identification of the gauge group. In the standard localization measure, with the constant gauge volume and canceled Gaussian factors normalized, the remaining measure is

$$d s. \tag{O04.3}$$

We use the generator with unit charge, not a rescaled generator whose charge lattice has silently changed. Equation (O04.3) fixes an overall convention as well as the variable; an arbitrary multiplicative constant would change the unnormalized free energy.

Add masses and derive the classical source factor

Give both chiral scalars the same conserved trial R-charge $r$. An axial flavor symmetry assigns charge $+1$ to both multiplets. Its real mass comes from a fixed vector multiplet with scalar $m_A$ and auxiliary $D_A=i m_A/\ell$. Set the central charge parameter $z=0$. The effective signed physical masses and their dimensionless versions are

$$m_\pm=\pm\sigma+m_A,\qquad x_\pm=\pm s+\mu,\qquad \mu=\ell m_A. \tag{O04.4}$$

$m_A$ and $\sigma$ have mass dimension one. The variables $s,\mu$ are dimensionless. The opposite gauge charges do not make the axial contributions opposite.

An FI coupling can likewise be encoded by a fixed unit topological source with scalar $\nu$ and auxiliary $D_\nu=i\nu/\ell$. The previous action lesson fixed $\xi=-\nu/(2\pi)=-\zeta$, where $\xi$ multiplies the Euclidean FI density. Define

$$\eta=\ell\nu=-2\pi\ell\xi=2\pi\ell\zeta. \tag{O04.5}$$

Here $\xi,\zeta,\nu$ have dimension one; $\eta$ is dimensionless. The curved FI density is $\xi(D-\sigma H)$ on this background. At (O04.2), its two terms add:

$$\begin{aligned} S_{\rm FI} &=(2\pi^2\ell^3)\,\xi \left(\frac{is}{\ell^2}+\frac{is}{\ell^2}\right)\\ &=4\pi^2i\ell\xi s=-2\pi i\eta s, \qquad e^{-S_{\rm FI}}=e^{2\pi i\eta s}. \end{aligned} \tag{O04.6}$$

Keeping only $\xi D$ would miss a factor of two. The Yang–Mills deformation vanishes on the locus; it does not supply a Gaussian $e^{-c s^2/g_{\rm YM}^2}$. The protected answer is independent of its coefficient under the regulated Ward-identity assumptions of localization. This does not assert that every observable is independent of the gauge coupling.

Assemble the normalized Cartan integral

Recall the determinant prescription already derived:

$$Z_\chi(r,x)=e^{\mathcal L(1-r-ix)},\qquad \mathcal L'(u)=-\pi u\cot(\pi u),\qquad \mathcal L(0)=0.$$

The derivative has a removable singularity at zero, and the logarithm is continued from there in $-1<\operatorname{Re}u<1$. Each field of charge $q$ contributes this determinant with $x=qs+\mu$. Multiplying both factors and the classical FI factor gives

$$\boxed{ Z_{\rm SQED}(r;\mu,\eta)= \int_{-\infty}^{\infty}ds\, e^{2\pi i\eta s} e^{\mathcal L(1-r-i(s+\mu)) +\mathcal L(1-r-i(-s+\mu))}.} \tag{O04.7}$$

This is a one-dimensional matrix integral: the “matrix” is the single real Cartan entry of $U(1)$. The measure has been normalized, but $Z_{\rm SQED}$ is still a partition function, not a normalized expectation value.

The general mass and gauge-measure construction is W16, (3.30)–(3.33). The source’s variables translate as $\widehat\sigma_{\rm W16}=-s$, $\widehat m_{\rm W16}=-\mu$, and $\widehat\zeta_{\rm W16}=\eta$. Its determinant and FI exponent then reproduce (O04.7). This translation is consistent with the opposite mass argument established in the determinant lesson; copying the source’s scalar symbol without that change would conjugate individual matter factors.

The quantum prescription is part of (O04.7). We retain the symmetric chiral determinants and the compact, unit-charge, spin-theory regulator convention of the SQED action lesson, with total gauge level zero. The two Dirac fermions allow this integer level choice: $\tfrac12\sum q_s^2=1$. No extra integer background CS factors are appended to (O04.7). This fixes a representative for this sphere family; it is not a claim that every background contact coefficient vanishes. A different prescribed background counterterm can multiply the answer by a source-dependent phase. A different dynamical gauge level changes the theory and its integrand.

Check the integral before evaluating it

At fixed real $s,\mu$, the finite chiral Gaussians converge for $0<r<2$. That does not ensure convergence of the remaining integral over $s$.

Set $a=1-r$. For real $x$ with large magnitude, the cotangent equation gives

$$\log|Z_\chi(r,x)| =-\pi a|x|+O\!\left((1+|x|)e^{-2\pi|x|}\right). \tag{O04.8}$$

For example, at $x>0$, $\cot[\pi(a-ix)]\to i$ and $\partial_x\mathcal L(a-ix)=i\pi(a-ix)\cot[\pi(a-ix)]$ has real part tending to $-\pi a$. To fix the real constant, use oddness and complex conjugation of $\mathcal L$: $\mathcal L(ix)$ is purely imaginary. Integrating $\operatorname{Re}\mathcal L'(b+ix)$ from $b=0$ to $b=a$ then gives $-\pi ax$ plus an exponentially small remainder, which establishes (O04.8).

For $|s|>|\mu|$, $|s+\mu|+|-s+\mu|=2|s|$. The product therefore decays as $e^{-2\pi(1-r)|s|}$. A sufficient working domain for (O04.7) is

$$0<r<1,\qquad \mu,\eta\in\mathbb R. \tag{O04.9}$$

The FI factor has unit magnitude. At $r=1,\mu=0$, oddness of $\mathcal L$ makes the determinant product exactly one, so the remaining Fourier integral is not an ordinary convergent partition function. The local matter operators can be nonsingular while the integral along the gauge scalar still fails. Analytic continuation or a distributional prescription must be stated separately.

Work an exact trial-charge example

Take $r=1/2$ and $\mu=0$, retaining arbitrary real $\eta$. The opposite-mass identity derived in the free-chiral lesson gives

$$Z_{\rm SQED}\left(\frac12;0,\eta\right) =\int_{\mathbb R}\frac{e^{2\pi i\eta s}}{2\cosh\pi s}\,ds. \tag{O04.10}$$

The same paired factor appears in KWY, (3.68) and W16, (3.28). Its appearance here does not add an adjoint chiral multiplet or enhance the supersymmetry of our fixed theory.

To evaluate the integral, integrate $f(w)=e^{2\pi i\eta w}/(2\cosh\pi w)$ around the rectangle with vertices $-R,R,R+i,-R+i$, counterclockwise. Since $f(w+i)=-e^{-2\pi\eta}f(w)$, the two horizontal segments sum to $(1+e^{-2\pi\eta})Z$ as $R\to\infty$. The vertical segments vanish exponentially for each fixed real $\eta$. The only enclosed pole is $w=i/2$, with residue $e^{-\pi\eta}/(2\pi i)$. Hence

$$(1+e^{-2\pi\eta})Z=e^{-\pi\eta},\qquad \boxed{Z_{\rm SQED}\left(\frac12;0,\eta\right) =\frac1{2\cosh\pi\eta}.} \tag{O04.11}$$

At zero FI source, $Z=1/2$ and $F=-\log Z=\log2$. This is an exact answer for the declared trial R-charge and prescription. It is not an identification of $r=1/2$ with the interacting theory’s superconformal R-charge.

The absolute truncation error from integrating only $|s|\le S$ is bounded uniformly in real $\eta$:

$$\left|\int_{|s|>S}\frac{e^{2\pi i\eta s}}{2\cosh\pi s}\,ds\right| \le \frac{2}{\pi}e^{-\pi S}. \tag{O04.12}$$

This follows from $1/(2\cosh\pi s)\le e^{-\pi|s|}$. It turns a numerical agreement into a controlled test. At large $|\eta|$, the exact answer is exponentially small and cancellation makes a small absolute error insufficient for high relative accuracy.

Practice

1. Separate an axial mass from a shift of the gauge scalar

Replace the masses in (O04.7) by $x_+=s+\alpha$ and $x_-=-s+\beta$, with real dimensionless $\alpha,\beta$ and $0<r<1$. Reduce the integral to the common-axial-mass form. Evaluate it when $r=1/2$ and $\beta=-\alpha$. Explain which phase must be retained.

Hint 1

Use the average and half-difference of $\alpha,\beta$.

Hint 2

A real translation preserves the integration contour and its measure, but changes the FI exponential.

Solution

Let $\mu=(\alpha+\beta)/2$ and $d=(\alpha-\beta)/2$. Put $s'=s+d$. Then $s+\alpha=s'+\mu$ and $-s+\beta=-s'+\mu$, while $e^{2\pi i\eta s}=e^{-2\pi i\eta d}e^{2\pi i\eta s'}$. Absolute convergence permits the translation, so

$$Z(r;\alpha,\beta,\eta) =e^{-2\pi i\eta d}Z_{\rm SQED}(r;\mu,\eta).$$

For $\beta=-\alpha$, $d=\alpha,\mu=0$, and at $r=1/2$,

$$Z=\frac{e^{-2\pi i\eta\alpha}}{2\cosh\pi\eta}.$$

The average mass cannot be removed by shifting one gauge scalar, because it enters both charged fields with the same sign. The difference can be absorbed into that scalar in this zero-level integral, but the topological-source phase remains. It records the fixed source convention; equality of absolute values does not justify deleting it. At $\eta=0$ it disappears, as the change of variables requires. Introducing a gauge CS term would also transform its quadratic classical exponent, so this result cannot be copied unchanged into a different-level theory.

2. Use the FI parameter to measure the Cartan distribution

At $r=1/2,\mu=\eta=0$, define the normalized positive density $p(s)=Z(0)^{-1}/(2\cosh\pi s)$. Find $\langle s^2\rangle$ and $\langle s^4\rangle$ using (O04.11). Is this density Gaussian? Relate the second moment to $F''(0)$, where the derivative is with respect to $\eta$.

Hint 1

The ratio $Z(\eta)/Z(0)$ is the characteristic function $\langle e^{2\pi i\eta s}\rangle$.

Hint 2

Expand $\operatorname{sech}(\pi\eta)$ through fourth order and retain the powers of $2\pi i$.

Solution

Since $Z(0)=1/2$, $p(s)=\operatorname{sech}(\pi s)$ and its characteristic function is

$$\frac{Z(\eta)}{Z(0)} =1-\frac{\pi^2\eta^2}{2} +\frac{5\pi^4\eta^4}{24}+O(\eta^6).$$

Odd moments vanish by reflection symmetry. Comparing the second and fourth derivatives with $(2\pi i)^n\langle s^n\rangle$ gives

$$\langle s^2\rangle=\frac14,\qquad \langle s^4\rangle=\frac5{16}.$$

A centered Gaussian with variance $1/4$ would instead have fourth moment $3/16$. Localization need not leave a Gaussian measure.

Here $F=\log(2\cosh\pi\eta)$, so $F''(0)=\pi^2=4\pi^2\langle s^2\rangle$. The positive sign follows from the imaginary FI insertion in $Z$; it should not be guessed from a real Boltzmann source. These are moments of the localized integration variable in this prescription, not automatically the separated-point correlators of a local field on the sphere.

3. Decide whether a mode cutoff may be integrated first

The determinant lesson defined a finite subtracted approximation

$$\mathcal L_K(u)= \sum_{j=1}^{K}j\log\frac{j+u}{j-u}-(2K+1)u.$$

For $a=1-r\in(0,1)$, $\mu=\eta=0$, form $w_K(s)=\exp[\mathcal L_K(a-is)+\mathcal L_K(a+is)]$. Find its large-$|s|$ limit at fixed $K$. Can one obtain (O04.7) by integrating $w_K$ over the whole real line and only then sending $K$ to infinity? Give a valid order of approximations for a numerical calculation.

Hint 1

Multiply the conjugate factors before taking the large-$|s|$ limit.

Hint 2

Pointwise convergence does not by itself supply an integrable bound independent of $K$.

Solution

The finite expression is positive:

$$w_K(s)=e^{-2(2K+1)a} \prod_{j=1}^{K} \left[\frac{(j+a)^2+s^2}{(j-a)^2+s^2}\right]^j.$$

Every ratio tends to one at fixed $K$, so $w_K(s)\to e^{-2(2K+1)a}>0$. Its whole-line integral diverges for every finite $K$, even though its pointwise infinite-$K$ limit has the exponentially decaying tail (O04.8). The proposed interchange is invalid.

One valid approach is to evaluate the already regulated function $\mathcal L$ and control the remaining Cartan tail. Alternatively, fix a compact interval $[-S,S]$, take the determinant limit there with a uniform error estimate, and only then enlarge $S$ using a bound for the limiting integrand. A small nonzero plateau at finite cutoff cannot be ignored merely because its height is exponentially small: the real line has infinite length.

The partition function is now a finite, testable integral with masses, FI parameter and normalization specified. R-mixing and F-maximization determines which conserved R-current should be used at an interacting fixed point; the convenience of the hyperbolic example does not answer it.

Paper guides

Connect this lesson to the original papers and their notation.

• Reading three-dimensional localization with matter