Gauge projection and an index check of SQED-XYZ

Integrating a flavor fugacity can impose electric neutrality. It cannot create the magnetic sectors of a compact gauge theory. We will see both operations at work: the neutral sector supplies the SQED meson, while nonzero flux supplies its two monopole operators. Together they reproduce a flavored index of the interacting XYZ theory.

Use the free-chiral counting method and the SQED–XYZ operator dictionary. Our electric theory is three-dimensional $\mathcal N=2$ SQED: compact $U(1)$ of rank one, chirals $Q,\widetilde Q$ of gauge charges $+1,-1$, Chern–Simons level zero, and no superpotential, mass or FI deformation. Its proposed infrared partner has no gauge field and has $W=XYZ$. The fixed point has four real Poincaré and four real conformal supercharges. The index selects one Poincaré component $q$ and its radial adjoint.

We use $\mathcal H_q=D-R-J_3$ and the commuting exponent $K=D+J_3$. Fermion parity supplies the signs. The axial generator $A$ gives both electric chirals charge $+1$; the topological generator is $J$. Their dimensionless fugacities are $a,v$. Do not confuse $J$ with the spatial rotation component $J_3$.

Project within each magnetic sector

On the sphere in radial quantization, let $\mathcal a$ be the dynamical gauge connection. Its integer magnetic flux and compact holonomy variable are

$$m=\frac1{2\pi}\int_{S^2}d\mathcal a\in\mathbb Z, \qquad u=e^{ih},\qquad h\sim h+2\pi. \tag{O12.1}$$

Here $h$ is the gauge-transformation angle inserted in the trace; the circle path integral realizes it as a holonomy. An excitation of electric charge $g$ carries $u^g$. Therefore

$$\oint_{|u|=1}\frac{du}{2\pi i u}\,u^g=\delta_{g,0}. \tag{O12.2}$$

This is Gauss-law projection, with a normalized measure. Perform it in every sector $m$, then sum those sectors with $v^m$, since $J=m$. A monopole sector is a sector of states on the sphere, or equivalently a local disorder insertion at the origin. It is not obtained by adding an electrically charged oscillator to the zero-flux vacuum.

For $U(1)$ there is no Weyl division and no root determinant. The vector multiplet’s nonzero contributions cancel in the localized index; its gauge holonomy and flux remain. These statements specialize the localization formula in Kapustin–Willett, (2.4) and (2.12). We use that formula as an established input, rather than claiming to derive gauge-theory localization from the free Fock space.

The vacuum also carries charges

Write the electric scalar R-charge as $r$. The physical symmetric fixed point has $r=1/3$; keeping $r$ symbolic briefly will show how flavor mixing enters. The charged fields themselves are not gauge-invariant local primaries, so this notation does not assign them physical local-operator dimensions.

In flux $m$, a charge-$g$ scalar has monopole angular momentum starting at $|gm|/2$. The surviving highest-weight scalar and conjugate-fermion towers of the localized quadratic complex have respective weights

$$u^g a\,x^{|gm|+r+2n}, \qquad -u^{-g}a^{-1}x^{|gm|+2-r+2n}, \qquad n=0,1,\ldots . \tag{O12.3}$$

The fermion tower starts at angular momentum $(|gm|+1)/2$; its R-weight is $1-r$. The remaining $|gm|$ unpaired angular fermion modes determine the vacuum charge below. These are equivariant weights in the localization complex, given by source (2.3), (2.9)–(2.12). They are not a claim that an interacting gauge theory obeys the free equations of motion with $1/2$ replaced by $r$.

For a fermion mode of charge $f$ under any generator, symmetric ordering assigns $f(c^\dagger c-\tfrac12)$. Thus each empty unpaired mode contributes $-f/2$. The two matter fermions have charges

Fermion	Gauge	Axial	R
$\psi_Q$	$+1$	$1$	$r-1$
$\psi_{\widetilde Q}$	$-1$	$1$	$r-1$

Each has $p=|m|$ unpaired modes. In the paired regulator with level zero their vacuum charges are consequently

$$g_0=-\frac p2(1-1)=0,\qquad A_0=-p,\qquad R_0=-\frac p2\,2(r-1)=(1-r)p. \tag{O12.4}$$

The bare BPS vacuum is rotationally scalar, so its index exponent is $K=R_0$. The Abelian gaugino adds no charged-root contribution. The required prefactor is therefore $v^m a^{-p}x^{(1-r)p}$. Dropping it would give every flux sector a constant term, producing an immediate divergent sum and the wrong monopole charges.

We fix symmetric zero-point ordering, the paired gauge regulator and the axial/topological charge normalization above. We add no further integer background Chern–Simons counterterms relative to the matched prescription of Kapustin–Willett, and normalize the zero-flux identity state to one. This does not set every fractional contact coefficient to zero. Integer counterterms can shift source-dependent phases, whereas fractional contact data cannot generally be removed: see Closset et al., (1.3)–(1.7) and (4.5). Both flavor background fluxes are zero for the main calculation. That slice alone cannot determine all contact data on other bundles; Exercise 3 probes a larger slice.

Assemble the gauge-theory index

Define $(z;x^2)_\infty=\prod_{n\geq0}(1-zx^{2n})$. Applying the Bose and Fermi products to (O12.3), and then (O12.2), gives

$$\boxed{ \mathcal I_{\rm SQED} =\sum_{m\in\mathbb Z}v^m (a^{-1}x^{1-r})^{|m|} \oint_{|u|=1}\frac{du}{2\pi i u} \prod_{\epsilon=\pm1} \frac{(a^{-1}u^{-\epsilon}x^{|m|+2-r};x^2)_\infty} {(a u^\epsilon x^{|m|+r};x^2)_\infty}. } \tag{O12.5}$$

We will set $r=1/3$. For numerical evaluation take $0<x<1$. A sufficient domain for the displayed contour expansion and flux sum is

$$|a|x^r<1,\qquad |a|^{-1}x^{1-r}|v|<1,\qquad |a|^{-1}x^{1-r}|v|^{-1}<1. \tag{O12.6}$$

Unit-modulus $a,v$ satisfy these conditions for $0<r<1$. Outside the convergence domain, an analytically continued expression need not be a convergent trace. Alternatively, the low-order calculation below is a formal series with Laurent-polynomial coefficients in $a,v$.

The original source uses half-integer flux $s$, with $m=2s$, and begins its SQED formula with R-label $r_0=1/2$. To use its (3.17) and (3.29) at our fixed point, make the exact substitution

$$\alpha_{\rm source}=a\,x^{-1/6},\qquad w_{\rm source}=v. \tag{O12.7}$$

It changes the vacuum factor as well as the oscillator weights. This is a flavor-fugacity redefinition, not a change of the physical SCFT.

Compare with XYZ

The map $X=Q\widetilde Q$, $Y=V_+$, $Z=V_-$ gives

Field	$A$	$J$	R-charge	Flavor weight
$X$	$2$	$0$	$2r$	$w_X=a^2$
$Y$	$-1$	$+1$	$1-r$	$w_Y=a^{-1}v$
$Z$	$-1$	$-1$	$1-r$	$w_Z=a^{-1}v^{-1}$

The cubic has flavor weight one and R-charge two. These charges agree with the original SQED discussion, (3.1)–(3.2), after the R-mixing and harmless superpotential rephasing explained in the duality lesson. Kapustin–Willett’s XYZ axial generator has the opposite sign to the electric one; their identification $\widetilde\alpha=\alpha^{-1}$ implements that translation.

At the symmetric fixed point all three XYZ fields have physical R-charge and dimension $2/3$. The localized expression is

$$\mathcal I_{\rm XYZ} =\prod_{w\in\{a^2,a^{-1}v,a^{-1}v^{-1}\}} \frac{(w^{-1}x^{4/3};x^2)_\infty} {(w x^{2/3};x^2)_\infty}. \tag{O12.8}$$

Although written as three chiral factors, this is the index of the interacting theory with the allowed flavors and R-symmetry fixed by $W=XYZ$. It is not the index of three free SCFT chirals, whose R-charge would be $1/2$. Kapustin–Willett derive the exact equality with (O12.5), at zero axial background flux, by contour residues and basic hypergeometric identities in (3.19)–(3.31). We will reproduce terms that test its operator interpretation without assuming the equality.

A calculation that sees the monopoles

Let $y=x^{2/3}$. In the zero-flux sector, the lowest scalar weights are $a u^{\pm1}x^{1/3}$. Electric neutrality pairs them into mesons. Through $x^2=y^3$, the only fermionic corrections are the two neutral scalar–conjugate-fermion combinations, each of weight $-x^2$. No derivative mode contributes yet. Thus

$$\mathcal I_{m=0} =1+a^2y+a^4y^2+(a^6-2)y^3+O(y^4). \tag{O12.9}$$

For $p=|m|\geq1$, the bare vacuum contributes $a^{-p}v^m y^p$. The first electrically neutral oscillator correction needs one scalar of each charge. Its full weight starts at

$$x^{p(1-r)+2p+2r}=y^{4p+1} \quad\text{when }r=\frac13. \tag{O12.10}$$

It cannot affect the order we are computing. Only $|m|\leq3$ can contribute through $y^3$, so summing their vacua gives

$$\begin{aligned} \mathcal I_{\rm SQED} ={}&1+\bigl[a^2+a^{-1}(v+v^{-1})\bigr]y\\ &+\bigl[a^4+a^{-2}(v^2+v^{-2})\bigr]y^2\\ &+\bigl[a^6+a^{-3}(v^3+v^{-3})-2\bigr]y^3 +O(y^4). \end{aligned} \tag{O12.11}$$

In particular, the $m=\pm1$ terms supply $Y$ and $Z$. Integrating only the $m=0$ sector would miss them.

Now expand XYZ independently. Write $w_1,w_2,w_3$ for its three flavor weights, so $w_1w_2w_3=1$. The lowest bosons have weights $w_i y$, the fermions contribute $-w_i^{-1}y^2$, and derivative corrections start at $y^4$. Let $h_n$ be the sum of all degree-$n$ bosonic monomials in the three $w_i$. The first coefficients are

$$\begin{aligned} [y]\mathcal I_{\rm XYZ}&=\sum_i w_i,\\ [y^2]\mathcal I_{\rm XYZ} &=h_2-\sum_i w_i^{-1}=\sum_i w_i^2,\\ [y^3]\mathcal I_{\rm XYZ} &=h_3-\Bigl(\sum_i w_i\Bigr)\Bigl(\sum_jw_j^{-1}\Bigr) =\sum_i w_i^3-2. \end{aligned} \tag{O12.12}$$

At order $y^2$, the inverse weights cancel the mixed bosonic monomials because $w_i^{-1}=w_jw_k$. At order $y^3$, the product in the subtraction contains the six mixed cubic weights and three copies of one, while $h_3$ contains those six and only one copy of one. This explains the $-2$ and reproduces every term of (O12.11).

The cancellation of mixed quadratic monomials is consistent with the F-term relations $XY=XZ=YZ=0$. It does not make the whole index a chiral-ring Hilbert series: that series has no $-2y^3$. An index coefficient is a signed protected count, and matching it alone does not prove equality of all observables or reconstruct an unprotected spectrum.

There is also a useful numerical check beyond the displayed truncation. At $x=1/8$ and $a=v=1$, gauge projection with the full flux sum and the full XYZ product both give

$$\mathcal I=1.970684111387770949326\ldots . \tag{O12.13}$$

One can evaluate the electric side by expanding each holonomy factor, extracting its neutral coefficient and then summing the fluxes as geometric series. Direct unit-circle quadrature checks that coefficient extraction independently. Keeping flavor phases nonzero tests more than the unflavored number; convergence bounds must accompany all truncated products and sums.

Exercises

1. Extend the first comparison

Find the coefficient of $y^4$ at $r=1/3$. Explain which fluxes must be retained and why the zero-flux derivative modes cannot simply be omitted.

Hint 1

The $|m|=1$ oscillator correction starts at $y^5$. The $|m|=4$ vacuum is already of order $y^4$.

Hint 2

At zero flux, pair a scalar with its first derivative mode, and compare with a conjugate fermion dressed by three lowest scalars.

Solution

The meson fourth power contributes $a^8y^4$. There are two neutral derivative–scalar combinations, each $+a^2x^{8/3}$. There are also two neutral combinations of one conjugate fermion and three lowest scalars, each $-a^2x^{8/3}$. These cancel, leaving $a^8$ in the zero-flux coefficient.

For $1\leq |m|\leq3$, the vacuum occurs below $y^4$ but the next contribution starts at $y^{4|m|+1}$, above it. The $m=\pm4$ vacua supply the remaining terms:

$$[y^4]\mathcal I=a^8+a^{-4}(v^4+v^{-4}). \tag{O12.14}$$

The XYZ check is also instructive. Its coefficient is $h_4-h_2\sum_iw_i^{-1}+\sum_{i<j}(w_iw_j)^{-1}+\sum_iw_i$. The last term is the first scalar derivative tower. Using $w_1w_2w_3=1$, this reduces to $\sum_iw_i^4$, exactly (O12.14). Keeping only unit flux is enough for the leading monopole terms, not for an arbitrarily long series.

2. Change the R-label without changing the theory

Replace the reference electric R-label $r$ by $r+\delta$. Show that this can be implemented by $a\mapsto a x^\delta$ everywhere in (O12.5), including its zero-point factor. Find the induced XYZ R-labels. Why does this not move the physical symmetric fixed point away from $r=1/3$?

Hint 1

Track one scalar factor, one conjugate-fermion factor and one bare monopole vacuum separately.

Hint 2

The three XYZ axial charges are $2,-1,-1$. A flavor redefinition changes the presentation of the trace, not its physical Hilbert space.

Solution

The scalar weight gains $x^\delta$, the conjugate-fermion weight gains $x^{-\delta}$, and the vacuum gains $x^{-\delta|m|}$. They become precisely the weights of (O12.5) with $r+\delta$:

$$\mathcal I_{r+\delta}(x,a,v) =\mathcal I_r(x,a x^\delta,v),\qquad (R_X,R_Y,R_Z)=(2r+2\delta,1-r-\delta,1-r-\delta). \tag{O12.15}$$

Their sum remains two. Forgetting the vacuum shift would spoil the monopole match. Starting with the source’s $r_0=1/2$ requires $\delta=-1/6$, giving (O12.7).

This is a reparametrization by the conserved axial generator. The physical superconformal $R$ in $\mathcal H_q$ remains the one fixed by the interacting theory. A differently labeled function with the same numerical $a$ weights states differently; it does not assign new scaling dimensions to the same fixed-point operators. Analytic substitutions should also remain in a convergence domain or be identified as analytic continuation.

3. A fixed topological background flux

Turn on a fixed integer topological background flux $N=1$, keeping the axial background flux zero. Use the index BF convention of Kapustin–Willett (2.16): the electric integrand acquires $u^N$, while its flux weight remains $v^m$. Here $N=2n_{\rm source}$, and neither $N$ nor $v$ is summed or integrated.

At $r=1/3$, put $t=x^{1/3}$. Calculate the index through $t^6$ in both descriptions. Then define an extra integer mixed axial/topological source counterterm by its index factor $a^{\kappa N}$. What changes for $\kappa=2$? Could the ordinary zero-background-flux comparison detect it?

Hint 1

The holonomy integral now selects oscillator electric charge $-1$. In zero dynamical flux, the lightest such mode is $\widetilde Q$.

Hint 2

On XYZ, $Y,Z$ see fluxes $+1,-1$, while $X$ sees zero. Compute their combined zero-point factor before expanding their products.

Solution

The $m=0$ sector contributes $a t+a^3t^3+(a^5-a^{-1})t^5$: the first three bosonic states are $\widetilde Q$ times $1,M,M^2$, and the negative term is the charge-$-1$ conjugate fermion of $Q$. In $m=\pm1$, the vacuum has weight $a^{-1}v^{\pm1}t^2$, and the lowest charge-$-1$ scalar has weight $a t^4$. Higher flux cannot contribute yet. Thus

$$\mathcal I_{N=1} =a t+a^3t^3+(a^5-a^{-1})t^5 +(v+v^{-1})t^6+O(t^7). \tag{O12.16}$$

On XYZ, the two flux-carrying chirals have R-charge $2/3$, axial charge $-1$ and opposite topological charges. Applying symmetric zero-point ordering gives their combined factor $a x^{1/3}=a t$. The $X$ product begins $1+a^2t^2+(a^4-a^{-2})t^4$. The first $Y,Z$ scalar weights are $a^{-1}v^{\pm1}t^5$. Multiplying gives (O12.16), including its two nonzero electric-flux contributions.

The extra counterterm multiplies the entire answer by $a^2$, so the leading term becomes $a^3t$. At $N=0$ its factor is one: the ordinary comparison cannot detect this source choice. The sign of $\kappa$ here is defined by the stated index BF factor, rather than inferred from an unspecified Euclidean connection convention. We have used integral background bundles for the charge normalization above; this example does not classify extensions to every possible quotient of the global symmetry group. The matched prescription used in the lesson has no such additional term, $\kappa=0$.

The fixed-background result is a generalized index. It is not a sum over a new dynamical gauge field, and its charged vacuum need not have the ordinary zero-flux identity contribution.

The essential test is now reproducible: fix the vacuum and source prescription, project within each flux sector, and only then compare the two descriptions. The source’s exact identity explains why this controlled comparison continues beyond the low orders computed here; the individual terms explain why magnetic sectors are indispensable.

For further protected traces, compare a separately sourced duality index or boundary conditions and their half-indices. The boundary problem has its own preserved algebra and reference R charge.