Count a free chiral multiplet by hand

The index of a free multiplet is already more than a count of its component fields. A field creates infinitely many sphere modes; its equation of motion removes some descendants; and arbitrarily many bosons can occupy one surviving mode. We will do these three steps separately, beginning with states that can be listed by hand.

Use the definition of the index and the free chiral action. Our theory is a massless, gauge-neutral three-dimensional $\mathcal N=2$ chiral multiplet with no superpotential. It has four real Poincaré and four real conformal supercharges, although the trace selects only one complex Poincaré component and its radial adjoint. There is no dynamical gauge group, gauge projection or magnetic sector in this calculation. The flavor background also has zero flux.

Count physical modes, not auxiliary fields

Let $\phi$ have flavor charge $+1$. The unitary free SCFT fixes its scaling dimension and superconformal R-charge to $\Delta_\phi=r_\phi=1/2$. Its fermion has $\Delta_\psi=1$ and $r_\psi=-1/2$. Conjugate fields have opposite R and flavor charges and the same dimensions.

Here a dagger denotes the physical conjugate field in the unitary radial theory, with the spinor index converted to the usual spin-$1/2$ representation when necessary. It does not impose a new conjugation rule on the independent Euclidean tilded variables used to prove the action’s invariance. That action gives, away from other insertions,

$$\partial^2\phi=0,\qquad \gamma^\mu\partial_\mu\psi=0,\qquad F_\Phi=\widetilde F_\Phi=0, \tag{O11.1}$$

and the conjugate equations. The auxiliaries have no kinetic term and no independent oscillator states. Counting them as additional bosons would count an off-shell description instead of the Hilbert space.

A derivative of a local field raises $\Delta$ by one. At derivative order $n$, commuting derivatives give a symmetric rank-$n$ spatial tensor. For the scalar, the equation of motion removes its traces. There are

$$\binom{n+2}{2}-\binom n2=2n+1 \tag{O11.2}$$

independent components, with the second term absent for $n<2$. These form the spin-$j=n$ rotation representation. Equivalently, the descendants are polynomials in three translation generators modulo their quadratic equation $P^2\phi=0$.

For the spinor there are initially $2\binom{n+2}{2}$ components. The derivatives of its first-order equation remove $2\binom{n+1}{2}$, leaving $2n+2$, the dimension of spin $j=n+1/2$. There is no further relation among these equation-of-motion descendants: multiplication by $\gamma\cdot P$ is injective on polynomial spinors, because its square is $P^2$ and a nonzero polynomial cannot annihilate another polynomial. This explains the subtraction without treating all derivatives as unrelated letters.

Under the state–operator correspondence, these are precisely the free one-particle sphere modes. Their cylinder energies above the vacuum are $\Delta/L$ on a sphere of radius $L$; $n$ is dimensionless. The result is

Field family	$\Delta$	$j$	$r$	Flavor $f$	Parity
Scalar $\phi$	$n+1/2$	$n$	$1/2$	$+1$	Even
Conjugate scalar $\phi^\dagger$	$n+1/2$	$n$	$-1/2$	$-1$	Even
Fermion $\psi$	$n+1$	$n+1/2$	$-1/2$	$+1$	Odd
Conjugate fermion $\psi^\dagger$	$n+1$	$n+1/2$	$1/2$	$-1$	Odd

Every row has $n=0,1,\ldots$ and $J_3$ eigenvalues $m=-j,-j+1,\ldots,j$. These are on-shell free representations. An interacting multiplet with a different scalar dimension cannot be counted by simply retaining these free equations of motion.

Which modes survive?

Use the previously chosen charge and weights:

$$\mathcal H_q=\{q,q^\dagger\}=D-R-J_3,\qquad K=D+J_3,\qquad \mathcal I(x,z)=\operatorname{Tr}\Gamma e^{-\beta\mathcal H_q}x^Kz^{F_{\rm fl}}. \tag{O11.3}$$

Here $\Gamma$ is fermion parity, $F_{\rm fl}$ is the flavor generator, and $z$ is its fixed fugacity. The charge $q$ changes $(\Delta,r,m)$ by $(1/2,1,-1/2)$ and leaves flavor unchanged. Its radial adjoint is a conformal charge. The positive eigenvalue to test is $\delta=\Delta-r-m$, not the cylinder energy.

For $\phi$, $\delta=n-m$, which vanishes only at $m=n$. For $\phi^\dagger$ it is $n+1-m\geq1$. For $\psi$ it is $n+3/2-m\geq1$. Finally, for $\psi^\dagger$ it is $n+1/2-m$, zero only at $m=n+1/2$. We retain one scalar and one fermion at each $n$.

One can also see the discarded pairs explicitly. Up to normalized nonzero coefficients, the on-shell component transformations give

$$\begin{aligned} q:\quad \psi_{n,m}&\longmapsto\phi_{n+1,m-1/2},\qquad &&-n-\tfrac12\leq m\leq n+\tfrac12,\\ q:\quad (\phi^\dagger)_{n,m}&\longmapsto(\psi^\dagger)_{n,m-1/2}, &&-n\leq m\leq n. \end{aligned} \tag{O11.4}$$

The first image contains every scalar mode of order $n+1$ except its highest-weight component; the second contains every conjugate-fermion mode of order $n$ except its highest-weight component. The free transformations have the structures $q\psi\sim\partial\phi$ and $q\phi^\dagger\sim\psi^\dagger$; their other images vanish after eliminating $F_\Phi$. On each displayed pair the coefficient can be chosen $\sqrt\delta$ in normalized states, since $\{q,q^\dagger\}=\delta$. Both states have the same $K$ and flavor and opposite parity. This verifies the cancellation, including its angular-momentum endpoints.

Write $P_+$ for the translation component with $J_3$ charge $+1$. The surviving modes are

$$B_n=P_+^n\phi,\qquad \Psi_n=P_+^n\psi^\dagger_+,\qquad n\geq0, \qquad b_n=z x^{1/2+2n},\quad f_n=z^{-1}x^{3/2+2n}. \tag{O11.5}$$

The subscript $+$ on the fermion means $m=+1/2$. Each $P_+$ raises both $D$ and $J_3$ by one, hence raises $K$ by two. The symbols $b_n,f_n$ are weights; the fermionic contribution to the trace is $-f_n$. These towers are non-null: the surviving highest-weight derivatives are not traces or Dirac-equation descendants.

The single-particle signed sum is therefore

$$i_{\rm ch}(x,z)=\sum_{n\geq0}(b_n-f_n) =\frac{z x^{1/2}-z^{-1}x^{3/2}}{1-x^2}. \tag{O11.6}$$

It agrees with the neutral, zero-flux, $\Delta_\Phi=1/2$ specialization of Kapustin–Willett (2.3) and (2.9). The source’s $t_a$ is our $z$ for this flavor; its $\Delta_\Phi$ is the scalar R-charge. Equation (O11.6) counts single-particle modes, not the full index.

A first count before the product

Set $t=x^{1/2}$, so all powers are integers. The first letters are $B_0$ of weight $zt$, $\Psi_0$ of weight $z^{-1}t^3$, and $B_1$ of weight $zt^5$. Products mean normally ordered free-field states. Bosonic powers are unrestricted; an identical fermionic mode can occur only once.

Power of $t$	Even states	Odd states	Signed coefficient
$0$	Vacuum	None	$1$
$1$	$B_0$	None	$z$
$2$	$B_0^2$	None	$z^2$
$3$	$B_0^3$	$\Psi_0$	$z^3-z^{-1}$
$4$	$B_0^4$	$B_0\Psi_0$	$z^4-1$
$5$	$B_0^5,\ B_1$	$B_0^2\Psi_0$	$z^5+z-z=z^5$
$6$	$B_0^6,\ B_0B_1$	$B_0^3\Psi_0$	$z^6+z^2-z^2=z^6$

In particular, the vanishing coefficient of $zt^5$ hides two surviving states. Both are annihilated by $q$ and $q^\dagger$; they are not one of the positive-$\delta$ pairs already removed. Their identical weights and opposite parity cancel only in the signed trace. Also, $B_1$ and $B_0B_1$ are descendants, and are intentionally counted. Radial states related by a translation are not identified as they would be in an integrated action modulo a boundary term.

This is already different from the free scalar chiral ring, whose monomials $1,\phi,\phi^2,\ldots$ have generating function $1/(1-zt)$. The index includes the fermion at $t^3$ and the derivative towers as well.

Assemble the free Fock space

In a free theory $\mathcal H_q$ is additive on oscillator occupations. Since every one-particle $\delta$ is nonnegative, a many-particle zero state can contain only the surviving modes. The full zero space is the bosonic symmetric and fermionic exterior algebra of (O11.5). Thus a single bosonic mode contributes $1+b_n+b_n^2+\cdots$, while a single fermionic mode contributes $1-f_n$.

$$\boxed{ \mathcal I_{\rm ch}(x,z) =\prod_{n=0}^{\infty} \frac{1-z^{-1}x^{3/2+2n}}{1-zx^{1/2+2n}}.} \tag{O11.7}$$

Multiplying the first factors reproduces the table:

$$\mathcal I_{\rm ch} =1+zt+z^2t^2+(z^3-z^{-1})t^3+(z^4-1)t^4 +z^5t^5+z^6t^6+O(t^7). \tag{O11.8}$$

There is one normalized vacuum and no flux-dependent zero-point factor here. A background contact-term completion of a path integral on general bundles is additional data; the expression above is the specified radial Hilbert-space trace. The flavor fugacity is not integrated: the flavor symmetry is global.

Only now is compact notation useful. Define $(a;Q)_\infty=\prod_{n\geq0}(1-aQ^n)$, with $Q$ a number unrelated to the supercharge $q$. Then

$$\mathcal I_{\rm ch} =\frac{(z^{-1}x^{3/2};x^2)_\infty}{(zx^{1/2};x^2)_\infty} =\exp\!\left[\sum_{m=1}^{\infty}\frac1m i_{\rm ch}(x^m,z^m)\right]. \tag{O11.9}$$

The last equality follows by expanding $-\log(1-b)=\sum_m b^m/m$ and $\log(1-f)=-\sum_m f^m/m$. It is called a plethystic exponential. Its fermionic minus sign stays negative for every $m$: parity is already included in $i_{\rm ch}$. These are the product manipulations in (2.7)–(2.11), now derived from the modes and their statistics.

For the literal absolutely convergent zero-space trace, take $0<x<1$, $z\ne0$ and $|z|x^{1/2}<1$. The fermion product converges for every fixed nonzero $z$ because $\sum_n|f_n|<\infty$. The unreduced trace (O11.3) additionally needs $\beta>0$ and $|z|^{-1}x^{1/2}e^{-\beta}<1$, so that conjugate-scalar occupations also converge; a sufficiently large $\beta$ ensures this. For the logarithmic series in (O11.9), instead require $|z|^{-1}x^{3/2}<1$ in addition to the zero-space conditions: both elementary logarithmic expansions then converge. In particular, a unit-modulus flavor fugacity is safe. Formal power series in $t$ with Laurent-polynomial coefficients in $z$ give another precise interpretation. Equality after analytic continuation is not a license to replace a divergent occupation trace by a finite answer.

Exercises

1. Two more coefficients

List all states through orders $t^7$ and $t^8$, keeping their flavor charges. Find the two signed coefficients. Are states containing two fermions needed yet?

Hint 1

The new mode is $\Psi_1$ of weight $z^{-1}t^7$. The next bosonic mode first appears at $t^9$.

Hint 2

At order $t^8$, include $B_1\Psi_0$ as well as $B_0\Psi_1$. The same-mode square $\Psi_0^2$ is zero.

Solution

At $t^7$ the even states are $B_0^7$ and $B_0^2B_1$, with weights $z^7,z^3$. The odd states are $B_0^4\Psi_0$ and $\Psi_1$, with weights $z^3,z^{-1}$. The coefficient is $z^7-z^{-1}$.

At $t^8$ the even states are $B_0^8,B_0^3B_1$, with weights $z^8,z^4$. The odd states are $B_0^5\Psi_0,B_1\Psi_0,B_0\Psi_1$, with weights $z^4,1,1$. The coefficient is $z^8-2$. The first two-fermion state is $\Psi_0\Psi_1$ at $t^{10}$, with positive sign and flavor $-2$. There is no $\Psi_0^2$ state at $t^6$. Expanding the factors with exponents $1,3,5,7$ in (O11.7) checks both results.

2. Change the number of free fields

Add a second independent free chiral with flavor charge $-1$ under the same global $U(1)$; keep both scalar R-charges $1/2$ and add no interaction. Derive its total index through $t^3$. Explain why the neutral scalar product at $t^2$ is counted and why charged states are still present.

Hint 1

The two Hilbert spaces tensor together. Replace $z$ by $z^{-1}$ in the second copy of the one-chiral answer.

Hint 2

At cubic order, first count all four scalar monomials, then the two single-fermion states. Being neutral is not an additional rule for a global symmetry.

Solution

The answer is $\mathcal I_{\rm ch}(x,z)\mathcal I_{\rm ch}(x,z^{-1})$, hence

$$\mathcal I_{2\rm ch} =1+(z+z^{-1})t+(z^2+1+z^{-2})t^2 +(z^3+z^{-3})t^3+O(t^4).$$

If the scalars are $A,B$, their cubic states $A^3,A^2B,AB^2,B^3$ have weights $z^3,z,z^{-1},z^{-3}$. The conjugate fermions subtract $z^{-1}+z$, leaving the displayed coefficient. The neutral $AB$ is an ordinary allowed product state. No Gauss law has been imposed, so $A$ and $B$ themselves remain as charged states. Integrating over the flavor fugacity would change this counting problem; it is not part of adding another free field.

3. A zero and a pole

Fix $0<x<1$. Compare the choices $z=x^{3/2}$ and $z=x^{-1/2}$. Does either define a convergent zero-space trace? Interpret a zero or divergence of the answer without calling it a count of supersymmetric vacua.

Hint 1

Examine the occupation sum of $B_0$ and the two allowed occupations of $\Psi_0$ separately.

Hint 2

A unit fermionic weight gives $1-1$ and a finite absolute sum $1+1$. A unit bosonic weight gives an infinite geometric series.

Solution

At $z=x^{3/2}$, $b_n=x^{2+2n}$ and $f_n=x^{2n}$. The bosonic product converges; the absolute fermionic product also converges, including its first factor $2$. But $f_0=1$, so the signed factor $1-f_0$ vanishes. Every state built from the other surviving modes is cancelled by the state with $\Psi_0$ added. This is a cancellation between zero-space states with the same specialized weight, not their removal from the Hilbert space. The index is zero, and the vacuum still exists. The logarithmic series for this first fermionic factor fails at its endpoint, while the product and the absolutely convergent trace remain meaningful.

At $z=x^{-1/2}$, $b_0=1$. All $B_0^k$ have unit weight, so the absolute trace and the bosonic occupation sum diverge. Now $f_0=x^2<1$; no zero numerator cancels the divergence. The product has a pole at precisely this occupation boundary. Neither result is a statement about how many flat-space supersymmetric vacua the theory has.

The useful checkpoint is to recover both surviving towers and the $z^8-2$ coefficient without a memorized determinant. The next physical change is to make a symmetry dynamical: then Gauss law and allowed magnetic sectors must be included. That is a new calculation, not a modification of Bose or Fermi statistics.

Carry out that next step in Gauge projection and an index check of SQED–XYZ. The monopole vacuum factors are as important as the neutral oscillator products.