What survives in a supersymmetric index?

A supersymmetric spectrum can change while a signed trace stays fixed. To understand what such a trace remembers, we must specify more than a symbol $(-1)^F$: which states are traced, which supercharge pairs them, and which weights agree on each pair?

Use R-symmetry versus flavor symmetry and the radial norm argument in What BPS and protected mean. We work in a unitary three-dimensional $\mathcal N=2$ superconformal field theory, or SCFT, in its conformal vacuum. Its full algebra has four real Poincaré and four real conformal supercharges. We will select one nilpotent complex component and its radial adjoint; selecting them does not change the theory’s full supersymmetry. No gauge group or rank is fixed by the definition below.

Put local operators on a sphere

The Hilbert space is the physical space $\mathcal H(S^2)$ of radial quantization, not a space of massive one-particle scattering states. Recall the cylinder construction. With radial distance $\rho=L e^{t/L}$,

$$ds_{\rm flat}^2=e^{2t/L} \left(dt^2+L^2d\Omega_2^2\right), \qquad H_{\rm cyl}-E_{\rm vac}=\frac DL.$$

The Weyl transformation removes the overall factor. A dilation becomes cylinder time translation. The dimensionless generator $D$ has eigenvalue $\Delta$ on the state created by an operator of scaling dimension $\Delta$. The radius $L$ converts it to an energy gap. We use the radial normalization $D|0\rangle=0$; a partition function with additional vacuum-energy or local-source factors would require its own prescription.

Let $R$ be the actual superconformal R-generator, $J_3$ one Hermitian sphere rotation generator, and $F_a$ mutually commuting flavor generators. Denote their eigenvalues by $r,m,f_a$. The subscript on $F_a$ matters: flavor charge is not fermion parity. Write the latter as $\Gamma=(-1)^F$, with eigenvalues $+1$ and $-1$ on even and odd states.

In a gauge theory, $\mathcal H(S^2)$ contains physical, Gauss-law-invariant states. The theory’s global gauge group determines which magnetic sectors are allowed; local monopole insertions can create states in those sectors. They cannot be discarded by calling all states “polynomials in elementary fields.” Here the flavor background has zero magnetic flux. Turning on prescribed flavor flux gives a different, generalized index, while dynamical gauge flux must be summed when constructing the physical trace. This distinction is explicit in Kapustin–Willett, section 2.2, (2.13)–(2.14). We will not need a model-specific flux measure yet.

Choose the charge before choosing the weights

Choose $q$ from the R-charge-$+1$ Poincaré family, called $\widetilde Q$ in the component lessons, with

$$q^2=0,\qquad [D,q]=\frac12q,\qquad [R,q]=q,\qquad [J_3,q]=-\frac12q,\qquad [F_a,q]=0. \tag{O10.1}$$

Its adjoint is a conformal supercharge in the radial inner product. It is not the flat-space Poincaré adjoint whose anticommutator gives momentum. Normalize it as in the prerequisite:

$$\boxed{\mathcal H_q=\{q,q^\dagger\}=D-R-J_3.} \tag{O10.2}$$

This is the source’s (2.2): its energy symbol $\Delta$ is our $D$, and its $H$ is our $\mathcal H_q$. We use dimensionless radial generators, with the sphere radius setting the units of $q$. In particular, $\mathcal H_q$ is not $L(H_{\rm cyl}-E_{\rm vac})$.

For a normalized state of eigenvalue $\delta=\Delta-r-m$,

$$\delta=\|q|v\rangle\|^2+\|q^\dagger|v\rangle\|^2\geq0. \tag{O10.3}$$

The zero space is annihilated by both charges. It can contain states with positive cylinder energy: a scalar chiral primary has $\Delta=r>0$, $m=0$, and hence $\delta=0$. Calling this a zero of $\mathcal H_q$ does not make it the cylinder vacuum.

Now ask which continuous chemical potentials respect this pairing. For $G=aD+bJ_3+cR+\sum_i\nu_i F_i$, its commutator is

$$[G,q]=\left(\frac{a-b}{2}+c\right)q. \tag{O10.4}$$

Thus the allowed linear combinations satisfy $a-b+2c=0$. A useful basis is

$$K=D+J_3,\qquad \mathcal H_q=D-R-J_3,\qquad F_a. \tag{O10.5}$$

Indeed, any such $G$ is $\tfrac12(a+b)K-c\mathcal H_q+\sum_i\nu_iF_i$. Weighting by $\mathcal H_q$ changes the regulator; the remaining nontrivial weights are $K$ and flavor. Both $K$ and $\mathcal H_q$ commute with $q$, since their charge shifts are respectively $1/2-1/2=0$ and $1/2-1+1/2=0$. The same is true of their commutators with $q^\dagger$.

Calculate one cancellation completely

For an eigenspace with $\delta>0$, set $c=q/\sqrt\delta$. Then $c^2=0$ and $\{c,c^\dagger\}=1$. It decomposes into two-state blocks. Choose a unit vector $|b\rangle$ with $q^\dagger|b\rangle=0$ in one such block and define $|f\rangle=q|b\rangle/\sqrt\delta$. It has unit norm because $\langle b|q^\dagger q|b\rangle=\delta$. The inverse map is $q^\dagger/\sqrt\delta$. The names can be interchanged if the initial vector is odd.

For a concrete block take the charges

State	Parity	$\Delta$	$r$	$m$	One flavor charge $f$	$\delta$	$K$
$b$	Even	$2$	$0$	$0$	$1$	$2$	$2$
$f=q b/\sqrt2$	Odd	$5/2$	$1$	$-1/2$	$1$	$2$	$2$

These are the charges of a selected supercharge block, not a claimed complete SCFT spectrum. In its ordered basis,

$$q=\begin{pmatrix}0&0\\\sqrt2&0\end{pmatrix},\qquad \mathcal H_q=2I_2.$$

With a dimensionless regulator $\beta>0$, a weight $x^K$ and flavor weight $z^f$, the two contributions are

$$x^2z\,e^{-2\beta}-x^2z\,e^{-2\beta}=0. \tag{O10.6}$$

Their ordinary dimensions differ, but their allowed weights agree. If we instead insert $x^D$, the answer is $z e^{-2\beta}(x^2-x^{5/2})$, which does not vanish for generic $x$. Conservation of $D$ alone is not enough: it fails to commute with $q$.

The three-dimensional superconformal index is

$$\boxed{\mathcal I(x,z_a)= \operatorname{Tr}_{\mathcal H(S^2)} \left[\Gamma e^{-\beta\mathcal H_q}x^K \prod_a z_a^{F_a}\right].} \tag{O10.7}$$

This is (2.1) with the additional regulator translation $\beta_{\rm source}=-\beta$: the source writes $e^{+\beta_{\rm source}H}$, while we use $e^{-\beta\mathcal H_q}$ with $\beta>0$. A fugacity is the multiplicative weight, such as $z=e^{i\alpha}$ for a compact flavor rotation; one can also continue it into a convergence domain. Initially take $0<x<1$ and flavor phases, and require the weighted trace to exist with finite multiplicities. More general complex fugacities need their own domain and choice of logarithm. An expansion has the actual allowed $K$ weights, which need not be integer powers.

Positive-$\delta$ blocks cancel at each common fugacity weight, so

$$\mathcal I= \operatorname{Tr}_{\ker\mathcal H_q} \left[\Gamma x^K\prod_a z_a^{F_a}\right]. \tag{O10.8}$$

This establishes independence of $\beta$ under the stated spectral and convergence assumptions. It does not assert independence of $x$ or $z_a$. Nor does $e^{-\beta\mathcal H_q}$ suppress an infinite tower already at $\delta=0$; the remaining weights are essential.

What survives is a cohomology class

A state is closed if $q|v\rangle=0$, and exact if $|v\rangle=q|w\rangle$. Nilpotence makes every exact state closed. The quotient $\ker q/\operatorname{im}q$ is called $q$-cohomology: two closed states represent the same class when they differ by an exact state.

On a positive-$\delta$ eigenspace, a closed state obeys

$$|v\rangle=\frac1\delta\{q,q^\dagger\}|v\rangle =q\left(\frac{q^\dagger|v\rangle}{\delta}\right). \tag{O10.9}$$

It is exact. Conversely a zero-$\delta$ state is orthogonal to every exact state, since $q^\dagger|v\rangle=0$. State by state, or in finite regulated charge spaces, this identifies cohomology with $\ker\mathcal H_q$. An extension to arbitrary infinite superpositions needs domain control for division by $\delta$; it cannot be assumed when positive eigenvalues accumulate at zero. The trace argument above retains its spectral and convergence assumptions. It remains a signed trace: an even and an odd class at identical weights cancel despite both being present. Equality of two indices is consequently a useful test of a proposed equivalence, not a proof that their full spectra or dynamics agree.

The index can include derivative descendants. Let $P_+$ be the highest-spin component of translations, with $D$-charge one, $J_3$-charge one, and R- and flavor charges zero. Then $[\mathcal H_q,P_+]=0$ and $[K,P_+]=2P_+$. If $P_+^n|\mathcal O\rangle$ is nonzero, it stays in the zero space and has weight $x^{\kappa+2n}\prod_a z_a^{f_a}$, where $\kappa$ was the original $K$ weight. A non-null independent tower contributes a geometric series, not only its first state. Null relations must be removed before counting. This explains why an index calculation is not a list of scalar chiral primaries.

For comparison, a chiral ring organizes scalar operators annihilated by both components of the chiral supercharge family, with products and relations. Its usual Hilbert series counts those classes without this fermion-parity subtraction or the full derivative tower. A supersymmetric $S^3$ partition function is a path integral on a different background and topology; it is not (O10.7). The index has a path-integral interpretation on $S^2\times S^1$, with the fermion-parity insertion giving periodic fermionic gluing and the fugacities imposing additional symmetry twists. It is not an ordinary thermal partition function with antiperiodic fermions. The free sphere determinant and its source prescription belong to the sphere observable, not to this state count.

Which changes preserve the answer?

A continuous supersymmetric deformation can change dimensions and let short representations join into a long one. A pair that leaves the zero space contributes opposite signs at equal commuting weights. Invariance requires that the chosen charge, grading, flavor data and trace prescription can be transported consistently, with no lost states, continuum boundary contribution or failure of convergence. A mass deformation taking a theory away from conformality is not literally a continuous change of the same SCFT dilation generator; using an index along an RG flow requires the corresponding preserved symmetries and supersymmetric background construction.

There is also a harmless change of labels that should not be confused with a physical change of the radial algebra. For one flavor generator, write $R_t=R+tF$. The exact positive operator remains

$$\mathcal H_q=D-R_t-J_3+tF. \tag{O10.10}$$

On its kernel, $K=R+2J_3=R_t+2J_3-tF$. Thus a weight written with $R_t$ can be returned to the physical one by a flavor-fugacity substitution. It cannot change an operator’s actual dimension. The source’s footnote 1 explicitly warns that a background construction with arbitrary R assignments need not retain the direct superconformal interpretation; it is not permission to replace $R$ in (O10.2) and discard the compensating term.

Quantum-mechanics comparison

This optional comparison uses What the Witten index can tell us. There the positive operator was the physical Hamiltonian, with $\{Q,Q^\dagger\}=2H_{\rm QM}$. Its convergent graded trace counted the signed number of zero-energy vacua. Here $\mathcal H_q=D-R-J_3$ is a dimensionless combination, and its zero states can have positive cylinder energy. The index weights those states by $x^K$ and flavor charges. A nonzero coefficient certifies a nonzero signed population at those weights, not that an ordinary excited-state degeneracy equals that coefficient.

The same lesson about convergence survives the change of Hilbert space. One cannot infer deformation invariance merely by writing a formal anticommutator. In particular, taking a fugacity to a boundary where infinitely many protected descendants have equal weight may destroy the trace even though all positive-$\delta$ pairs still cancel.

Practice

1. Another conserved weight

Test $G_1=2D-R$ and $G_2=D-J_3$ as continuous index weights. Express any allowed combination in terms of $K$ and $\mathcal H_q$. For the positive pair in the table, compute the two $G_2$ weights and its uncanceled contribution. Does including a factor $x^{\mathcal H_q}$ change the final cohomology trace?

Hint 1

Use the charge shifts in (O10.1), not only conservation of the generators.

Hint 2

$K+\mathcal H_q=2D-R$. On the kernel, $\mathcal H_q$ has eigenvalue zero.

Solution

$[G_1,q]=(1-1)q=0$, whereas $[G_2,q]=(1/2+1/2)q=q$. The first is allowed and equals $K+\mathcal H_q$. Multiplication by $x^{\mathcal H_q}$ shifts the regulator from $\beta$ to $\beta-\log x$; where the trace converges it leaves the kernel trace unchanged.

For the displayed pair, $G_2(b)=2$ and $G_2(f)=5/2-(-1/2)=3$. Its contribution becomes $z e^{-2\beta}(x^2-x^3)$, which is generically nonzero. A different conserved Hamiltonian weight is not automatically a supersymmetric fugacity.

2. Two states can disappear without changing the index

Consider a finite model of the selected charge algebra, not a complete SCFT. It has an even state $b$, an odd state $f$, and an additional even state $a$. For a real $u\geq0$, set $q f=\sqrt u\,b$ and let $q$ annihilate $a,b$. The nonzero adjoint action is $q^\dagger b=\sqrt u\,f$. Give $b,f$ flavor charge one and charges

$$(\Delta,r,m)_b=(2+u,2,0),\qquad (\Delta,r,m)_f=(3/2+u,1,1/2).$$

The state $a$ has $(\Delta,r,m)=(3,3,0)$ and flavor charge $-1$. Compute the index, the number of zero states at $u=0$ and at $u>0$, and the unsigned weighted zero-state count. What does a missing coefficient at weight $x^2z$ mean at $u=0$?

Hint 1

Both $b$ and $f$ have $\delta=u$ and $K=2+u$. Check the charge differences when $q$ acts from $f$ to $b$.

Hint 2

At $u=0$ the charge matrix vanishes, but the grading remains. Count states before subtracting signs.

Solution

The charge shifts are $(1/2,1,-1/2)$, as required. The pair contributes $e^{-\beta u}x^{2+u}z-e^{-\beta u}x^{2+u}z=0$ for every $u$. The remaining answer is $\mathcal I=x^3z^{-1}$.

At $u=0$ there are three zero states, with unsigned weighted count $2x^2z+x^3z^{-1}$. At $u>0$ only $a$ remains in the zero space, with count $x^3z^{-1}$. The signed answer never changes. Thus the absent $x^2z$ coefficient at $u=0$ hides two classes; it does not prove their absence. The finite matrices demonstrate the selected algebra and its cancellation, without asserting that every such deformation embeds in a unitary SCFT.

3. Change an R label and approach a convergence boundary

Suppose an even scalar chiral primary has $\Delta=r$, flavor charge one, and a non-null independent tower $P_+^n|\mathcal O\rangle$ for $n\geq0$. Compute this tower’s index contribution. Reexpress the same weights using $R_t=R+tF$. If $C_t(x,z)$ denotes the tower weighted by $x^{R_t+2J_3}z^F$, relate it to the physical contribution $C(x,z)$. What fails at $x=1$ with $z=1$? Has the relabeling changed $\Delta$?

Hint 1

At level $n$, the three charges are $(\Delta,r,m)=(r+n,r,n)$, so $\delta=0$.

Hint 2

Sum the geometric series only in its convergence domain. Then compare the flavor powers before changing the fugacity.

Solution

The tower gives

$$C(x,z)=\frac{x^r z}{1-x^2},\qquad C_t(x,z)=\frac{x^{r+t}z}{1-x^2} =C(x,zx^t).$$

Equivalently $C(x,z)=C_t(x,zx^{-t})$. This is an equality of weights on the same kernel. The physical dimension is still $r$, and the compensating $tF$ in (O10.10) keeps the norm unchanged. A trace formula with arbitrary R data and a different background would need a separate construction.

For $|x|<1$ the displayed sum converges at fixed finite $z$. At $x=z=1$ every state contributes one, and the sum diverges. The $\mathcal H_q$ regulator cannot help because all these states have $\delta=0$. The computation concerns one stipulated independent tower; it is not a full free-multiplet index and includes no claim about additional towers or null relations.

The next task is dynamical: determine which actual scalar, fermion and derivative states survive in a particular theory, then implement gauge invariance and flux sectors when required. The definition already tells us which weights to compare and which cancellations those calculations must respect.

Count a free chiral multiplet by hand carries out that first dynamical calculation: remove equation-of-motion descendants, identify the surviving towers, and assemble their Bose and Fermi occupations.