BPS Indices, Absolute Degeneracies, and Wall Crossing

A protected BPS index and an absolute degeneracy answer different questions. The index is a signed, often spin-weighted trace that can survive continuous deformations; the degeneracy is the positive dimension of a specified state space. An index becomes an absolute black-hole count only after the broken-supercharge multiplet, exterior hair, multicenter states, chamber, and possible cancellations have all been controlled. Across a marginal-stability wall, even the protected index of a fixed total charge can jump.

Required background. The Witten Index, Vacuum Counting, and Its Failure Modes supplies the protected-trace logic; Microscopic Black-Hole Entropy: Claim and Ensemble Contract distinguishes the asymptotic trace from the horizon count.

Helpful background. Marginal Stability, Chambers, and Wall Crossing develops the general wall-crossing machinery; Index Inversion, Recombination, and Protected-Spectrum Limits explains why a protected trace usually cannot reconstruct a spectrum.

We will first separate complete multiplets from reduced states, then apply that distinction to torsion-one $\mathcal N=4$ dyons, and finally stress-test the inference with wall crossing and large cancellations.

Helicity supertraces and spin multiplets

Fix a charge $\Gamma$ and a chamber $\mathcal C$ of the asymptotic moduli space. After any stated removal of translational center-of-mass motion and continuum contributions, decompose the internal BPS Hilbert space into spatial-rotation multiplets,

$$\mathcal H^{\mathrm{BPS,int}}_{\Gamma,\mathcal C} =\bigoplus_j \mathbb C^{\,n_j}\otimes V_j,$$

where $V_j$ is the spin-$j$ irreducible representation of $SU(2)$ and $n_j$ is its multiplicity. The multiplet-completed absolute count in this convention is

$$d_{\mathrm{abs}}^{\mathrm{multiplet}}(\Gamma,\mathcal C) =\sum_j(2j+1)n_j.$$

Let $J_3$ be the Cartan generator of spatial rotations. The even helicity supertrace is

$$B_{2k}(\Gamma,\mathcal C) =\frac{1}{(2k)!} \operatorname{Tr}_{\Gamma,\mathcal C} \left[(-1)^{2J_3}(2J_3)^{2k}\right].$$

Equivalently,

$$B_{2k} =\frac{1}{(2k)!} \sum_j n_j(-1)^{2j} \sum_{r=-j}^{j}(2r)^{2k}.$$

The sign is constant within a spin multiplet because $r-j$ is an integer. The weights, however, are not bounded by one. If $j_{\max}$ is the largest spin present, the triangle inequality gives

$$\lvert B_{2k}\rvert \le \frac{(2j_{\max})^{2k}}{(2k)!} d_{\mathrm{abs}}^{\mathrm{multiplet}}.$$

For fixed $k$ and $j_{\max}>0$ this implies

$$\log d_{\mathrm{abs}}^{\mathrm{multiplet}} \ge \log\lvert B_{2k}\rvert -2k\log(2j_{\max}) +\log((2k)!).$$

Here $S_0$ denotes the candidate leading entropy scale. Exponential growth of $\lvert B_{2k}\rvert$ gives the same leading lower bound on the absolute entropy only if the helicity weight grows subexponentially, for example $\log j_{\max}=o(S_0)$ along the chosen charge sequence. This is a lower bound, not an equality. A small or vanishing index gives no upper bound on the number of states because positive and negative contributions may nearly cancel.

Broken supersymmetry and the reduced count

Suppose every state in the sector breaks exactly $4k$ real supercharges. The corresponding $2k$ complex fermion zero modes generate a universal Clifford module,

$$\mathcal H_{\Gamma,\mathcal C} \simeq \mathcal F_{4k}\otimes\mathcal H_{\mathrm{red}}, \qquad \dim\mathcal F_{4k}=2^{2k}.$$

The $(2J_3)^{2k}$ insertion saturates these zero modes. In the normalization above,

$$B_{2k}=(-1)^k\Omega_{\mathrm{red}}, \qquad \Omega_{\mathrm{red}} =\operatorname{Tr}_{\mathcal H_{\mathrm{red}}}(-1)^{2J_3},$$

while

$$d_{\mathrm{abs}}^{\mathrm{multiplet}} =2^{2k}d_{\mathrm{red}}, \qquad d_{\mathrm{red}} =\operatorname{Tr}_{\mathcal H_{\mathrm{red}}}\mathbf 1.$$

Consequently $\lvert B_{2k}\rvert\le d_{\mathrm{red}}$, with equality precisely when every nonzero reduced contribution has the same helicity sign. The selection rule is one-sided: a multiplet breaking more than $4k$ supercharges vanishes in $B_{2k}$; one breaking exactly $4k$ gives the clean factor above; multiplets breaking fewer supercharges can contribute unless the charge orbit excludes them or they are separately removed Dabholkar, Gomes, Murthy, and Sen 2011, § 2, eqs. (2.1)–(2.2), pp. 11–12.

An exact index–degeneracy comparison

The half-BPS Dabholkar–Harvey sector of heterotic string theory on $T^6$ provides a finite comparison before any black-hole asymptotics are taken. Its reduced left-moving oscillator degeneracies are generated by

$$\frac{1}{\Delta_{\mathrm{mod}}(\tau)} =\frac{1}{\eta(\tau)^{24}} =q^{-1}+24+324q+3200q^2+\cdots,$$

where the subscript distinguishes the modular discriminant $\Delta_{\mathrm{mod}}$ from the charge discriminant used below. For a primitive electric charge with $Q^2/2=1$, the coefficient of $q$ gives

$$d_{\mathrm{red}}=324.$$

A half-BPS state in four-dimensional $\mathcal N=4$ supersymmetry breaks eight real supercharges, so $k=2$ and the first saturated trace is $B_4$. The reduced oscillator states—the Clifford vacua after the broken-supersymmetry multiplet is stripped off—have the same helicity sign, and therefore

$$B_4=324, \qquad d_{\mathrm{abs}}^{\mathrm{multiplet}} =2^4d_{\mathrm{red}} =5184.$$

Here the index equals the reduced absolute multiplicity exactly, whereas the complete supermultiplet count differs by the known Goldstino factor $16$. This low-charge result establishes the exact index–multiplet relation; it is not a regular macroscopic two-derivative black hole Dabholkar, Murthy, and Zagier 2012, § 3.2, eqs. (3.16)–(3.24), pp. 20–22, Dijkgraaf, Verlinde, and Verlinde 1997, § 1 after eq. (1.3), p. 2.

From a single-center index to a horizon count

A quarter-BPS state in four-dimensional $\mathcal N=4$ supersymmetry preserves four of sixteen real supercharges and breaks twelve. Its first saturated helicity trace is therefore $B_6$. If the single horizon carries zero spatial angular momentum and the complete exterior sector consists only of the six universal fermion-zero-mode pairs—so there are no additional neutral or charged hair states—then

$$B_6^{\mathrm{single}}=-d_{\mathrm{hor}}, \qquad -B_6^{\mathrm{single}}=d_{\mathrm{hor}}>0.$$

The minus sign is the zero-mode factor $(-1)^3$. Spherical symmetry of a single supersymmetric four-dimensional horizon motivates the zero-angular-momentum condition; the resulting sign is a necessary test of the horizon interpretation Sen 2011, § 1, especially eq. (1.1), pp. 2–4. It does not prove that cancellations elsewhere are small.

The simple equality must be replaced whenever additional exterior hair contributes, even if that hair is neutral and spinless. Charged or spinning hair additionally requires the corresponding charge and helicity convolution,

$$B_6^{\mathrm{single}}(\Gamma) =\sum_{\Gamma_{\mathrm{hor}}+\Gamma_{\mathrm{hair}}=\Gamma} d_{\mathrm{hor}}(\Gamma_{\mathrm{hor}}) B_6^{\mathrm{hair}}(\Gamma_{\mathrm{hair}}).$$

At the generating-function level this becomes a product. One may divide by a hair factor only when the model supplies that factor, the charge variables and boundary conditions match, and the inverse exists in the chosen expansion. Single-center extraction and horizon–hair deconvolution are distinct operations Dabholkar, Gomes, Murthy, and Sen 2011, § 2, eqs. (2.4)–(2.9), pp. 12–14.

Torsion-one N=4 dyons

Now specialize to type II string theory on $K3\times T^2$, or equivalently heterotic string theory on $T^6$. Let $Q$ and $P$ be the electric and magnetic charge vectors. Their squares and inner product use the compactification’s integral charge-lattice bilinear. We restrict to the torsion-one orbit

$$I=\gcd(Q\wedge P)=1,$$

where $\gcd(Q\wedge P)$ is the greatest common divisor of the antisymmetric components $Q_iP_j-Q_jP_i$ in an integral basis. Dyons with $I>1$ require a modified divisor-sum formula. Define the T-duality invariants

$$n=\frac{Q^2}{2}, \qquad m=\frac{P^2}{2}, \qquad \ell=Q\!\cdot\!P, \qquad \mathscr D=4mn-\ell^2.$$

The symbol $\mathscr D$ denotes the charge discriminant. The two-derivative entropy of a regular large single-center solution is

$$S_{\mathrm{BH}}^{(0)}=\pi\sqrt{\mathscr D}.$$

The asymptotic coefficient and its contour

Introduce

$$\Omega= \begin{pmatrix} \tau & z\\ z & \sigma \end{pmatrix}, \qquad Z(\Omega)=\frac{1}{\Phi_{10}(\Omega)},$$

where $\Phi_{10}$ is the weight-ten Igusa cusp form. In the standard physical convention, the complete quarter-BPS index in an asymptotic chamber $\mathcal C$ is

$$\begin{aligned} d_{\mathrm{ind}}^{\mathrm{asymp}}(Q,P;\mathcal C) &\equiv-B_6(Q,P;\mathcal C)\\ &=(-1)^{\ell+1} \int_{\mathcal K_{\mathcal C}}\mathrm d\tau\,\mathrm dz\,\mathrm d\sigma\, \frac{ e^{-2\pi i(n\tau+\ell z+m\sigma)} }{\Phi_{10}(\Omega)}. \end{aligned}$$

The real parts of $\tau$, $z$, and $\sigma$ run over unit periods. Their imaginary parts specify how the contour passes the poles, and therefore specify the chamber. For a regular charge with $m>0$, $n>0$, and $\mathscr D>0$, a useful attractor representative is

$$\operatorname{Im}\tau=\frac{2m}{\epsilon}, \qquad \operatorname{Im}\sigma=\frac{2n}{\epsilon}, \qquad \operatorname{Im}z=-\frac{\ell}{\epsilon}, \qquad \epsilon\to0^+.$$

The interchange of $m$ and $n$ between the exponent and these two imaginary parts, as well as the sign of $\operatorname{Im}z$, follows from

$$\widetilde\Omega= \begin{pmatrix} \sigma & -z\\ -z & \tau \end{pmatrix}, \qquad \operatorname{Im}\widetilde\Omega =\frac{1}{\epsilon} \begin{pmatrix} 2n & \ell\\ \ell & 2m \end{pmatrix}.$$

This is the attractor-contour prescription, not a matrix inversion Dabholkar, Murthy, and Zagier 2012, § 6, eqs. (6.1), (6.8), and (6.10)–(6.14), pp. 37–40.

Finite and polar parts

Because $1/\Phi_{10}$ is meromorphic, changing the contour can change a Fourier coefficient without changing the meromorphic function. At fixed $m$ write

$$\frac{1}{\Phi_{10}(\tau,z,\sigma)} =\sum_{m=-1}^{\infty} \psi_m(\tau,z)e^{2\pi i m\sigma}, \qquad \psi_m=\psi_m^{\mathrm F}+\psi_m^{\mathrm P}.$$

For $m>0$, $\psi_m^{\mathrm P}$ is the polar Appell–Lerch part. Its chamber-dependent Fourier expansion carries the two-center residues responsible for wall crossing. The holomorphic finite part $\psi_m^{\mathrm F}$ is a mock Jacobi form. With $q=e^{2\pi i\tau}$ and $y=e^{2\pi iz}$, define

$$d_{\mathrm{ind}}^{\mathrm F}(Q,P) \equiv (-1)^{\ell+1} [q^ny^\ell]\,\psi_m^{\mathrm F}(\tau,z).$$

For $m>0$, $n>0$, $\mathscr D>0$, and a regular single-center attractor, these coefficients are interpreted as the chamber-independent, or “immortal,” single-center indexed degeneracies in this torsion-one model. They are not absolute degeneracies and do not equal the horizon count without the separate hair analysis. The finite part can also have negative-discriminant coefficients, which must not be interpreted as regular horizons merely because they survived the subtraction Dabholkar, Murthy, and Zagier 2012, pp. 4–5 and § 11.1, eqs. (11.17)–(11.22), pp. 117–119.

These objects should not be interchanged:

Counting objects separated by chamber behavior and black-hole meaning

Object	Chamber behavior	Black-hole interpretation
Asymptotic dindasymp(Q, P; C) = −B6(Q, P; C)	Changes when the contour crosses a pole	Index of the complete asymptotic charge sector, including chamber-supported bound states
Polar coefficient from ψmP	Its Fourier expansion changes with the chamber	Two-center contribution that carries the primitive wall jump in this model
Finite coefficient from ψmF	Independent of the asymptotic chamber after the decomposition is fixed	Protected single-center coefficient in the regular positive-discriminant domain, before any separate hair analysis
dhor from the fixed-charge horizon theory	Has no asymptotic chamber	Absolute single-horizon count only under the stated horizon boundary conditions

A reproducible comparison

Before comparing the finite coefficient with a black-hole entropy, record each item below.

Reproducibility requirements for the finite-coefficient entropy comparison

Item	Choice in this application	Why it matters
Theory	Type II on K3 × T2, equivalently heterotic on T6	Fixes the charge lattice, supersymmetry, and generating function.
Charge orbit	I = 1 with declared m, n, and ℓ	Excludes the nonprimitive divisor-sum formula.
Trace	dind = −B6 with the factor (−1)ℓ+1	Fixes the physical sign and the zero-mode normalization.
Chamber	The stated contour, with the attractor contour used for the single-center comparison	Selects which asymptotic Fourier expansion is meant.
Multicenter treatment	Extract the finite part ψmF rather than the polar part ψmP	Separates the torsion-one two-center wall contribution.
Exterior treatment	Identify and, when possible, deconvolve the model's hair generating function	Prevents an asymptotic trace from being mislabeled as a horizon count.
Regularity domain	m > 0, n > 0, 𝒟 > 0, plus a regular single-center attractor	Positive discriminant alone is not a complete horizon test.
Scaling	A primitive torsion-one sequence with m, n, ℓ = O(Λ2) at fixed ratios and 𝒟 ∼ Λ4	States which terms are leading and preserves the declared orbit.
Macroscopic object	The single-horizon fixed-charge quantity, not a sum over asymptotic multicenters	Matches the microscopic single-center object.
Failure test	A sign violation, chamber dependence after finite-part extraction, or exponential mismatch after hair deconvolution	Identifies which claimed inference would fail.

The scaling sequence must keep $\gcd(Q_\Lambda\wedge P_\Lambda)=1$; uniformly replacing both charge vectors by $\Lambda$ times fixed vectors would instead increase the torsion. Under the full set of conditions in the table, the licensed leading statement is

$$\log\lvert d_{\mathrm{ind}}^{\mathrm F}(Q_\Lambda,P_\Lambda)\rvert =\pi\sqrt{\mathscr D_\Lambda} +o(\sqrt{\mathscr D_\Lambda}).$$

After excluding multicenter sectors and accounting for all exterior degrees of freedom, the quantity compared with the fixed-charge horizon theory is $d_{\mathrm{hor}}$. Rotational invariance of the horizon sector gives

$$B_{0;\mathrm{hor}} \equiv \operatorname{Tr}_{\mathcal H_{\mathrm{hor}}}(-1)^{2J_3} =d_{\mathrm{hor}}.$$

If the only relevant exterior contribution consists of the six universal Goldstino pairs, then

$$-B_6^{\mathrm{single}} =d_{\mathrm{hor}}, \qquad \log d_{\mathrm{hor}} =\pi\sqrt{\mathscr D} +o(\sqrt{\mathscr D}).$$

This is the zero-mode and exterior factorization of Dabholkar, Gomes, Murthy, and Sen 2011, § 2, eqs. (2.4)–(2.9), pp. 12–14.

The expected sign is necessary evidence for this interpretation, not evidence that cancellations are mild Chattopadhyaya and David 2021, §§ 2 and 4.1.

A wall-crossing experiment with an absolute count

Import the stable-minus-empty convention from the prerequisite page. Let a primitive total charge split as $\Gamma=\Gamma_1+\Gamma_2$ and define

$$\kappa=\langle\Gamma_1,\Gamma_2\rangle.$$

Write $\Omega_{\mathrm{BPS}}(\Gamma_i)$ for the reduced protected index of constituent $i$. Call the side containing the two-center bound state the stable chamber and the adjacent side without it the empty chamber. Assume that the charges are primitive, only this two-body decay becomes marginal, the constituent indices stay fixed, and no scaling solution or additional aligned charge ray participates. The imported relative-quantum-mechanics result is

$$j_{\mathrm{rel}}=\frac{\lvert\kappa\rvert-1}{2}, \qquad d_{\mathrm{rel}}=2j_{\mathrm{rel}}+1=\lvert\kappa\rvert.$$

and the stable-minus-empty jump is

$$\Delta_{\mathrm{st}}\Omega_{\mathrm{BPS}}(\Gamma) =(-1)^{\kappa-1}\lvert\kappa\rvert \Omega_{\mathrm{BPS}}(\Gamma_1) \Omega_{\mathrm{BPS}}(\Gamma_2).$$

The derivation, stability condition, and convention checks remain with the primitive-jump discussion and Brennan, Moore, and Royston 2018, § 3.3, eqs. (3.28)–(3.29).

For an exact $\mathcal N=4$ worked charge sector, choose torsion-one charges with

$$m=n=1, \qquad \ell=\kappa=1, \qquad \mathscr D=3,$$

and let the pure electric and pure magnetic half-BPS constituents each have $B_4=324$. The relative spin is zero, so its absolute reduced degeneracy and index both have magnitude one. In the convention $d_{\mathrm{ind}}=-B_6$,

$$\Delta_{\mathrm{st}}d_{\mathrm{ind}} =d_{\mathrm{red,stable}}-d_{\mathrm{red,empty}} =324^2 =104\,976.$$

The same $104\,976$ new reduced states that appear in the stable chamber account for the protected-index jump; there is no cancellation in this one-multiplet relative sector. The empty chamber contains none of them. This comparison is exact, but it counts a two-center bound state—not a single horizon. It also proves that $\mathscr D>0$ does not make the complete asymptotic index chamber-independent. In the generating function, the contour crosses the $z=0$ pole and the polar contribution changes Sen 2008, § 1 and eqs. (2)–(4), pp. 3–6.

Cancellation can change the leading entropy

Wall crossing and cancellation are independent failure modes. A minimal algebraic stress test has $n_+=N+K$ positive-sign reduced states and $n_-=K$ negative-sign reduced states. Then

$$\Omega_{\mathrm{BPS}}=n_+-n_-=N, \qquad d_{\mathrm{red}}=n_++n_-=N+2K.$$

Choose $N_L=2^{\lfloor\alpha L\rfloor}$ and $K_L=2^L$ with $0<\alpha<1$. The leading logarithms are

$$\log\lvert\Omega_{\mathrm{BPS},L}\rvert =\alpha L\log2+O(1), \qquad \log d_{\mathrm{red},L} =L\log2+O(1).$$

The index stays positive and grows exponentially, yet it misses the leading absolute entropy. This example is a diagnostic, not a claim that a particular dyon spectrum realizes these multiplicities.

A physical example shows why the large-charge trajectory matters. Let $\mathcal C_2$ denote the zero-mode-saturated second helicity trace used in the cited analysis. For the five-dimensional D1–D5–momentum system at fixed $Q_1Q_5$ and $n\to\infty$, both formulas below use the ensemble summed over the $SU(2)_L$ angular momentum $J$:

$$\log\lvert\mathcal C_2\rvert \sim2\pi\sqrt{Q_1Q_5n}, \qquad \log d_{\mathrm{abs}} \sim2\pi\sqrt{(Q_1Q_5+2)n}.$$

Their ratio is exponentially small in $\sqrt n$ on that trajectory. This is a microscopic asymptotic chosen to expose cancellation, not automatically a controlled macroscopic-gravity limit. If $Q_1Q_5$ also becomes large, the added $2$ is subleading and the leading entropies agree. “Index and degeneracy have the same leading growth” is therefore incomplete until the scaling limit is named Dabholkar, Gomes, Murthy, and Sen 2011, § 5.1, eq. (5.20) and the following discussion, pp. 39–40.

Common pitfalls

“Protected” means chamber-independent. Protection applies under deformations that do not cross a locus where states enter or leave the Hilbert space. The fixed-total-charge index generally jumps across a marginal-stability wall.

The expected sign proves equality with a degeneracy. A positive remainder can arise from exponentially large positive and negative sectors. The sign tests compatibility with a nonrotating single horizon; equality still needs a same-sign theorem or an independent absolute count.

Positive discriminant proves that every coefficient is a single horizon. The regularity inequalities are necessary, not sufficient. The charge orbit, contour, finite-part extraction, attractor, and exterior sector still have to be specified.

The contour is only a computational choice. For a meromorphic generating function, the contour selects a Fourier expansion and hence a physical chamber.

Removing the polar part produces an absolute count. It produces a protected single-center coefficient in this model. Hair deconvolution and cancellation control are separate steps.

The primitive jump applies to every wall. Nonprimitive constituents, simultaneous aligned rays, and scaling solutions require the more general wall-crossing formulas.

A full asymptotic index is automatically the quantum entropy function. The former may include exterior and multicenter states; the latter is a fixed-charge single-horizon quantity. Their boundary conditions and state spaces must first be matched.

Exercises

1. Bound a weighted trace

Starting from the spin decomposition, derive the $j_{\max}$ bound on $B_{2k}$. What extra condition makes the difference between $\log\lvert B_{2k}\rvert$ and the resulting lower bound subleading compared with an entropy scale $S_0$?

Solution

For every state, $\lvert2J_3\rvert\le2j_{\max}$. Applying the triangle inequality to the trace gives

$$\lvert B_{2k}\rvert \le\frac{(2j_{\max})^{2k}}{(2k)!} d_{\mathrm{abs}}^{\mathrm{multiplet}}.$$

Rearranging and taking logarithms produces a correction $2k\log(2j_{\max})-\log((2k)!)$. At fixed $k$, it is $o(S_0)$ if $\log j_{\max}=o(S_0)$. This proves only a lower bound on the absolute count.

2. Saturate the quarter-BPS zero modes

A single-center quarter-BPS $\mathcal N=4$ sector has $d_{\mathrm{hor}}=700$ states of zero horizon angular momentum and no charged or spinning hair beyond the universal broken-supercharge modes. Compute $B_6^{\mathrm{single}}$ and the multiplet-completed absolute count including the Goldstino module.

Solution

Twelve broken real supercharges give $k=3$, so the zero-mode factor is $(-1)^3=-1$ and the Clifford module has dimension $2^6=64$. Therefore

$$B_6^{\mathrm{single}}=-700, \qquad d_{\mathrm{abs}}^{\mathrm{multiplet}}=64\times700=44\,800.$$

The result uses every hypothesis in the question. Charged hair would replace the simple product by a convolution.

3. Cross a primitive wall

Take $\kappa=3$ and two constituents with one reduced state each and $\Omega_{\mathrm{BPS}}(\Gamma_1)=\Omega_{\mathrm{BPS}}(\Gamma_2)=1$. Find the relative spin, the new absolute reduced degeneracy, and the stable-minus-empty index jump.

Solution

The relative spin is $j_{\mathrm{rel}}=(3-1)/2=1$, so the multiplet contains three states. Its helicity sign is $(-1)^{3-1}=+1$. Hence

$$d_{\mathrm{stable}}-d_{\mathrm{empty}}=3, \qquad \Delta_{\mathrm{st}}\Omega_{\mathrm{BPS}}=+3.$$

Magnitude equality holds because this single relative multiplet has one sign. Both quantities nevertheless change across the wall.

4. Separate the finite coefficient

In chamber A, suppose the complete asymptotic indexed coefficient is $37$ and the polar contribution is $12$. In chamber B, the corresponding values are $29$ and $4$. Find the finite coefficient and explain what has—and has not—been established.

Solution

Subtracting the polar contribution gives

$$d^{\mathrm F}_{\mathrm A}=37-12=25, \qquad d^{\mathrm F}_{\mathrm B}=29-4=25.$$

The finite protected coefficient is chamber-independent in this example, while the asymptotic and polar pieces jump together. The calculation has not shown that $25$ is an absolute horizon degeneracy; that requires the horizon–hair relation and cancellation control.

What the protected calculation establishes

The implication has four separate rungs:

1. A chamber-specified helicity trace is a protected signed observable.
2. Finite–polar separation can isolate a chamber-independent single-center index in the torsion-one $\mathcal N=4$ model.
3. A known exterior factor can relate that index to a horizon index.
4. A same-sign result or an independent absolute count can promote the horizon index to an absolute degeneracy.

Failure of a later rung does not invalidate the earlier ones. It narrows the conclusion. For the general protected-trace logic, continue to what vacuum counting can establish. For the geometry and algebra of jumps, continue to marginal-stability wall crossing. For reconstruction limits, see what an index can determine. The next black-hole application is D-Brane Bound States and the Strominger–Vafa Count.

The chapter overview contains the structure diagram and validity and failure diagram. They are embedded there once so that their shared chapter-level context is not repeated on every article.

For the chapter-wide comparison of assumptions, counterevidence, falsifiers, and claim limits, see the claim-domain table.

References

• Brennan, T. Daniel, Gregory W. Moore, and Andrew B. Royston. “Wall Crossing from Dirac Zeromodes.” Journal of High Energy Physics 2018, 9 (2018): 038. DOI. Open PDF.
• Chattopadhyaya, Aradhita, and Justin R. David. “Horizon States and the Sign of Their Index in N=4 Dyons.” Journal of High Energy Physics 2021, 3 (2021): 106. DOI. Open PDF.
• Dabholkar, Atish, João Gomes, Sameer Murthy, and Ashoke Sen. “Supersymmetric Index from Black Hole Entropy.” Journal of High Energy Physics 2011, 4 (2011): 034. DOI. Open PDF.
• Dabholkar, Atish, Sameer Murthy, and Don Zagier. “Quantum Black Holes, Wall Crossing, and Mock Modular Forms.” arXiv:1208.4074 [hep-th] (2012; revised 2014). arXiv.
• Dijkgraaf, Robbert, Erik Verlinde, and Herman Verlinde. “Counting Dyons in N=4 String Theory.” Nuclear Physics B 484 (1997): 543–561. DOI. Open PDF.
• Sen, Ashoke. “How Do Black Holes Predict the Sign of the Fourier Coefficients of Siegel Modular Forms?” General Relativity and Gravitation 43 (2011): 2171–2183. DOI. Open PDF.
• Sen, Ashoke. “Wall Crossing Formula for N=4 Dyons: A Macroscopic Derivation.” Journal of High Energy Physics 2008, 7 (2008): 078. DOI. Open PDF.