not apply to it. And indeed, (160) is satisfied along a gradient descent
trajectory because $M_{\textrm{Pl}}|\nabla V|/V=2\sqrt{2/3}\geq\sqrt{2}$,
satisfying (161). In fact, precisely this system appears in the dilaton-radion
field space of heterotic string theory compactified to four dimensions
Rudelius:2021oaz .

It is also important to note that our bound forbids indefinite periods of
accelerated expansion in asymptotic regimes of scalar field space, but it does
not restrict accelerated expansion of the universe in the interior of field
space, nor does it forbid finite periods of accelerated expansion. For
instance, certain compactifications of supercritical string theories support
periods of accelerated expansion Dodelson:2013iba , and even in a theory that
satisfies (161), one can attain a short burst of accelerated expansion by
giving the scalar field a kick of kinetic energy towards the top of its
potential.999We thank Thomas Van Riet for pointing this out to us. Instead,
our bound simply implies that any period of accelerated expansion in
asymptotic regimes of string theory must end after a finite time. Indeed, it
is widely believed that de Sitter vacua in string theory are at best
metastable Dyson:2002pf ; Goheer:2002vf ; Bousso:2011aa , so it seems quite
plausible that any period of accelerated cosmological expansion in quantum
gravity can persist for only a finite length of time.

This does not imply that eternal inflation is forbidden, however. In the
standard picture of eternal inflation Starobinsky:1986fx ; Vilenkin:1999pi ;
Linde:2007fr ; Guth:2007ng , a given observer will, with probability 1
Dyson:2002pf ; Bousso:2011aa , experience a phase transition after a finite
period of time from a parent accelerating phase to a daughter
decelerating/Minkowski/AdS phase. Inflation is nonetheless “eternal” in this
scenario because bubbles of the daughter phase continue to inflate away from
one another due to the accelerated expansion in the parent phase, and some
nonzero fraction of the universe at comoving time $t$ remains in the parent
accelerating phase even in the limit $t\rightarrow\infty$ (see e.g.
Rudelius:2019cfh  for further explanation). This means that eternal inflation
may occur even if every de Sitter vacuum is metastable and every period of
accelerated expansion in quantum gravity can persist for only a finite length
of time. Thus, while our results suggest that quintessence, like de Sitter,
cannot exist indefinitely in quantum gravity, they do not necessarily point
towards a ban on eternal inflation.

It is interesting that the coefficient $c_{\textrm{min}}=2/\sqrt{d-2}$ we have
derived in (161) is twice the Distance Conjecture coefficient
$\lambda_{\textrm{min}}=1/\sqrt{d-2}$ we have proposed throughout this paper.
Such a relationship between $c_{\textrm{min}}$ and $\lambda_{\textrm{min}}$
has been discussed previously Hebecker:2018vxz ; Andriot:2020lea ;
Bedroya:2020rmd , and it is encouraging here that the resulting value of
$c_{\textrm{min}}$ matches precisely with a value that is already supported by
several independent lines of evidence Rudelius:2021oaz ; Rudelius:2021azq .

###  6.2 Large-Field Inflation

Specializing to four dimensions, the bounds developed here may lead to
important consequences for large-field inflation. A scalar field displacement
$\Delta\phi$ in four dimensions implies a tower of light states beginning at
the mass scale $m\lesssim\exp(-|\Delta\phi|/\sqrt{2})$ in Planck units, which
for $\Delta\phi\gg M_{\textrm{Pl}}$ implies species bound cutoff no larger
than

$\Lambda_{\textrm{UV}}\lesssim\exp\left(-\frac{|\Delta\phi|}{3\sqrt{2}M_{\textrm{Pl}}}\right)M_{\textrm{Pl}}\,,$
(162)

assuming that the tower begins at or below the Planck scale $M_{\textrm{Pl}}$
when the field starts to roll. For a large-field model of inflation with
$|\Delta\phi|=10M_{\textrm{Pl}}$, this leads to a bound
$\Lambda_{\textrm{UV}}\lesssim 10^{-1}M_{\textrm{Pl}}$, which yields a narrow
window for a hierarchy of scales between the IR cutoff $H\sim
10^{-4}M_{\textrm{Pl}}$ and the UV cutoff $\Lambda_{\textrm{UV}}$.

In fact, the window for large-field inflation may be even tighter than this.
Invoking the Emergent String Conjecture once again, we assume that every
infinite-distance limit in scalar field space is either an emergent string
limit or a decompactification limit. This leads to a cutoff

$\Lambda_{\textrm{UV}}\equiv\min(M_{\textrm{string}},m_{\textrm{KK}})\lesssim\exp\left(-\frac{|\Delta\phi|}{\sqrt{2}M_{\textrm{Pl}}}\right)M_{\textrm{Pl}}\,.$
(163)

Here, we have made the seemingly reasonable assumption Baumann:2014nda  that
the string scale and the Kaluza-Klein are bounded above by the Planck scale
when the field starts to roll. For $|\Delta\phi|=10M_{\textrm{Pl}}$, this
yields a string scale cutoff $\Lambda_{\textrm{UV}}\sim
10^{-3}M_{\textrm{Pl}}$, and for $|\Delta\phi|=15M_{\textrm{Pl}}$, it yields a
string scale cutoff $\Lambda_{\textrm{UV}}\sim 10^{-5}M_{\textrm{Pl}}$.

Thus, at best, the UV cutoff is only one order of magnitude above the IR
cutoff $H\approx 10^{-4}M_{\textrm{Pl}}$ when $|\Delta\phi|\sim
10M_{\textrm{Pl}}$, effectively ruling out any semblance of theoretical
control over the effective field theory. Models with $|\Delta\phi|\gtrsim
15M_{\textrm{Pl}}$, such as $m^{2}\phi^{2}$ inflation, are excluded entirely
for $H\approx 10^{-4}M_{\textrm{Pl}}$. A detailed investigation into our
proposed bound and its applicability beyond the supersymmetric context could
therefore have important consequences for the viability of large-field
inflation (and relaxion models Graham:2015cka ) in string theory.

A couple of important caveats are in order here: first, we have assumed that
string scale and the Kaluza-Klein scale begin below the Planck scale, so the
tower of particles implied by the Distance Conjecture appears almost
immediately as the scalar field starts to roll. Although the Distance
Conjecture is, strictly speaking, a statement about asymptotic behavior of
scalar field space, various works (see e.g. Klaewer:2016kiy ; Erkinger:2019umg
) have found in examples that the exponential towers required by the Distance
Conjecture appear rather quickly, when $|\Delta\phi|\lesssim
1\,M_{\textrm{Pl}}$, so our assumption is likely a reasonable one.

Perhaps a more serious difficulty is the question of whether or not our
argument here may be applied to axion models of inflation, which have
historically represented the most popular models of large-field inflation
Freese:1990rb ; McAllister:2008hb ; Silverstein:2008sg . We will consider this
question next.

###  6.3 Axions and a Tower Scalar Weak Gravity Conjecture

As discussed in the introduction, the Distance Conjecture deals with infinite-
distance limits of moduli space and so does not directly constrain axions,
which have a compact field space. On the other hand, the Scalar Weak Gravity
Conjecture does constrain couplings to axion fields, but it does not imply the
existence of an infinite tower of light particles: a finite number of
particles may be sufficient to satisfy the convex hull condition.

The evidence we have seen here for both the Distance Conjecture and the Scalar
Weak Gravity Conjecture–along with the fact that neither conjecture is quite
strong enough to imply the other–suggests a third, stronger conjecture which
implies both of them. This led the authors of Calderon-Infante:2020dhm  to
propose the Convex Hull Distance Conjecture, which we call the tower Scalar
Weak Gravity Conjecture:

###### The Tower Scalar Weak Gravity Conjecture.

Given a massless scalar field modulus $\phi$ in a quantum gravity theory in
$d$ spacetime dimensions, there necessarily exists an infinite tower of
particles of mass $m_{n},n\in\mathbb{Z}$ satisfying

$\frac{(\partial_{\phi}m_{n})^{2}}{g_{\phi\phi}m_{n}^{2}}\geq{\lambda_{\text{min}}^{2}}{\kappa_{d}^{2}}\equiv\frac{\kappa_{d}^{2}}{d-2}\,,~{}~{}~{}\text{
with }\partial_{\phi}m_{n}<0\,,$ (164)

where $g_{\phi\phi}$ is the $\phi\phi$ component of the metric on scalar
moduli space.

In other words, the Scalar Weak Gravity Conjecture should be satisfied by a
whole tower of particles. When $\phi$ is a non-compact field, this tower is
the usual Distance Conjecture tower, but this conjecture goes beyond the
Distance Conjecture in that it also requires a tower if $\phi$ is a compact
scalar field.

One way to justify the tower Scalar Weak Gravity Conjecture qualitatively
(i.e., up to $O(1)$ factors) is by the emergence proposal Grimm:2018ohb ;
Heidenreich:2018kpg ; Heidenreich:2017sim . Large field traversals in quantum
gravity generally come from weakly coupled scalar fields, which in turn come
from integrating out towers of exponentially light particles. This is true
even for axion fields Heidenreich:2018kpg : a large-field model of natural
inflation requires a large axion decay constant, which by the emergence
proposal requires a tower of exponentially light particles, which points
strongly towards the tower Scalar Weak Gravity Conjecture. This emergence
argument only works, however, when the mass scale of the tower of particles in
question is far below the quantum gravity scale $\Lambda_{\textrm{QG}}$, where
gravity becomes strongly coupled.

Thus, it seems likely that the arguments of the previous subsection do apply
to large-field models of natural inflation as well, which may help explain why
these models have so far resisted an embedding in quantum gravity Banks:2003sx
; Rudelius:2015xta ; delaFuente:2014aca ; Heidenreich:2015wga ;
Montero:2015ofa ; Brown:2015iha ; Bachlechner:2015qja ; Conlon:2016aea . It is
less clear, however, that the tower Scalar Weak Gravity Conjecture will place
important constraints on axion monodromy models of inflation McAllister:2008hb
; Silverstein:2008sg , since these do not require a large axion decay
constant, and towers of particles may remain heavy even as the axion traverses
its fundamental domain many times (see however Baume:2016psm ). A more
thorough investigation of axion monodromy in light of our results in this
paper could be a worthwhile direction for future study.

###  6.4 Black Holes and Repulsive Forces

Consider a Einstein-Maxwell-dilaton action in $d$ dimensions for a 1-form
gauge field:

$S=\frac{1}{2\kappa_{d}^{2}}\int\textrm{d}^{d}x\sqrt{-g}\left({\cal
R}-\frac{1}{2}(\nabla\phi)^{2}\right)-\frac{1}{2e^{2}}\int
d^{d}x\sqrt{-g}{\mathrm{e}}^{-\alpha\phi}F_{2}^{2}\,.$ (165)

The extremality bound for a black hole of quantized charge $Q$ and mass $M$
takes the form

$\frac{e^{2}Q^{2}}{M^{2}}\geq\left[\frac{\alpha^{2}}{2}+\frac{d-3}{d-2}\right]\kappa_{d}^{2}\,.$
(166)

For $\phi\rightarrow-\infty$, the gauge field becomes weakly coupled, and by
the tower Weak Gravity Conjecture we expect a tower of particles whose mass
scales as

$m\sim\exp(-\alpha\phi/2)\sim\exp(-\alpha\kappa_{d}\hat{\phi}/\sqrt{2})\,,$
(167)

where $\hat{\phi}=\phi\sqrt{2}\kappa_{d}$ is the canonically normalized scalar
field. For large charge, we expect that this tower will approach the black
hole extremality bound, so that (167) applies to both quantum mechanical
particles at small charge and semiclassical black holes at large charge and
smoothly interpolates between them.

Two copies of a given particle in this theory will repel each other if their
gauge repulsion overwhelms the attraction due to the gravitation force and the
scalar force–in other words, if

$\frac{e^{2}Q^{2}}{M^{2}}\geq\left[2\frac{(\partial_{\phi}M)^{2}}{M^{2}}+\frac{d-3}{d-2}\right]\kappa_{d}^{2}\,.$
(168)

Plugging in $\partial_{\phi}M=\alpha M/2$ from (167), this is precisely the
bound (166): two copies of a particle in the tower will repel each other at
long distances precisely if they are superextremal, so the Repulsive Force
Conjecture is equivalent to the Weak Gravity Conjecture for this tower of
particles.

So far, this discussion is simply a review (see e.g. Palti:2017elp ;
Lee:2018spm ; Heidenreich:2019zkl ; Heidenreich:2020upe ). The novelty comes
when we impose our bound $\lambda\geq 1/\sqrt{d-2}$, which by (167) implies
$\alpha\geq\sqrt{2/(d-2)}$. As a result, both the tower Weak Gravity
Conjecture and the tower Repulsive Force Conjecture imply the bound

$\frac{e^{2}Q^{2}}{M^{2}}\geq\kappa_{d}^{2}$ (169)

for charged particles in the tower. Note that this bound only applies for
exactly massless scalar fields, since massive scalar fields do not mediate
long range forces. The consequences and validity of this bound are problems
for future study.

###  6.5 The Emergent String Conjecture and an Upper Bound on $\lambda$

The Emergent String Conjecture has played an important role throughout this
paper. If every infinite-distance limit in scalar moduli space is either an
emergent string limit or a decompactification limit, then our analysis in §2
is not merely one illustrative example of what may happen in an asymptotic
limit of scalar field space. Rather, it is more or less the _only_ example of
what may happen. The Emergent String Conjecture therefore provides strong
evidence that our proposed lower bound on $\lambda$ is correct. It also offers
a compelling explanation for why towers with $\lambda=1/\sqrt{(d-1)(d-2)}$ are
so often accompanied by other towers with $\lambda=\sqrt{(d-1)/(d-2)}$, since
these values occur readily in dimensional reduction.

In turn, our analysis provides strong evidence for the Emergent String
Conjecture. We have found in all of our examples that the limits which
saturate our proposed bound $\lambda\geq 1/\sqrt{d-2}$ are emergent string
limits, and the states which saturate this bound are oscillator modes of
fundamental strings. The other towers we have encountered have the scaling
properties expected of Kaluza-Klein towers, with
$\lambda_{\textrm{KK}}>1/\sqrt{d-2}$.

When our results are combined with the Emergent String Conjecture, further
interesting results emerge. For one thing, if we assume that the _lightest_
tower of particles in a given infinite-distance limit is either a Kaluza-Klein
tower or a tower of string oscillator modes, then we conclude that any
generator of the convex hull in $\vec{\zeta}$-vector space must have length

$|\vec{\zeta}^{(d)}|=\frac{1}{\sqrt{d-2}}~{}~{}\text{ or
}~{}~{}|\vec{\zeta}^{(d)}|=\sqrt{\frac{n+d-2}{n(d-2)}}\,,$ (170)

for some $n\geq 1$. These values correspond to string oscillator modes and
Kaluza-Klein towers, respectively. As a corollary to this, we also deduce that
the parameter $\lambda$ is upper-bounded as

$\lambda\leq\lambda_{\textrm{max}}=\sqrt{(d-1)/(d-2)}\,.$ (171)

This bound is satisfied in the examples we have encountered in this paper.
Further exploration of this bound–as well as a more thorough investigation of
the Emergent String Conjecture–is warranted in light of the results of this
paper.

## Acknowledgements

We thank Matthew Reece for collaboration in the early stages of this work. We
thank David Andriot, Matthew Reece, and Irene Valenzuela for comments on a
previous version of this manscript. We thank David Andriot, Daniel Klaewer,
Miguel Montero, Lisa Randall, Matthew Reece, Cumrun Vafa, and Irene Valenzuela
for useful discussions. The work of SK and TR was supported in part by the
Berkeley Center for Theoretical Physics; by the Department of Energy, Office
of Science, Office of High Energy Physics under QuantISED Award DE-SC0019380
and under contract DE-AC02-05CH11231; and by the National Science Foundation
under Award Number 2112880. The work of BH and ME was supported by NSF grant
PHY-1914934. The work of YQ was supported by NSF grants PHY-1820675 and
PHY-2112800.

##  Appendix A Top Down Evidence in Maximal Supergravity for Six, Five, and
Four dimensions

We now investigate the value $\lambda_{\text{min}}$ for the Scalar Weak
Gravity Conjecture for maximal supergravity in six, five, and four dimensions,
making extensive use of U-duality to simplify the calculation. In all cases,
we obtain $\lambda_{\text{min}}=1/\sqrt{d-2}$.

###  A.1 Symplectic Conventions

The unitary symplectic group $\text{USp}(2n)$ is the subgroup of $U(2n)$ that
preserves the symplectic form $\Omega^{ab}$, which is an antisymmetric tensor
with the property that

$\Omega^{ac}\Omega^{\star}_{bc}=\delta^{a}_{b}.$ (172)

Note that an arbitrary antisymmetric tensor $A^{ab}$ of $U(2n)$ can be put in
the block-diagonal form

$A^{ab}=\text{diag}\left[\left(\begin{array}[]{cc}0&\lambda_{1}\\\
-\lambda_{1}&0\end{array}\right),\ldots,\left(\begin{array}[]{cc}0&\lambda_{n}\\\
-\lambda_{n}&0\end{array}\right)\right],\qquad\text{for $\lambda_{i}$ real and
non-negative}$ (173)

by a unitary transformation $A\rightarrow UAU^{\top}$. In the case of the
symplectic form, the constraint (172) implies
$\lambda_{1}=\cdots=\lambda_{n}=1$, hence all choices of $\Omega^{ab}$ are
equivalent up to a unitary transformation. Thus,

$\Omega^{ab}=\left(\begin{array}[]{ccc}\mathbf{\varepsilon}_{2\times
2}&\mathbf{0}_{2\times 2}&\mathbf{0}_{2\times 2}\\\ \mathbf{0}_{2\times
2}&\ddots&\mathbf{0}_{2\times 2}\\\ \mathbf{0}_{2\times 2}&\mathbf{0}_{2\times
2}&\mathbf{\varepsilon}_{2\times
2}\end{array}\right),\qquad\text{or}\qquad\Omega^{ab}=\left(\begin{array}[]{cc}\mathbf{0}_{n\times
n}&\mathbf{1}_{n\times n}\\\ -\mathbf{1}_{n\times n}&\mathbf{0}_{n\times
n}\end{array}\right),$ (174)

in two closely related bases, where $\mathbf{\varepsilon}_{2\times
2}=i\sigma_{2}=\left(\begin{array}[]{cc}0&1\\\ -1&0\end{array}\right)$. These
two bases manifest the important subgroups
$\text{SU}(2)^{n}\subseteq\text{USp}(2n)$ (with equality for $n=1$) and
$U(n)\subset\text{USp}(2n)$.

It is convenient to define
$\Omega_{ab}=(\Omega^{ab})^{-1}=(\Omega^{ba})^{\star}$, so that
$\Omega_{ab}\Omega^{bc}=\delta_{a}^{c}$. We raise and lower indices with the
symplectic form $\Omega_{ab}$ acting on the left, as follows:

$V^{a}=\Omega^{ab}V_{b}\qquad\Leftrightarrow\qquad V_{a}=\Omega_{ab}V^{b}.$
(175)

Note that this implies

$V^{a}W_{a}=(\Omega^{ab}V_{b})W_{a}=-V_{b}(\Omega^{ba}W_{a})=-V_{b}W^{b},$
(176)

so it is important to keep track of index position.

###  A.2 U-Duality Results

Per Obers:1998fb , the 1/2 BPS particles have a mass formula that is quadratic
in the central charge

$M^{2}=M_{IJ}(\phi)Z^{I}Z^{J},$ (177)

where $M_{IJ}(\phi)$ is some moduli-dependent matrix. Moreover, the $1/2$ BPS
condition takes the form

$f^{\alpha}=S_{IJ}^{\alpha}Z^{I}Z^{J}=0,$ (178)

where $S_{IJ}^{\alpha}$ is some $\phi$-independent, U-duality invariant such
that $\alpha$ sits in the string representation of the U-duality group.

We are interested in

$\zeta_{a}=\frac{1}{M}\frac{\partial
M}{\partial\phi^{a}}=\frac{1}{2M^{2}}\frac{\partial
M_{IJ}}{\partial\phi^{a}}Z^{I}Z^{J},$ (179)

where $\phi^{a}$ are the independent scalar fluctuations at a given point in
the moduli space, with accompanying metric $G^{ab}(\phi)$. Let us write this
as

$\zeta^{a}=G^{ab}\zeta_{b}=\zeta^{a}_{IJ}\hat{Z}^{I}\hat{Z}^{J},\qquad\hat{Z}^{I}\equiv\mathcal{N}\frac{Z^{I}}{\sqrt{M_{IJ}Z^{I}Z^{J}}},$
(180)

where $\mathcal{N}$ is some normalization factor that depends on conventions.

The moduli space has a coset structure $\mathcal{M}=G/H$. Focusing on a single
point, the subgroup $H$ plays the role of a “symmetry” group in the low-energy
effective theory, albeit not an exact symmetry for the same reasons that,
e.g., $\text{SO}(2)\subset\text{SL}(2,\mathbb{R})$ is not an exact symmetry of
type IIB string theory: the charge lattice, massive spectrum, etc., are not
invariant, but the two-derivative effective action _is_ invariant at the
classical level.

This means that $M_{IJ}$, $S_{IJ}^{\alpha}$, and $\zeta^{a}_{IJ}$ must be $H$
invariants. Combined with representation theory, that is enough to fix the
formula for $\zeta^{a}$ in terms of the central charges (again, ignoring
charge quantization) up to a change of basis. In particular, to do so we need
to know what representation of $H$ the scalars $\phi^{a}$ transform under
(call it $R_{\text{mod}}$), as well as the representation $R_{\text{part}}$
under which the central charges $Z^{I}$ transform and the representation
$R_{\text{str}}$ under which the 1/2 BPS conditions $f^{\alpha}=0$ transform.
For $4\leq d\leq 9$, we get:

$\begin{array}[]{c|cccc}d&H&R_{\text{mod}}&R_{\text{part}}&R_{\text{str}}\\\
\hline\cr 9&\text{SO}(2)&\ydiagram{2}\oplus\mathbf{1}&\ydiagram
1\oplus\mathbf{1}&\ydiagram 1\\\
8&\text{SO}(3)\times\text{SO}(2)&\left(\ydiagram{2},\mathbf{1}\right)\oplus\left(\mathbf{1},\ydiagram{2}\right)&(\ydiagram
1,\ydiagram 1)&(\ydiagram 1,\mathbf{1})\\\
7&\text{SO}(5)&\ydiagram{2}&\ydiagram{1,1}&\ydiagram 1\\\
6&\text{Spin}(5)\times\text{Spin}(5)&(\ydiagram 1,\ydiagram
1)&(S,S)&(\ydiagram 1,1)+(1,\ydiagram 1)\\\
5&\text{USp}(8)&\ydiagram{1,1,1,1}&\ydiagram{1,1}&\ydiagram{1,1}\\\ \\\
4&\text{SU}(8)&\ydiagram{1,1,1,1}&\ydiagram{1,1}_{\mathbb{C}}&\text{Adj}\oplus\ydiagram{1,1,1,1}\end{array}$
(181)

where $S$ is the four-dimensional (pseudoreal) spinor of $\text{Spin}(5)$,
which is the $\ydiagram 1$ of $\text{USp}(4)$. All the reps are real except
the explicitly indicated $\ydiagram{1,1}_{\mathbb{C}}$ for particles in 4d,
where the reality condition on the vectors will relate $F$ and $\star F$,
i.e., the complexity of the representation is related to the presence of
magnetic charge. Note that the 6d line can also be written in terms of
$H=\text{USp}(4)\times\text{USp}(4)$, whence
$R_{\text{mod}}=\left(\ydiagram{1,1},\ydiagram{1,1}\right)$,
$R_{\text{part}}=(\ydiagram 1,\ydiagram 1)$, and
$R_{\text{str}}=\left(\ydiagram{1,1},1\right)\oplus\left(1,\ydiagram{1,1}\right)$.
This makes the embedding into $\text{USp}(8)$ in one lower dimension more
obvious.

To see where (181) comes from, note that $R_{\text{mod}}$ can be fixed from
the coset structure $\mathcal{M}=G/H$, where $G$ and $H$ are well known (see,
e.g., table B.3 of Polchinski:1998rr ). Moreover, $R_{\text{part}}$ and
$R_{\text{str}}$ descend from the irreps $R_{\text{part}}^{(G)}$ and
$R_{\text{str}}^{(G)}$ of $G$ that are listed in hep-th/9809039, so one only
needs to understand the $G\rightarrow H$ branching rules. In particular, we
have the following:

$\begin{array}[]{c|cccc}d&G&H&R_{\text{part}}^{(G)}&R_{\text{str}}^{(G)}\\\
\hline\cr 9&\text{SL}(2,\mathbb{R})\times\text{SO}(1,1)&\text{SO}(2)&\ydiagram
1\oplus\mathbf{1}&\ydiagram 1\\\
8&\text{SL}(3,\mathbb{R})\times\text{SL}(2,\mathbb{R})&\text{SO}(3)\times\text{SO}(2)&(\ydiagram
1,\ydiagram 1)&(\ydiagram 1,\mathbf{1})\\\
7&\text{SL}(5,\mathbb{R})&\text{SO}(5)&\ydiagram{1,1}&\ydiagram 1\\\
6&\text{Spin}(5,5)&\text{Spin}(5)\times\text{Spin}(5)&S&\ydiagram 1\\\
5&E_{6(6)}&\text{USp}(8)&\mathbf{2}\mathbf{7}&\mathbf{2}\mathbf{7}^{\prime}\\\
4&E_{7(7)}&\text{SU}(8)&\mathbf{5}\mathbf{6}&\text{Adj}=\mathbf{1}\mathbf{3}\mathbf{3}\end{array}$
(182)

Here $S$ denotes the minimal, 16-dimensional Majorana-Weyl spinor rep of
$\text{Spin}(5,5)$, $E_{6(6)}$ and $E_{7(7)}$ denote the split real forms of
the $E_{6}$ and $E_{7}$ algebras (and their corresponding groups) and
$\mathbf{2}\mathbf{7}$ and $\mathbf{2}\mathbf{7}^{\prime}$ are the two
different fundamental representations of $E_{6(6)}$ related by the
$\mathbb{Z}_{2}$ outer automorphism of the Dynkin diagram. The non-trivial
point now is to understand the branching rules in the exceptional cases (which
depend only on the Lie algebras, not on the choice of real forms); these can
be computed using SAGE sagemath . One finds that

$\displaystyle E_{6}\rightarrow
C_{4}:\qquad\mathbf{2}\mathbf{7},\mathbf{2}\mathbf{7}^{\prime}\rightarrow\ydiagram{1,1},\qquad\text{Adj}\rightarrow\text{Adj}\oplus\ydiagram{1,1,1,1},$
(183) $\displaystyle E_{7}\rightarrow
A_{7}:\qquad\mathbf{5}\mathbf{6}\rightarrow\ydiagram{1,1}\oplus\overline{\ydiagram{1,1}},\qquad\text{Adj}\rightarrow\text{Adj}\oplus\ydiagram{1,1,1,1}.$
(184)

Noting that $\text{Adj}_{G}\rightarrow\text{Adj}_{H}\oplus R_{\text{mod}}$,
this is enough to deal with the cases of interest in the table.

###  A.3 Six Dimensions

In 6d, $H=\text{USp}(4)\times\text{USp}(4)$ and the central charge $Z^{a;b}$
transforms in the bivector representation $(\ydiagram 1,\ydiagram 1)$. While
the rep $\ydiagram 1$ of $\text{USp}(4)$ is pseudoreal, the $(\ydiagram
1,\ydiagram 1)$ rep is real. Specifically, the reality condition

$(Z^{a;b})^{\star}=Z_{a;b}=\Omega_{ac}\Omega_{bd}Z^{c;d},$ (185)

is consistent, since

$Z^{\star}=\Omega
Z\Omega^{\top}\qquad\Rightarrow\qquad(Z^{\star})^{\star}=\Omega^{\star}Z^{\star}\Omega^{{\dagger}}=\Omega^{\star}\Omega
Z(\Omega^{\star}\Omega)^{\top}=Z$ (186)

using $\Omega^{\star}\Omega=-1$.

Noting that

$(\ydiagram 1,\ydiagram 1)\otimes_{S}(\ydiagram 1,\ydiagram
1)=\left(\ydiagram{2},\ydiagram{2}\right)\oplus\left(\ydiagram{1,1}\oplus\mathbf{1},\ydiagram{1,1}\oplus\mathbf{1}\right),$
(187)

we see that the 1/2 BPS shortening condition, in the
$\left(\ydiagram{1,1},\mathbf{1}\right)\oplus\left(\mathbf{1},\ydiagram{1,1}\right)$
rep, is uniquely fixed by $H$ invariance to be

$Z^{a;c}Z_{b;c}=Z^{c;a}Z_{c;b}=\frac{1}{4}\delta^{a}_{b}Z^{c;d}Z_{c;d}.$ (188)

Likewise, due to $H$ invariance, the mass formula must take the form:

$M^{2}=Z^{a;b}(Z^{a;b})^{\star}=Z^{a;b}Z_{a;b},$ (189)

up to an overall normalization factor that we absorb into the definition of
$Z^{a;b}$.

Last but not least, the scalar charge-to-mass ratio vector $\zeta^{ab;cd}$
sits in the $\left(\ydiagram{1,1},\ydiagram{1,1}\right)$ representation. Note
that the representation $\ydiagram{1,1}$ of $\text{USp}(4)$ (the $\ydiagram 1$
of $\text{SO}(5)$) is itself real, i.e., the condition

$(A^{ab})^{\star}=A_{ab}=\Omega_{ac}\Omega_{bd}A^{cd}$ (190)

is consistent for much the same reason as in (186). Likewise, $\zeta$
satisfies

$(\zeta^{ab;cd})^{\star}=\zeta_{ab;cd}.$ (191)

As there is a unique singlet in
$\left(\ydiagram{1,1},\ydiagram{1,1}\right)\otimes_{S}\left(\ydiagram{1,1},\ydiagram{1,1}\right)$,
i.e., written in terms of $\text{SO}(5)\times\text{SO}(5)$ reps:

$(\ydiagram 1,\ydiagram 1)\otimes_{S}(\ydiagram 1,\ydiagram
1)=\left(\ydiagram{2}\oplus\mathbf{1},\ydiagram{2}\oplus\mathbf{1}\right)\oplus\left(\ydiagram{1,1},\ydiagram{1,1}\right)=\mathbf{1}\oplus(\cdots),$
(192)

the metric on $\zeta$-space is also fixed by $H$ invariance to be

$\zeta^{2}=\zeta^{ab;cd}(\zeta^{ab;cd})^{\star}=\zeta^{ab;cd}\zeta_{ab;cd},$
(193)

up to an overall factor that we absorb into the definition of $\zeta^{ab;cd}$.

In the case of a $1/2$ BPS state, we should have an expression schematically
of the form

$\zeta=\frac{1}{M^{2}}ZZ$ (194)

with some index structure. Because $(\ydiagram 1,\ydiagram
1)\otimes_{S}(\ydiagram 1,\ydiagram 1)$ has a unique copy of
$\left(\ydiagram{1,1},\ydiagram{1,1}\right)$ in its decomposition, the index
structure is again fixed by $H$ invariance to be

$\zeta^{ab;cd}=\mathcal{N}\left(\hat{Z}^{a;c}\hat{Z}^{b;d}-\hat{Z}^{a;d}\hat{Z}^{b;c}-\frac{1}{8}\Omega^{ab}\Omega^{cd}\right),\qquad\hat{Z}^{a;b}=\frac{Z^{a;b}}{\sqrt{Z^{a;b}Z_{a;b}}},$
(195)

where the last term is required to ensure $\Omega_{ab}\zeta^{ab;cd}=0$ and
$\zeta^{ab;cd}\Omega_{cd}=0$.

The normalization factor $\mathcal{N}$ can no longer be absorbed by
redefinitions. To determine it, first note that the unit-normalized central
charge $\hat{Z}^{a;b}$ satisfies

$\hat{Z}^{a;b}\hat{Z}_{a;c}=\frac{1}{4}\delta^{b}_{c},\qquad\hat{Z}^{a;b}\hat{Z}_{c;b}=\frac{1}{4}\delta^{a}_{c},$
(196)
$\Rightarrow\qquad\Omega_{ab}\hat{Z}^{a;c}\hat{Z}^{b;d}=-\frac{1}{4}\Omega^{cd},\qquad\hat{Z}^{a;c}\hat{Z}^{b;d}\Omega_{cd}=-\frac{1}{4}\Omega^{ab},$
(197)

upon imposing the 1/2 BPS condition. Thus, for a $1/2$ BPS state

$\zeta^{2}=2\mathcal{N}^{2}\left(\hat{Z}^{a;c}\hat{Z}^{b;d}-\hat{Z}^{a;d}\hat{Z}^{b;c}-\frac{1}{8}\Omega^{ab}\Omega^{cd}\right)\hat{Z}_{a;c}\hat{Z}_{b;d}=2\mathcal{N}^{2}\left(1-\frac{1}{4}-\frac{1}{8}\right)=\frac{5}{4}\mathcal{N}^{2}.$
(198)

However, we know that Kaluza Klein modes are $1/2$ BPS states, which moreover
satisfy $\zeta^{2}=\frac{d-1}{d-2}=\frac{5}{4}$. Thefore, $\mathcal{N}=1$.

Next, note that reality condition combined with the $1/2$ BPS condition gives
$\hat{Z}^{a;b}(\hat{Z}^{c;b})^{\star}=\frac{1}{4}\delta^{a}_{c}$, i.e,
$\hat{Z}\hat{Z}^{{\dagger}}=\frac{1}{4}$, so $2\hat{Z}$ is a unitary matrix.
We can therefore fix $\hat{Z}^{a;}\,_{b}=\frac{1}{2}\delta^{a}_{b}$ by a
unitary rotation on the right-hand factor of
$\text{USp}(4)\times\text{USp}(4)$. Thus, we obtain

$\Omega^{R}_{bc}\hat{Z}^{a;c}=\frac{1}{2}\delta^{a}_{b}\qquad\Rightarrow\qquad\hat{Z}^{a;b}=-\frac{1}{2}\Omega_{R}^{ab},$
(199)

where we are now careful to distinguish the symplectic form $\Omega_{R}$ for
the right-hand factor from the one $\Omega_{L}$ for the left-hand factor.
However, the reality condition on $\hat{Z}$ now gives:

$(\hat{Z}^{a;b})^{\star}=\hat{Z}_{a;b}=\Omega^{L}_{ac}\Omega^{R}_{bd}\hat{Z}^{c;d}\quad\Rightarrow\quad-\frac{1}{2}\Omega_{ba}^{R}=-\frac{1}{2}\Omega^{L}_{ac}\Omega^{R}_{bd}\Omega_{R}^{cd}\quad\Rightarrow\quad\Omega_{ab}^{R}=\Omega^{L}_{ab},$
(200)

using $(\Omega^{ab})^{\star}=\Omega_{ba}$, so they are in fact the same.

Thus, we obtain:

$\zeta^{ab;cd}=\frac{1}{4}\left[\Omega^{ac}\Omega^{bd}-\Omega^{ad}\Omega^{bc}-\frac{1}{2}\Omega^{ab}\Omega^{cd}\right],$
(201)

in this basis for a $1/2$ BPS state. As a check this satisfies
$\Omega_{ab}\zeta^{ab;cd}=0$, $\zeta^{ab;cd}\Omega_{cd}=0$, and

$\zeta^{2}=\frac{1}{8}\left[\Omega^{ac}\Omega^{bd}-\Omega^{ad}\Omega^{bc}-\frac{1}{2}\Omega^{ab}\Omega^{cd}\right]\Omega_{ac}\Omega_{bd}=\frac{1}{8}[16-4-2]=\frac{5}{4}.$
(202)

Now we want to write $\zeta$ and $\zeta^{I;J}$, a bivector of
$\text{SO}(5)\times\text{SO}(5)$ with metric

$\zeta^{2}=\zeta^{I;J}\zeta_{I;J}=\text{Tr}(\zeta\zeta^{\top}).$ (203)

By choosing a $\hat{Z}^{a;b}$ above, we picked a particular identification
between the two factors of the group, and having done so $\zeta$ could be
written purely in terms of invariants of the diagonal $\text{USp}(4)$ that
remained. Likewise, if we consider the diagonal $\text{SO}(5)$ there is a
unique invariant matrix, i.e., the identity matrix. Thus, we conclude that, in
this basis

$\zeta_{I;J}=\frac{1}{2}\delta_{IJ},$ (204)

where the normalizing factor is chosen to ensure that

$\zeta^{2}=\frac{1}{4}\text{Tr}(1)=\frac{5}{4},$ (205)

as before.

Thus, in an arbitrary basis we have the $1/2$ BPS state $\zeta$ vectors

$\zeta_{I;J}=\frac{1}{2}O_{IJ},$ (206)

where $O_{IJ}$ is any element of $\text{SO}(5)$.

We want to understand the convex hull of these $\zeta$ vectors. Note that it
is actually $O(5)\times O(5)$ invariant. Consider some arbitrary bivector
$P_{I;J}$. Using the singular value decomposition, we can set

$P_{I;J}=\text{diag}(\lambda_{1},\ldots,\lambda_{5}),\qquad\lambda_{I}\geq 0$
(207)

after an $O(5)\times O(5)$ transformation. Thus, for arbitrary
$\zeta_{I;J}=\frac{1}{2}O_{IJ}$,

$\text{Tr}(\zeta P^{\top})=\frac{1}{2}\sum_{I=1}^{5}\lambda_{I}O_{II}.$ (208)

We have the constraints

$\sum_{I}O_{IJ}O_{IK}=\delta_{JK}.$ (209)

Clearly this requires $O_{II}\leq 1$, and the constraint can be saturated by
taking $O_{IJ}=\delta_{IJ}$. Thus,

$\text{Tr}(\zeta P^{\top})\leq\frac{1}{2}\sum_{I=1}^{5}\lambda_{I}=1,$ (210)

and we should take $\sum_{I=1}^{5}\lambda_{I}=2$ to enfore $\zeta\cdot P\leq
1$ with at least one $1/2$ BPS state saturating the constraint.

With this constraint, we seek to maximize

$P\cdot P=\text{Tr}(PP^{\top})=\sum_{I}\lambda_{I}^{2},$ (211)

subject to the constraints $\sum_{I=1}^{5}\lambda_{I}=2$ and $\lambda_{I}\geq
0$. As before, for fixed $x+y=t$, $x^{2}+y^{2}$ increases as the difference
between $x$ and $y$ increases. Thus, if more than one $\lambda_{I}$ is
positive we can increase $P^{2}$ by reducing the value of the smaller one and
increasing the value of the larger one. As a result, we achieve the maximum
value when all but one vanishes, e.g., $\lambda_{1}=2$ and
$\lambda_{2}=\lambda_{3}=\lambda_{4}=\lambda_{5}=0$. Thus,

$P^{2}_{\text{max}}=\sum_{I}\lambda_{I}^{2}=4.$ (212)

This gives
$\lambda_{\min}=\frac{1}{P_{\text{max}}}=\frac{1}{\sqrt{4}}=\frac{1}{2}$,
which indeed verifies (and saturates) the bound
$\lambda_{\text{min}}\geq\frac{1}{\sqrt{d-2}}$.

###  A.4 Five Dimensions

In 5d, $H=\text{USp}(8)$ and the central charge $Z^{ab}$ transforms in the
traceless antisymmetric tensor representation $\ydiagram{1,1}$. This
representation is real, with the reality condition

$(Z^{ab})^{\star}=Z_{ab}\equiv\Omega_{ac}\Omega_{bd}Z^{cd}.$ (213)

Since
$\ydiagram{1,1}\otimes_{S}\ydiagram{1,1}=\ydiagram{1,1,1,1}\oplus\ydiagram{2,2}\oplus\ydiagram{1,1}\oplus
1$, there is (1) a unique mass formula, coming from the $1$ component, (2) a
unique $1/2$ BPS constraint, coming from the $\ydiagram{1,1}$ component, and
(3) a unique $\zeta\propto\hat{Z}^{2}$, coming from the $\ydiagram{1,1,1,1}$
component. We examine these in turn. The mass formula is

$M^{2}=\frac{1}{2}Z^{ab}(Z^{ab})^{\star}=\frac{1}{2}Z^{ab}Z_{ab},$ (214)

up to an overall normalization that we absorb into the definition of $Z^{ab}$.
The $1/2$ BPS constraint is

$Z^{ac}Z_{bc}=\frac{1}{8}\delta^{a}_{b}Z^{cd}Z_{cd},$ (215)

transforming in the traceless antisymmetric tensor rep as required.

Likewise, the charge-to-mass ratio $\zeta^{abcd}$ transforms in the traceless
four-form representation $\ydiagram{1,1,1,1}$. This representation is real,
with the reality condition

$(\zeta^{abcd})^{\star}=\zeta_{abcd}\equiv\Omega_{ae}\Omega_{bf}\Omega_{cg}\Omega_{dh}\zeta^{efgh}.$
(216)

The metric is

$\zeta^{2}=\frac{1}{4!}\zeta^{abcd}(\zeta^{abcd})^{\star}=\frac{1}{4!}\zeta^{abcd}\zeta_{abcd},$
(217)

up to an overall normalization absorbed into $\zeta^{abcd}$, which is unique
because

$\ydiagram{1,1,1,1}\otimes_{S}\ydiagram{1,1,1,1}=\ydiagram{2,2,2,2}\oplus\ydiagram{2,2}\oplus
1,$ (218)

contains only one singlet.

For $1/2$ BPS particles, we must have

$\zeta^{abcd}=3\mathcal{N}\left[\hat{Z}^{[ab}\hat{Z}^{cd]}+\frac{1}{12}\Omega^{[ab}\Omega^{cd]}\right],\qquad\hat{Z}^{ab}=\frac{Z^{ab}}{\sqrt{\frac{1}{2}Z^{ab}Z_{ab}}},$
(219)

for a still-to-be determined normalization factor $\mathcal{N}$, where the
second factor is chosen to ensure tracelessness:

$\displaystyle\begin{aligned}
\Omega_{ab}\zeta^{abcd}=&\mathcal{N}\Omega_{ab}\left[\hat{Z}^{ab}\hat{Z}^{cd}+\hat{Z}^{ac}\hat{Z}^{db}+\hat{Z}^{ad}\hat{Z}^{bc}+\frac{1}{12}\Omega^{ab}\Omega^{cd}+\frac{1}{12}\Omega^{ac}\Omega^{db}+\frac{1}{12}\Omega^{ad}\Omega^{bc}\right]\\\
=&\mathcal{N}\left[0+\frac{1}{4}\Omega^{cd}+\frac{1}{4}\Omega^{cd}-\frac{2}{3}\Omega^{cd}+\frac{1}{12}\Omega^{cd}+\frac{1}{12}\Omega^{cd}\right]=0.\end{aligned}$
(220)

We then find

$\displaystyle\begin{aligned}
\zeta^{2}=&\frac{\mathcal{N}^{2}}{8}\hat{Z}_{ab}\hat{Z}_{cd}\left[\hat{Z}^{ab}\hat{Z}^{cd}+\hat{Z}^{ac}\hat{Z}^{db}+\hat{Z}^{ad}\hat{Z}^{bc}+\frac{1}{12}\Omega^{ab}\Omega^{cd}+\frac{1}{12}\Omega^{ac}\Omega^{db}+\frac{1}{12}\Omega^{ad}\Omega^{bc}\right]\\\
=&\frac{\mathcal{N}^{2}}{8}\left[4-\frac{1}{2}-\frac{1}{2}+0-\frac{1}{6}-\frac{1}{6}\right]=\frac{\mathcal{N}^{2}}{3}.\end{aligned}$
(221)

Since Kaluza Klein modes are $1/2$ BPS and satisfy
$\zeta^{2}=\frac{d-1}{d-2}=\frac{4}{3}$, we conclude that $\mathcal{N}=2$,
i.e.,

$\displaystyle\begin{aligned}
\zeta^{abcd}=&6\left[\hat{Z}^{[ab}\hat{Z}^{cd]}+\frac{1}{12}\Omega^{[ab}\Omega^{cd]}\right]\\\
=&2\left[\hat{Z}^{ab}\hat{Z}^{cd}+\hat{Z}^{ac}\hat{Z}^{db}+\hat{Z}^{ad}\hat{Z}^{bc}+\frac{1}{12}\Omega^{ab}\Omega^{cd}+\frac{1}{12}\Omega^{ac}\Omega^{db}+\frac{1}{12}\Omega^{ad}\Omega^{bc}\right].\end{aligned}$
(222)

Now consider the conditions on $\hat{Z}^{ab}$. Written in terms of
$\hat{Z}^{a}\,_{b}\equiv\Omega_{bc}\hat{Z}^{ac}$, we find (recalling that
$(\Omega_{ab})^{\star}=\Omega^{ba}=-\Omega^{ab}$):

$\hat{Z}^{a}\,_{a}=0,\qquad\left(\hat{Z}^{a}\,_{b}\right)^{\star}=\hat{Z}^{b}\,_{a},\qquad\hat{Z}^{a}\,_{c}\hat{Z}^{c}\,_{b}=\frac{1}{4}\delta^{a}_{b},$
(223)

i.e., $2\hat{Z}^{a}\,_{b}$ is a traceless Hermitian matrix that squares to
$1$. Thus, after a unitary change of basis:

$\hat{Z}^{a}\,_{b}=\frac{1}{2}\text{diag}(1,1,1,1,-1,-1,-1,-1).$ (224)

In this basis, we observe that $\Omega^{ab}$ and
$\hat{Z}^{ab}=\hat{Z}^{a}\,_{c}\Omega^{bc}$ are both antisymmetric matrices.
This requires $\Omega^{ab}$ to be block diagonal in the $4\times 4$ blocks
defined by $\hat{Z}^{a}\,_{b}$, i.e., the stabilizer subgroup of $\hat{Z}$ is
$\text{USp}(4)\times\text{USp}(4)$, where

$\hat{Z}^{ab}=\frac{1}{2}(-\Omega_{1}^{ab}+\Omega_{2}^{ab}),\qquad\Omega^{ab}=\Omega_{1}^{ab}+\Omega_{2}^{ab},\qquad\Longrightarrow\qquad\hat{Z}_{ab}=\frac{1}{2}([\Omega_{1}]_{ab}-[\Omega_{2}]_{ab}),$
(225)

in terms of the symplectic forms $\Omega_{1}^{ab}$ and $\Omega_{2}^{ab}$ for
the two $\text{USp}(4)$ blocks. Then:

$\zeta^{abcd}=2(\Omega_{1}^{[ab}\Omega_{1}^{cd]}+\Omega_{2}^{[ab}\Omega_{2}^{cd]}-\Omega_{1}^{[ab}\Omega_{2}^{cd]}).$
(226)

Now choose an arbitary traceless four-form $P^{abcd}$ satisfying the reality
condition (216). We seek a 1/2 BPS central charge $\hat{Z}^{ab}_{(0)}$ that
globally maximizes $P^{abcd}\zeta_{abcd}^{(0)}$ with $\zeta_{abcd}$ given by
(222). Since all $\hat{Z}^{ab}$’s are related by $\text{USp}(8)$
transformations, we can rephrase this problem by going to the basis where
$\hat{Z}^{ab}_{(0)}$ takes the form (225), then requiring that
$P^{abcd}\zeta^{(0)}_{abcd}$ is maximized under $\text{USp}(8)$ rotations of
$P^{abcd}$ with $\zeta^{(0)}_{abcd}$ held fixed in its canonical form (226).
In other words, fixing
$\Omega_{1}=\Omega_{2}=\text{diag}\left[\left(\begin{array}[]{cc}0&1\\\
-1&0\end{array}\right),\left(\begin{array}[]{cc}0&1\\\
-1&0\end{array}\right)\right]$ we maximize

$P\cdot\zeta=\frac{1}{3}(2P^{1234}+2P^{5678}-P^{1256}-P^{1278}-P^{3456}-P^{3478}).$
(227)

The traceless condition on $P^{abcd}$ implies that

$\displaystyle\begin{aligned} P^{1234}+P^{1256}+P^{1278}=0,\qquad
P^{1234}+P^{3456}+P^{3478}=0,\\\ P^{1256}+P^{3456}+P^{5678}=0,\qquad
P^{1278}+P^{3478}+P^{5678}=0.\end{aligned}$ (228)

These equations can be rewritten as

$P^{5678}=P^{1234},\qquad P^{3478}=P^{1256},\qquad P^{3456}=P^{1278},\qquad
P^{1234}+P^{1256}+P^{1278}=0,$ (229)

where the reality condition on $P^{abcd}$ implies that all of these are real.
Thus,

$P\cdot\zeta=P^{1234}+P^{5678}=2P^{1234}.$ (230)

Defining the four-form $\Pi_{abcd}$ such that $\Pi_{1234}=+1$ with other
components vanishing, we see $P\cdot\zeta=\frac{2}{4!}P^{abcd}\Pi_{abcd}$ and
the condition for a critical point is

$\delta(P\cdot\zeta)\propto iT^{a}_{e}P^{ebcd}\Pi_{abcd}=0,$ (231)

where $T^{a}_{e}$ is an arbitrary generator of $\text{USp}(8)$, i.e., a
Hermitian matrix satisfying $T^{a}_{c}\Omega^{cb}+T^{b}_{c}\Omega^{ac}=0$. In
particular,

$T^{a}\,_{b}=\left(\begin{array}[]{cc}T_{1}&X\\\
\Omega_{1}^{-1}X\Omega_{2}&T_{2}\end{array}\right),$ (232)

for any $T_{1},T_{2}$ generators of $\text{USp}(4)_{1,2}$ and $X$ any $4\times
4$ matrix. This means that (231) becomes

$P^{ebcd}\Pi_{abcd}=0,$ (233)

i.e., $P^{abcd}$ has no components with 3 legs on the first $4\times 4$ block
and 1 leg on the second. By an analogous argument $P^{abcd}$ has no components
with 1 leg on the first block and 3 on the second either. Thus, we can write

$P^{abcd}=\frac{3}{2}\lambda(\Omega_{1}^{[ab}\Omega_{1}^{cd]}+\Omega_{2}^{[ab}\Omega_{2}^{cd]}-\Omega_{1}^{[ab}\Omega_{2}^{cd]})+\hat{P}^{abcd},$
(234)

where $\lambda=2P^{1234}$ and $\hat{P}^{abcd}$ only has components with 2 legs
on each $4\times 4$ block. Combined with the traceless condition, this implies
that $\hat{P}^{abcd}(\Omega_{1})_{cd}=\hat{P}^{abcd}(\Omega_{2})_{cd}=0$, and
so $\hat{P}^{abcd}$ transforms in the
$\left(\ydiagram{1,1},\ydiagram{1,1}\right)$ irrep of
$\text{USp}(4)\times\text{USp}(4)$, which is the $(\ydiagram 1,\ydiagram 1)$
irrep of $\text{Spin}(5)\times\text{Spin}(5)$. Notice that (234) implies

$\displaystyle\begin{aligned}
\frac{1}{2}P^{abcd}\hat{Z}_{cd}=&\frac{3}{8}\lambda(\Omega_{1}^{[ab}\Omega_{1}^{cd]}+\Omega_{2}^{[ab}\Omega_{2}^{cd]}-\Omega_{1}^{[ab}\Omega_{2}^{cd]})([\Omega_{1}]_{cd}-[\Omega_{2}]_{cd})\\\
=&\frac{\lambda}{2}(-\Omega_{1}^{ab}+\Omega_{2}^{ab})=\lambda\hat{Z}^{ab}.\end{aligned}$
(235)

using (225). We recognize this is an eigenvalue equation for the $28\times 28$
matrix $P^{ab}_{\hskip 8.4ptcd}$

$\frac{1}{2}P^{ab}_{\hskip 8.4ptcd}\hat{Z}^{cd}=\lambda\hat{Z}^{ab}.$ (236)

We’ve shown that this is a necessary condition to maximize $P\cdot\zeta$. In
fact, it is also a sufficient condition, since

$P\cdot\zeta=\frac{1}{4}\hat{Z}_{ab}\hat{Z}_{cd}P^{abcd},$ (237)
$\Rightarrow\qquad\delta(P\cdot\zeta)\propto
iT^{a}_{e}\hat{Z}_{ab}\hat{Z}_{cd}P^{ebcd}=2\lambda
iT^{a}_{e}\hat{Z}_{ab}\hat{Z}^{eb}=\frac{i\lambda}{2}T^{a}_{a}=0,$ (238)

since the generators of $\text{USp}(8)$ are traceless. The reformulation (236)
of the condition for extremizing $P\cdot\zeta$ will be useful again later.

So far, we have imposed the condition that $P\cdot\zeta$ is extremized, but
not that it is maximized, much less that it realizes its global maximum value.
To do so, we first use the fact that $\hat{P}^{abcd}$ reduces to the
$(\ydiagram 1,\ydiagram 1)$ irrep of $\text{Spin}(5)\times\text{Spin}(5)$ to
put it into a canonical form. First, we need to determine the map between
$\ydiagram{1,1}$ of $\text{USp}(4)$ and $\ydiagram 1$ of $\text{Spin}(5)$,
which is related to the $\Gamma$ matrices for $\text{Spin}(5)$. In terms of
the $2\times 2$ Pauli matrices

$\mathbf{\sigma}^{1}=\left(\begin{array}[]{cc}0&1\\\
1&0\end{array}\right),\qquad\mathbf{\sigma}^{2}=\left(\begin{array}[]{cc}0&-i\\\
i&0\end{array}\right),\qquad\mathbf{\sigma}^{3}=\left(\begin{array}[]{cc}1&0\\\
0&-1\end{array}\right),$ (239)

we choose a basis where

$(\Gamma^{i})^{a}\,_{b}=\left(\begin{array}[]{cc}\mathbf{0}&-i\mathbf{\sigma}^{i}\\\
i\mathbf{\sigma}^{i}&\mathbf{0}\end{array}\right),\qquad(\Gamma^{4})^{a}\,_{b}=\left(\begin{array}[]{cc}\mathbf{0}&\mathbf{1}\\\
\mathbf{1}&\mathbf{0}\end{array}\right),\qquad(\Gamma^{5})^{a}\,_{b}=\left(\begin{array}[]{cc}\mathbf{1}&\mathbf{0}\\\
\mathbf{0}&-\mathbf{1}\end{array}\right).$ (240)

Choosing the symplectic form

$\Omega^{ab}=\left(\begin{array}[]{cc}\mathbf{\varepsilon}&\mathbf{0}\\\
\mathbf{0}&\mathbf{\varepsilon}\end{array}\right),\qquad\mathbf{\varepsilon}=i\sigma^{2}=\left(\begin{array}[]{cc}0&1\\\
-1&0\end{array}\right),\qquad\Rightarrow\qquad\Omega_{ab}=\left(\begin{array}[]{cc}-\mathbf{\varepsilon}&\mathbf{0}\\\
\mathbf{0}&-\mathbf{\varepsilon}\end{array}\right),$ (241)

we find that $\Gamma^{m}_{ab}\equiv\Omega_{ac}(\Gamma^{m})^{c}\,_{b}$ is given
by

$\Gamma^{i}_{ab}=\left(\begin{array}[]{cc}\mathbf{0}&i\mathbf{\varepsilon}\mathbf{\sigma}^{i}\\\
-i\mathbf{\varepsilon}\mathbf{\sigma}^{i}&\mathbf{0}\end{array}\right),\qquad\Gamma^{4}_{ab}=\left(\begin{array}[]{cc}\mathbf{0}&-\mathbf{\varepsilon}\\\
-\mathbf{\varepsilon}&\mathbf{0}\end{array}\right),\qquad\Gamma^{5}_{ab}=\left(\begin{array}[]{cc}-\mathbf{\varepsilon}&\mathbf{0}\\\
\mathbf{0}&\mathbf{\varepsilon}\end{array}\right).$ (242)

Since $\mathbf{\varepsilon}^{\top}=-\mathbf{\varepsilon}$ and

$i\mathbf{\varepsilon}\mathbf{\sigma}^{1}=\left(\begin{array}[]{cc}i&0\\\
0&-i\end{array}\right),\qquad
i\mathbf{\varepsilon}\mathbf{\sigma}^{2}=\left(\begin{array}[]{cc}-1&0\\\
0&-1\end{array}\right),\qquad
i\mathbf{\varepsilon}\mathbf{\sigma}^{3}=\left(\begin{array}[]{cc}0&-i\\\
-i&0\end{array}\right),$ (243)

are all symmetric, the $\Gamma^{m}_{ab}$ are all antisymmetric, with
$\Omega^{ab}\Gamma^{m}_{ab}=0$ because the $(\Gamma^{m})^{a}\,_{b}$ are
traceless. Thus, $(\Omega^{-1}\Gamma^{m})^{\top}=-\Omega^{-1}\Gamma^{m}$, or
$\Gamma^{m}\Omega=\Omega(\Gamma^{m})^{\top}$, from which one can readily check
that $\Omega$ is indeed a $\text{Spin}(5)$ invariant.

The $\Gamma_{ab}^{m}$ matrices (242) provide the desired dictionary for
translating between the traceless antisymmetric tensor representation of
$\text{USp}(4)$ and the vector representation of $\text{Spin}(5)$. Each
$\Gamma_{ab}^{m}$ has non-zero components with indices in disjoint pairs,
i.e., $\Gamma_{13}^{1}=-\Gamma_{24}^{1}=i$,
$\Gamma^{2}_{13}=\Gamma_{24}^{2}=-1$, $\Gamma^{3}_{14}=\Gamma_{23}^{3}=-i$,
$\Gamma^{4}_{14}=-\Gamma^{4}_{23}=-1$ and
$\Gamma^{5}_{12}=-\Gamma^{5}_{34}=-1$, with all other components vanishing
except those related by antisymmetry. Thus, if we write $\hat{P}^{abcd}$ as a
$(\ydiagram 1,\ydiagram 1)$ of $\text{Spin}(5)\times\text{Spin}(5)$ and
diagonalize it, then in 4-form langauge its non-vanishing components are

$\displaystyle\begin{aligned}
\hat{P}^{1357}&=\hat{P}^{2468},\qquad\hat{P}^{1368}=\hat{P}^{2457},\qquad\hat{P}^{1458}=\hat{P}^{2367},\hat{P}^{1467}=\hat{P}^{2358},\\\
\hat{P}^{1256}&=\hat{P}^{3478}=-\hat{P}^{1278}=-\hat{P}^{3456},\end{aligned}$
(244)

with all of them real. Working backwards, $P^{abcd}$ therefore has real
components

$\displaystyle\begin{aligned} P^{1357}&=P^{2468},\qquad
P^{1368}=P^{2457},\qquad P^{1458}=P^{2367},\qquad P^{1467}=P^{2358},\\\
P^{1256}&=P^{3478},\qquad P^{1278}=P^{3456},\qquad
P^{1234}=P^{5678},\end{aligned}$ (245)

subject to the tracelessness condition $P^{1234}+P^{1256}+P^{1278}=0$. We
define

$\displaystyle\begin{aligned}
\lambda_{1}&=P^{1234},\quad\lambda_{2}=P^{1256},\quad\lambda_{3}=P^{1278},\quad\lambda_{4}=P^{1357},\\\
\lambda_{5}&=P^{1368},\quad\lambda_{6}=P^{1458},\quad\lambda_{7}=P^{1467}.\end{aligned}$
(246)

with $\lambda_{1}+\lambda_{2}+\lambda_{3}=0$. The matrix $P^{ab}\,_{cd}$ is
now block diagonal. First, there is $4\times 4$ block consisting of the
$12,34,56$ and $78$ components:101010Note that raising the index pairs using
the symplectic form takes $12\rightarrow 12,34\rightarrow 34,56\rightarrow
56,78\rightarrow 78$.

$12,34,56,78:\left(\begin{array}[]{cccc}0&\lambda_{1}&\lambda_{2}&\lambda_{3}\\\
\lambda_{1}&0&\lambda_{3}&\lambda_{2}\\\
\lambda_{2}&\lambda_{3}&0&\lambda_{1}\\\
\lambda_{3}&\lambda_{2}&\lambda_{1}&0\end{array}\right),\qquad\lambda_{1}+\lambda_{2}+\lambda_{3}=0.$
(247)

One can easily check that the non-zero eigenvalues $\lambda$ for this block
are $2\lambda_{1},2\lambda_{2}$ and $2\lambda_{3}$ (the zero eigenvalue is
associated to the trace component and should be ignored). There are $6$
remaining $4\times 4$ blocks, e.g.,111111In this case, raising the index pairs
takes $13\leftrightarrow 24$, $57\leftrightarrow 68$, $14\leftrightarrow-23$,
$58\leftrightarrow-67$.

$13,24,57,68:\left(\begin{array}[]{cccc}-\lambda_{1}&0&\lambda_{5}&\lambda_{4}\\\
0&-\lambda_{1}&\lambda_{4}&\lambda_{5}\\\
\lambda_{5}&\lambda_{4}&-\lambda_{1}&0\\\
\lambda_{4}&\lambda_{5}&0&-\lambda_{1}\end{array}\right),\qquad
14,23,58,67:\left(\begin{array}[]{cccc}-\lambda_{1}&0&-\lambda_{7}&-\lambda_{6}\\\
0&-\lambda_{1}&-\lambda_{6}&-\lambda_{7}\\\
-\lambda_{7}&-\lambda_{6}&-\lambda_{1}&0\\\
-\lambda_{6}&-\lambda_{7}&0&-\lambda_{1}\end{array}\right).$ (248)

The eigenvalues in the first case are
$\lambda=-\lambda_{1}\pm_{1}\lambda_{4}\pm_{2}\lambda_{5}$, and likewise
$\lambda=-\lambda_{1}\pm_{1}\lambda_{6}\pm_{2}\lambda_{7}$ in the second.
Similarly,121212Now raising the index pairs takes $15\leftrightarrow 26$,
$37\leftrightarrow 48$, $16\leftrightarrow-25$, $38\leftrightarrow-47$,
$17\leftrightarrow 28$, $35\leftrightarrow 46$, $18\leftrightarrow-27$,
$36\leftrightarrow-45$.

$15,26,37,48:\left(\begin{array}[]{cccc}-\lambda_{2}&0&-\lambda_{6}&-\lambda_{4}\\\
0&-\lambda_{2}&-\lambda_{4}&-\lambda_{6}\\\
-\lambda_{6}&-\lambda_{4}&-\lambda_{2}&0\\\
-\lambda_{4}&-\lambda_{6}&0&-\lambda_{2}\end{array}\right),\qquad
16,25,38,47:\left(\begin{array}[]{cccc}-\lambda_{2}&0&\lambda_{7}&\lambda_{5}\\\
0&-\lambda_{2}&\lambda_{5}&\lambda_{7}\\\
\lambda_{7}&\lambda_{5}&-\lambda_{2}&0\\\
\lambda_{5}&\lambda_{7}&0&-\lambda_{2}\end{array}\right),$ (249)

with eigenvalues $\lambda=-\lambda_{2}\pm_{1}\lambda_{4}\pm_{2}\lambda_{6}$
and $\lambda=-\lambda_{2}\pm_{1}\lambda_{5}\pm_{2}\lambda_{7}$, and finally

$17,28,35,46:\left(\begin{array}[]{cccc}-\lambda_{3}&0&\lambda_{7}&\lambda_{4}\\\
0&-\lambda_{3}&\lambda_{4}&\lambda_{7}\\\
\lambda_{7}&\lambda_{4}&-\lambda_{3}&0\\\
\lambda_{4}&\lambda_{7}&0&-\lambda_{3}\end{array}\right),\qquad
18,27,36,45:\left(\begin{array}[]{cccc}-\lambda_{3}&0&-\lambda_{6}&-\lambda_{5}\\\
0&-\lambda_{3}&-\lambda_{5}&-\lambda_{6}\\\
-\lambda_{6}&-\lambda_{5}&-\lambda_{3}&0\\\
-\lambda_{5}&-\lambda_{6}&0&-\lambda_{3}\end{array}\right),$ (250)

with eigenvalues $\lambda=-\lambda_{3}\pm_{1}\lambda_{4}\pm_{2}\lambda_{7}$
and $\lambda=-\lambda_{3}\pm_{1}\lambda_{5}\pm_{2}\lambda_{6}$.

Using (236), we see that the eigenvalues found above are precisely the
critical values of $P\cdot\zeta$ as we vary $\hat{Z}^{ab}$ with $P^{abcd}$
held fixed.131313We must be careful here, as we have not yet imposed the
conditions (223) on the $\hat{Z}^{ab}$ eigenvectors. However, it is
straightforward to check that they are satisfied for all the eigenvectors in
question, up to a (complex) overall factor that can be freely chosen. The
critical value we found initially is $2\lambda_{1}=2P^{1234}$; now we can
write down the necessary and sufficient conditions for this to be a global
maximum of $P\cdot\zeta$:

$\displaystyle\begin{aligned}
\lambda_{1}\geq&\;\lambda_{2},\quad\lambda_{1}\geq\lambda_{3},\quad
2\lambda_{1}\geq-\lambda_{1}+|\lambda_{4}|+|\lambda_{5}|,\quad
2\lambda_{1}\geq-\lambda_{1}+|\lambda_{6}|+|\lambda_{7}|,\\\
2\lambda_{1}\geq&\;-\lambda_{2}+|\lambda_{4}|+|\lambda_{6}|,\quad
2\lambda_{1}\geq-\lambda_{2}+|\lambda_{5}|+|\lambda_{7}|,\\\ \quad
2\lambda_{1}\geq&\;-\lambda_{3}+|\lambda_{4}|+|\lambda_{7}|,\quad
2\lambda_{1}\geq-\lambda_{3}+|\lambda_{5}|+|\lambda_{6}|.\end{aligned}$ (251)

Using $\lambda_{1}+\lambda_{2}+\lambda_{3}=0$, the conditions on the second
line simply to

$\displaystyle\begin{aligned}
\lambda_{1}-\lambda_{3}\geq&\;|\lambda_{4}|+|\lambda_{6}|,\qquad\lambda_{1}-\lambda_{3}\geq|\lambda_{5}|+|\lambda_{7}|,\\\
\lambda_{1}-\lambda_{2}\geq&\;|\lambda_{4}|+|\lambda_{7}|,\qquad\lambda_{1}-\lambda_{2}\geq|\lambda_{5}|+|\lambda_{6}|,\end{aligned}$
(252)

from which the conditions on the first line automatically follow. Thus,
setting $\lambda_{1}=\frac{1}{2}$ (so that $P\cdot\zeta\leq 1$) we get

$\displaystyle\begin{aligned}
\lambda_{3}+|\lambda_{4}|+|\lambda_{6}|\leq&\;\frac{1}{2},\quad\lambda_{3}+|\lambda_{5}|+|\lambda_{7}|\leq\frac{1}{2},\quad\lambda_{2}+|\lambda_{4}|+|\lambda_{7}|\leq\frac{1}{2},\\\
\lambda_{2}+|\lambda_{5}|+|\lambda_{6}|\leq&\;\frac{1}{2},\quad\lambda_{2}+\lambda_{3}=-\frac{1}{2}.\end{aligned}$
(253)

Subject to these constraints, we wish to maximize:

$P^{2}=2\lambda_{1}^{2}+2\lambda_{2}^{2}+2\lambda_{3}^{2}+2\lambda_{4}^{2}+2\lambda_{5}^{2}+2\lambda_{6}^{2}+2\lambda_{7}^{2}.$
(254)

Clearly we need to saturate at least two of the inequalities in (253) to do
so, since otherwise we can increase at least one of $\lambda_{4,5,6,7}$ for
free. In particular, we must either saturate the first two or the second two,
since if we only saturate one from the first pair and one from the second then
there is still one of $\lambda_{4,5,6,7}$ that can be increased for free.
Because the problem is symmetric under the simultaneous exchange
$\lambda_{2}\leftrightarrow\lambda_{3}$,
$\lambda_{4}\leftrightarrow\lambda_{5}$, we can assume that the saturated pair
is

$\lambda_{2}+|\lambda_{4}|+|\lambda_{7}|=\lambda_{2}+|\lambda_{5}|+|\lambda_{6}|=\frac{1}{2},$
(255)

without loss of generality. Summing the inequalities for $\lambda_{3}$, this
implies that

$2\lambda_{2}+|\lambda_{4}|+|\lambda_{5}|+|\lambda_{6}|+|\lambda_{7}|=1\geq
2\lambda_{3}+|\lambda_{4}|+|\lambda_{5}|+|\lambda_{6}|+|\lambda_{7}|\quad\Longrightarrow\quad\lambda_{2}\geq\lambda_{3}.$
(256)

Thus, for fixed $\lambda_{2}$ the sums $|\lambda_{4}|+|\lambda_{7}|$ and
$|\lambda_{5}|+|\lambda_{6}|$ are fixed. Maximizing $P^{2}$ with this
constraint, it is optimal to maximize the _difference_ between $|\lambda_{4}|$
and $|\lambda_{7}|$ and between $|\lambda_{5}|$ and $|\lambda_{6}|$, i.e., to
set one of each pair to zero. If, e.g., we set
$|\lambda_{6}|=|\lambda_{7}|=0$, then the inequalities

$\lambda_{3}+|\lambda_{4}|+|\lambda_{6}|=\lambda_{3}+|\lambda_{4}|\leq\frac{1}{2},\quad\lambda_{3}+|\lambda_{5}|+|\lambda_{7}|=\lambda_{3}+|\lambda_{5}|\leq\frac{1}{2},$
(257)

follow from (255) and $\lambda_{2}\geq\lambda_{3}$, so we can do so consistent
with the $\lambda_{3}$ inequalities.141414We cannot necessarily set, e.g.,
$|\lambda_{5}|=|\lambda_{7}|=0$, since the $\lambda_{3}$ inequality
$\lambda_{3}+|\lambda_{4}|+|\lambda_{6}|\leq\frac{1}{2}$ might then be
violated; we could keep track of the point where we first violate the
$\lambda_{3}$ inequality in this direction, but the resulting $P^{2}$ is
submaximal, so we can just ignore this case altogether. Thus, for fixed
$\lambda_{2}\geq\lambda_{3}$, the maximum $P^{2}$ occurs when

$|\lambda_{4}|=|\lambda_{5}|=\frac{1}{2}-\lambda_{2},\qquad\lambda_{6}=\lambda_{7}=0.$
(258)

Specifically

$\displaystyle\begin{aligned}
P^{2}&=2\left[\frac{1}{4}+\lambda_{2}^{2}+\left(-\frac{1}{2}-\lambda_{2}\right)^{2}+\left(\frac{1}{2}-\lambda_{2}\right)^{2}+\left(\frac{1}{2}-\lambda_{2}\right)^{2}+0+0\right],\\\
\quad-\frac{1}{4}\leq&\;\lambda_{2}\leq\frac{1}{2},\end{aligned}$ (259)

where the latter constraint comes from
$\lambda_{3}\leq\lambda_{2}\leq\lambda_{1}$. Since this is a quadratic
function of $\lambda_{2}$ with a positive coefficient for $\lambda_{2}^{2}$,
it is maximized at one boundary or the other. In fact, the value is the _same_
at both boundaries:

$P^{2}_{\max}=3.$ (260)

This indeed yields $\lambda_{\min}=\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{d-2}}$ as
expected.

###  A.5 Four Dimensions

In 4d, $H=\text{SU}(8)$ and the central charge $Z^{ab}$ transforms in the
complex antisymmetric tensor representation $\ydiagram{1,1}$. Let us define
$Z_{ab}\equiv(Z^{ab})^{\star}$. $Z$ can alternately be thought of as living in
the real representation $\ydiagram{1,1}\oplus\overline{\ydiagram{1,1}}$ with
twice the dimension, composed of components $Z^{ab}$ and $Z_{ab}$ with the
reality condition $Z_{ab}=(Z^{ab})^{\star}$.

Note that

$\left(\ydiagram{1,1}\oplus\overline{\ydiagram{1,1}}\right)\otimes_{S}\left(\ydiagram{1,1}\oplus\overline{\ydiagram{1,1}}\right)=\ydiagram{1,1,1,1}\oplus\ydiagram{2,2}\oplus\overline{\ydiagram{1,1,1,1}}\oplus\overline{\ydiagram{2,2}}\oplus\ydiagram{2,2,1,1,1,1}\oplus\text{Adj}\oplus
1.$ (261)

This means that there is a unique mass formula

$M^{2}=\frac{1}{2}Z^{ab}(Z^{ab})^{\star}=\frac{1}{2}Z^{ab}Z_{ab},$ (262)

up to an overall normalizing factor that we set to $1$ by redefining the
central charge. However, since
$\ydiagram{1,1,1,1}\cong\overline{\ydiagram{1,1,1,1}}$ and
$R_{\text{str}}=\text{Adj}\oplus\ydiagram{1,1,1,1}$ the half-BPS condition at
first seems ambiguous. In particular,

$X^{abcd}=Z^{[ab}Z^{cd]},$ (263)

is complex, but the BPS condition should involve a _real_ four-form, i.e., one
satisfying

$\omega^{\star}_{abcd}=\frac{1}{4!}\varepsilon_{abcdefgh}\omega^{efgh},$ (264)

which is a consistent condition since

$[(\omega^{\star})^{\star}]^{abcd}=\frac{1}{4!}\varepsilon^{abcdefgh}\omega_{efgh}^{\star}=\frac{1}{4!^{2}}\varepsilon^{abcdefgh}\varepsilon_{efghijkl}\omega^{ijkl}=\omega^{abcd}.$
(265)

In general, the 1/2 BPS condition could involve a linear combination of the
real and imaginary parts of $X^{abcd}$, i.e., schematically
$\text{Im}[e^{i\theta}ZZ]=0$, but by redefining $Z^{ab}$ by a phase (leaving
the mass formula invariant) we can set $\theta=0$. Thus, after this
redefinition the $1/2$ BPS conditions become

$Z^{ac}Z_{bc}=\frac{1}{8}\delta^{a}_{b}Z^{cd}Z_{cd},\qquad
Z_{[ab}Z_{cd]}=\frac{1}{4!}\varepsilon_{abcdefgh}Z^{ef}Z^{gh}.$ (266)

The scalar charge-to-mass ratio $\zeta^{abcd}$ also sits in the
$\ydiagram{1,1,1,1}$ rep, satisfying the reality condition

$\zeta^{\star}_{abcd}=\frac{1}{4!}\varepsilon_{abcdefgh}\zeta^{efgh}.$ (267)

Because

$\ydiagram{1,1,1,1}\otimes_{S}\ydiagram{1,1,1,1}=\ydiagram{2,2,2,2}\oplus\ydiagram{2,2,1,1,1,1}\oplus
1,$ (268)

the metric is uniquely determined to be

$\zeta^{2}=\frac{1}{4!}\zeta^{abcd}(\zeta^{abcd})^{\star}=\frac{1}{4!}\zeta^{abcd}\zeta_{abcd}=\frac{1}{4!^{2}}\varepsilon_{abcdefgh}\zeta^{abcd}\zeta^{efgh},$
(269)

up to an overall normalization that we can absorb into the definition of
$\zeta$, where $\zeta_{abcd}\equiv(\zeta^{abcd})^{\star}$.

Given the $1/2$ BPS condition, there is a unique $\ydiagram{1,1,1,1}$
component inside
$\left(\ydiagram{1,1}\oplus\overline{\ydiagram{1,1}}\right)\otimes_{S}\left(\ydiagram{1,1}\oplus\overline{\ydiagram{1,1}}\right)$,
so for $1/2$ BPS states we must have

$\zeta^{abcd}=3\mathcal{N}\hat{Z}^{[ab}\hat{Z}^{cd]}=\mathcal{N}(\hat{Z}^{ab}\hat{Z}^{cd}+\hat{Z}^{ac}\hat{Z}^{db}+\hat{Z}^{ad}\hat{Z}^{bc}),\qquad\hat{Z}^{ab}\equiv\frac{Z^{ab}}{\sqrt{\frac{1}{2}Z^{cd}Z_{cd}}},$
(270)

up to an unknown normalizing factor $\mathcal{N}$. To fix $\mathcal{N}$, we
compute $\zeta^{2}$ for a $1/2$ BPS state:

$\zeta^{2}=\frac{3}{4!}\mathcal{N}^{2}(\hat{Z}^{ab}\hat{Z}^{cd}+\hat{Z}^{ac}\hat{Z}^{db}+\hat{Z}^{ad}\hat{Z}^{bc})\hat{Z}_{ab}\hat{Z}_{cd}=\frac{1}{8}\mathcal{N}^{2}\left(4-\frac{1}{2}-\frac{1}{2}\right)=\frac{3\mathcal{N}^{2}}{8}.$
(271)

Since Kaluza Klein modes are $1/2$ BPS with
$\zeta^{2}=\frac{d-1}{d-2}=\frac{3}{2}$, we conclude that $\mathcal{N}=2$,
i.e.,

$\zeta^{abcd}=6\hat{Z}^{[ab}\hat{Z}^{cd]}=2(\hat{Z}^{ab}\hat{Z}^{cd}+\hat{Z}^{ac}\hat{Z}^{db}+\hat{Z}^{ad}\hat{Z}^{bc}),\qquad\hat{Z}^{ab}\equiv\frac{Z^{ab}}{\sqrt{\frac{1}{2}Z^{cd}Z_{cd}}}.$
(272)

Note that we can solve the $1/2$ BPS conditions

$\hat{Z}^{ac}\hat{Z}_{bc}=\frac{1}{4}\delta^{a}_{b},\qquad\hat{Z}_{[ab}\hat{Z}_{cd]}=\frac{1}{4!}\varepsilon_{abcdefgh}\hat{Z}^{ef}\hat{Z}^{gh},$
(273)

as follows. After an $\text{SU}(8)$ transformation we can fix

$\hat{Z}^{ab}=e^{i\phi}\text{diag}\left[\left(\begin{array}[]{cc}0&\lambda_{1}\\\
-\lambda_{1}&0\end{array}\right),\ldots,\left(\begin{array}[]{cc}0&\lambda_{4}\\\
-\lambda_{4}&0\end{array}\right)\right],$ (274)

where $\lambda_{1},\ldots,\lambda_{4}$ are real and non-negative and the
overall phase factor is needed because we can only perform a _special unitary_
change of basis, not a general unitary change of basis. The first equation of
(273) then implies that
$\lambda_{1}=\lambda_{2}=\lambda_{3}=\lambda_{4}=\frac{1}{2}$. The second one
implies that

$\displaystyle\begin{aligned}
\zeta^{2}&=\frac{3}{2}=\frac{3}{2}\hat{Z}^{[ab}\hat{Z}^{cd]}\hat{Z}_{[ab}\hat{Z}_{cd]}=\frac{3}{2}\cdot\frac{1}{4!}\varepsilon_{abcdefgh}\hat{Z}^{ab}\hat{Z}^{cd}\hat{Z}^{ef}\hat{Z}^{gh}=\frac{3}{2}\text{Pf}(2\hat{Z}^{ab})\\\
\Rightarrow\qquad&\text{Pf}(2\hat{Z}^{ab})=1,\end{aligned}$ (275)

where

$\text{Pf}(M^{ab})=\frac{1}{2^{n}n!}\varepsilon_{a_{1}\ldots
a_{n}}M^{a_{1}a_{2}}\cdots M^{a_{n-1}a_{n}}$ (276)

is the Pfaffian. Thus, we obtain $\text{Pf}(2\hat{Z}^{ab})=e^{4i\phi}=1$, so
$e^{i\phi}$ is a fourth root of unity. However, the $\text{SU}(8)$
transformation $\text{diag}(e^{\pi i/4},\ldots,e^{\pi i/4})$ imparts an
overall phase $e^{i\pi/2}$ to $\hat{Z}^{ab}$, so we can set $e^{i\phi}=1$ by
such transformations, leaving

$\hat{Z}^{ab}=\text{diag}\left[\left(\begin{array}[]{cc}0&1\\\
-1&0\end{array}\right),\ldots,\left(\begin{array}[]{cc}0&1\\\
-1&0\end{array}\right)\right],$ (277)

in this basis. In other words, all the $1/2$ BPS choices of $\hat{Z}^{ab}$ are
equivalent up to $\text{SU}(8)$ transformations, and a particular choice of
$\hat{Z}^{ab}$ is left invariant by a $\text{USp}(8)\subset\text{SU}(8)$
subgroup.

Now choose an arbitary four-form $P^{abcd}$ satisfying the reality condition
(267). We seek a 1/2 BPS central charge $\hat{Z}^{ab}_{(0)}$ that globally
maximizes $P^{abcd}\zeta_{abcd}^{(0)}$ with $\zeta_{abcd}$ given by (272).
Since all $\hat{Z}^{ab}$’s are related by $\text{SU}(8)$ transformations, we
can rephrase this problem by going to the basis where $\hat{Z}^{ab}_{(0)}$
takes the form (277), then requiring that $P^{abcd}\zeta^{(0)}_{abcd}$ is
maximized under $\text{SU}(8)$ rotations of $P^{abcd}$ with
$\zeta^{(0)}_{abcd}$ held fixed in its canonical form (277). In other words,
we maximize

$P^{1234}+P^{1256}+P^{1278}+P^{3456}+P^{3478}+P^{5678}=2\text{Re}[P^{1234}+P^{1256}+P^{1278}],$
(278)

under $\text{SU}(8)$ rotations, where $P^{5678}=(P^{1234})^{\star}$,
$P^{3478}=(P^{1256})^{\star}$ and $P^{3456}=(P^{1278})^{\star}$ by the reality
condition (267).

Maximizing with respect to the Cartan subgroup, it is evident that $P^{1234}$,
$P^{1256}$ and $P^{1278}$ must be real and non-negative. More generally, the
condition to obtain an extremum of $P\cdot\zeta$ is

$\delta(P\cdot\zeta)=\frac{1}{6}iT^{a}_{e}P^{ebcd}\zeta_{abcd}=0$ (279)

for any generator $T^{a}_{b}$ of $\text{SU}(8)$, i.e., for any traceless
Hermitian matrix $T^{a}_{b}$. In fact, since the condition is
$\mathbb{C}$-linear and any traceless matrix can be written as $T_{1}+iT_{2}$
for traceless Hermitian matrices $T_{1}$ and $T_{2}$, the condition holds for
_any_ traceless matrix $T^{a}_{b}$, so that

$\frac{1}{3}P^{ebcd}\zeta_{abcd}=\delta^{e}_{a}(P\cdot\zeta).$ (280)

In terms of $\hat{Z}_{ab}$, this becomes:

$2P^{ebcd}\hat{Z}_{ab}\hat{Z}_{cd}=\delta^{e}_{a}(P\cdot\zeta)\qquad\Leftrightarrow\qquad\frac{1}{2}P^{abcd}\hat{Z}_{cd}=\hat{Z}^{ab}(P\cdot\zeta),$
(281)

using (273). Thus, decomposing

$P^{abcd}=4\lambda\hat{Z}^{[ab}\hat{Z}^{cd]}+\hat{P}^{abcd},\qquad\lambda=P\cdot\zeta,$
(282)

we find that $\hat{P}^{abcd}\hat{Z}_{cd}=0$. In other words $\hat{P}^{abcd}$
transforms in the traceless 4-form representation $\ydiagram{1,1,1,1}$ of the
residual $\text{USp}(8)$ preserving $\hat{Z}^{ab}$. Moreover, because any two
top-forms are proportional

$\frac{1}{2^{4}\cdot
4!}\hat{Z}_{[ab}\hat{Z}_{cd}\hat{Z}_{ef}\hat{Z}_{gh]}=\frac{1}{8!}\varepsilon_{abcdefgh}(\text{Pf}\hat{Z})=\frac{1}{2^{4}\cdot
8!}\varepsilon_{abcdefgh},$ (283)

where the overall normalization can be fixed by contracting with
$\varepsilon^{abcdefgh}$. Thus, the reality condition (267) for $P^{abcd}$
becomes

$\hat{P}^{\star}_{abcd}=\frac{8!}{4!^{2}}\hat{Z}_{[ab}\hat{Z}_{cd}\hat{Z}_{ef}\hat{Z}_{gh]}\hat{P}^{efgh}=2^{4}\hat{Z}_{ae}\hat{Z}_{bf}\hat{Z}_{cg}\hat{Z}_{dh}\hat{P}^{efgh},$
(284)

i.e., $\hat{P}^{abcd}$ is real in the sense of (216) with symplectic form
$\Omega_{ab}=2\hat{Z}_{ab}$. Thus, per the discussion in §A.4, we can put
$\hat{P}^{abcd}$ into canonical form with non-vanishing real components

$\displaystyle\begin{aligned}
\hat{P}^{1357}&=\hat{P}^{2468},\quad\hat{P}^{1368}=\hat{P}^{2457},\quad\hat{P}^{1458}=\hat{P}^{2367},\quad\hat{P}^{1467}=\hat{P}^{2358},\\\
\hat{P}^{1256}&=\hat{P}^{3478},\quad\hat{P}^{1278}=\hat{P}^{3456},\quad\hat{P}^{1234}=\hat{P}^{5678},\end{aligned}$
(285)

subject to the tracelessness condition
$\hat{P}^{1234}+\hat{P}^{1256}+\hat{P}^{1278}=0$. Working backwards, we see
that $P^{abcd}$ has real components

$\displaystyle\begin{aligned} P^{1357}&=P^{2468},\quad P^{1368}=P^{2457},\quad
P^{1458}=P^{2367},\quad P^{1467}=P^{2358},\\\ P^{1256}&=P^{3478},\quad
P^{1278}=P^{3456},\quad P^{1234}=P^{5678},\end{aligned}$ (286)

satisfying no further constraints.

To see that (286) is not, in fact, some random collection of directions, note
that exactly one in each pair has a leg along any given direction, e.g., the
$1$ direction. Stripping off this leg, we find the combinations

$234,\quad 256,\quad 278,\quad 357,\quad 368,\quad 458,\quad 467.$ (287)

These are nothing but the lines on a Fano plane with points $2,\ldots,8$ (up
to convention-dependent relabling). To present it more symmetrically,
represent each number with 3 digit binary code, in this case the binary digits
of that number minus 1:

$2\rightarrow 001,\quad 3\rightarrow 010,\quad 4\rightarrow 011,\quad
5\rightarrow 100,\quad 6\rightarrow 101,\quad 7\rightarrow 110,\quad
8\rightarrow 111,\quad$ (288)

where the codes are chosen such that 3 points are colinear iff the sums of
their digits $\text{mod}2$ are all zero (e.g., $001+011+010=000$ with _no
carrying_ , so $001-011-010$ forms the line 234). The symmetries of the Fano
plane are then $\text{GL}(3;\mathbb{Z}_{2})$ acting on these digits. Note that
this group is 2-transitive—any pair of points can be mapped to any other pair
of points—and the stabilizer subgroup fixing any pair of points is
$\mathbb{Z}_{2}\times\mathbb{Z}_{2}$; e.g., if the fixed points are $100$ and
$010$ then one $\mathbb{Z}_{2}$ is generated by adding the third digit to the
first ($001\leftrightarrow 101,011\leftrightarrow 111$) and the other by
adding the third digit to the second ($001\leftrightarrow
011,101\leftrightarrow 111$).

Using these facts, one can also show that $\text{GL}(3;\mathbb{Z}_{2})$ acts
2-transitively on the set of lines, with the stabilizer subgroup fixing a pair
of lines equal to $\mathbb{Z}_{2}\times\mathbb{Z}_{2}$; e.g., if the fixed
lines are $001-010-011$ and $001-100-101$ then one $\mathbb{Z}_{2}$ is
generated by adding the first digit to the third ($100\leftrightarrow
101,110\leftrightarrow 111$) and the other is generated by adding the second
digit to the third ($010\leftrightarrow 011,110\leftrightarrow 111$). In fact,
the Fano plane is self-dual under the exchange of lines and points, so this
similarity between the action on points and lines is no accident. Explicitly,
on such point-line duality is

$(001,010,011)\rightarrow 100,\quad(100,011,111)\rightarrow
011,\quad(110,101,011)\rightarrow 111,$ (289)

with the other elements of the dictionary given by permuting the three digits.
With this in mind, we assign labels:

$\displaystyle\begin{aligned}
\lambda_{001}&=P^{1357}=P^{2468},\quad\lambda_{010}=P^{1256}=P^{3478},\quad\lambda_{011}=P^{1458}=P^{2367},\\\
\lambda_{100}&=P^{1234}=P^{5678},\quad\lambda_{101}=P^{1368}=P^{2457},\quad\lambda_{110}=P^{1278}=P^{3456},\\\
\quad\lambda_{111}&=P^{1467}=P^{2358}.\end{aligned}$ (290)

Now we study the “eigenvalue” problem:

$\frac{1}{2}P^{abcd}\hat{Z}_{cd}=\lambda(\hat{Z}_{ab})^{\star},$ (291)

which is a necessary and sufficient condition for a given $\hat{Z}_{ab}$
satisfying (273) to extremize $P\cdot\zeta$, where the eigenvalue
$\lambda\in\mathbb{R}$ is equal to $P\cdot\zeta$. In general,

$Mv=\lambda
v^{\star}\qquad\Leftrightarrow\qquad\left(\begin{array}[]{cc}0&M\\\
M^{\star}&0\end{array}\right)\left(\begin{array}[]{c}v\\\
v^{\star}\end{array}\right)=\lambda\left(\begin{array}[]{c}v\\\
v^{\star}\end{array}\right),$ (292)

so this becomes a standard eigenvalue problem for a matrix twice as large,
supplemented by a reality condition on the eigenvectors

$\left(\begin{array}[]{cc}0&1\\\
1&0\end{array}\right)\left(\begin{array}[]{c}v\\\
v^{\star}\end{array}\right)=\left(\begin{array}[]{c}v\\\
v^{\star}\end{array}\right)^{\star}.$ (293)

In particular, if $M$ is real and symmetric then it has a basis of real
eigenvectors with real eigenvalues

$Mv_{i}=\lambda_{i}v_{i},\qquad v_{i}=v_{i}^{\star}.$ (294)

From each such eigenvector we obtain a _pair_ of
$\left(\begin{array}[]{cc}0&M\\\ M&0\end{array}\right)$ eigenvectors
satisfying the reality condition:

$\left(\begin{array}[]{cc}0&M\\\
M&0\end{array}\right)\left(\begin{array}[]{c}v_{i}\\\
v_{i}\end{array}\right)=\lambda_{i}\left(\begin{array}[]{c}v_{i}\\\
v_{i}\end{array}\right),\qquad\left(\begin{array}[]{cc}0&M\\\
M&0\end{array}\right)\left(\begin{array}[]{c}iv_{i}\\\
-iv_{i}\end{array}\right)=-\lambda_{i}\left(\begin{array}[]{c}iv_{i}\\\
-iv_{i}\end{array}\right).$ (295)

Thus the eigenvalues in the equation $Mv=\lambda v^{\star}$ are those of $M$
together with those of $-M$.

Fortunately, the real symmetric matrix $P^{abcd}$ (viewed as a $28\times 28$
matrix of index pairs) is block diagonal. For example, in the $12,34,56,78$
block we find:

$M=\left(\begin{array}[]{cccc}0&\lambda_{100}&\lambda_{010}&\lambda_{110}\\\
\lambda_{100}&0&\lambda_{110}&\lambda_{010}\\\
\lambda_{010}&\lambda_{110}&0&\lambda_{100}\\\
\lambda_{110}&\lambda_{010}&\lambda_{100}&0\end{array}\right).$ (296)

The eigenvalues of this matrix are
$\lambda_{100}+\lambda_{010}+\lambda_{110}$,
$\lambda_{100}-\lambda_{010}-\lambda_{110}$,
$-\lambda_{100}+\lambda_{010}-\lambda_{110}$ and
$-\lambda_{100}-\lambda_{010}+\lambda_{110}$, so in combination with those of
$-M$, we obtain
$\pm_{1}\lambda_{100}\pm_{2}\lambda_{010}\pm_{3}\lambda_{110}$, the largest
eigenvalue of which is

$\lambda_{\max}=|\lambda_{100}|+|\lambda_{010}|+|\lambda_{110}|.$ (297)

It is straightforward to check that the resulting eigenvectors $\hat{Z}_{ab}$
satisfy (273) after fixing their overall normalization as required.

Fortunately, due to the symmetries of the Fano plane, there is no need to
consider the other six $4\times 4$ blocks of $P^{abcd}$, which are related by
permutations on the indices $1,\ldots,8$ that correspond to these symmetries.
Thus, the requirement $P\cdot\zeta\leq 1$ for all half-BPS central charges
$\hat{Z}_{ab}$ translates to the seven conditions,

$|\lambda_{ijk}|+|\lambda_{i^{\prime}j^{\prime}k^{\prime}}|+|\lambda_{(i+i^{\prime})(j+j^{\prime})(k+k^{\prime})}|\leq
1,$ (298)

one for each line on the Fano plane with points $100$, $010$, etc. Meanwhile,

$P^{2}=2\sum_{i,j,k}|\lambda_{ijk}|^{2}.$ (299)

We seek to maximize $P^{2}$ while satisfying all the constraints. To do so,
first note that we can take $\lambda_{ijk}\geq 0$ without loss of generality.
If any given $\lambda_{ijk}$ does not show up in a constraint (298) that is
saturated then we can increase it at no cost, increasing $P^{2}$ at no cost.
Thus, the lines corresponding to the saturated constraints must intersect
every point on the Fano plane. Therefore, there must be a least one point for
which _all_ the lines intersecting it are saturated
constraints.151515Otherwise, since any two lines on the place intersect at a
point, we can form a chain of saturated lines. Once we have three such lines
we will have a loop of saturated lines but this only covers six of the points,
leaving one out. Choose this to be $\lambda_{111}$, so that

$\lambda_{111}+\lambda_{101}+\lambda_{010}=\lambda_{111}+\lambda_{100}+\lambda_{011}=\lambda_{111}+\lambda_{001}+\lambda_{110}=1.$
(300)

If there were no other constraints, then for fixed $\lambda_{111}$ the maximum
value of $P^{2}$ would occur when one of each of the other pairs vanishes,
e.g., $\lambda_{101}=\lambda_{110}=\lambda_{011}=0$, so that

$\lambda_{100}=\lambda_{010}=\lambda_{001}=1-\lambda_{111}.$ (301)

This satisfies the remaining constraints provided that
$\lambda_{111}\geq\frac{1}{2}$, and since

$P^{2}=2(\lambda_{111}^{2}+3(1-\lambda_{111})^{2}),$ (302)

the maximum values are achieved at the boundaries $\lambda_{111}=1$ (implying
$\lambda_{100}=\lambda_{010}=\lambda_{001}=0$) and $\lambda_{111}=\frac{1}{2}$
(implying $\lambda_{100}=\lambda_{010}=\lambda_{001}=\frac{1}{2}$), with
$P^{2}=2$ in both cases.

Thus, by process of elimination we consider the case
$\lambda_{111}<\frac{1}{2}$. Now after maximizing $P^{2}$ the constraint
associated to at least one other line in the Fano plane must be saturated.
Using the symmetries of the Fano plane, we can fix this to be

$\lambda_{110}+\lambda_{101}+\lambda_{011}=1,$ (303)

without loss of generality. Ignore the other constraints for the time being.
Applying the symmetries of the Fano plane, we may assume that
$\lambda_{110}\geq\lambda_{101}$ and $\lambda_{110}\geq\lambda_{011}$ (without
loss of generality). Then taking

$\lambda_{110}\rightarrow\lambda_{110}+\Delta,\qquad\lambda_{101}\rightarrow\lambda_{101}-\Delta,\qquad\lambda_{001}\rightarrow\lambda_{001}-\Delta,\qquad\lambda_{010}\rightarrow\lambda_{010}+\Delta,$
(304)

preserves the constraints (300) and (303), whereas

$P^{2}\rightarrow P^{\prime
2}=P^{2}+4(\lambda_{110}+\lambda_{010}-\lambda_{101}-\lambda_{001})\Delta+8\Delta^{2}=P^{2}+8(\lambda_{110}-\lambda_{101})\Delta+8\Delta^{2}>P^{2},$
(305)

since $\lambda_{110}\geq\lambda_{101}$ by assumption, so $P^{2}$ increases.
This can be continued until either $\lambda_{101}=0$ or $\lambda_{001}=0$. If
the former occurs, then we exchange the roles of $\lambda_{101}$ and
$\lambda_{011}$ and carry out the same process until either $\lambda_{011}$
also vanishes or $\lambda_{001}=0$. If the former, then (303) implies that
$\lambda_{110}=1$, but then (300) implies that $\lambda_{001}=0$, so we
ultimately obtain $\lambda_{001}=0$ regardless. Next, we again apply the
symmetries of the Fano plane to fix $\lambda_{101}\geq\lambda_{011}$, and then
carry out an analogous process to increase $\lambda_{101}$ at the expense of
$\lambda_{011}$ while steadily increasing $P^{2}$. This terminates when either
$\lambda_{011}=0$ or $\lambda_{010}=0$. If the former occurs, then our Fano
plane now looks like

$\begin{array}[]{ccccc}&&\lambda_{111}&&\\\
&\lambda_{001}&\lambda_{010}&\lambda_{100}&\\\
\lambda_{110}&&\lambda_{101}&&\lambda_{011}\end{array}=\begin{array}[]{ccccc}&&x&&\\\
&0&1-2x&1-x&\\\ 1-x&&x&&0\end{array},$ (306)

but then $\lambda_{110}+\lambda_{010}+\lambda_{100}=1-x+1-2x+1-x=3-4x>1$ since
$x=\lambda_{111}<\frac{1}{2}$, so we reach a contradiction. If the latter, the
our Fano plane now looks like:

$\begin{array}[]{ccccc}&&\lambda_{111}&&\\\
&\lambda_{001}&\lambda_{010}&\lambda_{100}&\\\
\lambda_{110}&&\lambda_{101}&&\lambda_{011}\end{array}=\begin{array}[]{ccccc}&&x&&\\\
&0&0&2-3x&\\\ 1-x&&1-x&&2x-1\end{array},$ (307)

but now since $x=\lambda_{111}<\frac{1}{2}$, $\lambda_{011}=2x-1<0$ and we
again reach a contradiction.

Thus, we conclude that another line must be saturated. Using the symmetries of
the Fano plane, we can fix it to be

$\lambda_{001}+\lambda_{010}+\lambda_{011}=1.$ (308)

Imposing all the constraints (300), (303), (308), the Fano plane looks like:

$\begin{array}[]{ccccc}&&\lambda_{111}&&\\\
&\lambda_{001}&\lambda_{010}&\lambda_{100}&\\\
\lambda_{110}&&\lambda_{101}&&\lambda_{011}\end{array}=\begin{array}[]{ccccc}&&x&&\\\
&y&1-x-y&1-2x&\\\ 1-x-y&&y&&x\end{array},$ (309)

and the remaining two lines impose the constraints

$1-2x\leq y\leq x.$ (310)

As usual, the maximum value of $P^{2}$ will be found on a boundary, either
where $y=x$ or $y=1-2x$, where in either case we obtain
$P^{2}=2(4x^{2}+3(1-2x)^{2})$. Since $x\geq\frac{1}{3}$ to satisfy the
previous inequality and $x\leq\frac{1}{2}$ by prior assumption, we obtain
$P^{2}=2$ for $x=\frac{1}{2}$ (as before) and $P^{2}=\frac{14}{9}$ for
$x=\frac{1}{3}$. The former is larger of course, so we finally obtain

$P^{2}_{\text{max}}=2,$ (311)

which corresponds to
$\lambda_{\text{min}}=\frac{1}{\sqrt{d-2}}=\frac{1}{\sqrt{2}}$ precisely as
expected.

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