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For a complex antisymmetric tensor $A\in {\wedge }^2{\mathbb{C}}^4$, the SU(4) action is $A\mapsto UA{U}^T$. The unitary skew-diagonalization of an antisymmetric matrix gives $A=\mathrm{diag}(aϵ,bϵ)$ in suitable coordinates, where $ϵ=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. If $|a|\ne |b|$, preserving $A{A}^{†}$ forces a stabilizing unitary matrix to preserve the two coordinate planes. The identity $VϵV^T=(\det V)ϵ$ then forces both two-by-two blocks to have determinant one, using $\det U=1$ also when one of $a,b$ vanishes. The exact stabilizer is $SU(2)\times SU(2)$. If $|a|=|b|\ne 0$, the stabilizer is the larger compact symplectic group $USp(4)$. If $a=b=0$, it is all of $SU(4)$. The real structure of the SU4 exterior square imposes $b=\bar{a}$ in this form, so a nonzero real six-dimensional field has the larger symplectic stabilizer, whereas unequal paired singular values are generic for a complex six-dimensional field.

Use $\{Q_{\alpha }^A,Q_{\beta }^B\}={ϵ}_{\alpha \beta }{Z}^{AB}$. Unitary skew-diagonalization of an antisymmetric matrix gives two-by-two blocks with entries $Z_r=2z_r$. At rest, the paired supercharges have anticommutators $2(M\pm |z_r|)$. The squared norms of an operator and its adjoint sum to the expectation of their anticommutator, so positive norm requires $M\ge |z_r|$ for every block. A convention with $2Z$ in the algebra instead writes $M\ge \max |Z_r|$.

The fundamental case. Set $⟨\varphi ⟩=v{e}_4$, where $v\ne 0$. Fixing this vector forces the fourth column of a unitary matrix to be $e_4$, and hence also fixes its fourth row. The remaining three-by-three block has determinant one. ThereforeEvery nonzero complex vector can be brought to this form by an SU(4) transformation: complete its normalized direction to a unitary basis and adjust the phase of a complementary basis vector to obtain determinant one. Thus a generic nonzero fundamental vacuum expectation value has a conjugate of $H_3$ as its exact unbroken subgroup. The particular embedded $H_3$ of the question is fixed only when the vacuum expectation value lies along $e_4$.

The antisymmetric case. With $ϵ=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, takeSince $VϵV^T=ϵ$ for $V\in SU(2)$, this antisymmetric tensor is fixed by $H_2$ for every $a,b$. Its transformation is $A\mapsto UA{U}^T$, so any stabilizing $U$ also preservesIf $|a|\ne |b|$, its two eigenspaces force $U=\mathrm{diag}(V,\tilde{V})$ with unitary two-by-two blocks. For each nonzero block, $VϵV^T=(\det V)ϵ$ forces the block determinant to be one. If one block is zero, the SU(4) determinant condition forces the other block determinant to be one as well. HenceThe unitary skew-diagonalization of an antisymmetric matrix shows that a generic complex $6$ has two distinct paired singular values; choosing an SU(4) rather than a $U(4)$ basis merely leaves an allowed overall phase in $a,b$. Thus its generic stabilizer is conjugate to $H_2$.

If instead $|a|=|b|\ne 0$, the form is proportional, up to unitary coordinates and phases, to a nondegenerate standard symplectic form. Its stabilizer is the compact symplectic group $USp(4)$, of dimension ten, containing $H_2$ properly. The explicit choice $a=b=1$ is a concrete counterexample to inferring an exact $H_2$ stabilizer from the existence of two singlets alone. For a real $6$, the real structure of the SU4 exterior square imposes $b=\bar{a}$ in this form, so these equal magnitudes are compulsory. Every nonzero real $6$ therefore has the larger compact symplectic group stabilizer. The zero vacuum expectation value preserves all of $SU(4)$.

The adjoint cases. The $H_2$-fixed and $H_3$-fixed traceless matrices have respectively the formsFor a Hermitian adjoint scalar field, choose $a$ real. More generally a complex $a$ gives the same stabilizers. The repeated eigenvalues implyHere $S$ means that the product of the block determinants is one. These groups contain respectively $H_2$ and $H_3$, together with an additional continuous $U(1)$. Explicit extra transformations are $\mathrm{diag}(e^{it}{I}_2,e^{-it}{I}_2)$ and $\mathrm{diag}(e^{it}{I}_3,e^{-3it})$. They commute with $X_2$ and $X_3$ for every $t$, proving that neither exact endpoint $H_2$ or $H_3$ occurs for a single adjoint vacuum expectation value.

Nor do these larger subgroups survive for a generic Hermitian traceless matrix: four distinct eigenvalues leave only the maximal torus $U(1{)}^3$. For a generic complex adjoint matrix, there need not even be a continuous stabilizer; generically its unitary centralizer is just the centre of a group ${\mathbb{Z}}_4$. Thus the displayed degenerate adjoint directions are special. All these statements concern the stabilizer of a possible vacuum expectation value; no unspecified scalar potential can be assumed to select a particular direction or a particular reality condition.

Choose the extended supersymmetry normalizationThe central charge in supersymmetry is antisymmetric in $A,B$, because the full anticommutator is symmetric and ${ϵ}_{\alpha \beta }$ is antisymmetric. In a massive one-particle state rest frame, $P_{\mu }=(M,0,0,0)$, and the first anticommutator becomes $2M{\delta }_{\alpha \beta }{\delta }^A{}_B$.

A unitary skew-diagonalization of an antisymmetric matrix puts $Z$ into blocks $\left(\begin{array}{cc}0& z_r\\ -z_r& 0\end{array}\right)$, with a remaining zero direction if $\mathcal{N}$ is odd. For one block write $z=|z|e^{i\phi }$, and label its two supersymmetries $1,2$. The combinationshaveFor example the mixed contribution to the first expression is $\pm (e^{-i\phi }z+e^{i\phi }\bar{z})/2=\pm |z|$. This is the phase-aligned version of taking a supercharge minus a unitary combination of its adjoints, as suggested by the hint.

For every operator $X$ and normalized state $|v⟩$,Applying this to $A_{-},B_{-}$ in every block proves the BPS boundIf the algebra is instead written with $2Z$ on its right side, the same result is $M\ge {\max }_r|Z_r|$; the factor is purely a central-charge convention.

A BPS state saturates at least one of these inequalities. The corresponding zero-norm combinations of supercharges and their adjoints annihilate the representation, so part of supersymmetry is preserved. Its massive supermultiplet is shortened: each saturated block removes two complex fermionic oscillators, giving $(2j+1)2^{2\mathcal{N}-2k}$ states from a spin-$j$ Clifford vacuum when $k$ blocks saturate, before any necessary CPT completion of a supermultiplet. It preserves $4k$ of the $4\mathcal{N}$ real supercharges. Its mass is tied to the central charges in supersymmetry and remains protected while it stays on the same short branch. Stability is not unconditional: a wall of marginal stability can permit a charged BPS state to decay into other BPS states with the same total charge.

For the unheaded extended-algebra request, fix the central-charge normalization explicitly:The central charges are scalar numbers on an irreducible massive representation. A unitary skew-diagonalization of an antisymmetric matrix puts $Z$ into two-by-two blocks $\left(\begin{array}{cc}0& z_r\\ -z_r& 0\end{array}\right)$ and, for odd $\mathcal{N}$, a leftover zero entry. Since this change of supercharge basis is unitary, it preserves the mixed anticommutator. Go to the rest frame $P^{\mu }=(M,0,0,0)$.

For one block set $q_{\alpha }=Q_{\alpha }^1$ and $r_{\alpha }={ϵ}_{\alpha \gamma }(Q_{\gamma }^2{)}^{†}$, with ${ϵ}_{12}=1$. Direct use of the algebra givesWrite $z=|z|e^{i\phi }$ and form the combinations suggested by the hint:ThenEach diagonal anticommutator has nonnegative expectation because it is the sum of squared norms of an operator and its adjoint. Therefore $M-|z_r|\ge 0$ for every block, provingIf the algebra is instead written with $\{Q,Q\}=ϵZ$ without the factor two, the same proof gives $M\ge \frac{1}{2}{\max }_r|Z_r|$. This is the normalization used by some definitions of the BPS bound in supersymmetry; the physical bound is unchanged by the naming of the central charge.

A massive BPS state saturates at least one block bound. At saturation $\{B_{\alpha },B_{\alpha }^{†}\}=0$, so both $B_{\alpha }$ and $B_{\alpha }^{†}$ annihilate every state in its multiplet. The two complex inactive oscillators are four preserved real supersymmetries per saturated block. If $r$ blocks saturate, the preserved fraction is $r/\mathcal{N}$ and only $2\mathcal{N}-2r$ complex fermionic oscillators remain active. For a spin-$j$ Clifford vacuum,compared with $(2j+1)2^{2\mathcal{N}}$ for a long massive multiplet. Thus BPS states have an exact mass-charge relation, preserve part of the supersymmetry, and belong to shortened representations. A physical CPT-invariant spectrum may additionally require the conjugate-charge multiplet. The short representation protects the saturated mass relation while it exists; it does not prohibit a decay at a threshold where central-charge phases align and the sum of daughter masses equals the bound.

A BPS state is a state whose mass saturates a lower bound set by conserved charges in the Super-Poincaré algebra. Saturation leaves some linear combinations of supercharges unbroken and produces a shortened massive supermultiplet. Both statements follow from the positivity of the physical Hilbert-space norm, rather than from an assumed classical field configuration.

For four-dimensional extended supersymmetry, choose the normalizationThe conjugate algebra is obtained by Hermitian conjugation. The antisymmetry of the central charge in supersymmetry matrix follows because the anticommutator is symmetric in the combined labels $(\alpha ,A)$ while ${ϵ}_{\alpha \beta }$ is antisymmetric. These central charges commute with momentum and supercharges, so their eigenvalues are fixed in an irreducible massive representation. In the rest frame, $P_{\mu }=(M,0,0,0)$ and $\{Q_{\alpha }^A,(Q_{\beta }^B{)}^{†}\}=2M{\delta }^{AB}{\delta }_{\alpha \beta }$.

For any operator $S$ and physical state $|v⟩$,Apply this to $S_{\alpha }^A=Q_{\alpha }^A-{ϵ}_{\alpha \beta }{U}^{AB}(Q_{\beta }^B{)}^{†}$, where $U$ is unitary. This is the rest-frame version of the positive operator constructed from $Q-\Gamma$. Summing over spinor and internal labels givesThe two diagonal terms each contribute $4\mathcal{N}M$, because $U^{†}U=I$; each mixed term contains ${\sum }_{\alpha \beta }{ϵ}_{\alpha \beta }^2=2$. For $\mathcal{N}=2$, write $Z^{12}=|Z|e^{i\phi }$ and choose $U=e^{i\phi }\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$. Then $\mathcal{H}=16M-8|Z|$, so $M\ge |Z|/2$. The factor of one-half reflects the stated convention with $Z$, rather than $2Z$, in the equal-chirality anticommutator.

To obtain the sharp general bound and exhibit the shortening, use unitary skew-diagonalization of an antisymmetric matrix to put $Z$ into two-by-two blocks $Z_r\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, with a possible zero row and column for odd $\mathcal{N}$. In one block, relabel its two internal indices $1,2$ and write $Z_r=|Z_r|e^{i{\phi }_r}$. DefineDirect substitution, using ${ϵ}_{12}=1$, givesand all annihilator-annihilator anticommutators vanish. The minus operators must have nonnegative norm in every block, hence the BPS bound in supersymmetry isEquivalently, defining $z_r=Z_r/2$ gives $M\ge {\max }_r|z_r|$. Applying the positive $Q-\Gamma$ construction to each block gives the same bound; applying only its full trace need not isolate the largest block. For $M>0$ and no central charge saturation, each operator can be normalized by $\sqrt{2M\pm |Z_r|}$ to an ordinary fermionic annihilation operator. The unmatched internal index, if any, contributes two more such operators.

There are consequently $2\mathcal{N}$ complex creation operators in a generic massive supermultiplet. Starting with a spin-$j$ Clifford vacuum, each creation operator can be occupied zero or one times, so the long representation contains $(2j+1)2^{2\mathcal{N}}$ states. If precisely $k$ central-charge blocks saturate $2M=|Z_r|$, then the two $a_{r,i}^{-}$ and their adjoints have zero anticommutator. The norm identity forces each of them to annihilate every physical state in that representation. Removing these $2k$ complex oscillators givesThus a massive $\mathcal{N}=2$ BPS supermultiplet is half-BPS. For $j=0$, its four states consist of two spin-zero states and a spin-one-half doublet, instead of the sixteen states of a long representation. A charged representation generally needs its conjugate for CPT completion of a supermultiplet, doubling this four-state count; this yields the eight states of a full charged hypermultiplet. For $\mathcal{N}=4$, one saturated block gives a quarter-BPS representation and two give a half-BPS representation. This rest-frame argument assumes $M>0$; massless multiplets follow instead from the null-momentum algebra.

The same bound has a geometric realization for solitons. Consider a static magnetic monopole in an $SU(2)$ gauge theory with an adjoint Higgs field ${\varphi }^a$, asymptotic magnitude $v$, vanishing scalar potential and magnetic field $B_i^a$. For magnetic charge $n_m$, the energy can be writtenThe second equality uses the covariant magnetic Bianchi identity to turn the cross term into the surface integral $\oint {\varphi }^a{B}_i^{a}d{S}_i$. Choosing its sign proves the Bogomolny bound $\boxed{E\ge 4\pi v|n_m|/g_{\mathrm{YM}}}$. Equality gives the first-order Bogomolny equations $B_i^a=\pm D_i{\varphi }^a$. In a suitable $\mathcal{N}=2$ supersymmetric embedding, the surface charge is the appropriate central charge and the soliton obeys precisely the algebraic BPS bound in supersymmetry. Its fermionic variation vanishes for the preserved supersymmetry parameters, while the broken supercharges supply its fermionic zero modes.

Shortening is useful because an isolated short multiplet cannot continuously acquire the additional states of a long representation. While the state exists and stays short, its exact mass is determined by the exact central charge, so masses of strongly interacting solitons can be found without computing a generic interacting-particle self-energy. This does not mean that its numerical mass is independent of coupling or vacuum expectation values: those can enter the central charge itself. Supersymmetric indices and the reduced fermionic state content provide ways to count protected states, subject to possible recombination or continuum effects.

For example, in rank-one Seiberg–Witten theory the electric and magnetic charges $(n_e,n_m)$ enter through periods $a,a_D$. In the normalization above, $Z^{12}=2\sqrt{2}(n_{e}a+n_m{a}_D)$, so a BPS state has $\boxed{M=\sqrt{2}|n_{e}a+n_m{a}_D|}$. A magnetic monopole can become massless where $a_D=0$, even though the original electric description is strongly coupled. Including this light magnetic hypermultiplet supplies a useful local description there; states with both electric and magnetic charge are dyons. Thus the protected spectrum identifies important singularities of the vacuum space and tests electric-magnetic dual descriptions.

The charge bound also controls stability. For any proposed decay with additive complex charges $Z={\sum }_i{Z}_i$, each daughter satisfies $M_i\ge |Z_i|/2$. A BPS parent satisfiesEnergy conservation permits decay only if both inequalities are equalities: all daughters must be BPS and their nonzero complex charges must have the same phase. With nonaligned phases the parent lies strictly below the multiparticle threshold. A wall of marginal stability is a locus where such phases align, and a short bound state can disappear across it. Therefore BPS saturation protects a mass-charge relation and short representation, while stability and existence still depend on the spectrum and parameters.

Fix a positive-mass unitary irreducible representation at rest, $P^{\mu }=(M,0,0,0)$ with $M>0$. Use the central charge in supersymmetry conventionThe conjugate relation is $\{{\bar{Q}}_{\dot{\alpha }A},{\bar{Q}}_{\dot{\beta }B}\}={ϵ}_{\dot{\alpha }\dot{\beta }}\overline{Z^{AB}}$, with the dotted epsilon tensor the conjugate of the undotted one. The supercharges commute with $P_{\mu }$ and the central charges. Their Lorentz transformation is that of a Weyl spinor; together with the ordinary Poincare algebra, these relations specify the Super-Poincaré algebra. Take ${ϵ}_{12}=1$ and write $Z^{12}=2z$. This makes the normalization of the mass bound explicit: a convention that writes $2{ϵ}_{\alpha \beta }{Z}^{AB}$ instead has $z=Z^{12}$.

When $Z=0$, define four fermionic annihilation operators and their fermionic creation operators byIn the rest frame they obey the canonical anticommutation relations $\{a_i,a_j^{†}\}={\delta }_{ij}$ and $\{a_i,a_j\}=0$. Choose a spin-zero Clifford vacuum annihilated by all four $a_i$. Acting with each distinct creator at most once constructs the fermionic Fock spaceThe five levels have the binomial coefficient multiplicities $1,4,6,4,1$, so the massive supermultiplet contains $2^4=16$ states. The four creators are two copies of the spin-one-half SU(2) representation. Their second exterior power isHere the first factor carries physical spin and the second counts the two supersymmetries. Thus level two contains one spin-one triplet and three spin-zero singlets; levels one and three each contain two spin-one-half doublets. The final spin content isEven levels give eight boson states and odd levels eight fermion states, exhibiting boson-fermion degeneracy in a supermultiplet.

For nonzero $z=|z|e^{i\phi }$, pair the two supersymmetries before normalizing. DefineThe rest-frame algebra gives $\{q_{\alpha },r_{\beta }^{†}\}=2z{\delta }_{\alpha \beta }$ and consequentlyFor $M>|z|$, divide $b^{\pm }$ by $\sqrt{2(M\pm |z|)}$ to recover four normalized oscillators. Positivity of the Hilbert space norm proves the BPS bound in supersymmetry, sinceThereforeFor a BPS state saturating this bound, both $b_{\alpha }^{-}$ and their adjoints annihilate every state of the representation. Four of the eight real supercharges act trivially. Only the two $b^{+}$ oscillators remain, so a scalar Clifford vacuum gives a shortened massive supermultiplet with two spin-zero states and one spin-one-half doublet:This is the irreducible count at fixed central charge. A CPT completion of a supermultiplet can require the charge-conjugate multiplet as well, doubling it to eight states; the full hypermultiplet count must not be confused with this four-state count. Saturation is a representation-theoretic relation between mass and conserved charge: shortening persists while the state remains BPS, although its existence can change across parameter space.

For even $\mathcal{N}=2n$, unitary skew-diagonalization of an antisymmetric matrix brings the central-charge matrix to blocks $2z_r\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)$, $r=1,\dots ,n$. Repeat the oscillator construction separately for each block. Every block supplies two oscillators with norm $2(M+|z_r|)$ and two with norm $2(M-|z_r|)$, givingIf $k$ blocks saturate, $2k$ complex oscillators disappear, leaving $2\mathcal{N}-2k$. Starting with spin $j_0$, the irreducible state count and preserved fraction areThe largest massive shortening, $k=n$, preserves one-half. The ordinary massive representation has $k=0$. This rest-frame discussion excludes $M=0$, where the massless supermultiplet construction uses a null momentum instead.