# Coleman-Mandula theorem

 Jeffrey E. Mandula (2015), Scholarpedia, 10(2):7476. doi:10.4249/scholarpedia.7476 revision #147717 [link to/cite this article]
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Curator: Jeffrey E. Mandula

The Coleman-Mandula theorem states what Lie group symmetries are possible in a relativistic theory of interacting particles.

## Statement of the Coleman-Mandula theorem

The Coleman-Mandula Theorem concerns what symmetries are possible in a relativistic theory of interacting particles (Coleman and Mandula, 1967). It states that the only possible such Lie group symmetries are direct products of the Poincaré group and an internal symmetry group. If we denote the state of a single particle by $\left| i\lambda p \right\rangle$, where $i$ denotes the particle type, $\lambda$ its intrinsic spin, and $p$ its momentum, an internal symmetry transformation is represented by a matrix acting only on the particle type index while a Poincaré transformation leaves the particle type unchanged and only affects its spin and momentum. The sorts of transformations excluded by the Coleman-Mandula theorem are those whose action on the $i\lambda p$ labels cannot be factored into the product of two terms, one acting only on the particle type index and the other acting only on the space-time indices. We would call such transformations hybrid symmetries.

Two important properties of a symmetry group of a relativistic particle theory are that single particle states form representations of the group, and that multiparticle states transform like direct products of single-particle representations. We will assume that the elements of a symmetry group are unitary operators acting on the space of states.

Corresponding to the unitarity of the group elements, the generators of the symmetry group are assumed Hermitean. The eigenvalues of the generators are the quantum numbers carried by the particles of the theory. The statement that multiparticle states transform like direct products of single particle representations is equivalent to the additive conservation of these quantum numbers. All of the generators of an internal symmetry group commute with each of the generators of the Poincaré group.

The theorem is phrased purely in terms of observable properties of particles and their interactions. Neither the statement of the theorem nor its proof use the full apparatus of quantum field theory. This pedagogical orientation is an artifact of the state of particle theory when the questions that led to its discovery arose. This will be reviewed in the next section of this article.

## Context of the theorem

In the 1950's, experiments at particle accelerators had discovered many new particles and scattering resonances. The masses and spins of these particles were measured, and the patterns of allowed and forbidden decays were observed. These discoveries were new information about the strong, nuclear force, at much higher energies than in previous experiments. They invigorated the attempt to build a theory of the strong interactions.

At that time the only successful theory of elementary particle interactions known was quantum electrodynamics, which seemed to describe all electromagnetic phenomena. This was the natural model from which to start, but it was immediately realized that trying to describe the strong interactions as a quantum field theory was not straightforward. Among the host of obstacles was the fact that there were no reliable methods of calculation in strongly coupled field theory, and that there was no method for choosing a set of fundamental fields.

Without a clear path forward, progress came from attempts to exploit general properties of relativistic quantum field theories. The results that came from such work could be true properties of the strong interactions. One approach was the use of dispersion relations, which express the principle of causality in terms of the analytic properties of scattering amplitudes as functions of complex momenta. Dispersion relations, supplemented by quite reasonable simplifying assumptions, gave good descriptions of many aspects of high energy scattering, such as the electromagnetic form factors of nucleons. Attempts to build a complete theory of the strong interactions were not immediately successful, but the ideas developed in this context have had a powerful and continuing effect on elementary particle theory.

A particularly successful approach was to look for symmetry principles to organize the data. Of course, symmetries alone could not give a complete description of strong interactions, but it was plausible that one could discover the symmetries of the theory underlying the resonances and stable particles before having found the true theory. Furthermore, discovering the symmetries of the theory could be a major step in finding the theory itself.

Symmetry groups had been successfully applied in nuclear physics. The paradigm of an internal symmetry was isotopic spin, which related the nuclei of isotopes of different elements, but with the same atomic number. Isotopic spin was seen to be a symmetry of the strong interactions as well. Resonances with different charges but (almost) identical mass could be grouped into multiplets that formed representations of the isospin group, SU(2), and their allowed decays followed the patterns expected from group representation theory.

Besides isospin, a new internal quantum number was discovered. It was observed that many possible decay modes that were allowed by conservation of energy, charge, and isotopic spin, did not occur at anything like the expected natural strong interaction rate. This puzzle was solved by assigning to each particle a new quantum number, called "strangeness", and positing that it had to be conserved in strong interaction decays (Nakano and Nishijima, 1953; Gell-Mann, 1956). Particles in the same isotopic spin multiplet had the same value of this new quantum number, that is, strangeness commuted with isotopic spin.

The 1960's saw a great expansion of the use of symmetry groups in particle physics. A spectacular development was the observation that resonances with similar masses but different charge, isospin, and strangeness could be collected into multiplets that formed representations of a larger group, SU(3) (Gell-Mann, 1961; Ne'eman, 1961). This was called the "unitary symmetry" group at the time, and is called the "flavor" SU(3) group today. It was curious that, although the simplest representations of SU(3) have dimensions 3 and 6, only higher representations seemed to occur in nature. The lightest bosonic particles and resonances formed two 8-dimensional representations of SU(3), one consisting of pseudoscalar mesons and the other of vector mesons. The lightest spin $1/2$ baryons also formed an 8-dimensional representation while the lightest spin $3/2$ baryons formed a 10-dimensional representation. The assumption that the interactions that violated unitary symmetry transformed in a specific simple way under the symmetry group led to many relations among particle masses that were accurately obeyed (Coleman and Glashow, 1961; Gell-Mann, 1961; Okubo 1962).

Naturally, the success and utility of unitary symmetry led to searches for further symmetries of the strong interactions, and for models that would give the new abstract symmetries a natural dynamical setting.

An especially appealing generalization of the internal symmetry group SU(3) was the hybrid of internal SU(3) symmetry and intrinsic spin based on the group SU(6) (Sakita, 1964; Gürsey and Radicati, 1964; Pais, 1964; Gürsey et al., 1964). As a method of grouping particles into multiplets it was strikingly successful. The ensemble of pseudoscalar and vector mesons formed a single 35-dimensional representation of SU(6) and the spin $1/2$ and spin $3/2$ baryons formed a single 56-dimensional representation. A similar hybrid symmetry had been successfully used in nuclear physics, but there was an conceptual difference between particle physics and nuclear physics. The theory of nuclei was framed in a non-relativistic context, and nuclear reactions could conserve angular momentum and intrinsic spin separately. However, relativity is inherent in particle physics, and intrinsic spin and orbital angular momentum are not separately conserved. The only conserved rotational quantum number in relativistic quantum mechanics is the total angular momentum.

Soon after SU(6) was proposed, several papers explored the problems associated with formulating SU(6) symmetry, and other hybrid symmetries, in a relativistic context (Coleman, 1964; Michel, 1964; Michel and Sakita, 1964; Bég and Pais, 1964; McGlinn, 1964; Weinberg, 1965; Jordan, 1965). The Coleman-Mandula theorem expressed clearly the reasons that hybrid symmetries could not be invariances of particle physics, and that the only possible Lie groups that can be symmetries of a relativistic particle theory are (locally) isomorphic to the direct product of the Poincaré group and an internal symmetry group.

## Proof of the theorem

As we remarked in the introduction, the Coleman-Mandula theorem rests on the incompatibility of Poincaré invariance and the conservation of hybrid quantum numbers that involve spin. Because the result involves the interplay of relativistic scattering theory and group representation theory, the proof is quite convoluted. The logical structure of the argument is to begin with an arbitrary symmetry group generator and whittle its structure down to the sum of a translation, a pure Lorentz transformation, and an internal symmetry generator.

The basic assumptions about the theory are:

1. The theory is Poincaré invariant, so its symmetry group includes the Poincaré group as a subgroup. In particular, any sum of translations or homogeneous Lorentz transformations of a generator is also in the generator algebra.
2. The number of particle types whose masses are less than any given value is finite. The states of each particle type transform according to some positive mass representation of the Poincaré subgroup of the full symmetry group of the theory. There is no generalization of the Coleman-Mandula theorem to theories with massless particles.
3. The $S$ matrix of the theory has the canonical analyticity properties. For a physical elastic scattering process $p + q \to p' + q'$, the momenta are on shell, ${{p}^{2}}={{{p}'}^{2}}=m_{p}^{2}$ and ${{q}^{2}}={{{q}'}^{2}}=m_{q}^{2}$, and the $T$ matrix, defined as $S - 1$ with the energy-momentum conservation $\delta$ function factored out, is an analytic function of the center-of-mass energy and momentum transfer variables \begin{align}\tag{1} & s={{(p+q)}^{2}} \\ & t={{(p-{p}')}^{2}} \\ \end{align}
4. The theory admits non-trivial scattering. The $T$ matrix does not vanish, except possibly at a discrete set of exceptional values of the center-of mass energy and momentum transfer.
5. An "ugly technical assumption", that the matrix elements of the group generators are distributions in momentum space.

To express these assumptions mathematically requires some notation. We denote the generators of translations (the 4-momenta) by ${{P}_{\mu}}$ and the generators of pure Lorentz transformations by ${M_ {\mu \nu} }$. We will call a general element of the algebra of the generators of the symmetry group $A$, and work with representations in which the 4-momentum is diagonal. On single particle states, the elements of the representation matrices are the matrix elements of $A$ between states of fixed momentum.

$\tag{2}{{A}_{{i}'{\lambda}';i\lambda}}({p}',p)=\left\langle {i}'{\lambda}'{p}'\left| A \right. | i\lambda p \right\rangle$

As before, $i({i}')$, $\lambda ({\lambda}')$, and $p(p')$ denote the initial(final) particle type, spin, and momentum. We will suppress the discrete indices and regard $A(p',p)$ as a matrix valued distribution in the momenta. Two particle states transform under the symmetry group like the direct product of their single particle components. In terms of group generator representations, this is expressed as

$\tag{3}A({p}'{q}',p\,q)=A({p}',p)\otimes I({q}',q)+I({p}',p)\otimes A({q}',q)$

The unit symbol $I({p}',p)$ is ${{\delta}^{4}}({p}'-p)$ times the identity matrix in the discrete indices.

It should be remarked that assumption 5, the "ugly technical assumption" is needed because the apparatus of local field theory is not assumed. In particular, there is no assumption that the generators are integrals of the time components of local currents. That assumption would, of course, imply that all generators are distributions in momentum space. Assumption 5 seems to be the minimum necessary to prove the theorem in the broader context considered here.

The proof of the Coleman-Mandula theorem can be separated into a series of intermediate results as follows:

#### Step 1: Show that $A({p}',p)$ is non-zero only when ${p}'=p$.

To this end we define a momentum projection of $A$ by

$\tag{4}{{A}^{[r]}}=\int{{{d}^{4}}}a\ {{e}^{i{{a}_{\mu}}{{P}^{\mu}}}}\,A\,\,{{e}^{-i{{a}_{\mu}}{{P}^{\mu}}}}\,\hat{r}(a)$

Here $r(\delta)$ is an arbitrary real function of momentum, and $\hat{r}(a)$ is its Fourier transform. Since translations are elements of the full symmetry group, ${{A}^{[r]}}$ is also an element of the group generator algebra. By construction, ${{A}^{[r]}}$ has non-zero matrix elements only between states whose momenta differ by a vector in the support of $r$, since

$\tag{5}{{A}^{[r]}}({p}',p)=r({p}'-p)\ A({p}',p)$

The following trivial observation is the basis of the first part of the proof: If an arbitrary momentum vector $\delta$ is added to an on-shell momentum vector $p$, the result, in general, is not on shell. The values of $\delta$ for which $(p+\delta)$ is on shell lie on a hyperboloid in $\delta$ space parallel to the $p$ mass shell hyperboloid, but centered at $(-p)$ rather than at the origin.

To apply this to a physical elastic scattering process $p+q\to {p}'+{q}'$, with all momenta fixed and on shell, choose the support of $r$ to consist of all momenta $\delta$ such that $(p+\delta)$ is on shell but $(q+\delta)$, $({p}'-\delta)$, and $({q}'-\delta)$ are all off shell. This region is the mass hyperboloid in $\delta$ space ${{(p+\delta)}^{2}}=m_{p}^{2}$, but excluding those points that are also on the ${{(q+\delta)}^{2}}=m_{q}^{2}$, ${{({p}'-\delta)}^{2}}=m_{p}^{2}$, or ${{({q}'-\delta)}^{2}}=m_{q}^{2}$ mass hyperboloids. The excluded points form a dimension 2 subset of the dimension 3 hyperboloid ${{(p+\delta)}^{2}}=m_{p}^{2}$, so they are a subset of measure $0$. The origin in $\delta$ space is always an excluded point, i.e., $r(0)=0$.

Since ${{A}^{[r]}}$ is an element of the symmetry generator algebra, it commutes with the $S$ matrix. The two particle matrix element of the symmetry relation $S{{A}^{[r]}}={{A}^{[r]}}S$ is

$\tag{6}\left\langle {p}'{q}' \right|S\,\left({{A}^{[r]}}\left| p \right\rangle \otimes \left| q \right\rangle +\left| p \right\rangle \otimes {{A}^{[r]}}\left| q \right\rangle \right)=\left(\left\langle {{p}'} \right|{{A}^{[r]}}\otimes \left\langle {{q}'} \right|+\left\langle {{p}'} \right|\otimes \left\langle {{q}'} \right|{{A}^{[r]}} \right)S\,\left| pq \right\rangle$

By construction of ${{A}^{[r]}}$, all terms but the first on the left hand side vanish, leaving

$\tag{7}\left\langle {p}'{q}' \right|S\left({{A}^{[r]}}\left| p \right\rangle \otimes \left| q \right\rangle \right)=0$

There are two possible ways this condition can be satisfied: Either the $S$ matrix is zero or the generator ${{A}^{[r]}}$ annihilates the state $\left|p \right\rangle$. While there is nothing amiss if the $S$ matrix vanishes at some isolated values of the kinematic variables $s$ and $t$, by a suitable choice of $q$, ${p}'$, and ${q}'$ for fixed $p$, the kinematic variables $s$ and $t$ can be given a continuous range of values. The assumed analyticity of the $S$ matrix and the existence of non-trivial scattering excludes the vanishing of $S$ as the explanation. Furthermore, the only value of $\delta$ that is generically excluded from the support of $r$ is the origin. The arbitrariness of $r(\delta)$ gives the conclusion of the first part of the proof, that $A({p}',p)$ vanishes when the momenta are unequal.

$\tag{8}A({p}',p)=0\quad \ \text{for}\;\text{all}\quad \ {p}'\ne p$

#### Step 2: Elucidate the structure of $A({p}',p)$ as a distribution in the momentum variables.

By the technical assumption that any matrix element of a group generator is a distribution in momentum space, it must be of a finite sum of derivatives of ${{\delta}^{4}}({p}'-p)$. The physical requirement that ${p}'$ and $p$ be on shell implies that the derivatives must be along directions tangential to the mass hyperboloid.

We can express these requirements for a generator whose matrix elements have derivatives up to order $N$ explicitly (suppressing the spin and particle type indices of the state vectors) as

$\tag{9}{{A}^{(N)}}({p}',p)=\left\langle {{p}'} \right|{{A}^{(N)}}\left| p \right\rangle =\sum\limits_{n=0}^{N}{{{A}^{(n)}}{{(p)}_{{{\mu}_{1}}{{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{n}}}}\frac{\partial}{\partial {{p}_{{{\mu}_{1}}}}}}\frac{\partial}{\partial {{p}_{{{\mu}_{2}}}}}\cdot \cdot \cdot \frac{\partial}{\partial {{p}_{{{\mu}_{n}}}}}\ \ {{\delta}^{4}}({p}'-p)$

The factors ${{A}^{(n)}}{{(p)}_{{{\mu}_{1}}{{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{n}}}}$ are finite matrices whose elements are labeled by the spin and particle type indices. The leading coefficient is isolated by multiple commutation with the momentum operator.

$\tag{10}~{{A}^{(N)}}{{(p)}_{{{\mu}_{1}}{{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}}}{{\delta}^{4}}({p}'-p)=\frac{1}{N!}\left\langle {{p}'} \right|\left[ {{P}_{{{\mu}_{1}}}},\left[ {{P}_{{{\mu}_{2}}}},\cdot \cdot \cdot \ \left[ {{P}_{{{\mu}_{N}}}},{{A}^{(N)}} \right] \right] \right]\left| p \right\rangle$

Note that since ${{P}_{\mu}}$ is an element of the symmetry algebra, the $N$-fold commutator of ${{A}^{(N)}}$ with ${{P}_{\mu}}$ is as well. It is an $N=0$ generator, whose matrix elements are given by Eq. (10).

Mathematically, the requirement that the derivatives are in directions tangent to the mass hyperboloid, so that the generators keep $p$ on shell, may be implemented as

$\tag{11}\left[ {{A}^{(N)}},\ {{P}^{\lambda}}{{P}_{\lambda}} \right]\ \ =\ \ 0$

#### Step 3: Isolate the momentum dependence of the $N=0$ representations.

In this step we factor out the energy-momentum $\delta$ function from the symmetry generators and the $S$ matrix, and are left with finite matrix functions of momentum. We use the same symbols, $S({p}'{q}',p\,q)$, ${{A}^{(0)}}(p)$, and $I(p)$, for the finite matrices corresponding to the indicated quantities.

The following argument shows that the trace of the single particle representation matrix of each $N=0$ generator is just a linear function of the momentum variable.

The two particle matrix element of the commutator of $S$ and ${{A}^{(0)}}$, which vanishes because ${{A}^{(0)}}$ is a symmetry generator, gives

\tag{12}\begin{align} & S({p}'{q}',p\,q)\left({{A}^{(0)}}(p)\ \otimes I(q)+I(p)\ \otimes {{A}^{(0)}}(q) \right) \\ & \quad \quad \quad =\left({{A}^{(0)}}({p}')\ \otimes I({q}')+I({p}')\ \otimes {{A}^{(0)}}({q}') \right)S({p}'{q}',p\,q) \\ \end{align}

Since the $S$ matrix is unitary, the representation matrices on each side of this equality are similarity transforms of each other, and so their traces are equal. This gives the remarkable functional relation

$\tag{13}\frac{Tr\,{{A}^{(0)}}({p}')}{N({{m}_{p}})}+\frac{Tr\,{{A}^{(0)}}({q}')}{N({{m}_{q}})}\quad =\quad \frac{Tr\,{{A}^{(0)}}(p)}{N({{m}_{p}})}+\frac{Tr\,{{A}^{(0)}}(q)}{N({{m}_{q}})}$

Here $N({{m}_{p}})=Tr\,I(p)$ is the number of states in the $p$ representation, which is finite by assumption 2. Since there is only one constraint among the momenta, ${p}'+{q}'=p+q$, this relation can only be satisfied if each term is the same linear function of its argument,

$\tag{14}Tr\ {{A}^{(0)}}(p)=\left({{a}_{\mu}}{{p}^{\mu}}+b \right)N({{m}_{p}})$

The coefficients ${{a}_{\mu}}$ and $b$ depend only on the element of the generator algebra. They are independent of the representations according to which the particles transform and of their momenta.

The distribution structure of symmetry generator representations with $N=0$ is

$\tag{15}{{A}^{(0)}}({p}',p)\ \ =\ \ {{A}^{(0)}}(p)\ {{\delta}^{4}}({p}'-p)$

Since there are no derivatives acting on the $\delta$ function, matrix elements of the commutator of ${{A}^{(0)}}$ with the momentum operator ${{P}_{\mu}}$ have a factor of $({{{p}_{\mu}}'}-{{p}_{\mu}})$ evaluated at ${p}'=p$. Thus all elements of the symmetry generator algebra whose representations have $N=0$ commute with the 4-momentum.

Since $Tr\,{{A}^{(0)}}(p)$ is the same for all representations, we can usefully define reduced symmetry generators $B$ by separating out the momentum and the constant term from $Tr\,{{A}^{(0)}}(p)$.

$\tag{16}{{A}^{(0)}}\ \,=\ \,{{a}_{\mu}}{{P}^{\mu}}+bI+\ B$

The numerical constants ${{a}_{\mu}}$ and $b$ are as in Eq. (14), and depend only on the generator ${{A}^{(0)}}$, and $I$ is the identity operator. The $B$ terms are themselves $N=0$ elements of the symmetry generator algebra. The traces of the coefficient matrices $B(p)$ multiplying the energy-momentum $\delta$ function vanish by construction. Furthermore, because the momenta commute with the $B$ generators and the trace of the commutator of any pair of finite matrices is zero, they close under commutation, and form a subalgebra of the algebra of $N=0$ symmetry generators.

#### Step 4: Show that the reduced $N=0$ operators generate only internal symmetries.

This argument uses the fact that $N=0$ generators commute with the momentum operators to show that they also commute with Lorentz transformations and do not act on spin indices, which is the definition of an internal symmetry. The following demonstration is much clearer than that given in the Coleman-Mandula paper. It is adapted from Volume III of Steven Weinberg's Quantum Field Theory treatise (Weinberg, 2000).

Denote by ${{B}_{\alpha}}$ a finite set of Hermitean generators that spans the reduced $N=0$ symmetry subalgebra. Their commutators are given by the real structure constants $C_{\alpha \beta}^{\gamma}$

$\tag{17}\left[{{B}_{\alpha}},\ {{B}_{\beta}}\right]=i\ C_{\alpha \beta}^{\gamma}{{B}_{\gamma}}$

Denote a finite Lorentz transformation by $\Lambda$ and the operator that effects this transformation on Hilbert space by $U(\Lambda)$. The Lorentz transforms of the generators, $U(\Lambda){{B}_{\alpha}}{{U}^{\mathsf{\dagger}}}(\Lambda)$, are also Hermitean reduced $N=0$ generators and also commute with ${{P}^{\mu}}$. Since we have taken the ${{B}_{\alpha}}$ to be a complete set of generators that span the reduced $N=0$ subalgebra, the transformed generators must be real linear combinations of the ${{B}_{\alpha}}$.

$\tag{18}U(\Lambda){{B}_{\alpha}}{{U}^{\mathsf{\dagger}}}(\Lambda)\ \ =\ \ D_{\alpha}^{{{\alpha}'}}(\Lambda)\,{{B}_{{{\alpha}'}}}$

The $D_{\alpha}^{{{\alpha}'}}(\Lambda)$ matrices form a real representation of the homogeneous Lorentz group. From the fact that the commutation relations of the similarity transformed generators have the same structure constants as the ${{B}_{\alpha}}$, there follows an invariance relation on the 3 index structure constant tensor

$\tag{19}C_{\alpha \beta}^{\gamma}\ \ =\ \ D_{\alpha}^{{{\alpha}'}}(\Lambda)\,D_{\beta}^{{{\beta}'}}(\Lambda)\,D_{{{\gamma}'}}^{\gamma}({{\Lambda}^{-1}})\,\ C_{{\alpha}'{\beta}'}^{{{\gamma}'}}$

and a similar relation follows for the Lie algebra metric $\ {{g}_{\alpha \beta}}\ =\ C_{\alpha \delta}^{\gamma}\,\ C_{\gamma \beta}^{\delta}$

$\tag{20}{{g}_{\alpha \beta}}\ \ =\ \ D_{\alpha}^{{{\alpha}'}}(\Lambda)\,\,D_{\beta}^{{{\beta}'}}(\Lambda)\,\,{{g}_{{\alpha}'{\beta}'}}$

We now reserve the indices $\mu$ and $\nu$ for the momenta and other Greek indices for all other quantum numbers. Since all $N=0$ generators commute with the momenta, the associated structure constants $C_{\mu \beta}^{\gamma}$ and $C_{\alpha \mu}^{\gamma}$vanish, as do ${{g}_{\mu \beta}}$ and ${{g}_{\alpha \mu}}$. This shows that the metric has a block structure: the elements that have an energy-momentum index, ${{g}_{\alpha \nu}}$, ${{g}_{\nu \beta}}$, and ${{g}_{\mu \nu}}$, are all $0$, while the remaining submatrix ${{g}_{\alpha \beta}}$ is positive definite and symmetric. Restricting ourselves to the ${{g}_{\alpha \beta}}$ submatrix, the invariance relation Eq. (20) can be rewritten as

$\tag{21}\left({{g}^{-1/2}}\,D\,{{g}^{1/2}} \right)\ \ {{\left({{g}^{-1/2}}\,D\,{{g}^{1/2}} \right)}^{\Tau}}=\ \ 1$

This shows that ${{g}^{-1/2}}\,D(\Lambda)\,\,{{g}^{1/2}}$ is a real orthogonal unitary representation of the homogeneous Lorentz group. However, because the Lorentz group is non-compact, it has no non-trivial finite dimensional unitary representations.

We arrived at this "contradiction" by making the tacit assumption that the representation $D(\Lambda)$ in Eq. (18) was non-trivial. In fact, we have just inferred that it must be the identity representation $D(\Lambda)=1$. Therefore, the entire algebra of $N=0$ generators ${{B}_{\alpha}}$ commutes with Lorentz transformations.

Consider the action of ${{B}_{\alpha}}$ on a single particle state with momentum $p$ and take the Lorentz transformation $\Lambda$ to be an arbitrary element of the little group of $p$, the subgroup of Lorentz transformations that leave the momentum $p$ of the state unchanged. The ensemble of $U(\Lambda)$ transformations effect arbitrary rotations of the spin of the state. The fact that all reduced $N=0$ generators ${{B}_{\alpha}}$ commute with all $U(\Lambda)$, Eq. (18) with $D(\Lambda)=1$, means that that they act trivially on the spin degrees of freedom. To sum up, the $N=0$ sector of the symmetry algebra is the direct sum of the algebra of the generators of translations, spanned by the 4-momenta ${{P}_{\mu}}$, and the algebra of internal symmetries, which are momentum independent and do not act on the spin indices.

#### Step 5: Show that there are no symmetry generators with $N>1$ derivatives.

To show that there are no generators with more than 1 derivative, consider the construction of ${{A}^{(N)}}{{(p)}_{{{\mu}_{1}}{{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}}}{{\delta}^{4}}({p}'-p)$ in Eq. (10). It is the single particle matrix element of an $N=0$ symmetry generator. From Step 3, this means that the trace of ${{A}^{(N)}}{{(p)}_{{{\mu}_{1}}{{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}}}$ is a linear function of the momentum

$\tag{22}Tr\ {{A}^{(N)}}{{(p)}_{{{\mu}_{1}}{{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}}}={{a}_{\nu {{\mu}_{1}}{{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}}}{{p}^{\nu}}+{{b}_{{{\mu}_{1}}{{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}}}$

and Eq. (9) shows that ${{A}^{(N)}}{{(p)}_{{{\mu}_{1}}{{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}}}$ is completely symmetric in its indices, and so the ${{a}_{\nu {{\mu}_{1}}{{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}}}$ coefficients are symmetric in the second through ${{(N+1)}^{\mathsf{st}}}$ indices $\{{{\mu}_{1}}{{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}\}$.

From Eqs. (10) and (14), the matrix trace of the multiple commutator of ${{P}^{\lambda}}{{P}_{\lambda}}$ and $N-1$ factors of ${{P}_{{{\mu}_{i}}}}$ with ${{A}^{(N)}}$ is

\tag{23}\begin{align} & \frac{1}{N!}Tr\ \left\langle {p}'\left| \left[ {{P}^{\lambda}}{{P}_{\lambda}},\,\,\left[ {{P}_{{{\mu}_{2}}}},\cdot \cdot \cdot \ \left[ {{P}_{{{\mu}_{N}}}},{{A}^{(N)}} \right] \right] \right] \right|p \right\rangle \\ & \quad \quad \quad \quad \quad = 2 {{p}^{\lambda}}\left({{a}_{\nu \lambda {{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}}}{{p}^{\nu}}+{{b}_{\lambda {{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}}} \right)\ {{\delta}^{4}}({p}'-p)\end{align}

This must vanish for any value of $p$ because it is the commutator of ${{P}^{\lambda}}{{P}_{\lambda}}$ with an element of the symmetry algebra, which keeps states on shell. Thus the ${{b}_{\lambda {{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}}}$ coefficients are zero and the ${{a}_{\nu \lambda {{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}}}$ coefficients must be antisymmetric in the $\lambda$ and $\nu$ indices. However, the latter is impossible because it is in contradiction with the complete symmetry of ${{a}_{\nu \lambda {{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}}}$ in the $\{\lambda {{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}\}$ indices. Explicitly,

\tag{24}\begin{align} {{a}_{\nu \lambda {{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}}} & =-\ {{a}_{\lambda \nu {{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}}}=-\,{{a}_{\lambda {{\mu}_{2}}\nu \cdot \cdot \cdot {{\mu}_{N}}}} = +\,{{a}_{{{\mu}_{2}}\lambda \nu \cdot \cdot \cdot {{\mu}_{N}}}} \\ & \quad \quad = +\,{{a}_{{{\mu}_{2}}\nu \lambda \cdot \cdot \cdot {{\mu}_{N}}}}=-\,{{a}_{\nu {{\mu}_{2}}\lambda \cdot \cdot \cdot {{\mu}_{N}}}}=-\,{{a}_{\nu \lambda {{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}}}\end{align}

Thus there are no symmetry generators with more than one momentum derivative acting on the momentum conservation $\delta$ function.

Lest the reader fear that we have excluded all generators, note that the above argument required that the coefficient ${{a}_{\nu {{\mu}_{1}}{{\mu}_{2}}\cdot \cdot \cdot {{\mu}_{N}}}}$ have at least three indices. That is, the foregoing says nothing about the cases $N=0$ or $N=1$.

#### Step 6: Identify the Poincaré transformation generators.

The $N=1$ generators have the form

$\tag{25}{{A}^{(1)}}({p}',p)=\left(\left(a_{\nu \mu}^{(1)}{{p}^{\nu}}+b_{\mu}^{(1)} \right)\frac{\partial}{\partial {{p}_{\mu}}}+A_{subleading}^{(1)}(p) \right)\ \ {{\delta}^{4}}({p}'-p)$

As shown in the previous Step, the coefficient $a_{\nu \mu}^{(1)}$ is antisymmetric in its indices and the coefficient $b_{\mu}^{(1)}$ vanishes. The first term, $\ a_{\nu \mu}^{(1)}{{p}^{\nu}}(\partial /\partial {{p}_{\mu}}){{\delta}^{4}}({p}'-p)$, is just the geometric part of a Lorentz generator. If we subtract the full Lorentz transformation generator from ${{A}^{(1)}}({p}',\ p)$

$\tag{26}{{A}^{(1)}}({p}',p)\ \ -\ \ a_{\nu \mu}^{(1)}\,{{M}^{\nu \mu}}=\ \ {{{A}'}^{(0)}}\ {{\delta}^{4}}({p}'-p)$

we obtain yet another $N=0$ element of the generator algebra. The part of the Lorentz generator $a_{\nu \mu}^{(1)}\,{{M}^{\nu \mu}}$ that acts on the spin indices is included in ${{{A}'}^{(0)}}$. However, in Step 4 we showed that any $N=0$ generator was the sum of a translation generator and an internal symmetry generator. Even though we have nominally incorporated the spin part of $a_{\nu \mu}^{(1)}\,{{M}^{\nu \mu}}$, into ${{{A}'}^{(0)}}$, the result of the analysis in Step 4 ensures that the operator so obtained does not affect the spin indices.

This completes the demonstration that the most general symmetry algebra of a Poincaré invariant theory with non-trivial scattering is a direct sum of the Lorentz algebra $(N=1)$ generated by the angular momentum and boost generators ${{M}_{\mu \nu}}$ , the algebra of translation generators $(N=0)$ generated by the linear momentum ${{P}_{\nu}}$, and an algebra of pure internal symmetry transformation generators $(N=0)$.

## Necessity of all the assumptions

The physics assumptions used in the proof of the Coleman-Mandula theorem are relativistic invariance and the existence of scattering. Both are needed. What fails in trying to implement a hybrid symmetry is not the possibility of grouping particles into representations of hybrid symmetry, but rather the conservation of the hybrid quantum numbers in scattering reactions or particle decays. To make the need for both assumptions clear, let us review examples of models that do display hybrid symmetries.

There is a well-known approximate hybrid symmetry in nuclear physics. It is the spin-isospin symmetry in the shell model introduced by Wigner (Wigner, 1937). The symmetry is represented on single nucleon states by arbitrary unitary mixing of the four basis states $\left| p\uparrow \right\rangle$, $\left| p\downarrow \right\rangle$, $\left| n\uparrow \right\rangle$, and $\left| n\downarrow \right\rangle$. The fact that in the real world this SU(4) symmetry group holds only approximately is not the point. It is a consistent model for the excitation level spectrum and, more relevantly, for nuclear decays. This existence of this example shows that there can be no analogue of the Coleman-Mandula theorem for non-relativistic systems.

There are also examples of relativistic theories with exact hybrid symmetries that only apply to the classification of bound states but not to the scattering of the bound states, but they are far less familiar (and less useful) than nuclear spin-isospin symmetry. The O(4) symmetry of the hydrogen atom carries over to the relativistic version of the same bound state problem: the Bethe-Salpeter equation for two massive scalar particles bound by the exchange of a massless scalar. Cutkosky solved this problem by Wick rotating the momenta to Euclidean space and mapping the relative momentum to the surface of a 4-sphere. The resulting equation was invariant under rotations on the 4-sphere, and so had the same O(4) symmetry as the non-relativistic hydrogen atom (Cutkosky, 1954).

In Minkowski space one may follow the same procedure, but there is a subtlety in identifying the physical symmetry transformations. It is necessary to enlarge the bound state symmetry group to O(4,C), the complex generalization of O(4). In any frame there is an O(4) subgroup of O(4,C) that is represented unitarily on the bound state wave functions. However, in each frame this physical O(4) subgroup is different. This structure is a reflection of the fact that the Lorentz group has no finite dimensional unitary representations. When each of the six generators of the physical O(4) subgroup in a given frame is boosted, it becomes a complex linear combination of the original set. The original set of six generators is thus effectively augmented by six new ones, each being $i$ times one of the original set. This set of twelve generators is closed under Lorentz transformations, showing that the full bound state symmetry group is O(4,C) (Mandula, 1969).

## Supersymmetry and the Haag-Łopuszański-Sohnius theorem

The Coleman-Mandula theorem deals only with symmetries expressed in terms of Lie groups, whose structure is described by the commutation relations between their generators. There are symmetries that cannot be so expressed, however. A class of such symmetries, called supersymmetries, involve transformations that change bosons into fermions and vice versa. These symmetries were discovered several years after the Coleman-Mandula theorem was proved.

Supersymmetries were first discovered in the context of string theory. Gervais and Sakita formulated an action for a theory with fermionic as well as bosonic variables, and observed that their action was invariant under set of transformations that interchanged the fermionic and bosonic world sheet fields (Gervais and Sakita, 1971). This was effectively a supersymmetry of a two dimensional field theory. A couple of years later, Wess and Zumino succeeded in extending the idea to four dimensional field theory (Wess and Zumino, 1974a). In a subsequent paper, they traced the reason that the Coleman-Mandula theorem does not apply to supersymmetries to the fact that the generators of supersymmetries are fermionic operators, and their structures are expressed by anticommutation relations (Wess and Zumino, 1974b).

Nonetheless, the possible supersymmetries are quite as restricted as ordinary symmetries. The restrictions on the possible supersymmetries were found by Haag, Łopuszański, and Sohnius, and is given by the theorem that bears their names (Haag et al., 1975). The proof of the Haag-Łopuszański-Sohnius theorem follows the same strategy as that of the Coleman-Mandula theorem. That is one begins with a completely general supersymmetry generator and, step by step, finds restrictions on the allowed generators and their anticommutators.

Unlike internal symmetries, supersymmetries are not Lorentz scalars, and their Lorentz transformation properties are relevant. Denote the generators of the commuting SU(2) factors of the Lorentz group by $J_{i}^{\pm}$ and denote a generic supersymmetry generator transforming according to the $({{j}^{+}},\,{{j}^{-}})$ representation of the Lorentz group by ${{Q}_{ar}}$. The index $a$ labels the Lorentz component $({{m}^{+}},\,\,{{m}^{-}})$ and the index $r$ distinguishes different generators with the same Lorentz transformation properties. The Hermitean adjoint operator $Q_{ar}^{\mathsf{\dagger}}$ transforms according to the conjugate representation $({{j}^{-}},\,{{j}^{+}})$.

In terms of the foregoing, the conclusions of the Haag-Łopuszański-Sohnius theorem are as follows:

1. The only possible fermionic supersymmetry generators ${{Q}_{ar}}$transform like the $(1/2, 0)$ representation and their conjugates ${{\bar{Q}}_{ar}}$ transform like $(0,1/2)$ in the standard basis. The linear relation between the conjugates ${{\bar{Q}}_{ar}}$ and the Hermitean adjoints of the ${{Q}_{ar}}$ is $\tag{27}{{\bar{Q}}_{ar}}=\ \ {{e}_{a{a}'}}\,Q_{{a}'r}^{\,\mathsf{\dagger}}$ where $e$ is an antisymmetric $2 \times 2$ matrix. The index $a$ takes the values $\pm 1/2$, and we fix the phase of $e$ by choosing ${\,{e}_{+{\scriptscriptstyle {}^1\!/\!{}_2}\,-{\scriptscriptstyle {}^1\!/\!{}_2}}}=+1$. The index $r$ distinguishes between different multiplets of supercharges.
2. The supercharges can be chosen so that their anticommutation relations are \tag{28}\begin{align} & \left\{{{Q}_{ar}},Q_{bs}^{\,\mathsf{\dagger}}\right\}=2\,{{\delta}_{rs}}\,\sigma _{ab}^{\,\mu}\,{{P}_{\mu}} \\ & \left\{{{Q}_{ar}},Q_{bs}^{\phantom{\,\mathsf{\dagger}}}\right\}={{e}_{ab}}\,{{Z}_{rs}} \\ \end{align} The matrices ${{\sigma}^{\,i}}\ (i=1,\,2,\,3)$ are the Pauli spin matrices, ${{\sigma}^{\,0}}$ is the $2 \times 2$ unit matrix, and $e$ is as above.
3. The supersymmetry generators ${{Q}_{ar}}$ and $Q_{as}^{\,\mathsf{\dagger}}$ commute with the translation generators ${{P}_{\mu}}$. The bosonic operators ${{Z}_{rs}}$ are antisymmetric in their indices and are "central charges" of the symmetry algebra, that is, they are symmetry generators that commute with all elements of the symmetry algebra. Therefore, the ${{Z}_{rs}}$ generators are singlets under Lorentz transformations and under the internal symmetry group.

Like the proof of the Coleman-Mandula Theorem, the proof of the Haag-Łopuszański-Sohnius theorem proceeds in several steps, as follows:

#### Step 1: Determine the Lorentz transformation properties of the supercharges.

Supersymmetry generators are fermionic charges, so they transform according to representations $({{j}^{+}},\,{{j}^{-}})$, where one of the labels ${{j}^{\,\pm}}$ is an integer and the other a half integer. In the usual basis, where ${{m}^{\pm}}$ label the elements of the representation by their eigenvalues of $J_{3}^{\,\pm}$, the adjoint of the supercharge $Q_{({{m}^{+}},\,{{m}^{-}})}^{({{\,j}^{\,+}},\,{{j}^{\,-}})}$ transforms exactly like the $(-{{m}^{-}},-{{m}^{+}})$ element of the $({{j}^{-}},\,{{j}^{+}})$ representation. We denote it by

$\tag{29}\bar{Q}_{(-{{m}^{\,-}},\,-{{m}^{\,+}})}^{\,({\,\,{j}^{\,-}}\,,\,\,{\,\,{j}^{\,+}})}=\ \ {{\left(Q_{({{m}^{\,+}},\,{{m}^{\,-}})}^{(\,{{j}^{\,+}},\,\,{{j}^{\,-}})} \right)}^{\mathsf{\dagger}}}$

From the rule for addition of ordinary angular momentum, one can show that the anticommutator of an extremal element of $Q_{({{m}^{\,+}},\,{{m}^{\,-}})}^{(\,{{j}^{\,+}},\,\,{{j}^{\,-}})}$ with its adjoint transforms as indicated by

$\tag{30}\left\{ Q_{({{j}^{\,+}},\,{-{j}^{\,-}})}^{(\,{{j}^{\,+}},\,\,{{j}^{\,-}})},\,\,\bar{Q}_{({{j}^{-}},\,{-{j}^{\,+}})}^{(\,{{j}^{\,-}},\,\,{{j}^{\,+}})} \right\}=B_{({{j}^{\,+}}+{{j}^{\,-}},\ -{{j}^{\,+}}-{{j}^{\,-}})}^{({{j}^{\,+}}+{{j}^{\ -}},\ \,{{j}^{\,+}}+\,{{j}^{\,-}})}$

The operator $B$ is an ordinary (bosonic) symmetry generator, since both of its representation labels are positive half integers. Furthermore, $B$ cannot vanish, because the anticommutator of any operator with its adjoint can only vanish if the operator vanishes itself. However, the Coleman-Mandula theorem shows that the only bosonic elements of the symmetry algebra are internal symmetries which are Lorentz scalars, the 4-momentum ${{P}_{\mu}}$, transforming like $(1/2, 1/2)$, and the generators of pure Lorentz transformations ${{M}_{\mu \nu}}$, which transform according to the reducible representation $(1,0) \oplus (0,1)$. The only solution to these constraints on the transformation properties of $B$ is that ${\,{j}^{+}}\!+{{j}^{-}}$ must be $1/2$, and so $B$ is proportional to the 4-momentum. Therefore, one of ${\,{j}^{\pm}}$ is $1/2$ and the other is $0$, and we conclude that the only allowed fermionic generators must transform like $(1/2, 0)$ or $(0, 1/2)$.

#### Step 2: Verify that the $\{{{Q}_{ar}},Q_{bs}^{\,\mathsf{\dagger}}\}$ anticommutator satisfies the first equality of Eq. (28).

The first anticommutator in Eq. (28) above follows from the Coleman-Mandula theorem: Since ${{P}_{\mu}}$ is the only allowed $(1/2, 1/2)$ symmetry generator, and the ${{\sigma}^{\,\mu}}$ matrices are the coefficients expressing the elements of a 4-vector in bi-spinor notation, the form of that anticommutator must be

$\tag{31}\left\{ {{Q}_{ar}},Q_{bs}^{\,\mathsf{\dagger}} \right\}\ \ =\ \ 2\,{{C}_{rs}}\sigma _{ab}^{\,\mu}\,{{P}_{\mu}}\ \ \equiv \ \ 2\,{{C}_{rs}}{{P}_{ab}}$

The matrix $C$ is positive definite, as can be seen as follows: Take $a=b$; the absence of massless particles implies that the diagonal elements of ${{P}_{ab}}$ are both positive. Contract the $r$ and $s$ indices above with an arbitrary row vector and its conjugate column vector. This gives the anticommutator of a supercharge and its adjoint, which must be positive. Since $C$ is positive definite, the given supercharges may be replaced by the real linear combinations given by the inverse square root of $C$. These new supercharges satisfy the first anticommutator of Eq. (28).

#### Step 3: Show that the supercharges commute with 4-momenta.

The argument that the supercharges commute with 4-momentum explicitly uses the transformation properties of the supercharges. The commutator of the momentum operator, transforming like $(1/2, 1/2)$, with a supercharge, transforming like $(1/2, 0)$, must be proportional to a conjugate supercharge, transforming like $(0,1/2)$, because $(0,1/2) \otimes (1/2, 1/2) = (1/2,0) \oplus (1/2,1)$, and there are no supersymmetry generators transforming like $(1/2,1)$. The detailed form of the relation is fixed by the Lorentz transformation properties of the supercharges and the 4-momentum:

\tag{32}\begin{align} \left[{{P}_{ab}},\,{{Q}_{cr}^{\phantom{\mathsf{\dagger}}}}\right] & = +{{C}_{rs}}\,{{e}_{ac}}\,Q_{bs}^{\mathsf{\dagger}} \\ \left[{{P}_{ab}},\,Q_{cs}^{\mathsf{\dagger}}\right] & = -C_{sr}^{*}\,{{e}_{bc}}\,{{Q}_{ar}} \\ \end{align}

Using the above it is straightforward to compute the triple commutator-anticommutator

$\tag{33}\left[ {{P}_{ab}},\,\left[ {{P}_{cd}},\left\{ {{Q}_{er}},\,Q_{fs}^{\mathsf{\dagger}} \right\} \right] \right]\ \ =\,\ 2\,{{\delta}_{rs}}\left[ {{P}_{ab}},\,\left[ {{P}_{cd}},\,{{P}_{ef}} \right] \right]$

The right hand side is zero for any values of the indices because translations form an Abelian group. However, the left hand side is not identically zero. To pick one example, Eq. (32) gives

$\tag{34}\sum\limits_{r}{\ \left[ {{P}_{+{\scriptscriptstyle {}^1\!/\!{}_2}\,+{\scriptscriptstyle {}^1\!/\!{}_2}}}\,,\,\left[ {{P}_{+{\scriptscriptstyle {}^1\!/\!{}_2}\,+{\scriptscriptstyle {}^1\!/\!{}_2}}}\,,\left\{ {{Q}_{-{\scriptscriptstyle {}^1\!/\!{}_2}\,r}},\,Q_{-{\scriptscriptstyle {}^1\!/\!{}_2}\,r}^{\mathsf{\dagger}} \right\} \right] \right]}\ \ =\,\ -4\left(Tr{{C}^{\mathsf{\dagger}}}C\right)\,{{P}_{+{\scriptscriptstyle {}^1\!/\!{}_2}\,+{\scriptscriptstyle {}^1\!/\!{}_2}}}$

These the two evaluations agree only if $Tr{\,{C}^{\mathsf{\dagger}}}C$, and hence $C$, vanishes. Thus we conclude that ${{P}_{\mu}}$ commutes with ${{Q}_{cr}}$ (and $Q_{cs}^{\mathsf{\dagger}}$).

$\tag{35}\left[{{P}_{ab}},\,\,{{Q}_{cr}^{\phantom{\mathsf{\dagger}}}}\right]\ \ =\ \,\left[{{P}_{ab}},\,Q_{cs}^{\mathsf{\dagger}}\, \right]\ \ =\,\ 0$

#### Step 4: Show that the $Z$ generators are central charges.

Note that the $Z$ operators commute with the 4-momenta because, per Eq. (28), they are comprised of two supercharges, and we have just shown that supercharges commute with the 4-momenta.

$\tag{36}\left[{{P}_{ab}},\,{{Z}_{rs}^{\phantom{\mathsf{\dagger}}}} \right]=\frac{1}{2}{{e}_{cd}}\,\left[ {{P}_{ab}},\left\{ {{Q}_{cr}},\,{{Q}_{ds}^{\phantom{\mathsf{\dagger}}}} \right\} \right]=0$

To show that the $Z$ operators commute with the $Q_{bs}^{\mathsf{\dagger}}$ supercharges, we use their definitions per Eq. (28) and evaluate

$\tag{37}\left[Q_{ct}^{\mathsf{\dagger}},\,\,{{Z}_{rs}} \right]=\frac{1}{2}{{e}_{ab}}\left[ Q_{ct}^{\mathsf{\dagger}},\left\{ {{Q}_{ar}^{\phantom{\mathsf{\dagger}}}},{{Q}_{bs}} \right\} \right]={{e}_{ab}}\left({{\delta}_{tr}}\left[ {{P}_{ac}},\,{{Q}_{bs}^{\phantom{\mathsf{\dagger}}}} \right]+{{\delta}_{ts}}\left[ {{P}_{bc}},{{Q}_{ar}^{\phantom{\mathsf{\dagger}}}} \right] \right)$

Both terms on the right vanish, again because the 4-momenta commute with all supercharges. Thus the $Z$ charges commute with the conjugate supercharges.

$\tag{38}\left[Q_{ct}^{\mathsf{\dagger}},\,\,{{Z}_{rs}} \right]=0$

To show that the $Z$ operators commute with the ${{Q}_{a\,r}}$ supercharges, expand the multiple commutator-anticommutator

$\tag{39}\left\{Q_{ct}^{\mathsf{\dagger}},\left[ {{Q}_{au}^{\phantom{\mathsf{\dagger}}}},\,\,{{Z}_{rs}} \right] \right\}=\left[ \left\{ Q_{ct}^{\mathsf{\dagger}},{{Q}_{au}} \right\},{{Z}_{rs}} \right]-\left\{ \left[ Q_{ct}^{\mathsf{\dagger}},{{Z}_{rs}} \right],{{Q}_{au}} \right\}$

Again, both terms on the right vanish. The first is the commutator of a momentum operator with a $Z$ operator (see Eq. (31)), which vanishes, and the second includes the commutator of a $Z$ operator with an adjoint supercharge, which we have just seen vanishes.

Since ${{Z}_{rs}}$ is the sum of products of two generators transforming according to the $(1/2, 0)$ representation of the Lorentz group, it must transform according to the $(0,0)$ or the $(1,0)$ representation, or a linear combination thereof. However, the Coleman-Mandula theorem shows that the only generators of the bosonic symmetry algebra that transform like $(1,0)$ are the Lorentz generators themselves, which do not commute with the 4-momenta. Therefore the $Z$ operators are Lorentz scalars and their commutators with the ${{Q}_{ar}}$ supercharges must transform according to the $(1/2, 0)$ representation. Since the only such symmetry generators are the supercharges themselves, the commutator must be a linear combination of those supercharges

$\tag{40}\left[{{Q}_{au}^{\phantom{\mathsf{\dagger}}}},\,\,{{Z}_{rs}} \right]=C_{uv}^{(rs)}{{Q}_{av}}$

and so the multiple commutator is

$\tag{41}\left\{Q_{ct}^{\mathsf{\dagger}},\left[ {{Q}_{au}^{\phantom{\mathsf{\dagger}}}},\,\,{{Z}_{rs}} \right] \right\}=\left\{ Q_{ct}^{\mathsf{\dagger}},C_{uv}^{(rs)}\,{{Q}_{av}} \right\}=2\,C_{ut}^{(rs)}{{P}_{ac}}\,$

We have seen above that this multiple commutator vanishes, which can only be true if the coefficients in the above linear combination, $C_{ut}^{(rs)}$, are all zero. This implies that the commutators of the ${{Z}_{rs}}$ operators with the ${{Q}_{at}}$ supercharges vanish.

To summarize the analysis so far, including the Hermitean conjugates of the foregoing, we have shown

\tag{42}\begin{align} \left[{{Q}_{at}^{\phantom{\mathsf{\dagger}}}},\,\,{{Z}_{rs}}\right] & = \left[{{Q}_{at}},\,\,Z_{rs}^{\mathsf{\dagger}}\right]=\left[Q_{bt}^{\mathsf{\dagger}},\,\,{{Z}_{rs}}\right]=\left[Q_{bt}^{\mathsf{\dagger}},\,\,Z_{rs}^{\mathsf{\dagger}}\right] \\ & = \left[{{Z}_{rs}^{\phantom{\mathsf{\dagger}}}},\,\,{{Z}_{tu}}\right]=\left[{{Z}_{rs}},\,\,Z_{tu}^{\mathsf{\dagger}}\right]=\left[Z_{rs}^{\mathsf{\dagger}},\,\,Z_{tu}^{\mathsf{\dagger}}\right]=0 \\ \end{align}

This shows that the $Z$ operators are central charges of the pure supersymmetry algebra.

Next consider the commutators of arbitrary bosonic internal symmetry generators ${{B}_{t}}$ with the supercharges and the $Z$ operators formed from them. The supercharges form a representation of the internal symmetry group, i.e.,

$\tag{43}\left[{{B}_{t}},\,\,{{Q}_{ar}^{\phantom{\mathsf{\dagger}}}}\right]\ \ =\ \ D_{r{r}'}^{\,t}\,{{Q}_{a{r}'}}$

where the matrix $D^{\,t}$ is a representation of ${{B}_{t}}$. Therefore the commutator of an internal symmetry with a $Z$ operator is

$\tag{44}\left[{{B}_{t}},\,\,{{Z}_{rs}^{\phantom{\mathsf{\dagger}}}}\right]\ \ =\ \ \left(D_{r{r}'}^{\,t}\,{{\delta}_{s{s}'}}+{{\delta}_{r{r}'}}D_{s{s}'}^{\,t}\right)\,{{Z}_{{r}'{s}'}}$

The $\,{{Z}_{rs}}$ charges are ordinary bosonic symmetries, and Eq. (44) says that they comprise an invariant subalgebra of the full internal symmetry algebra. The $\,{{Z}_{rs}}$ charges all commute with each other, so it is an Abelian invariant subalgebra. Since the $\,{{Z}_{rs}}$ charges are a subset of the ${{B}_{t}}$ charges, the coefficients in Eq. (44) are a subset of the structure constants of the internal symmetry group. If we lower the $t$ index with the group metric, we can write the structure constants in the standard form ${{f}_{t,rs,{r}'{s}'}}$. The index $t$ runs over the all the generators of the internal symmetry algebra while the oddly labeled indices $rs$ and ${r}'{s}'$ only run over the elements of the Abelian invariant subalgebra. The 3-index structure constant array is antisymmetric under exchange of any pair of indices, and the fact that the $\,{{Z}_{rs}}$ operators commute with each other means that the structure constants ${{f}_{rs,{r}'{s}',t}}$ are zero for all values of the indices $rs$, ${r}'{s}'$, and $t$. The antisymmetry of $f$ then implies that the ${{f}_{t,rs,{r}'{s}'}}$, which are the coefficients in Eq. (44), vanish. This shows that the $\,{{Z}_{rs}}$ charges are central charges of the full symmetry algebra, and so completes the proof of the Haag-Łopuszański-Sohnius theorem.