# Intuitive Meaning of the Coleman-Mandula Theorem

• A
Staff Emeritus
This theorem is summarized here: https://en.wikipedia.org/wiki/Coleman–Mandula_theorem

I sort of understand the mathematical content of the theorem, that

[Any reasonable theory with a mass gap] can only have a Lie group symmetry which is always a direct product of the Poincaré group and an internal group.
But what I don't understand is, intuitively, what sort of possibilities are ruled out. I've heard it said that flavor conservation laws such as conservation of lepton number and baryon number cannot be exact. Even if there are no single Feynman diagrams that show a violation of conservation of such quantities, they are not expected to hold in non-perturbative field theory. There's also the "no-hair" claim about black holes--black holes have definite charges and angular momentum and so-forth, but don't conserve anything else such as baryon number. Are these all related issues involving non-conservation?

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Haelfix
The Coleman Mandula theorem doesn't really say anything about whether global symmetries are exact in field theories. Provided of course that those gauge theories are allowed by the theorem and are of the trivial type (Spacetime * Internal).

The theorem is a kinematic consequence of the fact that the Poincare group has just enough symmetry to make for interesting theories, but anymore overconstrains the SMatrix and violates cherished physical postulates (like analyticity of scattering amplitudes).

Of course the list of loopholes to the theorem are large and interesting, but they tend to be of the same form. Namely that they preclude any mixing of (suitably generalized spacetime symmetries and internal gauge symmetries) at the algebraic level.

king vitamin
Staff Emeritus
Of course the list of loopholes to the theorem are large and interesting, but they tend to be of the same form. Namely that they preclude any mixing of (suitably generalized spacetime symmetries and internal gauge symmetries) at the algebraic level.
What, intuitively does "mixing" mean? Is it in the same sense as spin and orbital angular momentum mixing?

Haelfix
What, intuitively does "mixing" mean? Is it in the same sense as spin and orbital angular momentum mixing?
In which sense do you mean here? Like in the nonrelativistic quantum mechanics of a bound state?

I mean mixing in the sense that if there is a conserved quantity in your relativistic theory (with a mass gap) that transforms as a tensor under the Lorentz group, than that quantity is either the energy momentum vector P, the generators of Lorentz transformations J or completely internal scalars that commute with the former. The sort of 'hybrid' symmetries this precludes are probably not very familiar, but they were frequently written down in the early days of QFT when people were investigating QCD.

Alternatively you could point out that this also precludes 'unified' theories (with some caveats) where the full lie algebra of nature is enlarged to incorporate gravity, and where the breaking pattern is nontrivial.

Hans de Vries
Gold Member
This theorem is summarized here: https://en.wikipedia.org/wiki/Coleman–Mandula_theorem

Any reasonable theory with a mass gap] can only have a Lie group symmetry which is always a direct product of the Poincaré group and an internal group
The "direct product" in the theorem is is an unnecessary over-constraint. It is sufficient that the Poincaré group generators commute with the internal group generators.

A counter example is the maximal group structure of a universal fermion field with 4 spinors.
$$SO(4)\otimes SO(4) ~~~\cong~~~ (S^3_j\times S^3_i)\otimes(S^3_j\times S^3_i)$$
This is a direct product of two ##SO(4)\cong S^3_j\times S^3_i## groups. This is equivalent with four triplets of generators. Each triplet commutes with any other triplet. The right hand side can be reorganized as:
$$(S^3_j\otimes S^3_j)\times(S^3_i\otimes S^3_i) ~~~=~~~ (\mbox{Pointcaré generators})\times(\mbox{internal generators})$$
The Poincaré generators commute with the internal generators without the need for a direct product.

The separation of the Poincaré generators and the internal generators becomes clear if we organize all bilinear field components in a matrix (This entire matrix can be calculated with a single matrix multiplication).

The Poincaré generators define the columns and the internal generators define the rows.

With a 4 spinor fermion field we can define each fermion so that it has the correct coupling (given by the matrix above) to all electroweak vector bosons while also coming in three generations. So neutrinos are left-handed, transform like V-A particles giving rise to maximal parity violation. Anti neutrinos are right-handed. (Anti-)quarks have fractional charges ##\pm1/3## and ##\pm2/3## while, at the same time, have the correct coupling to the other vector bosons (for sin##^2\theta=0.25##)

With four spinors we can define more fermions than there are in the Standard Model but it turns out that all Standard Model fermions are the eigen functions of a single Operator expression with only the charge of the particle as input.

see my work here: https://thephysicsquest.blogspot.com/ and

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Hans: why do you pair off S3 factors from different SO(4)s in order to get the Poincare group and the internal symmetry group? What is the significance of rearranging the order of is and js, in the expression (S3xS3)x(S3xS3)?

Hans de Vries
Gold Member
Hans: why do you pair off S3 factors from different SO(4)s in order to get the Poincare group and the internal symmetry group? What is the significance of rearranging the order of is and js, in the expression (S3xS3)x(S3xS3)?
That's a good question Mitchell, thank you for asking it.

With regards to the thread's title: It will show that the "direct product" in the Coleman Mandula theorem is already circumvented at the single spinor level.

If we combine the Pauli Matrices ##\sigma^i## (acting on a single spinor) with the complex conjugate operator '*' (what we do all the time) then the imaginary and real parts become independent and the group increases from SU(2) to SO(4) . We obtain therefor an extra generator triplet. So, combining Pauli matrices with '*' gives additional generators which are generally not recognized as such.

- The extra generator triplet is the ##S_i^3## in ##SO(4)\cong S^3_j\times S^3_i##
- While in fact already part of the Standard Model, these generators are totally overlooked.
- The extra triplet ##S_i^3## can be associated with one of the "internal" triplets of the Standard Model.

------------------------ The two generator triplets of of SO(4) -------------------------

Let's express the complex SU(2) generators in the equivalent SO(4) form with the following standard substitutions:
$$a+ib ~\longleftrightarrow~ \left(\begin{array}{rr} a & -b \\ b & a \end{array}\right) ~~~~~~~~~~ * ~\longleftrightarrow~ \left(\begin{array}{rr} 1 & 0 \\ 0 & -1 \end{array}\right)$$

Note that the colored rectangles correspond with the three SO(3) generators for x, y and z. The standard Pauli matrices are ##\sigma^1##, ##\sigma^2## and ##\sigma^3##. The indices x, y and z in the ##\mathsf{j_x,j_y,j_z}##, the ##\mathsf{i_x,i_y,i_z}## and ##\sigma^x,\sigma^y,\sigma^z## are chosen so that they correspond with the SO(3) rotation matrices.

---------------------- The generators of local (internal) rotation --------------------

We see that the ##\mathsf{j_x,j_y,j_z}## are just the standard Lorentz rotation generators. In 3D jargon: They rotate the spinor in World coordinates.

If we do the calculations then it turns out that the generator triplet ##\mathsf{i_x,i_y,i_z}## rotates the spinor in the spinor's own, Local, coordinates. This corresponds with the rotation of ,for instance, an airplane around its Roll, Pitch and Yaw axis.

If we look at the triplet ##\mathsf{i_x,i_y,i_z}## in Pauli's notation then we see that we already know them.

The ##i\sigma^o## rotates the spinor around its own axis by 180##^o## regardless of the direction the spinor points in. We also know that the generators ##\sigma^2*## and ##i\sigma^2*## turn a spinor into the opposite direction, regardless of the direction it has. This corresponds with a 180##^o## degrees rotation around the other local axes. Parity inversion as defined by ##\sigma^2*## is an inversion with respect to a line, not a point as you might expect. Finally one may want to check that ##i\sigma^o##, ##\sigma^2*## and ##i\sigma^2*## behave like an anti-commuting triplet (##*## operates to the right).

-------------- The spinor base states in the real representation ---------------

The difference in behavior between ##S_j^3## and ##S_i^3## is also why the two ##S_j^3## combine to the Poincaré generators and the two ##S_i^3## combine to the "internal" generators. Lets have a further look at spinors in the real representation.

A spinor represents an orientation (unlike a vector which represents merely a direction). A spinor tells us how an object is rotated in space. Each spinor corresponds to a unique rotation matrix. We will show how to calculate this rotation matrix using only a single matrix multiplication using ##S_j^3## and ##S_i^3##.

A rotation matrix is defined by three orthogonal vectors. These vectors span up the local reference frame of the spinor. The four parameters of a spinor in the real representation give us the four base states as shown in the image above.

The base state (1,0,0,0) corresponds with an orientation were the local reference frame is aligned with the coordinate axes. x'=x, y'=y, z'=z. The other base states (0,1,0,0) (0,0,1,0) and (0,0,0,1) correspond with 180##^o## rotations around the x, y and z-axis.

Every spinor maps to a right handed local coordinate system. For left handed spinors we multiply the rotation matrix with -1 to obtain a left handed local coordinate system.

------------------ The local coordinate system of a spinor -----------------

For an arbitrary spinor (u,x,y,z) we can calculate its local coordinate system with the following matrix multiplication:

$$\check{\xi}\,\hat{\xi} ~=~ (\xi\cdot\mathsf{i})(\mathsf{j}\cdot\xi) ~=~ \big|\xi \big|^2\!\! \left( \begin{smallmatrix} 1~&0~&0~&0~\\ 0~&\mathsf{X}^{^x}&\mathsf{X}^{^y}&\mathsf{X}^{^z} \\ 0~&\mathsf{Y}^{^x}&\mathsf{Y}^{^y}&\mathsf{Y}^{^z} \\ 0~&\mathsf{Z}^{^x}&\mathsf{Z}^{^y}&\mathsf{Z}^{^z} \end{smallmatrix} \right)$$

With ##\hat{\xi}## and ##\check{\xi}## written out as:

\begin{array}{c}
\hat{\xi} ~~=~~ (\xi\cdot\mathsf{j})~~=~~(u\mathsf{j}_o+x\mathsf{j}_x+y\mathsf{j}_y+z\mathsf{j}_z) ~~=~~
\left(\begin{smallmatrix}
~u & ~x & ~y & ~z \\
\!\!-x & ~u &\!\!-z & ~y \\
\!\!-y & ~z & ~u &\!\!-x \\
\!\!-z &\!\!-y & ~x & ~u
\end{smallmatrix}\right)
\\ \\
\check{\xi} ~~=~~(\xi\cdot\mathsf{i})~~=~~(u\mathsf{i}_o+x\mathsf{i}_x+y\mathsf{i}_y+z\mathsf{i}_z) ~~=~~
\left(\begin{smallmatrix}
~u &\!\!-x &\!\!-y &\!\!-z \\
~x & ~u &\!\!-z & ~y \\
~y & ~z & ~u &\!\!-x \\
~z &\!\!-y & ~x & ~u
\end{smallmatrix}\right)
\end{array}

Although the matrix multiplication is mine, the result is exactly identical to the Euler-Rodrigues formula

------------------------- General spinor rotation formula ----------------------------

We can now define a general spinor rotation formula with both ##S_j^3## and ##S_i^3## triplets:
$$e^{~ \tfrac12(a\cdot\mathsf{i}+b\cdot\mathsf{j})}~\xi ~~\longrightarrow~~ e^{\,a\cdot\textbf{J}}\,(\check{\xi}\,\hat{\xi})~e^{-b\cdot\textbf{J}}$$

Where the left side operates on the spinor ##\xi## and the right side operates on the local coordinate system ##\check{\xi}\,\hat{\xi}## and the ##\mathsf{J}## contain the SO(3) rotation generators. To highlight the complete independence of the histories of rotations in world coordinates and internal coordinates we can write this formula using integrals over time.

$$e^{~ \tfrac12\int(a_{(t)}\cdot\,\mathsf{i}~+~b_{(t)}\cdot\,\mathsf{j})\,dt}~\xi ~~\longrightarrow~~ e^{+\int a_{(t)}\cdot\textbf{J}\,dt}\,(\check{\xi}\,\hat{\xi})~e^{-\int b_{(t)}\cdot\textbf{J}\,dt}$$

The triplet ##S_i^3## operates o the right while ##S_j^3## operates to the left. The same ordering occurs in the larger group:

$$SO(4)\otimes SO(4) ~~~\cong~~~ (S^3_j\otimes S^3_j)\times(S^3_i\otimes S^3_i)$$

Because of the left and right operating order the two ##S_j^3## are combined and form the Poincaré group while the two ##S_i^3## combine to the "internal" group.

------- Calculating the components of the precessing (spin 1/2) spinor ------

The 3 vectors that span the local reference frame of the spinor are in fact the same as the three spin vectors ##s_x##, ##s_y## and ##s_z## while the sum of the three is the total spin ##s##. The generator ##\mathsf{i}_x## makes the total spin precess around ##s_x##, the spinor's local x-axis. The other two generators make the total spin precess around the local y- and z-axis.

Compare this with rigid body dynamics where a body can rotate around any of its three principle axes.

We see thus that these overlooked generators, that are nevertheless already a part of the Standard Model, can describe three different, mutually orthogonal, types of fermions. The three generations of fermions did not have any generators associated with them in the Standard Model yet there is this overlooked triplet of generators that perfectly fits them. The CKM and PMNS matrices can be directly associated with this triplet of generators in order to mix generations.

See the sections 1.2, 1.3 and 1.4 of chapter 1 and the whole of chapter 2 here:

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haushofer