EDIT4: I think I was now able to track down where this dogma originally came from. Howard Georgi wrote in TOWARDS A GRAND UNIFIED THEORY OF FLAVOR

There is a deeper reason to require the fermion representation to be complex with respect to SU(3) × SU(2) × U(1). I am assuming that the grand unifying symmetry is broken all the way down to SU(3) × SU(2) × U(1) at a momentum scale of $10^{15}$ GeV. I would therefore expect any subset of the LH fermion representation which is real with respect to SU(3) X SU(2) X U(1) to get a mass of the order of $10^{15}$ GeV from the interactions which cause the spontaneous breakdown. As a trivial example of this, consider an SU(5) theory in which the LH fermions are a 10, a 5 and two $\bar 5$'s. In this theory there will be SU(3) × SU(2) X U(1) invariant mass terms connecting the 5 to some linear combination of the two $\bar 5$-'s. These ten (chiral) states will therefore correspond to 5 four-component fermions with masses of order 10 as GeV. The 10 and the orthogonal linear combination of the two 5-'s will be left over as ordinary mass particles because they carry chiral SU(2) X U(1).

Unfortunately I'm not able to put this argument in mathemtical terms. How exactly does the new, invariant mass term, combining the $5$ and the $\bar 5$ look like?

EDIT3: My current experience with this topic is summarized in chapter 5.1 of this thesis:

Furthermore the group should have complex representations necessary to accommodate the SU(3) complex triplet and the complex doublet fermion representation. [...] the next five do not have complex representations, and so, are ruled out as candidates for the GUT group. [...] It should be pointed out that it is possible to construct GUT's with fermions in the real representation provided we allow extra mirror fermions in the theory.

What? Groups without complex representations are ruled out. And a few sentences later everything seems okay with such groups, as long as we allow some extra particles called mirror fermions.

In almost every document about GUTs it is claimed that we need complex representations (=chiral representations) in order to be able to reproduce the standard model. Unfortunately almost everyone seems to have a different reason for this and none seems fully satisfactory to me. For example:

Witten says:

Of the five exceptional Lie groups, four ( G 2 , F 4 , E 7 , and E 8 ) only have real or pseu-doreal representations. A four-dimensional GUT model based on such a group will not give the observed chiral structure of weak interactions. The one exceptional group that does have complex or chiral representations is E6

This author writes:

Since they do not have complex representations. That we must have complex representations for fermions, because in the S.M. the fermions are not equivalent to their complex conjugates.

Another author writes:

Secondly, the representations must allow for the correct reproduction of the particle content of the observed fermion spectrum, at least for one generation of fermions. This requirement implies that G gut must possess complex representations as well as it must be free from anomalies in order not to spoil the renormalizability of the grand unified theory by an incompatibility of regularization and gauge invariance. The requirement of complex fermion representations is based on the fact that embedding the known fermions in real representations leads to diculties: Mirror fermions must be added which must be very heavy . But then the conventional fermions would in general get masses of order M gut . Hence all light fermions should be components of a complex representation of G gut .

And Lubos has an answer that does not make any sense to me:

However, there is a key condition here. The groups must admit complex representations - representations in which the generic elements of the group cannot be written as real matrices. Why? It's because the 2-component spinors of the Lorentz group are a complex representation, too. If we tensor-multiply it by a real representation of the Yang-Mills group, we would still obtain a complex representation but the number of its components would be doubled. Because of the real factor, such multiplets would always automatically include the left-handed and right-handed fermions with the same Yang-Mills charges!

So... what is the problem with real representations? Unobserved mirror fermions? The difference of particles and antiparticles? Or the chiral structure of the standard model?


I just learned that there are serious GUT models that use groups that do not have complex representations. For example, this review by Langacker mentions several models based on $E_8$. This confuses me even more. On the one hand, almost everyone seems to agree that we need complex representations and on the other hand there are models that work with real representations. If there is a really good why we need complex representations, wouldn't an expert like Langacker regard models that start with some real representation as non-sense?


Here Stech presents another argument

The groups E7 and E8 also give rise to vector-like models with $\sin^2 \theta = 3/4$. The mathematical reason is that these groups have, like G and F4, only real (pseudoreal) representations. The only exceptional group with complex... [...] Since E7 and Es give rise to vector-like theories, as was mentioned above, at least half of the corresponding states must be removed or shifted to very high energies by some unknown mechanism

  • $\begingroup$ Sketch of an answer: charge conjugation of spinors switches chirality: $(\psi_L)^C = (\psi^C)_R$. The SM is chiral - left- and right-handed fields have different charges. Thus fields and their charge conjugates must have different charges. If the GUT didn't have complex representations, that would be impossible. $\endgroup$
    – innisfree
    Commented Apr 27, 2015 at 20:13
  • 1
    $\begingroup$ @innisfree thanks for your comment. Unfortunately I'm not really convinced that it is correct. Charge conjugation flipping, for example, the isospin sign, works independently of if we have complex or real objects. The charge conjugate of an $SU(2)$ doublet is $\Psi_L \rightarrow \Psi_L^C= i \gamma_2 \Psi_L^\star $, where the spinors inside the doublet must be charge conjugated, too. The matrix $\gamma_2$ flips the upper and the lower component and this is what makes sure our charge conjugated fields inside the doublet have opposite isospin. $\endgroup$
    – jak
    Commented Apr 28, 2015 at 6:15
  • $\begingroup$ @innisfree In addition our spinors are complex, but this is independent of the gauge group and purely a result of studying the representations of the Poincare group and this is where the notions left-chiral and right-chiral come from. A left-chiral field becomes a right-chiral field with opposite sign of the isospin independent of which representation of the gauge group, for example $SU(2)$ (aren't they never complex?), we use. $\endgroup$
    – jak
    Commented Apr 28, 2015 at 6:19
  • 1
    $\begingroup$ @innisfree There is for every field a right-chiral companion with really different quantum number, which means that isospin does not differ merely by a sign. For example, we have the left-chiral electron with isospin $\frac {-1}{2}$, its charge conjugate which is right chiral and has isospin $\frac {+1}{2}$. In addition there is the right-chiral electron with isospin zero and its left-chiral charge conjugate. Nevertheless, I can't see why this should require complex representations of the gauge group $\endgroup$
    – jak
    Commented Apr 28, 2015 at 6:23
  • 2
    $\begingroup$ All the reasons in your quotes are the same - real representations would give vector-like representations of fermions. The $(\psi_L)^C$ has weak isospin 0 - it's right-handed $P_L (\psi_L)^C=0$. $\endgroup$
    – innisfree
    Commented Apr 29, 2015 at 8:17

4 Answers 4


Charge conjugation is extremely slippery because there are two different versions of it; there have been many questions on this site mixing them up (1, 2, 3, 4, 5, 6, 7, 8, 9), several asked by myself a few years ago. In particular there are a couple arguments in comments above where people are talking past each other for precisely this reason.

I believe the current answer falls into one of the common misconceptions. I'll give as explicit of an example as possible, attempting to make a 'Rosetta stone' for issues about chirality, helicity, and $\hat{C}$. Other discrete symmetries are addressed here.

A hypercharge example

For simplicity, let's consider hypercharge in the Standard Model, and only look at the neutrino, which we suppose has a sterile partner. For a given momentum there are four neutrino states:

$$|\nu, +\rangle \text{ has positive helicity and hypercharge } Y=0$$ $$|\nu, -\rangle \text{ has negative helicity and hypercharge } Y=-1/2$$ $$|\bar{\nu}, +\rangle \text{ has positive helicity and hypercharge } Y=1/2$$ $$|\bar{\nu}, -\rangle \text{ has negative helicity and hypercharge } Y=0$$

There are two neutrino fields: $$\nu_L \text{ is left chiral, has hypercharge } -1/2, \text{annihilates } |\nu, -\rangle \text{ and creates } |\bar{\nu}, + \rangle$$ $$\nu_R \text{ is right chiral, has hypercharge } 0, \text{annihilates } |\nu, +\rangle \text{ and creates } |\bar{\nu}, - \rangle$$ The logic here is the following: suppose a classical field transforms under a representation $R$ of an internal symmetry group. Then upon quantization, it will annihilate particles transforming under $R$ and create particles transforming under the conjugate representation $R^*$.

The spacetime symmetries are more complicated because particles transform under the Poincare group and hence have helicity, while fields transform under the Lorentz group and hence have chirality. In general, a quantized right-chiral field annihilates a positive-helicity particle. Sometimes, the two notions "right-chiral" and "positive-helicity" are both called "right-handed", so a right-handed field annihilates a right-handed particle. I'll avoid this terminology to avoid mixing up chirality and helicity.

Two definitions of charge conjugation

Note that both the particle states and the fields transform in representations of $U(1)_Y$. So there are two distinct notions of charge conjugation, one which acts on particles, and one which acts on fields. Acting on particles, there is a charge conjugation operator $\hat{C} $ satisfying $$\hat{C} |\nu, \pm \rangle = |\bar{\nu}, \pm \rangle.$$ This operator keeps all spacetime quantum numbers the same; it does not change the spin or the momentum and hence doesn't change the helicity. It is important to note that particle charge conjugation does not always conjugate internal quantum numbers, as one can see in this simple example. This is only true when $\hat{C}$ is a symmetry of the theory, $[\hat{C}, \hat{H}] = 0$.

Furthermore, if we didn't have the sterile partner, we would have only the degrees of freedom created or destroyed by the $\nu_L$ field, and there would be no way of defining $\hat{C}$ consistent with the definition above. In other words, particle charge conjugation is not always even defined, though it is with the sterile partner.

There is another notion of charge conjugation, which on classical fields is simply complex conjugation, $\nu_L \to \nu_L^*$. By the definition of a conjugate representation, this conjugates all of the representations the field transforms under, i.e. it flips $Y$ to $-Y$ and flips the chirality. This is true whether the theory is $\hat{C}$-symmetric or not. For convenience we usually define $$\nu_L^c = C \nu_L^*$$ where $C$ is a matrix which just puts the components of $\nu_L^*$ into the standard order, purely for convenience. (Sometimes this matrix is called charge conjugation as well.)

In any case, this means $\nu_L^c$ is right-chiral and has hypercharge $1/2$, so $$\nu_L^c \text{ is right chiral, has hypercharge } 1/2, \text{annihilates } |\bar{\nu}, +\rangle \text{ and creates } |\nu, - \rangle.$$ The importance of this result is that charge conjugation of fields does not give additional particles. It only swaps what the field creates and what it annihilates. This is why, for instance, a Majorana particle theory can have a Lagrangian written in terms of left-chiral fields, or in terms of right-chiral fields. Both give the same particles; it is just a trivial change of notation.

(For completeness, we note that there's also a third possible definition of charge conjugation: you could modify the particle charge conjugation above, imposing the additional demand that all internal quantum numbers be flipped. Indeed, many quantum field theory courses start with a definition like this. But this stringent definition of particle charge conjugation means that it cannot be defined even with a sterile neutrino, which means that the rest of the discussion below is moot. This is a common issue with symmetries: often the intuitive properties you want just can't be simultaneously satisfied. Your choices are either to just give up defining the symmetry or give up on some of the properties.)

Inconsistencies between the definitions

The existing answer has mixed up these two notions of charge conjugation, because it assumes that charge conjugation gives new particles (true only for particle charge conjugation) while reversing all quantum numbers (true only for field charge conjugation). If you consistently use one or the other, the argument doesn't work.

A confusing point is that the particle $\hat{C}$ operator, in words, simply maps particles to antiparticles. If you think antimatter is defined by having the opposite (internal) quantum numbers to matter, then $\hat{C}$ must reverse these quantum numbers. However, this naive definition only works for $\hat{C}$-symmetric theories, and we're explicitly dealing with theories that aren't $\hat{C}$-symmetric.

One way of thinking about the difference is that, in terms of the representation content alone, and for a $\hat{C}$-symmetric theory only, the particle charge conjugation is the same as field charge conjugation followed by a parity transformation. This leads to a lot of disputes where people say "no, your $\hat{C}$ has an extra parity transformation in it!"

For completeness, note that one can define both these notions of charge conjugation in first quantization, where we think of the field as a wavefunction for a single particle. This causes a great deal of confusion because it makes people mix up particle and field notions, when they should be strongly conceptually separated. There is also a confusing sign issue because some of these first-quantized solutions correspond to holes in second quantization, reversing most quantum numbers (see my answer here for more details). In general I don't think one should speak of the "chirality of a particle" or the "helicity of a field" at all; the first-quantized picture is worse than useless.

Why two definitions?

Now one might wonder why we want two different notions of charge conjugation. Charge conjugation on particles only turns particles into antiparticles. This is sensible because we don't want to change what's going on in spacetime; we just turn particles into antiparticles while keeping them moving the same way.

On the other hand, charge conjugation on fields conjugates all representations, including the Lorentz representation. Why is this useful? When we work with fields we typically want to write a Lagrangian, and Lagrangians must be scalars under Lorentz transformations, $U(1)_Y$ transformations, and absolutely everything else. Thus it's useful to conjugate everything because, e.g. we know for sure that $\nu_L^c \nu_L$ could be an acceptable Lagrangian term, as long as we contract all the implicit indices appropriately. This is, of course, the Majorana mass term.

Answering the question

Now let me answer the actual question. By the Coleman-Mandula theorem, internal and spacetime symmetries are independent. In particular, when we talk about, say, a set of fields transforming as a $10$ in the $SU(5)$ GUT, these fields must all have the same Lorentz transformation properties. Thus it is customary to write all matter fields in terms of left-chiral Weyl spinors. As stated above, this does nothing to the particles, it's just a useful way to organize the fields.

Therefore, we should build our GUT using fields like $\nu_L$ and $\nu_R^c$ where $$\nu_R^c \text{ is left chiral, has hypercharge } 0.$$ What would it have looked like if our theory were not chiral? Then $|\nu, + \rangle$ should have the same hypercharge as $|\nu, -\rangle$, which implies that $\nu_R$ should have hypercharge $-1/2$ like $\nu_L$. Then our ingredients would be $$\nu_L \text{ has hypercharge } -1/2, \quad \nu_R^c \text{ has hypercharge } 1/2.$$ In particular, note that the hypercharges come in an opposite pair.

Now let's suppose that our matter fields form a real representation $R$ of the GUT gauge group $G$. Spontaneous symmetry breaking takes place, reducing the gauge group to that of the Standard Model $G'$. Hence the representation $R$ decomposes, $$R = R_1 \oplus R_2 \oplus \ldots$$ where the $R_i$ are representations of $G'$. Since $R$ is real, if $R_i$ is present in the decomposition, then its conjugate $R_i^*$ must also be present. That's the crucial step.

Specifically, for every left-chiral field with hypercharge $Y$, there is another left-chiral field with hypercharge $-Y$, which is equivalent to a right-chiral field with hypercharge $Y$. Thus left-chiral and right-chiral fields come in pairs, with the exact same transformations under $G'$. Equivalently, every particle has an opposite-helicity partner with the same transformation under $G'$. That is what we mean when we say the theory is not chiral.

To fix this, we can hypothesize all of the unwanted "mirror fermions" are very heavy. As stated in the other answer, there's no reason for this to be the case. If it were, we run into a naturalness problem just as for the Higgs: since there is nothing distinguishing fermions from mirror fermions, from the standpoint of symmetries, there is nothing preventing matter from acquiring the same huge mass. This is regarded as very strong evidence against such theories; some say that for this reason, theories with mirror fermions are outright ruled out. For example, the $E_8$ theory heavily promoted in the press has exactly this problem; the theory can't be chiral.

  • $\begingroup$ From a second quantization field theory point of view, given $\mathcal{C} : \psi \mapsto \psi^c = C \psi^*$, in the Dirac basis, is the appropriate $C$ matrix $\gamma^0 \gamma^2$ or $\gamma^2$? References muddle this up as you mention, so I want to be sure I'm not making errors $\endgroup$
    – Craig
    Commented Mar 22, 2022 at 6:17

This can be explained by thinking about the coupling of fermions to the $SU(2)$ weak gauge field. Let's recap what we know

  1. Weyl fermions necessarily appear in two complex representations of the Lorentz group $L$ and $R$.
  2. Only fermions in the $L$ representation of the Lorentz group couple to the $SU(2)$ gauge field.
  3. CPT is a symmetry of the theory.

Now let's introduce the charge conjugation operator $C$. Consider a left-handed fermion field living in the fundamental representation $R$ of a gauge group $G$. Then the charge conjugation operator produces a left-handed anti-fermion field in the complex conjugate representation $\bar{R}$. If $R$ is a real representation then $R=\bar{R}$.

Why is this bad? Well if the left-handed anti-fermion lives in the same representation as the left-handed fermion, then it can couple to the gauge field in the same way. Indeed by the logic of effective field theory it must do, unless you invent some complicated new mechanism which prevents this from happening!

Now using CPT symmetry we can equivalently regard our left-handed anti-fermion as a right-handed fermion. But this means that you have a right-handed fermion coupling to the gauge field in the same way as your left-handed one did originally. In other words your theory is not chiral.

Are there any loopholes? Well, you could hypothesize that the right-handed fermions coupling to the weak field just haven't been observed yet! This is the idea of mirror matter. It is a necessary prediction of any theory using a Lie algebra which has no complex representations, such as $E_8$.

To conclude, I think that Witten has the clearest explanation, but it is a little terse! I agree that some of the arguments above are vague (as indeed was this answer originally). Do please keep asking questions in the comments and hopefully we can hone in on a really accessible explanation!

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – ACuriousMind
    Commented May 6, 2018 at 20:08

Trying to provide a short-winded answer: The Standard Model is chiral, and we define the chiral projection operator as $$ P_{RL} = \frac{1}{2}(1 \pm \gamma^5), $$ which involves $\gamma^5$ expressed as $$ \gamma^5 = i\gamma^0\gamma^1\gamma^2\gamma^3. $$ The imaginary number $i$ in above definition is crucial in keeping $\gamma^5$ hermitian $$ (\gamma^5)^\dagger = \gamma^5. $$ Given that the Standard model is chiral, the indispensable $i$ in the definition of chiral projection $P_{RL}$ behooves us to choose a complex representation.

That being said, a real representation is not strictly prohibited if you are innovative enough to come up with real chiral projection operator and real $\gamma^\mu$ representation.


The geometry of spacetime and of the gauge dimensions is fundamentally real. Many textbooks seem to suggest that the Dirac algebra is complexified so that a chiral decomposition is possible, but that's not right. You can't complexify something just because it leads to a result you want; that result won't show up in nature, which doesn't complexify.

What's really happening is that the gauge dimensions make the chiral decomposition possible. In Clifford-algebraic terms, the real even algebra of 3+1 dimensions has a pseudoscalar ($\mathbf{txyz}$) that squares to $-1$. If you add two extra dimensions (because $\mathrm{Spin}(2) \cong \mathrm{U}(1)$), the pseudoscalar of the internal dimensions (call it $\mathbf{uv}$) also squares to $-1$, but the pseudoscalar of the combined external-internal algebra ($\mathbf{txyzuv}$) squares to $-1\cdot-1=1$, and so you can decompose the algebra into two subalgebras with the projections $(1\pm\mathbf{txyzuv})/2$. This object $\mathbf{txyzuv}$ is the $γ^5$ matrix, and the $i$ in the usual complexification is $\mathbf{uv}$.

In terms of the matrix representations, the real even Clifford algebra of 3+1 dimensions is isomorphic to $\def\C{\mathbb C} \C^{2\times 2}$ (meaning 2×2 matrices over $\C$). The Clifford algebra of the $\mathrm{U}(1)$ gauge theory is isomorphic to $\C$. The real tensor product of the two is $(\C^{2\times 2})^2$, i.e., pairs of 2×2 matrices over $\C$, and that's the algebra that admits the chiral decomposition. The $\C^2$ vectors acted upon by these two matrices are the Weyl spinors. The $\C^2$ vector acted on by the $\C^{2\times 2}$ of the external dimensions is the Majorana spinor.

If you add 10 internal dimensions instead of 2 (because the Standard Model gauge group is embeddable in $\mathrm{Spin}(10)$) then the internal Clifford algebra is isomorphic to $\C^{16\times 16}$, and the tensor product is isomorphic to $(\C^{32\times 32})^2$.

If you try this with 4, 8, 12, ... internal dimensions, the internal Clifford algebra is isomorphic to pairs of matrices over $\def\R{\mathbb R} \R$ or $\def\H{\mathbb H} \H$. The tensor product still gives you pairs of matrices over $\C$, but the external dimensions aren't involved in the chiral decomposition; it's the same as if you'd discarded half of the internal algebra before the tensor product instead of after. So externally left- and right-handed particles have the same internal charges, which is not what's observed.

If the number of dimensions is odd then the even algebra has no pseudoscalar. The even algebras are isomorphic to single matrices over $\R$ or $\H$, and the tensor product gets you a single matrix over $\C$.

Complex numbers are genetically dominant in tensor products: $\C \otimes \R \cong \C,\; \C \otimes \C \cong \C^2,\; \C \otimes \H \cong \C^{2\times 2}$. Because of that they tend to infect everything once they show up, and one can be fooled into thinking that they were there from the start. But in this context they always come from a central element of a geometric algebra that has some geometric interpretation.

I think the quote from Luboš in the question (that made no sense to OP) is saying something similar to the above.

It's because the 2-component spinors of the Lorentz group are a complex representation, too.

In other words, because the external algebra is a matrix algebra over $\C$. (I think that the spinors are not technically complex, although there seem to be multiple inequivalent definitions of what a complex representation is and I've never quite understood it.)

If we tensor-multiply it by a real representation of the Yang-Mills group, we would still obtain a complex representation but the number of its components would be doubled.

I think he means $\C^k \otimes_\R \R^{2\ell} \cong \C^{2kl}$, whereas $\C^k \otimes_\R \C^\ell \cong \C^{kl} \oplus \C^{kl}$, and you can throw away one of those $\C^{kl}$ (postulate that it doesn't exist in nature), leaving you with half as many real components.


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