# Why chiral fermions don't exist in odd dimensions?

• synoe
In summary, in five dimensions, the chirality operator \gamma_5 can be written as a block diagonal matrix with a Lorentz-breaking term in the lower right block. This means that there is no Lorentz-invariant way to define chirality in five dimensions, and therefore the concept of chiral fermions does not exist in odd dimensions. The bilinear form \bar{\Psi}\Psi is also zero in any frame for 5D spinors due to the symplectic Majorana condition, which is a reality condition equivalent to the Majorana condition in 4D.
synoe
In four dimensions, left and right chiral fermion can be written as
$$\psi_L= \begin{pmatrix} \psi_+\\ 0 \end{pmatrix},\qquad \psi_R= \begin{pmatrix} 0\\ \psi_- \end{pmatrix},$$
respectively, where $\psi_+$ and $\psi_-$ are some two components spinors(Weyl spinors?). In this representation, the chirality operator $\gamma_5$ is written as
$$\gamma_5= \begin{pmatrix} \mathbb{1}&0\\ 0&-\mathbb{1} \end{pmatrix}.$$

In five dimensions, the fifth $\gamma$-matrix $\Gamma^4$ can coincide with the four dimensional chiral operator $\gamma_5$. The other matrices are same as four dimensional ones:
$$\Gamma^\mu=\gamma^\mu\qquad\mu=0,1,2,3\\ \Gamma^4=\gamma_5.$$
In this representation, a five dimensional fermion $\Psi=\psi_L$ seems to be "chiral" if I define the chiral operator as $\Gamma_6=\Gamma^4=\gamma_5$.

However in general, there is no notion of chirality in odd dimensions. Why the above $\Psi$ cannot be a chiral fermion?

Because ##\Gamma^4## is not Lorentz invariant in five dimensions, whereas ##\gamma_5## is Lorentz invariant in four dimensions.

Given gamma matrices that obey the anticommutation relations ##\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}##, we can define matrices ##S^{\mu\nu}={\textstyle{i\over 4}}[\gamma^\mu,\gamma^\nu]## that obey the Lorentz algebra. The question is then whether this representation of the Lorentz algebra is reducible or not. It is reducible if we can find a matrix that commutes with all the ##S^{\mu\nu}## but is not a multiple of the identity matrix. ##\gamma_5## has this property in four dimensions, but ##\Gamma^4## does not have this property in five dimensions.

vanhees71
Avodyne said:
Because ##\Gamma^4## is not Lorentz invariant in five dimensions, whereas ##\gamma_5## is Lorentz invariant in four dimensions.

Given gamma matrices that obey the anticommutation relations ##\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}##, we can define matrices ##S^{\mu\nu}={\textstyle{i\over 4}}[\gamma^\mu,\gamma^\nu]## that obey the Lorentz algebra. The question is then whether this representation of the Lorentz algebra is reducible or not. It is reducible if we can find a matrix that commutes with all the ##S^{\mu\nu}## but is not a multiple of the identity matrix. ##\gamma_5## has this property in four dimensions, but ##\Gamma^4## does not have this property in five dimensions.

I can't understand why a commutable matrix can decompose a representation into a block diagonal form. Could you tell me the concrete way of decomposition?

And I would like to realize the proposition in matrix form. I think the five dimensional spinors which is written as
$$\Psi= \begin{pmatrix} \psi_+\\ 0 \end{pmatrix}$$
can exist. In four dimensions, I regard that the spinors of this form as chiral. Is this not the case for five dimensions ?. Is it possible for this type of spinors have the mass term $\bar{\Psi}\Psi$ in five dimensions?

Taking what Avodyne said a slightly different way, in response to your question, a 5D Lorentz transformation acts on spinors as

$$\Psi \rightarrow \exp\left( i \alpha_{MN} S^{MN} \right) \Psi,$$

where ##S^{MN}## is related to the commutator of 5D##\gamma^M##s, as in Avodyne's post. Since ##\gamma^4\equiv \gamma^5##, we can see that there are Lorentz transformations that take

$$\begin{pmatrix} \psi_+ \\ 0 \end{pmatrix} \rightarrow \begin{pmatrix} \psi_+' \\ \psi_-' \end{pmatrix} .$$

Said in words, the chirality that we could define in 4D is no longer a Lorentz-invariant condition in 5D.

fzero said:
Taking what Avodyne said a slightly different way, in response to your question, a 5D Lorentz transformation acts on spinors as

$$\Psi \rightarrow \exp\left( i \alpha_{MN} S^{MN} \right) \Psi,$$

where ##S^{MN}## is related to the commutator of 5D##\gamma^M##s, as in Avodyne's post. Since ##\gamma^4\equiv \gamma^5##, we can see that there are Lorentz transformations that take

$$\begin{pmatrix} \psi_+ \\ 0 \end{pmatrix} \rightarrow \begin{pmatrix} \psi_+' \\ \psi_-' \end{pmatrix} .$$

Said in words, the chirality that we could define in 4D is no longer a Lorentz-invariant condition in 5D.

How about the Lorentz invariant bilinear form $\bar{\Psi}\Psi$? If $\Psi$ takes the form
$$\Psi= \begin{pmatrix} \psi_+\\ 0 \end{pmatrix}$$
in a frame, its Dirac adjoint takes the form
$$\bar{\Psi}= \begin{pmatrix} 0&\bar{\psi}_+ \end{pmatrix}.$$
Then, the bilinear $\bar{\Psi}\Psi$ equals to zero. This means $\Psi$ is massless in arbitrary frame if it takes the above form in a frame in five dimensions?

synoe said:
How about the Lorentz invariant bilinear form $\bar{\Psi}\Psi$? If $\Psi$ takes the form
$$\Psi= \begin{pmatrix} \psi_+\\ 0 \end{pmatrix}$$
in a frame, its Dirac adjoint takes the form
$$\bar{\Psi}= \begin{pmatrix} 0&\bar{\psi}_+ \end{pmatrix}.$$
Then, the bilinear $\bar{\Psi}\Psi$ equals to zero. This means $\Psi$ is massless in arbitrary frame if it takes the above form in a frame in five dimensions?

Well ##\bar{\Psi}\Psi =0## in any frame for 5D spinors, but the reason is a bit technical. The group ##\tex{Spin}(5)## is isomorphic to the symplectic group ##Sp(2)##, so there is a natural symplectic form ##\omega_{\alpha\beta}## that acts on the fundamental ##\mathbf{4}##, which is the spinor when we view the group as spin. Now, when we compute ##\bar{\Psi}##, since the ##\mathbf{4}## is an irreducible representation, ##\bar{\Psi}## must also transform in the ##\mathbf{4}##. In terms of components then ##\bar{\Psi}^\alpha = C^{\alpha\beta}\Psi_\beta## for some matrix ##C^{\alpha\beta}##. Because of the group theory, ##C^{\alpha\beta}## has to be related to the symplectic form and so it has to be antisymmetric. This is a reality condition equivalent to the Majorana condition in 4D, but here in 5D, because of the appearance of the symplectic group, it is called a symplectic Majorana condition.

The upshot of all that is ##\bar{\Psi}\Psi =C^{\alpha\beta} \Psi_\alpha \Psi_\beta=0## by symmetry. This is equivalent to what happens in 4D if you try to compute such a term for a Majorana fermion. The reason the mass term can be nonzero for a 4D Dirac fermion is that the 4-dimensional representation there is reducible to the Weyl spinors that transform in the ##(\mathbf{2},\mathbf{1})## and ##(\mathbf{1},\mathbf{2})## of ##SU(2)\times SU(2)##. So the part that survives is the cross term. But in 5D the structure is completely different.

fzero said:
The upshot of all that is ##\bar{\Psi}\Psi =C^{\alpha\beta} \Psi_\alpha \Psi_\beta=0## by symmetry.
In quantum field theory (which is the right context), the ##\Psi## field anticommutes. So ##C^{\alpha\beta} \Psi_\alpha \Psi_\beta## does not vanish. This is similar to what happens for a Weyl field ##\chi\sim(\mathbf{2},\mathbf{1})## in four dimensions, where ##\varepsilon^{ab}\chi_a\chi_b## (plus hermitian conjugate) is an allowed mass term; here ##\varepsilon^{ab}## is the Levi-Civita tensor of SU(2).

But this doesn't change the fact that there is no Lorentz-invariant chirality projection in 5D.

Avodyne said:
In quantum field theory (which is the right context), the ##\Psi## field anticommutes. So ##C^{\alpha\beta} \Psi_\alpha \Psi_\beta## does not vanish. This is similar to what happens for a Weyl field ##\chi\sim(\mathbf{2},\mathbf{1})## in four dimensions, where ##\varepsilon^{ab}\chi_a\chi_b## (plus hermitian conjugate) is an allowed mass term; here ##\varepsilon^{ab}## is the Levi-Civita tensor of SU(2).

Yes that's the Majorana mass term and what I ended up writing down in the 5d version. Obviously I have some sort of error in relating the Dirac mass term to that, so the Dirac term must vanish for some other reason (probably obvious that I'm not seeing).

I don't believe that the Dirac mass term vanishes in 5D either. A Dirac field can always be written as the sum of two Majorana fields (one with a factor of ##i##), and I don't see how you will get the Majorana terms to vanish (for anticommuting fields).

synoe said:
How about the Lorentz invariant bilinear form $\bar{\Psi}\Psi$? If $\Psi$ takes the form
$$\Psi= \begin{pmatrix} \psi_+\\ 0 \end{pmatrix}$$
in a frame, its Dirac adjoint takes the form
$$\bar{\Psi}= \begin{pmatrix} 0&\bar{\psi}_+ \end{pmatrix}.$$
Then, the bilinear $\bar{\Psi}\Psi$ equals to zero. This means $\Psi$ is massless in arbitrary frame if it takes the above form in a frame in five dimensions?
This form is not preserved by the Dirac equation (massless or massive) in 5D. That is, while it is Lorentz invariant, it is not time independent, so it doesn't matter if it happens to be zero at some particular time.

Avodyne said:
I don't believe that the Dirac mass term vanishes in 5D either.

Maybe the first part of my calculation is correct and I should just conclude that the Dirac term is isomorphic to the Majorana term in 5D. My mistake was in thinking I could say somethng more without extending the group theory to the case of anticommuting fields.

A Dirac field can always be written as the sum of two Majorana fields (one with a factor of ##i##), and I don't see how you will get the Majorana terms to vanish (for anticommuting fields).

In 5D the Dirac spinor ##\mathbf{4}## is irreducible, so it already satisfies the Majorana-type condition. In 6D the Dirac spinor ##\mathbf{8}## reduces to a pair of Majorana
spinor ##\mathbf{4}##s.

This form is not preserved by the Dirac equation (massless or massive) in 5D. That is, while it is Lorentz invariant, it is not time independent, so it doesn't matter if it happens to be zero at some particular time.

This seems to be the correct resolution.

## 1. Why can't chiral fermions exist in odd dimensions?

In physics, chiral fermions are particles that have a different spin and momentum depending on their direction of motion. In odd dimensions, the spin and momentum are not independent of each other, making it impossible for chiral fermions to exist.

## 2. What is the significance of odd dimensions in relation to chiral fermions?

Odd dimensions play a crucial role in the existence of chiral fermions. In even dimensions, the spin and momentum of a particle are independent of each other, allowing for the existence of chiral fermions. However, in odd dimensions, this independence does not hold, leading to the absence of chiral fermions.

## 3. Can chiral fermions exist in any odd dimension?

No, chiral fermions can only exist in odd dimensions that are higher than three. In three dimensions, chiral fermions do exist, as they correspond to the left and right-handed particles in the Standard Model of particle physics. However, in higher odd dimensions, the concept of chirality breaks down, making it impossible for chiral fermions to exist.

## 4. Are there any exceptions to the rule that chiral fermions don't exist in odd dimensions?

So far, there are no known exceptions to this rule. The theoretical framework of quantum field theory predicts that chiral fermions cannot exist in odd dimensions, and this has been supported by experimental evidence. However, there is ongoing research in this area to explore any potential loopholes or exceptions to this rule.

## 5. How does the absence of chiral fermions in odd dimensions affect our understanding of the universe?

The absence of chiral fermions in odd dimensions has a significant impact on our understanding of the universe at the most fundamental level. It is a crucial factor in the development of the Standard Model of particle physics, which explains the behavior of particles and their interactions. It also has implications for theories such as supersymmetry, which predicts the existence of additional particles, some of which may have chiral fermion-like properties.

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