Why chiral fermions don't exist in odd dimensions?

[synoe]

In four dimensions, left and right chiral fermion can be written as
<br /> \psi_L=<br /> \begin{pmatrix}<br /> \psi_+\\<br /> 0<br /> \end{pmatrix},\qquad<br /> \psi_R=<br /> \begin{pmatrix}<br /> 0\\<br /> \psi_-<br /> \end{pmatrix},<br />
respectively, where \psi_+ and \psi_- are some two components spinors(Weyl spinors?). In this representation, the chirality operator \gamma_5 is written as
<br /> \gamma_5=<br /> \begin{pmatrix}<br /> \mathbb{1}&amp;0\\<br /> 0&amp;-\mathbb{1}<br /> \end{pmatrix}.<br />

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:
<br /> \Gamma^\mu=\gamma^\mu\qquad\mu=0,1,2,3\\<br /> \Gamma^4=\gamma_5.<br />
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?

 

[Avodyne]

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.

 

[synoe]

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
<br /> \Psi=<br /> \begin{pmatrix}<br /> \psi_+\\<br /> 0<br /> \end{pmatrix}<br />
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?

 

[fzero]

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.

 

[synoe]

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
<br /> \Psi=<br /> \begin{pmatrix}<br /> \psi_+\\<br /> 0<br /> \end{pmatrix}<br />
in a frame, its Dirac adjoint takes the form
<br /> \bar{\Psi}=<br /> \begin{pmatrix}<br /> 0&amp;\bar{\psi}_+<br /> \end{pmatrix}.<br />
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?

 

[fzero]

How about the Lorentz invariant bilinear form \bar{\Psi}\Psi? If \Psi takes the form
<br /> \Psi=<br /> \begin{pmatrix}<br /> \psi_+\\<br /> 0<br /> \end{pmatrix}<br />
in a frame, its Dirac adjoint takes the form
<br /> \bar{\Psi}=<br /> \begin{pmatrix}<br /> 0&amp;\bar{\psi}_+<br /> \end{pmatrix}.<br />
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.

 

[Avodyne]

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.

 

[fzero]

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).

 

[Avodyne]

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).

How about the Lorentz invariant bilinear form \bar{\Psi}\Psi? If \Psi takes the form
<br /> \Psi=<br /> \begin{pmatrix}<br /> \psi_+\\<br /> 0<br /> \end{pmatrix}<br />
in a frame, its Dirac adjoint takes the form
<br /> \bar{\Psi}=<br /> \begin{pmatrix}<br /> 0&amp;\bar{\psi}_+<br /> \end{pmatrix}.<br />
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.

 

[fzero]

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.