### Orbifolding out of the 5D Chirality Problem

22Apr07

How to get left/right-handed 4D theories from an ambidexterous 5D theory… demote the 5th dimension from a manifold to an orbifold.

In a previous post I introduced the 5D chirality problem. The problem was that there is no such thing as a 5D parity operator, and hence 5D spinors are Dirac $(0,\frac 12)\oplus(\frac 12,0)$ spinors. In other words, there’s no clear way to get 4D chiral Weyl, $(0,\frac 12)$ or $(\frac 12,0)$, fermions that aren’t automatically paired up with another 4D fermion of opposite helicity and identical quantum numbers. The Standard Model is chiral, and hence any realistic theory that hopes to put the Standard Model fermions in a bulk 5D space would have to deal with this chirality problem.

A possible solution is to orbifold the the extra dimension. This means that we mod out the manifold (in this case $S^1$) by a discrete symmetry, which we will choose to be $\mathbb{Z}^2$. The resulting space is an interval or, more generally, a “manifold with boundary.” To impose this orbifold symmetry, we identify $\phi \leftrightarrow -\phi$ and impose that nature be invariant under this transformation.

Image from Professor Raman Sundrum’s TASI 2004 Lectures.

What does this do for us? The combination of the boundary conditions from circular compactification $\Psi(x,-\pi) = \Psi(x,\pi)$ and the orbifold symmetry $\mathcal L(x,\phi) = \mathcal L(x,-\phi)$ conspire to kill off one of the chiral zero modes, allowing for a low energy chiral theory. Let’s see how this works.

Without loss of generality we can write our Kaluza Klein decomposition in terms of functions that are even/odd under $\phi$ parity, we simply rewrite our exponentials in terms of sines and cosines:

$\Psi(x, \phi) = \frac{1}{\sqrt{\pi R}} \left[ \frac{1}{\sqrt{2}} \psi_0^{(+)}(x)+\sum \psi^{(+)}_n(x)\cos(n\phi)+ \sum \psi^{(-)}_n(x)\sin(n\phi)\right]$,

where the sums above run from $n=1$ to $\infty$. Recall that this decomposition comes from the boundary conditions of our $S^1$ compactification of the extra dimension.

Now consider the fermion kinetic term:

$\mathcal L \supset i\bar \Psi \gamma^M \partial_M \Psi$

Here $M$ is a bulk index running from 0 to 4, with 4 corresponding to the $\phi$ coordinate. The above expression includes the term $i\bar \Psi \gamma^\phi\partial_\phi\Psi$, where I’ve likely made an error of an overall sign or factor of $i$. This decomposes into

$i\bar \Psi_L \partial_5 \Psi_R + i \bar \Psi_R \partial_5 \Psi_L$.

Where I’ve explicitly written out the left-handed and right-handed components of the Dirac spinor $\Psi$. Now it is clear that $\partial_\phi = \partial/\partial \phi$ is odd under $\phi$ parity. Thus each term needs another odd term to cancel this sign and maintain $\mathcal L(x,\phi) = \mathcal L(x,-\phi)$. Hence in order for this term in the Lagrangian to respect the orbifold symmetry, exactly one of $\Psi_L$ or $\Psi_R$ must be $\phi$-odd while the other must be $\phi$-even. This is to say that the 4D Dirac spinor transforms as $\Psi(x) \rightarrow \gamma^5 \Psi(x)$ under $\phi$ parity.

Now we note that in the KK decomposition the odd chirality must be made up of only sines while the even chirality must be made up of only cosines (including the constant). In particular, the odd chirality cannot have a zero mode since $\sin(n\phi)|_{n=0} =0$.

And now we’ve done it. The low energy theory—that is the effective theory well below the mass of the first KK excitation—is chiral and we’ve finagled our way out of an awkward situation.

Some notes

There are a few notes I’d like to append to this brief discussion.

(1) First of all, one might still find the above situation a bit awkward because although we’ve written down chrial zero modes, there are higher KK modes of both chiralities. In general, orbifolding is a powerful tool for removing unwanted degrees of freedom. For the case of a scalar field, for example, the combination of the compactification boundary conditions and the orbifold symmetry forces half of the scalar modes to vanish. This is because the field must transform with a definite sign under $\phi$-parity. Thus the field must either be a scalar or pseudoscalar under $\phi \rightarrow -\phi$, and hence either the entire even or the entire odd KK tower vanishes.

(2) Secondly, though we have focused on a $S^1/\mathbb Z^2$ orbifold, the concept that orbifolding removes degrees of freedom is general for any orbifold.

(3) Note, also that a bulk fermion Dirac mass term, $m \bar \Psi \Psi$ is prohibited by the orbifold symmetry (just look at $\phi$-parity). Such a term does not transform properly under $\phi$-parity. To insert such a term, one must write $m \mathrm{sgn}(\phi) \bar \Psi\Psi$. This seems terribly odd, but it is suggestive of a higher dimensional Higgs mechanism where the 5D Higgs field is a $\phi$-pseudoscalar (i.e. it is $\phi$-odd).

(4) In my previous post I mentioned gauge-Yukawa unification using the example of a 5D bulk fermion with a 5D gauge field $A^M(x,\phi)$.

$S = \int d^4x \int (R d\phi) \bar\Psi i D_\mu\Gamma^\mu \Psi + \bar\Psi\gamma_5\partial_5\Psi + ig\bar\Psi A_5\gamma_5\Psi$

$\phantom{S} = 2\pi R\int d^4x \sum^{\infty}_{n=-\infty} \bar\psi^{(n)} i ( \gamma^\mu\partial_\mu - i\frac{n}{R}\gamma_5 )\psi^{(n)} + ig \bar\psi^{(n)}\gamma_5 A_5\psi^{(n)} +\cdots$

We see that the $A^5$ field, a 4D scalar, must be a $\phi$ pseudoscalar. Let uspick an `almost axial gauge’ where $A^5$ is taken to be independent of $\phi$, $A^5(x,\phi) = A^5(x)$. Thus, forcing $A^5$ to be a pseudoscalar eliminates it completely (in this gauge). Unfortunately, our gauge-Yukawa candidate Higgs boson was the sacrifice for imposing a chiral low energy theory.

References

For more information, see Professor Raman Sundrum’s 2004 TASI lecture on extra dimensions, in particular section 3 and 5, which I have followed closely. Also see Dienes’ lecture in the 2002 TASI proceedings (not on the arXiv, but available unofficially online if you look hard enough). Special thanks to M.R. for some clarification on the form of the 5D Dirac mass term.