# Apparent coincidence of spin-connection term and Higgs field?

In curved space time, there is a spin-connection term $$\overline{\psi}\gamma^\mu\sigma^{ab}\omega^{ab}_\mu\psi$$.

Here's my apparent problem. If there were no Higgs field and no gravity, all particles would be massless. And hence the left-handed and right handed electrons would uncouple and be seen as separate massless fermions with no connection to each other. It is only the higgs coupling that combines these in a pair with mass. As the higgs interaction turns the left-handed electron into a right handed one.

But the spin-connection term, for example, (as far as I can tell) mixes the left and right handed electrons. Even though, apparenly, these two particles have nothing to do with each other. (e.g. This predicts a left handed neutrino passing by a rotating black hole will experience some kind of torsion effects and can turn into a right-handed neutrino).

Worse still if we consider the Cabbibo mass mixing matrix, there is not pairing of the left-handed electron and right-handed but a pairing of the triplet of 3 generations of charged fermions with their counterparts.

e.g. we would get a term like $$\overline{\psi}^A(m^{AB}+\gamma^\mu\sigma^{ab}\omega^{ab}_\mu)\psi^B$$

Where $$m$$ is a mass mixing matrix e.g. for up-quarks.

Hence I don't see the justification for assuming that the spin-connection term pairs, say, left-handed up-quarks to right-handed up-quarks. Ignoring the Higgs term it could just as well pair left handed up-quarks to left-handed electrons. (e.g. this would predict a left handed quark passing by a rotating black hole would turn into a left-handed electron. Since we can't do the experiment how can we rule that out?)

So it seems to me there is a some unexplained coincidence occuring where the Higgs interactions are pairing up particles identically to the way the spin-connecton pairs up particles.

Or is there even an experiment to show how the chirality of a fermion is affected by gravitational torsion from a rotating gravitational mass? (Or perhaps some equivalent accelerating frame?)

Can you explain this?

My view is that a left-handed neturino in a gravitational field should stay a neutrino left-handed neutrino and there should be no mixing. Perhaps that could be achieved by replacing the term a pesudo-vector term: $$\overline{\psi}\gamma_5\gamma^\mu\sigma^{ab}\omega^{ab}_\mu\psi$$ (Just a guess)??

• A gamma and sigma make an odd number of gammas, insulating left from right, like a kinetic term, no? – Cosmas Zachos Mar 26 '20 at 2:38
• @Cosmas Zachos. I'm not sure. But that sounds promising. If that is true could you prove it in an answer? I tried working it out but I got a term involving $\gamma_5$ which switches the chiralities. So maybe I did it wrong? I would like to either prove that left and right are indeed insulated or if not there seems to be a problem. I would indeed expect a kinetic term since gravity is altering the momentum. – zooby Mar 26 '20 at 2:40
• Wait.. I think I understand. The gamma matrices are like this: $[[0,A],[A,0]]$ So an odd number of them would indeed insulate the result! Combined with the $\gamma_0$ which is like $[[0,1],[-1,0]]$ this would indeed uncouple the equations. Thanks! I wonder what I did wrong? – zooby Mar 26 '20 at 2:49

Frankly, I cannot follow every tangent in the discussion, nor should I attempt to comment on it. But the basic brutal fact that should consistently fit with all the rest is that odd numbers of visible gamma matrices insulate left from right chiralities inside spinor bilinears. This is a representation-independent statement and going to a specific representation might or might not be salutary.

Specifically, $$P_R = \frac{1 + \gamma^5}{2}, \qquad P_L = \frac{1 - \gamma^5}{2} ~,$$

so, e.g., the kinetic terms $$\overline{\psi_L} ~\gamma^\mu \partial_\mu \psi_L= \overline{\psi} P_R~\gamma^\mu \partial_\mu \psi_L = \overline{\psi} ~\gamma^\mu P_L~ \partial_\mu \psi \neq 0 , \hbox {whilst} \qquad \overline{\psi_R} ~\gamma^\mu \partial_\mu \psi_L= 0.$$

Likewise, covariant completions to them, $$\overline{\psi_L}\gamma^\mu\sigma^{ab}\omega^{ab}_\mu\psi_L= \overline{\psi}\gamma^\mu\sigma^{ab}\omega^{ab}_\mu P_L\psi \neq 0, \hbox {whilst} \qquad \overline{\psi_R}\gamma^\mu\sigma^{ab}\omega^{ab}_\mu\psi_L= 0.$$

So spin-connection completions are chirally coordinated with the gradients they complete, and much unlike mass terms.

• Yes. Thanks for your answer. I was counting the $\overline{\psi} = \psi^\dagger \gamma^0$ as an extra gamma matrix. So I was counting them as even instead of odd. But yes the same argument holds. Excellent answer thanks. – zooby Mar 26 '20 at 14:03
• OK, I lawyer-finessed by a "visible" to at least spook the reader about that. – Cosmas Zachos Mar 26 '20 at 14:39

From comment from Cosmas Zachos:

The gamma matrices in the Weyl representation are all of the form:

$$\gamma^\mu = \begin{bmatrix} 0 & A \\ B & 0 \end{bmatrix}$$

Thus an even power of them will be diagonal and the equations will decouple into left and right-handed fermions provided there is not mass terms. (Remembering the hidden gamma matrix $$\overline{\psi}=\psi^\dagger \gamma^0$$ )

Therefor a spinning black hole will not convert left-handed massless neutrinos into right-handed massless neutrinos after all!

In fact after some gamma matrix identities the term $$\overline{\psi}\gamma^\mu\sigma^{ab}\omega^{ab}_\mu\psi$$ should decompose into a vector and a pseudo-vector term: $$\overline{\psi}\gamma^\mu(\gamma_5 \Omega_\mu + i\Sigma_\mu)\psi$$. Not sure if both terms are non-zero. (I think $$\Sigma$$ must vannish as it has an $$i$$ so isn't Hermitian compatible).