Musings

We’ll work in Euclidean signature. Let $X$ be a Riemannian 4-manifold, and $V\to X$ a complex vector bundle with unitary connection. $$D: \Gamma(S^+\otimes V) \to \Gamma(S^- \otimes V)$$ is the usual chiral Dirac operator. Let’s also define $$\tilde{D}: \Gamma(S^-\otimes V^*) \to \Gamma(S^+ \otimes V^*)$$ and a nonderivative (scalar) operator, $\Phi\in\Gamma(Sym^2 V^*)$, or equivalently $$\Phi: \Gamma(V)\to \Gamma(V^*)$$ Putting these together, we can form $$\mathcal{D} = \begin{pmatrix}\Phi&\tilde{D}\\ D&\Phi^\dagger\end{pmatrix}: \Gamma(S^+\otimes V\oplus S^-\otimes V^*)\to \Gamma(S^+\otimes V^*\oplus S^-\otimes V)$$ and, finally, the self-adjoint operator $$\hat{\mathcal{D}} = \begin{pmatrix}0&\mathcal{D}\\ \mathcal{D}^\dagger & 0\end{pmatrix}$$

The Euclidean action for the fermions is

where $(\cdot,\cdot)$ denotes the natural skew-bilinear pairing, $S^\pm\otimes S^\pm \to \mathbb{C}$.

For the bosons one writes the nonlocal action

for some positive, but otherwise unspecified function $f$.

The particular case of interest is to take $V$ to be associated to a $G=SU(3)\times SU(2)\times U(1)$ principal bundle over $X$, via the representation $R=R_0^{\oplus 3}$, where

For $\Phi$, one takes $$\Phi = (\lambda'_d +\lambda'_e) \phi + (\lambda'_u {\color{red}+\lambda'_N}) \phi^* {\color{red}+M}$$ for $\phi\in \Gamma(U)$, where $U$ is the bundle associated to the representation $R_h=(1,2)_{-1/2}$. The $\lambda'_{u,d,e,N}$ are $G$-invariant elements $$\begin{gathered} \lambda'_d,\lambda'_e\in R_h^*\otimes Sym^2 R^* \\ \lambda'_u,{\color{red}\lambda'_N}\in R_h \otimes Sym^2 R^* \\ \end{gathered}$$ That is, they are Clebsch-Gordon coefficients² multiplied by coupling constants. ${\color{red}M}$ is a constant matrix, corresponding to the $G$-invariant subspace of $Sym^2 R^*$. It provides a Majorana mass term for the right-handed neutrinos, the ${\color{red} (1,1)_0 }\in R$.

Now, there’s a certain amount of mumbo-jumbo about interpreting $S^+\otimes V$ as the bundle of chiral spinors on a certain noncommutative spacetime and $\mathcal{D}$ as the associated chiral Dirac operator. But I don’t see that there’s much of anything to be gained thereby. Just take (1),(2) as the definition of the theory (at least at the classical level) and go from there.

Now, (2) is pretty uselessly nonlocal (not to mention under-specified). But you can do a heat-kernel expansion of it, and recover a local effective action, valid at low energies, $E\ll \Lambda$. The lowest-order terms in this expansion depend only on the first few moments of the function $f$. The complete expansion, up to dimension-4 operators, can be found in equation (3.41) of Chamseddine, Connes and Marcolli.

For instance, the Yang-Mills kinetic terms just depend on $f_0\equiv f(0)$. One reads off the $SU(3)\times SU(2)\times U(1)$ gauge couplings

The fact that the gauge couplings are “unified” is no surprise. $R$ consists of complete $SU(5)$ multiplets, so the “induced action”, (2) for the gauge fields is $SU(5)$-symmetric.

Slightly more interesting is that the kinetic term for the Higgs field is also proportional to $f_0$. The canonically-normalized Higgs fields $$\varphi= \sqrt{\tfrac{a f_0}{2\pi^2}}\phi$$ where $$a = tr(\lambda'^\dagger\lambda')\equiv 3(\lambda'_u^\dagger \lambda'_u+\lambda'_d^\dagger \lambda'_d)+ \lambda'_e^\dagger \lambda'_e{\color{red}+\lambda'_N^\dagger \lambda'_N}$$ So the physical Yukawa couplings $$\lambda = \frac{\lambda'}{\sqrt{a f_0/2\pi^2}}$$ satisfy

The Higgs potential depends, as well, on $f_2$.

where

The Einstein-Hilbert term also depends on $f_2$

There’s also a cosmological constant term, which depends on $f_4$, but let’s skip that for now.

What is interesting is that the heat-kernel expansion did not give us the most general possible gauge-invariant low energy effective action. Instead, a-priori independent couplings obeyed relations, given in (4),(5),(6)(7) and (8), for any choice of the function $f$.

While we have no clue how to quantize an action like (1)+(2), we do know how to treat an effective field theory. We should interpret above relations on the coupling constants as holding for the running coupling constant, evaluated at the cutoff scale, $\Lambda$. One might have hoped that renormalization would preserve the form of the action (1)+(2), and that divergences could be absorbed into a redefinition of the function $f$. Clearly, that does not happen in the effective theory. But perhaps that is just as well, because these relation clearly don’t hold in the real world at low energies.

Given the freedom that we do have, can we ensure that the restricted set of couplings, stemming from (1)+(2), give the correct physics, after being run down to low energies?

Somehow or other, by tuning $\Lambda f_2$ and $M$, we need to make (7) electroweak scale, while making (8) Planck-scale. That requires that $M$ be Planck scale (we shan’t worry about fine-tuning). But choosing $M$ be Planck scale would seem to preclude getting acceptable neutrino masses (see below) But let’s press on.

(4) suggests identifying $\Lambda$ as the GUT scale. Of course, that’s a little problematic, as the gauge couplings don’t actually unify (with just the SM degrees of freedom).

If the couplings did unify, then we could identify $g$ as the GUT coupling and (5) would give a sum rule on fermion masses-squared. Connes et al state this as an upper bound on the top mass, which is a rather odd way to state the result.

Similarly, (6) would yield a prediction for the Higgs mass. Again, eliding the small problem that the gauge couplings don’t actually unify, the claimed value for the Higgs mass is $$m_H =171.6\pm 5\text{GeV}$$

More fundamental, though, than the difficulty of getting the electroweak scale to come out right, or the lack of coupling constant unification, is that we don’t have a clue how the treat (1)+(2) as a quantum theory. Indeed, without a principle for choosing $f$, we don’t even know what theory we wish to quantize.

Update (8/4/2008): Uniqueness

I forgot to mention one really elementary point about their construction that confuses me. They start with the involutive algebra $$\mathcal{A}_{LR} = \mathbb{C}\oplus \mathbb{H}\oplus \mathbb{H}\oplus M_3(\mathbb{C})$$ The direct sum of all irreducible odd $\mathcal{A}_{LR}$ bimodules, $\mathcal{M}_F$, can be written as $$\mathcal{M}_F = \mathcal{E} \oplus \overline{\mathcal{E}}$$ where $\mathcal{E}$ is a bimodule³ of complex dimension 16, and $\mathcal{M}_F$ has a real structure, given by the antilinear map $$J(\xi,\overline{\eta}) = (\eta,\overline{\xi})$$ which satisfies $$J^2 =1,\quad u b = J b^* J u,\quad \forall\, u\in \mathcal{M}_F,\, b\in \mathcal{A}_{LR}$$

The way Connes et al read off the gauge group (and fermion representation) is to pick a certain subalgebra $\mathcal{A}_{F}\subset \mathcal{A}_{LR}$. The gauge group is then the group of unitaries, $G = SU(\mathcal{A}_{F})$ $$U(\mathcal{A}) = \{u\in \mathcal{A} | u u^* = u^* u =1\}$$ and $SU(\mathcal{A}_{F})\subset U(\mathcal{A}_{F})$ is the subgroup of determinant 1, when acting on $\mathcal{E}$.

But which subalgebra to choose? They take

embedded as $(\alpha, q, m)\mapsto (\alpha, q,\alpha,m)$, which yields $G= SU(3)\times SU(2)\times U(1)$, with $\mathcal{E}$ transforming as the representation $R_0$, defined above. But a different choice, which also seems to satisfy all of their criteria, is to take

embedded as

This leads to the gauge group $$G' = SU(3)\times SU(2)\times U(1) \times U(1)_{B-L}$$ So I don’t really understand the uniqueness theorem (proposition 2.11 of Chamseddine, Connes and Marcolli) that leads to (9).

Evidently, I’m missing something …

Update (8/5/2008): The Electroweak Scale

Perhaps it’s worth spelling out more explicitly the problem, alluded to above, with getting the electroweak scale and neutrino masses to work out. Start with (8), which says $$\frac{1}{8\pi G_N} = \frac{4 f_2 \Lambda^2}{\pi^2} - \frac{tr(M^\dagger M)}{48 g^2},$$ solve for $f_2 \Lambda^2$, and plug into (7):

We want $v\sim 250$GeV, while $\frac{1}{8\pi G_N}\sim (2.4\times 10^{18} \text{GeV})^2$. Now, the neutrino masses are $$m_\nu \sim \frac{\lambda_N^2 v^2}{M}$$ To get $m_\nu \sim 5\times 10^{-11}$ GeV, with $|\lambda_N|^2 \lesssim 1$ (for a perturbative analysis to be valid), we need $M \lesssim 10^{15}$ GeV. But that’s incompatible with (12), which requires $M\sim 10^{17}$ GeV.

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