A survey of spectral models of gravity coupled to matter

Abstract

This is a survey of the historical development of the spectral Standard Model and beyond, starting with the ground breaking paper of Alain Connes in 1988 where he observed that there is a link between Higgs fields and finite noncommutative spaces. We present the important contributions that helped in the search and identification of the noncommutative space that characterizes the fine structure of space-time. The nature and properties of the noncommutative space are arrived at by independent routes and show the uniqueness of the spectral Standard Model at low energies and the Pati–Salam unification model at high energies.

Dedicated to Alain Connes

Appendix: Pati–Salam model: potential analysis

We here include the scalar potential analysis for the composite Pati–Salam model, as described in Section 5.2 above.

If there is unification of lepton and quark couplings, then ρ = 1 so that the $\Sigma ^I_J$-field decouples. In that case we have

where we have absorbed some constant factors by redefining the couplings λ _H, λ _{H Σ}, and λ _Σ.

We choose unitarity gauge for the Δ and ϕ-fields, in the following precise sense.

Lemma 6

For each value of the fields $\{ \phi _{\dot a}^b, \Delta _{\dot a I}\}$ there is an element (u _R, u _L, u) ∈ SU(2)_R × SU(2)_L × SU(4) such that

$$\displaystyle \begin{aligned} u_R \begin{pmatrix} \phi_{\dot 1}^1 & \phi_{\dot 1}^2 \\ {} \phi_{\dot 2}^1 & \phi_{\dot 2}^2 \end{pmatrix} u_L^* & = \begin{pmatrix} h & 0 \\ 0 & \chi \end{pmatrix} \end{aligned} $$

and

$$\displaystyle \begin{aligned} u_R \begin{pmatrix} \Delta_{\dot 1 1}& \Delta_{\dot 1 2}& \Delta_{\dot 1 3}& \Delta_{\dot 1 4}\\ \Delta_{\dot 2 1}& \Delta_{\dot 2 2}& \Delta_{\dot 2 3}& \Delta_{\dot 2 4}\end{pmatrix} u^t &= \begin{pmatrix} 1+\delta_0 & 0& 0 & 0 \\ \delta_1 & \eta_1& 0 &0 \end{pmatrix} \end{aligned} $$

where h, δ ₀, δ ₁, η ₁ are real fields and χ is a complex field.

Proof

Consider the singular value decomposition of the 2 × 2 matrix $(\phi _{\dot a}^b)$:

$$\displaystyle \begin{aligned} \begin{pmatrix} \phi_{\dot 1}^1 & \phi_{\dot 1}^2 \\ {} \phi_{\dot 2}^1 & \phi_{\dot 2}^2 \end{pmatrix} = U \begin{pmatrix} h & 0 \\ 0 & k \end{pmatrix} V^* \end{aligned}$$

for unitary 2 × 2 matrices U, V  and real coefficients h, k. If we define

$$\displaystyle \begin{aligned} u_R &= \begin{pmatrix} 1 & 0 \\ 0 & \det U\end{pmatrix}U^* \in SU(2)_R\\ u_L &= \begin{pmatrix} 1 & 0 \\ 0 & \det V\end{pmatrix} V^*\in SU(2)_L \end{aligned} $$

it follows that

$$\displaystyle \begin{aligned} u_R \begin{pmatrix} \phi_{\dot 1}^1 & \phi_{\dot 1}^2 \\ {} \phi_{\dot 2}^1 & \phi_{\dot 2}^2 \end{pmatrix} u_L^* & = \begin{pmatrix} h & 0 \\ 0 & k \det UV^* \end{pmatrix} =: \begin{pmatrix} h & 0 \\ 0 & \chi \end{pmatrix}. \end{aligned} $$

Next, we consider $\Delta _{\dot aI}$ and write

$$\displaystyle \begin{aligned} \left(\Delta_{\dot aI}\right) = \begin{pmatrix} u_1^* \\ u_2^* \end{pmatrix},\qquad \text{with } u_a^* = \begin{pmatrix} \Delta_{\dot a 1}& \Delta_{\dot a 2}& \Delta_{\dot a 3}& \Delta_{\dot a 4} \end{pmatrix} \end{aligned}$$

for a = 1, 2. We may suppose that the vectors u ₁, u ₂ are such that their inner product $u_1^* u_2$ is a real number. Indeed, if this is not the case, then multiply $\Delta _{\dot aI}$ by a matrix in SU(2)_R as follows:

$$\displaystyle \begin{aligned} \begin{pmatrix} u_1^* \\ u_2^* \end{pmatrix}\mapsto \begin{pmatrix} \alpha & 0 \\ 0 & \alpha^* \end{pmatrix}\begin{pmatrix} u_1^* \\ u_2^* \end{pmatrix} = \begin{pmatrix} \alpha u_1^* \\ \alpha^* u_2^* \end{pmatrix}. \end{aligned}$$

Now the inner product is $(\alpha ^* u_1)^* \alpha u_2 = (\alpha )^2 u_1^* u_2$ and we may choose α so as to cancel the phase of $u_1^* u_2$. Moreover, this transformation respects the above form of $\phi _{\dot a}^b$ after a SU(2)_L-transformation of exactly the same form:

$$\displaystyle \begin{aligned} \begin{pmatrix} h & 0 \\ 0 & \chi \end{pmatrix} \mapsto \begin{pmatrix} \alpha & 0 \\ 0 & \alpha^* \end{pmatrix} \begin{pmatrix} h & 0 \\ 0 & \chi \end{pmatrix} \begin{pmatrix} \alpha & 0 \\ 0 & \alpha^* \end{pmatrix} ^* = \begin{pmatrix} h & 0 \\ 0 & \chi \end{pmatrix} . \end{aligned}$$

Thus let us continue with the vectors u ₁, u ₂ satisfying $u_1^* u_2 \in \mathbb {R}$. We apply Gramm–Schmidt orthonormalization to u ₁ and u ₂, to arrive at the following orthonormal set of vectors {e ₁, e ₂} in $\mathbb {C}^4$:

$$\displaystyle \begin{aligned} e_1 = \frac{u_1 }{\| u_1\|}; \qquad e_2 = \frac{u_2 - \frac{u_1^* u_2}{\|u_1\|} u_1}{\|u_2 - \frac{u_1^* u_2}{\|u_1\|} u_1\|}. \end{aligned}$$

We complete this set by choosing two additional orthonormal vectors e ₃ and e ₄ and write a unitary 4 × 4 matrix:

$$\displaystyle \begin{aligned} U = \begin{pmatrix} e_1 & e_2 & e_3 & e_4 \end{pmatrix} \end{aligned}$$

The sought-for matrix u ∈ SU(4) is determined by

$$\displaystyle \begin{aligned} u^t = U \begin{pmatrix} 1_3 & 0 \\ 0 & \det U^* \end{pmatrix} \end{aligned}$$

so as to give

$$\displaystyle \begin{aligned} \left(\Delta_{\dot aI}\right) u^t = \begin{pmatrix} u_1^* e_1 & 0 & 0 & 0 \\ u_2^* e_1 & u_2^* e_2 & 0 & 0 \end{pmatrix} =: \begin{pmatrix} 1+\delta_0 & 0 & 0 & 0 \\ \delta_1 & \eta_1 & 0 & 0 \end{pmatrix} \end{aligned} $$

□

Remark 7

Note that this is compatible with the dimension of the quotient of the space of field values by the group. Indeed, the fields $\phi _{\dot a}^b$ and $\Delta _{\dot a I}$ span a real 24-dimensional space (at each manifold point). The dimension of the orbit space is then $24- \dim P$ with P a principal orbit of the action of SU(2)_R × SU(2)_L × SU(4) on the space of field values. This dimension $\dim P$ is determined by the dimension of the group and of a principal isotropy group.

First, we see that up to conjugation there is always a SU(2)-subgroup of SU(4) leaving $\Delta _{\dot aI}$ invariant: it corresponds to SU(2)-transformations in the space orthogonal to the vectors $\Delta _{\dot 1 I}$ and $\Delta _{\dot 2I}$ in $\mathbb {C}^4$. Moreover, one can compute that the isotropy subgroup of the field values

$$\displaystyle \begin{aligned} \begin{pmatrix} \phi_{\dot a}^b \end{pmatrix} = \begin{pmatrix}1 & 0 \\ 0 & 0 \end{pmatrix}; \qquad \begin{pmatrix} \Delta_{\dot a I} \end{pmatrix} = \begin{pmatrix}1 & 0& 0 & 0 \\ 1 & 1 & 0 & 0 \end{pmatrix} \end{aligned}$$

is given by $\mathbb Z_2 \times SU(2)$. Hence, the dimension of the principal orbit is 21 − 3 = 18 so that the orbit space is six-dimensional. This corresponds to the 4 real fields h, δ ₀, δ ₁, η ₁ and the complex field χ.

We allow for the color SU(3)-symmetry not to be broken spontaneously, hence we only choose unitarity gauge in the SU(2)_R × SU(2)_L × U(1)-representations. That is, we retain the row vector $\Delta _{\dot 2 I}$ for I = 1, …, 4 as a variable and write

$$\displaystyle \begin{aligned} \begin{pmatrix} \Delta_{\dot a I} \end{pmatrix} = \begin{pmatrix}\sqrt{w}+\delta_0/\sqrt w & 0& 0 & 0 \\ \delta_1/\sqrt{w} & \eta_1/\sqrt{w} & \eta_2/\sqrt{w} & \eta_3/\sqrt{w} \end{pmatrix} \end{aligned}$$

so that (η _i) forms a scalar SU(3)-triplet field (so-called scalar leptoquarks). The reason for the rescaling with $\sqrt {w}$ is that it yields the right kinetic terms for δ ₀, δ ₁, and η. Indeed, from the spectral action we then have

$$\displaystyle \begin{aligned} \frac{1}{2}\partial_{\mu}H_{\overset{.}{a} I\overset{.}{b}J}\partial^{\mu}H^{\overset{.}{a}I\overset{.}{b}J} &=\frac{1}{2} \partial_\mu \left(\Delta_{\overset{.}{a}J}\Delta_{\overset{.}{b}I}\right) \partial^\mu \left(\Delta^{\overset{.}{a}J}\Delta^{\overset{.}{b}J}\right) \\ &\sim \sum_{a=0}^1 \partial_\mu \delta_a \partial^\mu \delta^a + \partial_\mu \eta \partial^\mu \eta^* + \text{ higher order} \end{aligned} $$

The scalar potential becomes in terms of the fields h, χ, δ ₀, δ ₁, η _i:

$$\displaystyle \begin{aligned} &\mathcal{L}_{pot}(h,\chi, \delta_0,\delta_1,\eta) = - \mu^2 ( h^2 + |\chi|{}^2) -\nu^2 \left( (w+ \delta_0)^2 + \delta_1^2 + |\eta|{}^2 \right)^2 /w^2 \\ & \quad + \lambda_{H\Sigma} \left( (w+\delta_0)^2 h^2 + (\delta_1^2 + |\eta|{}^2) |\chi|{}^2 \right)\left( (w+ \delta_0)^2 + \delta_1^2 + |\eta|{}^2 \right) /w^2 \\ & \quad + \lambda_H \left( (w+\delta_0)^4 + 2 (w+ \delta_0)^2 \delta_1^2+ (\delta_1^2 +|\eta|{}^2)^2 \right)^2/w^4 + \lambda_\Sigma (h^4+ |\chi|{}^4) \end{aligned} $$

As we are interested in the truncation to the Standard Model, we look for extrema with 〈δ ₁〉 = 〈η _i〉 = 0, while setting 〈h〉 = v, 〈δ ₀〉 = 0, 〈χ〉 = x. Note that the symmetry of these vevs is

$$\displaystyle \begin{aligned} &\left\{ \left(\begin{pmatrix} \lambda & 0 \\ 0 & \lambda^* \end{pmatrix}, \begin{pmatrix} \lambda^* & 0 \\ 0 & \lambda \end{pmatrix}, \begin{pmatrix} \lambda^* & 0 \\ 0 & m \end{pmatrix}\right): \lambda \in U(1) , m \in SU(3) \right\} \\ &\quad \subset SU(2)_R \times SU(2)_L \times SU(4) \end{aligned} $$

In other words, SU(2)_R × SU(2)_L × SU(4) is broken by the above vevs to U(1) × SU(3).

The first derivative of V  vanishes for these vevs precisely if

$$\displaystyle \begin{aligned} 2v (w^2 \lambda_{H \Sigma} +2v^2 \lambda_\Sigma - \mu^2 ) = 0,\\ 4 x^3 \lambda_\Sigma - 2x \mu^2 = 0,\\ 4w (2w^2 \lambda_H + v^2 \lambda_{H \Sigma} - \nu^2) =0. \end{aligned} $$

This gives rise to the fine-tuning of v, w as in [16]:

$$\displaystyle \begin{aligned} w^2 \lambda_{H \Sigma} +2v^2 \lambda_\Sigma - \mu^2 , \qquad 2w^2 \lambda_H + v^2 \lambda_{H \Sigma} - \nu^2 \end{aligned}$$

choosing μ and ν such that the solutions v, w are of the desired orders. Moreover, we find that the vev for χ either vanishes or is equal to $x=\sqrt {\mu ^2/2\lambda _\Sigma }$. Note that this latter vev appears precisely at the entry k ^dh (or k ^eh) of the finite Dirac operator, which we have disregarded by setting ρ = 1.

If 〈χ〉 = x = 0, then the Hessian is (derivatives with respect to h, χ, δ ₀, δ ₁, η):

where the 1 ₃ is the identity matrix in color space, corresponding to the η-field. This Hessian is not positive definite so we disregard the possibility that 〈χ〉 = 0.

If $x=\sqrt {\mu ^2/2\lambda _\Sigma }$, then the Hessian is

which is positive-definite if

$$\displaystyle \begin{aligned} \lambda_{H\Sigma}^2 \geq 8 \lambda_H \lambda_\Sigma. \end{aligned} $$

(56)

Note that this relation may hold only at high-energies. The masses for χ, δ ₁, and η are then readily found to be:

$$\displaystyle \begin{aligned} m_\chi^2 &= 4w^2 \lambda_{H \Sigma}+8v^2 \lambda_\Sigma,\\ m_{\delta_1}^2 &= w^2 \frac{\lambda_{H \Sigma}^2}{\lambda_\Sigma} ,\\ m_{\eta}^2 &=w^2 \frac{ \lambda_{H\Sigma}^2 - 8 \lambda_H \lambda_\Sigma}{\lambda_\Sigma} . \end{aligned} $$

Under the assumption that $v^2 \approx 10^2 \operatorname {\mathrm {GeV}}, w^2 \approx 10^{11} \operatorname {\mathrm {GeV}}$ we have $m_\chi ^2 \approx 10^{11} \operatorname {\mathrm {GeV}}$ and $m_{\delta _1}^2, m_{\eta } \approx 10^{11} \operatorname {\mathrm {GeV}}$.

The (non-diagonal) h and δ ₀ sector has mass eigenstates as in [16]:

$$\displaystyle \begin{aligned} m_\pm^2 &= 16 w^2\lambda_H +4 v^2 \lambda_\Sigma \\ &\quad \pm 4\sqrt{16 w^{4} \lambda_H^2 + v^4 \lambda_\Sigma^2 + 4v^2 w^2 \left(\lambda_{H \Sigma}^2 - 2 \lambda_H \lambda_\Sigma \right)} \end{aligned} $$

Under the assumption that v ² ≪ w ² we can expand the square root:

$$\displaystyle \begin{aligned} &4 \sqrt{ 16 \lambda_H^2 w^{4} \left(1 + \frac{\lambda_\Sigma^2}{\lambda_H^2}\frac{v^4}{w^{4}} + \frac{\lambda_{H \Sigma}^2 - 2 \lambda_H \lambda_\Sigma}{4\lambda_H^2} \frac{v^2}{w^2}\right)}\\ &\quad \approx 16 \lambda_H w^2 \left(1 +\frac{\lambda_{H \Sigma}^2 - 2 \lambda_H \lambda_\Sigma}{8\lambda_H^2} \frac{v^2}{w^2}\right) \\ &\quad = 16\lambda_H w^2 +\frac{2\lambda_{H \Sigma}^2}{\lambda_H} v^2 - 4\lambda_\Sigma v^2. \end{aligned} $$

Consequently,

$$\displaystyle \begin{aligned} m_+ &\approx 32 \lambda_H w^2 + 2 \frac{2\lambda_{H \Sigma}^2}{\lambda_H} v^2, \\ m_- &\approx 8 \lambda_\Sigma v^2 \left(1- \frac{\lambda_{H \Sigma}^2}{4\lambda_H \lambda_\Sigma} \right). \end{aligned} $$

which are of the order of 10¹¹ and $10^2 \operatorname {\mathrm {GeV}}$, respectively. This requires that we have at low energies

$$\displaystyle \begin{aligned} 4\lambda_H \lambda_\Sigma \geq \lambda_{H \Sigma}^2,\end{aligned} $$

(57)

which fully agrees with [16] when we identify δ ₀ ≡ σ and with the couplings related via

$$\displaystyle \begin{aligned} \lambda_H = \frac 14 \lambda_\sigma , \qquad \lambda_{H \Sigma} = \frac 12 \lambda_{h\sigma}, \qquad \lambda_\Sigma = \frac 14 \lambda_h\end{aligned} $$

Note the tension between Equations (57) and (56), calling for a careful study of the running of the couplings in order to guarantee positive mass eigenstates at their respective energies.

We have summarized the scalar particle content of the above model in Table 1. In terms of the original scalar fields $\phi _{\dot a}^b$ and $\Delta _{\dot aI}$ the vevs are of the following form:

$$\displaystyle \begin{aligned} \begin{pmatrix} \phi_{\dot a}^b \end{pmatrix} &= \begin{pmatrix} v & 0 \\ 0 & \sqrt{\mu^2/2 \Lambda_\Sigma}\end{pmatrix}\\ \begin{pmatrix} \Delta_{\dot a I} \end{pmatrix} &= \begin{pmatrix} w & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} . \end{aligned} $$

This shows that there are two scales of spontaneous symmetry breaking: at 10¹¹–$10^{12} \operatorname {\mathrm {GeV}}$ we have

$$\displaystyle \begin{aligned} SU(2)_R \times SU(2)_L \times SU(4) \to U(1)_Y \times SU(2)_L \times SU(3) \end{aligned}$$

and then at electroweak scale (both v and μ) we have

$$\displaystyle \begin{aligned} U(1)_Y \times SU(2)_L \times SU(3) \to U(1)_Q \times SU(3) \end{aligned}$$

Table 1 Scalar particle content with SM-representations