PHYSICS BY GPT-J

RESULT
"Three Fermion Generations from Octonions" Phys. Rev. D 76, 045015 (2007).

[^1]: E-mail: [email protected]

[^2]: E-mail: [email protected]

[^3]: E-mail: [email protected]

[^4]: The scalar curvature $R$ is a combination of the scalar curvature of the underlying space and of the other terms.

[^5]: The integral is evaluated over the whole space.

[^6]: The multiplicity of roots is not relevant here.

[^7]: A positive root of $M_0$ is given by $i\, {\cal M}$ and can be expressed as a combination of the roots $\pm i\, {\cal M}_k$, which are also positive, $\pm \Delta_k\, {\cal M}_k$.

[^8]: The parameter $\epsilon$ is related to the compactification scale $\Lambda_{c}$ and to the relation between $R_0$ and $\mu_0$.

[^9]: This makes sense only for the choice of a constant scale for the two vectors $Z$ and $\zeta$.

[^10]: Strictly speaking, we have to consider separately the case when $c_1<0$, because of the $\frac{1}{c_1}$ factor in the rhs of (\[rhos1\]). This could be in principle included in the calculations we perform.

[^11]: We actually have to normalize the integration measure in this case so that it is 1 at the maximum value of $\xi$ that we are interested in.

[^12]: This statement is true for a constant scale for the two vectors $Z$ and $\zeta$ as well.

[^13]: In general, one expects $p_4>0$ from unitarity bounds on the scattering of four massless scalar particles in flat Minkowski space.

[^14]: This statement is true for a constant scale for the two vectors $Z$ and $\zeta$ as well.

[^15]: More explicitly, we are interested in the leading logarithmic approximation in the limit of large $\Lambda_{c}$, with the relative corrections being subleading as $1/\Lambda_{c}$ in this limit.

[^16]: If $p_4<0$ this is not possible. The condition (\[n\]) is necessary for the problem to make sense. If it is not satisfied, one would have $r\le 0$ which is a contradiction.

[^17]: An alternative to this method to derive the constraint (\[rho\]) is to consider the contribution to $W_\mu W^\mu$ from the gauge sector.

[^18]: We thank G. Parisi for pointing this out.

[^19]: Note that the potential is logarithmically divergent if ${\cal M}^2=0$, i.e. for one of the roots. This does not arise in our analysis since we are only interested in the deep infrared limit of the theory.

[^20]: This is the reason why we use the minus sign for the condensate. It makes the eigenvalue equation for the modes to be fulfilled for $s>0$, instead of $s<0$.

[^21]: We thank S.S. Gubser for this comment.

[^22]: In particular, it can be shown that $\rho$ is constant in the deep infrared limit, which can be taken as a proof of absence of black hole formation in the dual field theory.

[^23]: Recall that the condensates are constrained by unitarity. In the case of $\Delta_4$ this constraint is the fact that the four particle scattering cross-section has to be finite. This is not the case of the condensates here.

[^24]: In the case $\sigma^\mu=0$ this corresponds to the solution where $f^\prime=1$ and $f$ is a constant. However, the contribution to the action from this solution is zero, due to the cancellation between the contributions from the first and from the last term in (\[etamax\]).

[^25]: This is not the case of the four-vector condensate $\Delta_4$ considered in this paper.

[^26]: We thank A. Fring for bringing this to our attention.

[^27]: One can alternatively perform the calculation in the eikonal approximation with respect to the angles between the momenta of the condensate and of the gauge bosons.

[^28]: If $g=0$, the string theory has an SU(2) symmetry. This is equivalent to a choice of a reference point in the $W$ plane. See also the footnote \[sec:su2\].

[^29]: Note that in the case of only the gauge fields, $\Delta_4$ is not equal to zero.

[^30]: We thank S. S. Gubser for this comment.

[^31]: The constant scale $g$ in (\[alpha\]) is not relevant for the calculation, since it can be included in the normalization of the integral measure.

[^32]: We thank J. Casalderrey-Solana for this comment.

[^33]: Note that the sum runs only over the positive roots.

[^34]: Note that $d_\alpha$ is the positive root of $M_0$ with respect to which we are normalizing the measure, so that the sum runs over the positive roots.

[^35]: We thank S. S. Gubser for this comment.

[^36]: This holds in particular when $R_0$ is chosen to be negative and smaller than the $AdS_5$ radius, as in our analysis of section \[sec:examples\].

[^37]: This is easily understood by considering the expression (\[eta4\]). Note that the ratio $\frac{1}{\sqrt{-g_5}}\frac{1}{\sqrt{-g_4}}$ is independent of the choice of the coordinates (since they are related only by a conformal transformation) and it is equal to $\sqrt{-g_5}\,\sqrt{-g_4}=\mu^2$ (we are taking the same normalization for the coordinates in (\[eta4\]) as in (\[etamax\])).

[^38]: For small values of $r$ (of the order of $\mu^2$), there is a contribution to the potential proportional to $r^2\log(r^2/\mu^2)$. However, this is a finite contribution and is therefore irrelevant for our discussion.

[^39]: In fact, in the small radius limit, $W_\mu W^\mu=0$, the expression (\[eta4\]) can be simplified to $$\eta = \frac{1}{2\, c_1} \, r\, (\, 4 \, \mu^4 - c_1 \, \Delta_2^2 + c_2 \, r^2 \, ) \,.
\label{smallr}$$ Note that for $r>0$ there are no contributions from $\Delta_4$ and $\Delta_6$ and that the only relevant contributions come from the first two terms. For this reason, in the above formula, we have included the term proportional to $r^2$. This term makes the r.h.s. of (\[smallr\]) be positive definite for $r>0$.

[^40]: See also the footnote \[sec:su2\].

[^41]: This is easily understood by considering the expression (\[eta4\]). Note that the ratio $\frac{1}{\sqrt{-g_5}}\frac{1}{\sqrt{-g_4}}$ is independent of the choice of the coordinates (since they are related only by a conformal transformation) and it is equal to $\sqrt{-g_5}\,\sqrt{-g_4}=\mu^2$ (we are taking the same normalization for the coordinates in (\[eta4\]) as in (\[etamax\])).

[^42]: For small values of $r$ (of the order of $\mu^2$), there is a contribution to the potential proportional to $r^2\log(r