The <AdS-CFT correspondence> maps the vacuum to empty anti-de Sitter spacetime and the stress-tensor conformal family to one-graviton normal modes. Therefore
$$
\boxed{n_g(E_0)=0,
\qquad n_g(E_1)=1,
\qquad n_g(E_2)=1.}
$$
The distinction between $E_1$ and $E_2$ is a different excitation within the same one-particle conformal family, not an additional graviton.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 354 / 1 / a / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-354/1/a/iv/solution