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--- advanced_tools:group_theory:desitter [2018/03/24 11:08]
jakobadmin [Why is it interesting?]
+++ advanced_tools:group_theory:desitter [2018/03/24 11:09]
jakobadmin [Student]
@@ Line 73: / Line 73: @@
 <tabbox Student>
-<note tip>
+<blockquote>
-In this section things should be explained by analogy and with pictures and, if necessary, some formulas.
+The group $SO(d,1)$ moves points on $dS^d$ around. We thus conclude that, just like the sphere, deSitter spacetime is maximally symmetric. So, according to the general theory of maximally symmetric spaces explained in chapter IX.6, the Riemann curvature tensor $R_{\mu\nu\lambda\sigma}$ must be equal to $(g_{\mu\lambda}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\lambda})$ up to an overall constant. [...] Then, by dimensional analysis, we must have
-</note>
+$$ R_{\mu\nu\lambda\sigma} =\frac{1}{L} (g_{\mu\lambda}g_{\nu\sigma}-g_{\mu\sigma}g_{\nu\lambda}) $$
+[...] de Sitter spacetime is a solution of Einstein's field equation $R_{\mu \nu}=8\pi G\Lambda g_{\mu\nu}$, with a positive cosmological constant given by
+$$ 8\pi G\Lambda = \frac{3}{L}$$
+[...]
+Topologically, de Sitter spacetime is $R \times S^3$, with a spatial section given by $S^3$, as just explained. In contrast, we know from chapters VI.2 and VI.5 that Einstein's field equation with a positive cosmological constant leads to $ds^2=-dt^2+e^{2Ht}(dx^2+dy^2+dz^2)$ with the Hubble constant given by $H= \left( \frac{8\pi G}{3}\Lambda\right)^{1/3}$, or in terms of the de Sitter length, $H=1/L$.
+<cite>page 627 in Einstein Gravity in a Nutshell - A. Zee</cite>
+</blockquote>
 <tabbox Researcher>

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