Given $n$-dimensional de Sitter or Anti-de Sitter space, $dS_n$ or $AdS_n$, what are the homotopy groups $\pi_m(dS_n)$ and $\pi_m(AdS_n)$ and how does one calculate such things?
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$\begingroup$ What's your interest on these homotopy groups? $\endgroup$– Fernando MuroCommented Sep 15, 2011 at 19:35
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De Sitter space $dS_n$ is homeomorphic to $\mathbb{R} \times S^{n - 1}$, so its homotopy groups are those of $S^{n - 1}$, which are intractably complicated when $n > 2$.
Anti de Sitter space $AdS_n$ is homeomorphic to $\mathbb{R}^{n - 1} \times S^1$, so its homotopy groups are those of $S^1$: $\pi_1(AdS_n) = \mathbb{Z}$, $\pi_m(AdS_n) = 0$ when $m > 1$.
The general rule for (finite dimensional) manifolds is that unless it has contractible universal cover (i.e., unless it is a $K(\pi, 1)$), there is no hope for computing its homotopy groups explicitly.
answered Sep 15, 2011 at 16:59
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2$\begingroup$ It may be that what the OP calls anti de Sitter space is actually the universal cover of what you have called anti de Sitter space (cf. Hawking and Ellis, for instance). If so, then it is contractible. $\endgroup$ Commented Sep 15, 2011 at 18:12