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Regarding the unit radius $AdS_{4}$ space, the metric in global coordinates, is given by:

$$ds^{2}_{AdS_{4}}=\frac{1}{\cos^{2}{\rho}}[dt^{2}-d\rho^{2}-\sin^{2}\rho d\Omega_{2}^{2}]$$

where $$d\Omega_{2}^{2}=d\theta^{2}+\sin^{2}\theta d\phi^{2}$$

Given the metric I tried to compute the volume form:

$$w^{0\rho\theta\phi}=\sin^{2}\rho \sin^{2}\theta d\rho\wedge d\theta\wedge d\phi$$

Am I mistaken to this point? Should I build it in another way?

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  • $\begingroup$ Actually I think I am indeed mistaken. Given the metric, it should rather be: $$\frac{sin^{4}\rho sin^{2}\theta}{cos^{8}\rho}dx^{0}\wedge d\rho\wedge d\theta\wedge d\phi$$ Can anyone enlighten me about this? Also, downvoting the question without a single comment has absolutely no meaning. $\endgroup$
    – Jordan
    Commented May 10, 2016 at 14:29
  • $\begingroup$ Hi Jordan. Not my downvote, but your question does look like a check my work question and this is off topic here. If you could expand your question a bit, e.g. explain why you think you might be mistaken, that might make people more receptive. $\endgroup$
    – John Rennie
    Commented May 10, 2016 at 15:42
  • $\begingroup$ Well, the mistake had to do with my lack of knowledge on differential forms. he correct answer is: $$ \frac{sin^{2}\rho sin\theta}{cos^{4}\rho}dx^{0}\wedge dx^{\rho}\wedge dx^{\theta}\wedge dx^{\phi}$$ given the fact that the volume form is given by: $$dV=\sqrt{\vert det(g)\vert}dx^{0}\wedge...\wedge dx^{d-1}$$ $\endgroup$
    – Jordan
    Commented May 11, 2016 at 21:41

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