**Question**

{ds}^2={d\psi}^2+\sin^2{\psi}\left({d\theta}^2+\sin^2{\theta}{d\phi}^2\right)

$$(a) Calculate the Christoffel connection coefficients. Use whatever method you like, but it is good practice to get the connection coefficients by varying the integral (3.49)

(b) Calculate the Riemann tensor ##R_{\ \ \ \sigma\mu\nu}^\rho## , Ricci tensor ##R_{\mu\nu}## and Ricci scalar ## R##.

(c) Show that (3.191) is obeyed by this metric, confirming that the 3-sphere is a maximally symmetric space (as you might expect.) 3.191 was$$

R_{\rho\sigma\mu\nu}=\frac{R}{n\left(n-1\right)}\left(g_{\rho\mu}g_{\sigma\nu}-g_{\rho\nu}g_{\sigma\mu}\right)

$$Calculating the Christoffel coefficients was easy because I could use the code I had written before. I did not take the advice "get good practice by varying the integral"! I forgot to use the result of exercise 3.3! Grrr. The Riemann tensor was gruelling. It has 81 components but in this case there are really only three which are independent and each of them generate only three more by index symmetries (as shown in the picture). However, every one of the 81 needs checking. We also notice that each component is ± a metric component and wonder if there is some relationship like the relation between these components similar to exercise 3.3. Perhaps a future challenge! The Ricci bits are almost trivial. For (c) I was able to write a simple bit of code to compare all 81 equations. It did the job.

### Answers

Note:A 1-sphere is a circle - the set of points in ##\mathbf{R}^2## at an equal distance from the origin.

A 2-sphere is the surface of a sphere - the set of points in ##\mathbf{R}^3## at an equal distance from the origin.

A 3-sphere is the set of points in ##\mathbf{R}^4## at an equal distance from the origin.

(a) Christoffel connection coefficients

\begin{align}

\Gamma_{\theta\theta}^\psi&=-\sin{\psi}\cos{\psi}&\phantom {10000}(5)\nonumber\\

\Gamma_{\phi\phi}^\psi&=-\sin{\psi}\cos{\psi}\sin^2{\theta}&\phantom {10000}(6)\nonumber\\

\Gamma_{\psi\theta}^\theta=\Gamma_{\theta\psi}^\theta&=\cot{\psi}&\phantom {10000}(7)\nonumber\\

\Gamma_{\phi\phi}^\theta&=-\sin{\theta}\cos{\theta}&\phantom {10000}(8)\nonumber\\

\Gamma_{\psi\phi}^\phi=\Gamma_{\phi\psi}^\phi&=\cot{\psi}&\phantom {10000}(9)\nonumber\\

\Gamma_{\theta\phi}^\phi=\Gamma_{\phi\theta}^\phi&=\cot{\theta}&\phantom {10000}(10)\nonumber

\end{align}(b) Riemann tensor components are

\begin{align}

R_{\ \ \ \theta\psi\theta}^\psi&=\sin^2{\psi}&\phantom {10000}(13)\nonumber\\

R_{\ \ \ \theta\theta\psi}^\psi&=-\sin^2{\psi}&\phantom {10000}(14)\nonumber\\

{R}_{\ \ \ {\phi\psi\phi}}^{\psi}&=\sin^2{\psi}\sin^2{\theta}&\phantom {10000}(15)\nonumber\\

R_{\ \ \ \phi\phi\psi}^\psi&=-\sin^2{\psi}\sin^2{\theta}&\phantom {10000}(16)\nonumber\\

R_{\ \ \ \psi\psi\theta}^\theta&=-1&\phantom {10000}(17)\nonumber\\

R_{\ \ \ \psi\theta\psi}^\theta&=1&\phantom {10000}(18)\nonumber\\

R_{\ \ \ \phi\theta\phi}^\theta&=\sin^2{\psi}\sin^2{\theta}&\phantom {10000}(19)\nonumber\\

R_{\ \ \ \phi\phi\theta}^\theta&=-\sin^2{\psi}\sin^2{\theta}&\phantom {10000}(20)\nonumber\\

R_{\ \ \ \psi\psi\phi}^\phi&=-1&\phantom {10000}(21)\nonumber\\

R_{\ \ \ \psi\phi\psi}^\phi&=1&\phantom {10000}(22)\nonumber\\

R_{\ \ \ \theta\theta\phi}^\phi&=-\sin^2{\psi}&\phantom {10000}(23)\nonumber\\

R_{\ \ \ \theta\phi\theta}^\phi&=\sin^2{\psi}&\phantom {10000}(24)\nonumber

\end{align}The Ricci tensor is twice the metric!

\begin{align}

R_{\mu\nu}=2g_{\mu\nu}&\phantom {10000}(35)\nonumber

\end{align}The Ricci scalar is 6

(c) was solved by VBA

Complete answer with more links in Ex 3.08 Vital statistics of a 3-sphere.pdf

Hi, this is the worst question I ever had to solve in my live and I am struggling with some points of it. How does your component of the Riemann tensor in the equation (19) has a sin^2(\psi) if the Christoffel symbols with \theta and \phi indices does not depend on \psi?

ReplyDeleteLook at equation (63) in the pdf where (19) is worked out. The last two terms of the expansion of ##R_{\ \ \ \phi\theta\phi}^\theta\ ## are ##\Gamma_{\lambda\theta}^\theta\Gamma_{\phi\phi}^\lambda-\Gamma_{\lambda\phi}^\theta\Gamma_{\phi\theta}^\lambda## both of which are summations over ##\lambda##. So Christoffel symbols with indices other than ##\theta,\phi## might be included. Inspecting the Christoffel symbols shows you that ##\Gamma_{\lambda\theta}^\theta\Gamma_{\phi\phi}^\lambda-\Gamma_{\lambda\phi}^\theta\Gamma_{\phi\theta}^\lambda=\Gamma_{\psi\theta}^\theta\Gamma_{\phi\phi}^\psi-\Gamma_{\phi\phi}^\theta\Gamma_{\phi\theta}^\phi##. The first two contain the index ##\psi##!

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