Ricci tensor equals a constant times the metric

[Pacopag]

Hello;
What does it mean physically if I have?
$$R_{a b} = Ag_{a b}$$

I think it means that my manifold is an n-sphere (i.e. if A is positive),
or it is AdSn (i.e. if A is negative). Is this correct?

 

[Pacopag]

I think I got it. What I wrote above is correct for some particular values of A;
$$n(n-1) \over r^2$$ where r is the radius of the n-sphere.
Put a minus sign in front of this for AdSn

 

[kdv]

I think I got it. What I wrote above is correct for some particular values of A;
$$n(n-1) \over r^2$$ where r is the radius of the n-sphere.
Put a minus sign in front of this for AdSn

Are you sure?

I thought that the equations were the following.

For maximally symmetric spaces, we have (n= number os spacetime dimensions)

$$R_{\rho \sigma \mu \nu} = C (g_{\rho \mu} g_{\sigma \nu} - g_{\rho \nu} g_{\sigma \mu})$$

By contracting, you can show that

$$C = \frac{R}{n(n-1)}$$
whre R is the Ricci scalar.
Therefore

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


Thsi agrees with eq 3.191 of Carroll

Now, this means

$$R_{\sigma \nu} = \frac{R}{n} g_{\sigma \nu}$$

 

[Pacopag]

To be honest, I don't really know what I'm talking about. But everything you wrote seems correct. But now when I find R for an n-sphere of radius r I find that
$$R={n(n-1)\over r^2}$$. So we get
$$R_{a b}={n-1\over r^2}g_{a b}$$
Oh wait! That's not what I had above. I had an extra factor of n on the right-hand side.

Thank you kdv