Directional Derivative of Ricci Scalar: Lev-Civita Connection?

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The discussion centers on the conditions required for the directional derivative of the Ricci scalar along a Killing Vector Field to equal zero. Participants debate whether the Levi-Civita connection is necessary, with some suggesting that the Lie derivative's fundamental nature may allow for alternative connections. It is noted that metric compatibility, expressed as the condition that the covariant derivative of the metric tensor vanishes, is a weaker requirement than the Levi-Civita connection. However, there is uncertainty about whether this condition alone ensures that the derivative of the curvature scalar along the Killing vector is zero. The conversation highlights the complexities of connections in relation to Killing fields and the Ricci scalar.
loops496
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I have a question about the directional derivative of the Ricci scalar along a Killing Vector Field. What conditions are necessary on the connection such that K^\alpha \nabla_\alpha R=0. Is the Levi-Civita connection necessary?
I'm not sure about it but I believe since the Lie derivative is more 'fundamental' than the covariant derivative it might be not necessary to have a Levi-Civita connection, but maybe I'm just conjecturing nonsense. Hope anyone can help me find an answer.
 
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I think you need ##\nabla_a g_{bc}=0##. If the Lie derivative of a field is zero then ##g_{ab}## is preserved on the integral curves. You do not need a connection for the Lie derivative.
 
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Hey Mentz114, thank for replying! Since you don't need a connection for the Lie derivative, and Killing Vector Fields depend upon the Lie derivative I suspect you don't need the Levi-Civita Connection, However for the derivation of such identity I used the Bianchi identities which rely on a torsion free connection. So I don't know wether you acually don't need it.
 
##\nabla_c g_{ab}=0## implies the LC connection.

I'm not sure if I'm answering your question ...
 
I think metric compatibility is a weaker condition, i.e. you can have various Riemannian connections without any being the LC. But that still does not guarantee that the derivative along the Killing of the curvature scalar is 0, or does it?
 
In an inertial frame of reference (IFR), there are two fixed points, A and B, which share an entangled state $$ \frac{1}{\sqrt{2}}(|0>_A|1>_B+|1>_A|0>_B) $$ At point A, a measurement is made. The state then collapses to $$ |a>_A|b>_B, \{a,b\}=\{0,1\} $$ We assume that A has the state ##|a>_A## and B has ##|b>_B## simultaneously, i.e., when their synchronized clocks both read time T However, in other inertial frames, due to the relativity of simultaneity, the moment when B has ##|b>_B##...

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