Are Christoffel Symbols Considered Tensors?

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I'm sort of confused about Christofel symbols being called tensors. I thought that to be considered a tensor, the tensor had to obey the standard component transformation law. For example: Ga'b' = Lca'Ldb'Gcd
But the Christofel symbols don't obey this transformation law; their components transform in a different way (I would like to show it, but I don't know how; see exercise 10.3 of MWT).
What gives?
 
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No, Christoffel symbols are not tensors.
 
Oh ... thought they were
 
Christoffel symbols make the covariant derivative a tensor. Try transforming one.
 
I haven't seen a single book on GR which introduces the Christoffel symbols without immediately pointing out that they're not tensors.
 
Karl G. said:
I'm sort of confused about Christofel symbols being called tensors. I thought that to be considered a tensor, the tensor had to obey the standard component transformation law. For example: Ga'b' = Lca'Ldb'Gcd
But the Christofel symbols don't obey this transformation law; their components transform in a different way (I would like to show it, but I don't know how; see exercise 10.3 of MWT).
What gives?

They transform inhomogeneously under general coordinates transformations, i.e., not tensors. However, the inhomogeneous term in the transformation law vanishes if the coordinate transformations are LINEAR. So, they do transform as tensors with respect to all linear coordinate transformations.

sam
 
I asked a question here, probably over 15 years ago on entanglement and I appreciated the thoughtful answers I received back then. The intervening years haven't made me any more knowledgeable in physics, so forgive my naïveté ! If a have a piece of paper in an area of high gravity, lets say near a black hole, and I draw a triangle on this paper and 'measure' the angles of the triangle, will they add to 180 degrees? How about if I'm looking at this paper outside of the (reasonable)...
From $$0 = \delta(g^{\alpha\mu}g_{\mu\nu}) = g^{\alpha\mu} \delta g_{\mu\nu} + g_{\mu\nu} \delta g^{\alpha\mu}$$ we have $$g^{\alpha\mu} \delta g_{\mu\nu} = -g_{\mu\nu} \delta g^{\alpha\mu} \,\, . $$ Multiply both sides by ##g_{\alpha\beta}## to get $$\delta g_{\beta\nu} = -g_{\alpha\beta} g_{\mu\nu} \delta g^{\alpha\mu} \qquad(*)$$ (This is Dirac's eq. (26.9) in "GTR".) On the other hand, the variation ##\delta g^{\alpha\mu} = \bar{g}^{\alpha\mu} - g^{\alpha\mu}## should be a tensor...

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