Graduate New Adler book on GR: Why do these coefficients go to zero?

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The discussion centers on a specific point from page 73 of a new Adler book regarding why mixed derivatives with affine connections vanish in a certain term. This vanishing is crucial for demonstrating that connections are not tensors. The terms involving second-order products of infinitesimal coordinate displacements are dropped, as only first-order terms are relevant for the analysis. The reasoning behind this simplification is clarified, leading to a better understanding of the mathematical concepts involved. The conversation concludes with appreciation for the explanation provided.
peasg
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This is page 73 of the book. As you can see, the mixed derivatives with the affine connections vanish in the second term. Why does that happen? This is used to prove that the connections are not a tensor, and i figured you could also reason it out even without making those terms vanish.

OBS: The derivatives are avaliated at P, for the reason that this is obtained via a taylor series of the transformation coefficients.
 
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peasg said:
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This is page 73 of the book. As you can see, the mixed derivatives with the affine connections vanish in the second term. Why does that happen?
The terms ##\left( \dfrac{\partial^2 \bar x^j}{\partial x^l \partial x^i } \right)_P \Gamma^i_{pq} V^q dx^l dx^p## have been dropped because they contain products ##dx^l dx^p##. Therefore, these terms are second-order in the infinitesimal coordinate displacements. Only terms up to first order need to be kept.
 
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TSny said:
The terms ##\left( \dfrac{\partial^2 \bar x^j}{\partial x^l \partial x^i } \right)_P \Gamma^i_{pq} V^q dx^l dx^p## have been dropped because they contain products ##dx^l dx^p##. Therefore, these terms are second-order in the infinitesimal coordinate displacements. Only terms up to first order need to be kept.
Oh, that makes perfect sense. Thank you for your time!
 
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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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