anonanon 152k1212 gold badges233233 silver badges398398 bronze badges $endgroup$ four $begingroup$ Which is perfect thanks a bunch! I also posted an analogous issue below: math.stackexchange.com/inquiries/2135276/… that You may have some Perception on? $endgroup$
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I am using the Einstein notation, so numerous indices are summed) or they offer two three-indices symbols with distinct signal
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This tensor shouldn't be bewildered With all the tensor density field outlined over. The presentation Within this part closely follows Carroll 2004.
In summary, the dialogue is about the Homes from the Levi-Civita symbol. The 1st component discusses a method that doesn't manage to match the envisioned answer, and the second part entails proving Yet another formulation. The conversation ends using a closing Option that satisfies the Attributes.
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On the pseudo-Riemannian manifold, one may possibly define a coordinate-invariant covariant tensor subject whose coordinate illustration agrees Together with the Levi-Civita image wherever the coordinate procedure is these kinds of that The premise in the tangent House is orthonormal with regard for the metric and matches a specific orientation.
The crucial element defining house on the symbol is full antisymmetry from the indices. When any two indices are interchanged, equal or not, the image is negated:
Do the thing is how this tends to have an effect on the value of $epsilon^ alphabetagammadelta epsilon_ alpha'beta'gamma'delta' $? $endgroup$