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Mathematics
Differential Geometry
Standard version of covariant derivative properties
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[QUOTE="Infrared, post: 6374206, member: 467682"] Why not? It's only defined in a coordinate neighborhood, but that's all you need anyway (since ##(\nabla_XT)(p)## depends only on the value of ##X## at ##p##, and ##T## in a neighborhood of ##p##). Item (3) in the first definition follows from the second definition because ##T(\omega,Y)## is a contraction of ##T\otimes\omega\otimes Y.## Item 4 in definition 1 just follows from ##\nabla T## being a tensor: ##\nabla_{fX+Z}T=(\nabla T)(fX+Z)=f \nabla T (X)+ \nabla T(Z)=f\nabla_XT+\nabla_ZT.## I wasn't able to tell from your post: do you have issues with the first two points in either definition? Also beware the slight shift in notation between the two sources. Your first definition doesn't consider the full tensor ##\nabla T##, only ##\nabla_XT## for a vector field ##X##. This corresponds to ##(\nabla T)(X)## in the second definition, which makes sense, because if ##T## is a ##(p,q)## tensor, then ##\nabla T## is a ##(p,q+1)## tensor, so it can be applied to a vector field ##X##, to give back another ##(p,q)## tensor. [/QUOTE]
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Standard version of covariant derivative properties
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