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was saying 'it's not clear' misinterpreted?
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octonion
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There is a chain rule if the object is a tensor formed by tensor products and contractions, for instance $$\nabla_c (A^{bde}B_{bdf})=B_{bdf}\nabla_c A^{bde}+A^{bde}\nabla_c B_{bdf}.$$ Something like $f(t_{ab})$ is in general not a tensor so it is not clear how you define the action of the connection coefficient part of the covariant derivative is undefined on it.

There is a chain rule if the object is a tensor formed by tensor products and contractions, for instance $$\nabla_c (A^{bde}B_{bdf})=B_{bdf}\nabla_c A^{bde}+A^{bde}\nabla_c B_{bdf}.$$ Something like $f(t_{ab})$ is in general not a tensor so it is not clear how you define the action of the connection coefficient part of the covariant derivative on it.

There is a chain rule if the object is a tensor formed by tensor products and contractions, for instance $$\nabla_c (A^{bde}B_{bdf})=B_{bdf}\nabla_c A^{bde}+A^{bde}\nabla_c B_{bdf}.$$ Something like $f(t_{ab})$ is in general not a tensor so the action of the connection coefficient part of the covariant derivative is undefined on it.

Source Link
octonion
  • 8.9k
  • 19
  • 35

There is a chain rule if the object is a tensor formed by tensor products and contractions, for instance $$\nabla_c (A^{bde}B_{bdf})=B_{bdf}\nabla_c A^{bde}+A^{bde}\nabla_c B_{bdf}.$$ Something like $f(t_{ab})$ is in general not a tensor so it is not clear how you define the action of the connection coefficient part of the covariant derivative on it.