Standard version of covariant derivative properties

[Throughout we're considering the intrinsic version of the covariant derivative. The extrinsic version isn't of any concern.]

I'm having trouble reconciling different versions of the properties to be satisfied by the covariant derivative. Essentially $\nabla$ sends $(p,q)$-tensors to $(p,q+1)$-tensors. I'll write down the required properties for $\nabla$ from the two sources.

This lecture (relevant timestamp linked)

If $X$ is a vector field,

1. $\nabla_Xf=Xf$, for a scalar field $f$
2. $\nabla_X(T+S)=\nabla_XT+\nabla_XS$
3. $\nabla_X(T(\omega,Y))=(\nabla_XT)(\omega,Y)+T(\nabla_X\omega,Y)+T(\omega,\nabla_XY)$
4. $\nabla_{fX+Z}\ T=f\nabla_XT+\nabla_ZT$

Core principles of special and general relativity (Luscombe):

1. $\nabla_if=\partial_if$
2. $\nabla(aT+bS)=a\nabla T+b\nabla S$ for real $a,b$
3. $\nabla(S\otimes T)=(\nabla S)\otimes T+S\otimes (\nabla T)$
4. $\nabla$ commutes with contractions, $\nabla_i(T^j_{\ \ jk})=(\nabla T)^j_{\ \ ijk}$

At least the second property is consistent. The first property from the book is a more restrictive version of the first property from the lecture. In fact, $\nabla_i$ means $\nabla_{\partial_i}$ and $\partial_i$ isn't even a vector field!

As for the last two properties from the two sources, I have no idea on how to relate them. Are these requirements incomplete for either of the sources?

If not, how can these two sets of requirements be shown to be equivalent?

1 Answer

The key point is that the covariant derivative is a local operator: If two tensors agree on an open set $U$, then their covariant derivatives agree on that open set. Moreover, if $X(p) = Y(p)$, then $(\nabla_X T)(p) = (\nabla_Y T)(p)$. So, given a vector field $X$ defined on an open set (in your case, $\partial_i$ on some coordinate chart $U$), using a bump function we can extend it to a globally defined vector field $\tilde X$ that agrees with $X$ on a pre-determined non-empty subset $V\subset U$. That means that your "local" definition of $\nabla$ on $V$ agrees with the "global" definition.