I'm studying about covariant differentiation from two different sources. One of them is the gravity and light lectures on youtube (lecture 7) in which a connection is defined as:
A connection $\nabla$ on a smooth manifold is a map that takes a pair consisting of a vector field and a $(p,q)$-tensor field, and sends them to a $(p,q)$-tensor field satisfying: (a bunch of conditions I'm not listing here, one of which is $\nabla_Xf=Xf$)
Now $f$ is a scalar field and the output $Xf$ is also a scalar field since it takes in a point $p$ and gives the directional derivative of $f$ in the direction $X(p)$ (please correct me if I'm wrong).
In the other reference (Core Principles of Special and General Relativity), the definition is:
The derivative operator $\nabla$ on a manifold $M$ is defined as an abstract mapping at a point $p\in M$ from type $(k,l)$ tensors to type $(k,l+1)$ tensors, $\nabla:\mathcal{T}^k_l\to\mathcal{T}^k_{l+1}$. […] The index on $\nabla_{\mu}$ is related to the variation of the tensor in the direction of the coordinate curve $x^{\mu}$.
So I'm guessing $\nabla_{\mu}$ is $\nabla_{\partial/\partial x^{\mu}}$, where $\frac{\partial}{\partial x^{\mu}}$ is a vector field that gives the basis vector at a point $p$ in the direction of the $\mu$-th coordinate curve. One of the requirements listed in the book is:
For any scalar field $\phi$, $\nabla_{\mu}\phi=\partial_{\mu}\phi$
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I'm having a hard time reconciling the definitions from the two sources. One maintains the tensor rank while the other changes it (adding an extra covariant component). In the last quote, $\phi$ is a scalar field. Isn't $\partial_{\mu}\phi$ also a scalar field? In which case the rank remains unchanged.
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I'm confused on how to interpret $\partial/\partial x^{\mu}$. It should take in a point $p$ and output a vector in $T_pM$. Naively it takes in $p$ and gives the "$\mu$-th basis vector of $T_pM$", but that depends on the chart selection, right? Output of a vector field shouldn't be chart-dependent.
Best Answer
Throughout, I'll define a tensor $T\in\mathcal{T}{}^k_lM$ as a function which takes $k$ covector fields and $l$ vector fields and returns a smooth function, and is $C^\infty$-linear in its arguments. $$ T:\left(\mathfrak{X}^*M\right)^k\times\left(\mathfrak{X}M\right)^l\to C^\infty M $$ $$ T(\omega^1,\dots,f\omega^i,\dots,\omega^k,V_1,\dots,V_l)=T(\omega^1,\dots,\omega^k,V_1,\dots,fV_i,\dots,V_l)=fT(\omega^1,\dots,\omega^k,V_1,\dots,V_k)\ \ \ \ \ \ \ f\in C^\infty M $$ The for a fixed vector field $X$ covariant derivative w.r.t. $X$, $\nabla_X T$, is also a $(k,l)$ tensor. Since $\nabla$ is $C^\infty$-linear in its first argument, we can define the "total" covariant derivative $\nabla T$ as $$ (\nabla T)(X,\omega^1,\dots,\omega^k,V_1,\dots,V_l)=(\nabla_X T)(\omega^1,\dots,\omega^k,V_1,\dots,V_l) $$ This is evidently a $(k,l+1)$ tensor, with the arguments in a nonstandard order. The relationship is similar to the relationship between directional derivative and gradient in $\mathbb{R}^n$: The gradient encodes all directional derivatives since we may dot it with the vector to produce the corresponding directional derivative.
In coordinates, we may write $\nabla_iT:=\nabla_{\partial_i}T$. One can see that the total covariant derivative has the coordinate representation $$ (\nabla T)_i{}^{n_1\dots n_k}{}_{m_1\dots m_l}=(\nabla_i T)^{n_1\dots n_k}{}_{m_1\dots m_l} $$
For your second question, it is true that $\partial_\mu$ is a (local) vector field, but it is dependent on the choice of chart. There's no canonical choice of basis for $TM$, but given a chart we can locally define a set of vector fields $\partial_1,\dots,\partial_n$ which form a basis at each point. Clearly this basis is chart dependent, but we can still make basis-independent statements (e.g. if $X^i=Y^i$ everywhere than $X=Y$, even though the components are chart dependent). For instance, $\nabla_i\phi=\partial_i\phi$ can be rewritten as $\nabla\phi=d\phi$, where $d\phi$ is the differential of $\phi$.