Standard version of covariant derivative properties

[Shirish]

[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.

Source 1: 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. $\nabla_{fX+Z}\ T=f\nabla_XT+\nabla_ZT$

Source 2: 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?

 

[Infrared]

$\partial_i$ isn't even a vector field!

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.

 

[Shirish]

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.

Thanks as usual! A couple of things that I'm in the dark about since they weren't explicitly mentioned in either source:

1. The definition of tensor that I've come across is that it's a multilinear map with $\mathbb{R}$ as the target space. Is there an alternative definition of tensor I'm missing out on? Confused because you mentioned in the last para that a $(p,q+1)$ tensor eats a vector to give a $(p,q)$ tensor.

2. Could you elaborate just a bit on what you mean by $T(\omega, Y)$ being a contraction of $T\otimes\omega\otimes Y$?

3. Applying covariant derivative on the above, we get
$$\nabla(T\otimes\omega\otimes Y)=(\nabla T)\otimes(\omega\otimes Y)+T\otimes(\nabla\omega\otimes Y)+T\otimes(\omega\otimes\nabla Y)$$ Then we can act the LHS and RHS on a vector field $X$ to get
$$\nabla_X(T(\omega,Y))=[(\nabla T)\otimes(\omega\otimes Y)](X)+[T\otimes(\nabla\omega\otimes Y)](X)+[T\otimes(\omega\otimes\nabla Y)](X)$$ How do I incorporate the $X$'s into the individual terms on the RHS?

 

[Infrared]

1. The definition of tensor that I've come across is that it's a multilinear map with $\mathbb{R}$ as the target space. Is there an alternative definition of tensor I'm missing out on? Confused because you mentioned in the last para that a $(p,q+1)$ tensor eats a vector to give a $(p,q)$ tensor.

If $T:(V^*)^{p}\times V^{q+1}\to\mathbb{R}$ is a $(p,q+1)$ tensor, and $X$ is a vector, then you can define a $(p,q)$ tensor by $(TX)(\varphi_1,\ldots,\varphi_p,Y_1,\ldots,Y_q)=T(\varphi_1,\ldots,\varphi_p,Y_1,\ldots,Y_q,X)$.

2. Could you elaborate just a bit on what you mean by $T(\omega, Y)$ being a contraction of $T\otimes\omega\otimes Y$?

I'm not sure what type of tensors $T,\omega,Y$ are supposed to be from looking at that expression, but
I think an example should make it clear: suppose $g=g_{ij}dx^i\otimes dx^j$ is $(0,2)$ form and $V=V^p\partial_p$ and $W=W^q\partial_q$. Then $g\otimes V\otimes W=g_{ij}V^pW^q dx^i\otimes dx^j\otimes\partial_p\otimes\partial_q.$ Contracting (i.e. taking the trace) the indices $i,p$ and $j,q$ gives $g_{ij}V^iW^j$, which is $g(V,W).$

3. Applying covariant derivative on the above, we get
$$\nabla(T\otimes\omega\otimes Y)=(\nabla T)\otimes(\omega\otimes Y)+T\otimes(\nabla\omega\otimes Y)+T\otimes(\omega\otimes\nabla Y)$$ Then we can act the LHS and RHS on a vector field $X$ to get
$$\nabla_X(T(\omega,Y))=[(\nabla T)\otimes(\omega\otimes Y)](X)+[T\otimes(\nabla\omega\otimes Y)](X)+[T\otimes(\omega\otimes\nabla Y)](X)$$ How do I incorporate the $X$'s into the individual terms on the RHS?

Your last line isn't correct. You contracted on the LHS but not the RHS.

Anyway, you should read rule 3 in the second definition as $\nabla_X(S\otimes T)=\nabla_XS\otimes T+S\otimes\nabla_XT.$ I think it's a little bit sloppy (if not wrong) to write it as they did, because you want the $1$-forms introduced by differentiation to appear at the end of the tensor, so that they act on a vector you apply to. See the answer in https://math.stackexchange.com/ques...ule-for-covariant-derivative-of-tensor-fields

With that in mind, start with $T\otimes \omega\otimes Y.$ Contracting first and then differentiating gives $\nabla_X(T(\omega,Y))$. Differentiating first, and then contracting gives $(\nabla_X T)(\omega,Y)+T(\nabla_X\omega,Y)+T(\omega,\nabla_XY).$ These must be equal.

 

[Shirish]

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 $p$).

One last doubt regarding the above. Even we if we want to calculate something like $\nabla_{\partial_i}T(p)$, $\partial_i$ will be the $i$-th element of the coordinate induced basis, so ultimately $\nabla_{\partial_i}T(p)$ will be chart dependent.

This problem doesn't occur with $\nabla_XT(p)$ since $X(p)$ is a well-defined vector in a unique direction at $p$. That was my reason for not calling $\partial_i$ a vector field (at least, not in the same sense as ordinary vector fields like $X$). Another reason being that $\partial_i$, being chart-dependent, I don't see how it is a well-defined section of the tangent bundle on $M$.

Does this mean that I should treat $\nabla_{\partial_i}T(p)$ as the chart-induced derivative of $T$ at $p$? (similar to how $X^i(p)$ represents a chart-induced component of $X(p)$)

 

[Infrared]

Another reason being that $\partial_i$, being chart-dependent, I don't see how it is a well-defined section of the tangent bundle on $M$.

Given a coordinate system on a neighborhood $U$, then $\partial_i$ is a well-defined section of the tangent bundle restricted to $U.$ For $p\in U$, it smoothly assigns $\partial_i\big\vert_p\in T_pU$. This is the definition of a section.

It does depend on the coordinate system, but this is fine. It just means that you can get different vector fields depending on which coordinate system you choose.

You'll want to use this in computations. If you want to compute $\nabla_XT$ in some example, you'll generally expand $X=X^i\partial_i$ in some coordinate system, and compute $\nabla_XT=X^i\nabla_iT,$ just like if you want to compute a directional derivative: $Xf=X^i\partial_if.$