Derivative Operators on a manifold

Question

I am having some trouble coming to terms with the notion of a derivative operator on a manifold. In Robert M. Wald's General Relativity, the definition in the textbook is given in terms of 5 stipulations:

1. Linearity: For all tensors $\mathit{A}, \mathit{B} \in \mathscr{T}(k,l)$ and $\alpha, \beta \in \mathbb{R},$ $$\nabla_c ( \alpha \mathit{A}^{a_1...c...a_k}_{b_1...c...b_l} + \beta \mathit{B}^{a_1...c...a_k}_{b_1...c...b_l}) = \alpha \nabla_c \mathit{A}^{a_1...c...a_k}_{b_1...c...b_l} + \beta \nabla_c \mathit{B}^{a_1...c...a_k}_{b_1...c...b_l} .$$

2. Leibnitz rule: For all $\mathit{A} \in \mathscr{T}(k,l),$ $\mathit{B} \in \mathscr{T}(k',l'),$ $$\nabla_e [\mathit{A}^{a_1...c...a_k}_{b_1...c...b_l} \mathit{B}^{c_1...c...c_k'}_{d_1...c...d_l'}] = \nabla_e [A]B + A \nabla_e[B] .$$

3. Commutativity with contraction: For all tensors $\mathit{A} \in \mathscr{T}(k,l)$, $\nabla_{d}(\mathcal{A}^{a_1...c...a_k}_{b_1...c...b_l}) = \nabla_{d}\mathcal{A}^{a_1...c...a_k}_{b_1...c...b_l}$.

4. Consistency with the notion of tangent vectors as directional derivatives on scalar fields: for all functions $f \in \mathscr{F}:M \rightarrow \mathbb{R}$ and all tangent vectors $t^a \in V_p$, it is required that $t(f)=t^a\nabla_a f.$ Note that this implies the action of a derivative operator will agree in the action on any scalar field.

5. (Optional) Torsion free: For all $f \in \mathcal{F}$ $$\nabla_a \nabla_b f = \nabla_b \nabla_a f.$$

Immediately thereafter, it is shown that derivative operators exist by considering the "ordinary derivative operator" $\partial_a$ in some coordinate system. However, I am not sure why ordinary derivatives will satisfy condition 4. That is, I am not sure why ordinary derivative operators will agree in their action scalar fields

For example, suppose our manifold is the plane $\mathbb{R}^2,$ with one coordinate system being the usual cartesian coordinate system, and the other coordinate system being the usual polar coordinate system.

Consider the function $f(x,y) = xy.$ Consider the tangent vector $\hat{y}$. The ordinary derivative operator we write $\frac{\partial}{\partial x} \hat{x} + \frac{\partial}{\partial y} \hat{y}.$ Then $t(f) = \hat{y} \cdot (y \hat{x} + x\hat{y}) = x.$

Now consider this in the polar coordinate system. $\tilde{f} (r, \theta) = r^2 \cos \theta \sin \theta.$ Consider the same tangent vector $\hat{y} = \sin\theta \hat{r} + \cos \theta \hat{ \theta}.$ Now consider the ordinary derivative operator $\frac{\partial}{\partial r} \hat{r} + \frac{\partial}{\partial \theta} \hat{\theta}.$ Note that this is different than what you would get if you wanted to consider what the gradient is in polar coordinates. Then $$t(\tilde{f}) = (\sin\theta \hat{r} + \cos \theta \hat{ \theta}) \cdot (2r \cos \theta \sin \theta \hat{r} + r^2(\cos^2 \theta - \sin^2 \theta) \hat{\theta}).$$ But note that $t(f) \neq t(\tilde{f}).$

So how do we have condition 4?

Answer 1

If I'm not mistaken, the problem is that you are not calculating the (covariant) derivative in the the coordinate basis ( $\boldsymbol{\hat{r}}$ and $\boldsymbol{\hat{\theta}}$ are the normalized version of the basis, but not the standard basis associated with polar coordinates per se) and you are also considering the basis vectors instead of the basis covectors (1-forms), which are not necessarily the same mathematical "entity" for a general curvilinear coordinate system. If the scalar field is $f=r^2\sin \theta \cos \theta$, then the gradient would be $\boldsymbol{\nabla}f=\frac{\partial f}{\partial x^{\alpha}}\boldsymbol{\omega}^{\alpha}=\frac{\partial f}{\partial r}\boldsymbol{\omega}^{r}+\frac{\partial f}{\partial \theta}\boldsymbol{\omega}^{\theta}=(2r\sin \theta \cos \theta)\boldsymbol{\omega}^{r}+(r^2\cos 2\theta)\boldsymbol{\omega}^{\theta}$. Where $\boldsymbol{\omega}^{r}$ and $\boldsymbol{\omega}^{\theta}$ are the polar (not necessarily unitary) basis covectors, and those can be written in the cartesian basis as: $ \boldsymbol{\omega}^{r}=\cos\theta \boldsymbol{\hat{x}} + \sin \theta \boldsymbol{\hat{y}} \\ \boldsymbol{\omega}^{\theta}=-\frac{1}{r}\sin \theta \boldsymbol{\hat{x}}+ \frac{1}{r}\cos \theta \boldsymbol{\hat{y}} $ While the basis polar vectors, $\boldsymbol{e}_r $ and $\boldsymbol{e}_{\theta}$ can be written in the same cartesian basis (this can be done only because cartesian basis covectors and cartesian basis vectors happen to be the same thing, but it's absolutely not true for more general spaces and coordinate systems and manifolds; curvature, wheter apparent, i.e. due to a "bad" choice of coordinates, or intrinsic, i.e. due to the manifold itself, always messes things up): $\boldsymbol{e}_r=\cos \theta \boldsymbol{\hat{x}}+\sin \theta \boldsymbol{\hat{y}} \\ \boldsymbol{e}_{\theta}=-r\sin \theta \boldsymbol{\hat{x}}+ r\cos \theta \boldsymbol{\hat{y}} $

Now, in order to calculate the directional derivative we have to take the inner product between the vector $\boldsymbol{\hat{y}}$ and the gradient (a covector): $\langle \boldsymbol{\hat{y}} , \boldsymbol{\nabla}f \rangle = \langle \sin \theta \boldsymbol{e}_{r} + \frac{\cos \theta}{r} \boldsymbol{e}_{\theta}, (2r\sin \theta \cos \theta)\boldsymbol{\omega}^{r}+(r^2\cos 2\theta)\boldsymbol{\omega}^{\theta} \rangle = 2r \sin^2\theta \cos \theta + r \cos \theta \cos 2\theta = r \cos \theta (2\sin ^2 \theta +\cos^2 \theta -\sin^2 \theta)=r\cos \theta (\cos^2 \theta + \sin ^2 \theta )= r\cos \theta = x $ Hence, the property is respected. Obviously the value of a directional derivative should not depend on the choice of coordinates.

Answer 2

I think you may be slightly confused by the notion of a tangent vector. This 'ordinary derivative operator' that you are using is slightly different to what he is talking about. You are most likely thinking of the Laplacian operator,

$$\nabla = \frac{\partial}{\partial x}+\frac{\partial}{\partial y}+\frac{\partial}{\partial z} +\dots$$

however as we are only interested in the derivative of the function in a certain direction, we use the partial derivative in say, the y direction, to be our tangent vector in that direction. I'm not good with words so I will use an example.

Say we are interested in the tangent vector in the Y direction, then the tangent vector is

$$ Y = \frac{\partial}{\partial y} $$

But we can write this more generally as a vector product, in this example we would write $$ Y= y^a \partial_a$$ where $y^a$ is a vector with a 1 in the y-position e.g. (0,1) in your example. Now:

$$Y(f) =y^a\partial_a(f)= \frac{\partial f}{\partial y} = x$$

If we swap into polar coordinates we transform our vector appropriately.

$$Y = \frac{\partial}{\partial y}=\frac{\partial r}{\partial y}\frac{\partial}{\partial r} + \frac{\partial \theta}{\partial y} \frac{\partial}{\partial \theta}$$

$$=sin\theta\frac{\partial}{\partial r} + \frac{1}{r}cos\theta \frac{\partial}{\partial \theta}$$

so then

$$\tilde{Y}(\tilde{f})=sin\theta\frac{\partial}{\partial r}(r^2cos\theta sin\theta) + \frac{1}{r}cos\theta \frac{\partial}{\partial \theta}(r^2cos\theta sin\theta)$$

$$=2rcos\theta(sin^2\theta)+rcos\theta(cos^2\theta - sin^2\theta))$$

$$=rcos\theta(sin^2\theta)+rcos\theta(cos^2\theta + sin^2\theta - sin^2\theta))$$

$$=rcos\theta(sin^2\theta)+rcos\theta(1 - sin^2\theta))$$

$$=rcos\theta = x$$

I was comparing the result of one tangent vector represented in two different bases. As such our answer agrees. You were comparing two different things in two different bases. Of-course they wont agree. I think you would benefit on reading an introduction to the basic Diff-geo used in GR, perhaps by Carrol or Tong. Wald is an incredibly dense and heavy book, it can be very difficult for a person new to the subject. Perhaps recap on some general coordinate transformations aswell, notably;

$$\frac{\partial}{\partial r}+\frac{\partial }{\partial \theta}\neq \frac{\partial}{\partial y} + \frac{\partial }{\partial x} $$