Gradient as a covariant vector in Lee smooth manifolds. Incorrect formula?

Question

From Lee's introduction to smooth manifolds

" In elementary calculus, the gradient of a smooth real-valued function ( f ) on ${R}^n$ is defined as the vector field whose components are the partial derivatives of ( f ). In our notation, this would read:

$$ \text{grad} \, f = \sum_{i} \frac{\partial f}{\partial x^i} \frac{\partial}{\partial x^i} $$

Unfortunately, in this form, the gradient does not make sense independently of coordinates (the fact that it violates our index conventions is a strong clue).

Exercise 11.17:

Let $f(x,y) = x^2$ on $\Bbb R^2$ and let $X$ be the vector field $X = \text{grad}\ f = 2x\frac{\partial}{\partial x}$. Compute the coordinate expression for $X$ in polar coordinates (on some open subset on which they are defined) and show that it is not equal to $\frac{\partial f}{\partial r}\frac{\partial}{\partial r}+\frac{\partial f}{\partial \theta}\frac{\partial}{\partial\theta}$.

I have two questions regarding this.

1. I do not understand what he means by "does not make sense independently of coordinates (the fact that it violates our index conventions is a strong clue)" What does he mean by "making sense independently of coordinates, why writing a vector in the form $ v^i{\partial x^i}$ or covariant vector $w^i {dx}_i$ makes sense "independently of coordinates"

2. Why does he want us to show the expression is not equal to $\frac{\partial f}{\partial r}\frac{\partial}{\partial r}+\frac{\partial f}{\partial \theta}\frac{\partial}{\partial\theta}$. I thought the correct expression for the gradiant in polar coordinates was given by $$ \text{grad} \, f = \frac{\partial f}{\partial r} \frac{\partial}{\partial r} + \frac{1}{r} \frac{\partial f}{\partial \theta} \frac{\partial}{\partial \theta} $$

Shouldn't this be the expression we expect ?

Answer 1

Prof. Lee is essentially making a point about the necessity for the invariance of geometric quantities. That is, the direction on a manifold in which a function is increasing the most rapidly should obviously not depend on the coordinates we are using to "describe" the underlying manifold. Now, we have two types of objects on our manifold; contravariant and covariant. Under a change of coordinates contravariant and covariant objects transform in precisely the opposite way of one another, that is;

$$e_{i'}=J^i_{i'}e_i$$

$$e^{i'}=J^{i'}_ie^i$$

So, if we wish for the expression that describes some object on our surface to remain intact regardless of coordinate transformations then we require both a contravariant and covariant index to be summed over. For example, a vector, $v$, can be expressed as;

$$v=v^ie_i=v^{i'}J^i_{i'}J^{i'}_{i}e_{i'}=v^{i'}e_{i'}$$

in every coordinate system on the manifold. The invariance of this expression is made obvious by the utilization of the Einstein summation convention where we have an implicit summation over upper and lower indices. Notationally, because we have one upper and one lower index, our expression for the object in question is entirely independent of coordinate transformations.

Now, naively, the gradient of a function $f$ is often thought of as being;

$$\nabla f=\bigg(\frac{\partial f}{\partial x^i}\bigg)e_i$$

which works perfectly well in cartesian coordiantes in $\mathbb{R}^n$. However, as we have just seen, we have a slight issue; This expression has two lower indices. Therefore, when we attempt a coordiante change we will have;

$$\nabla f = \bigg(\frac{\partial f}{\partial x^i}\bigg)e_i = \big(J^{i'}_i\big)^2\bigg(\frac{\partial f}{\partial x^{i'}}\bigg)e_{i'}$$

So, our expression is not invariant. Utilizing the clue given to us by the violation of our index summation rule makes the solution to our problem somewhat self-evident. If we want an invariant expression for the gradient we must have;

$$\nabla f = \bigg(\frac{\partial f}{\partial x^i}\bigg)e^i$$

So the gradient is, in some sense, most naturally a covector rather than a vector. This is the thrust of what Prof. Lee is getting at... For the specific example he makes use of, we have;

$$\nabla f = 2x \frac{\partial}{\partial x}$$

This is clearly the correct expression for the gradient of $f$ in cartesian coordinates. Converting this vector to polar coordinates then yields;

$$\nabla f = 2r\cos^2\theta \frac{\partial}{\partial r}-2\sin\theta\cos\theta\frac{\partial}{\partial \theta}$$

Which is just the same vector as before now in the new coordinate system. Therefore, this must be the correct expression for the gradient of our function in polar coordinates. However, if we were to try and naively compute this expression in the same way that we did in cartesian coordinates we would find that;

$$\nabla f = \bigg(\frac{\partial f}{\partial r}\bigg)\frac{\partial}{\partial r}+\bigg(\frac{\partial f}{\partial \theta}\bigg)\frac{\partial}{\partial \theta}$$

$$=2r\cos^2\theta \frac{\partial}{\partial r}-2\color{red}{r^2}\cos\theta\sin\theta\frac{\partial}{\partial \theta}$$

So, we have an extra incorrect factor of $r^2$ on the $\theta$-component of our gradient. This, of course, all gets taken care of by utilzing the more "natural" expression for the gradient;

$$df=\bigg(\frac{\partial f}{\partial r}\bigg)dr+\bigg(\frac{\partial f}{\partial \theta}\bigg)d\theta$$

Which, by utilizing the musical isomorphism, gives us the unique vector in our tangent space associated with this dual vector;

$$\nabla f = (df)^{\sharp} =g^{rr}\bigg(\frac{\partial f}{\partial r}\bigg)\frac{\partial}{\partial r}+g^{\theta\theta}\bigg(\frac{\partial f}{\partial \theta}\bigg)\frac{\partial}{\partial \theta}$$

$$=\bigg(\frac{\partial f}{\partial r}\bigg)\frac{\partial}{\partial r}+\frac{1}{r^2}\bigg(\frac{\partial f}{\partial \theta}\bigg)\frac{\partial}{\partial \theta}$$

$$=2r\cos^2\theta \frac{\partial}{\partial r}-2\sin\theta\cos\theta\frac{\partial}{\partial \theta}$$

Which yields the correct vector... Additionally, the proper expression for the gradient vector of some function in the orthonormal polar basis would then be;

$$\nabla f = \bigg(\frac{\partial f}{\partial r}\bigg)\hat{r}+\frac{1}{r}\bigg(\frac{\partial f}{\partial \theta}\bigg)\hat{\theta}$$

as expected.