Is this definition of the gradient using the exterior derivative consistent with calculus 1?

Question

I'm trying to get a better grasp on what the exterior derivative means on my own and I'm trying to connect the language of forms to my pre-existing knowledge. I came along the following formula for the gradient on wikipedia: $$\text{grad}f=\nabla f=(df)^\sharp$$ where $d$ is the exterior derivative and $^\sharp$ is a musical isomorphism. Since $(\omega_ie^i)^\sharp=\omega^ie_i$ this gives the following definition in component form $$\nabla f=e_i\partial^if=(\partial^if)\frac{\partial}{\partial x^i}.$$ From my calculus classes I remember the gradient being defined as a vector which is consistent with this formula. But what confuses me is that the vector component have raised indices. In Cartesian coordinates you wouldn't notice this but in polar coordinates, or any other coordinates for that matter, you would notice a difference. But again when I recall my calculus classes I remember always calculating the ordinary partial derivatives which correspond to lower indices $\partial_i f$. So is this definition consistent with elementary calculus intuition? So my elementary calculus definition would perhaps be $(\partial_i f)e_i$ even though I know that wouldn't make sense.

Answer 1

The answer to the question in the title is "yes," and here's why.

As the Wikipedia article on the musical isomorphism explains, the so-called "musical isomorphism" is equivalent to raising and lowering indices (sometimes called "index juggling") in component form. The process of raising and lowering indices is allowed due to the metric ($g_{ij}$) and its inverse ($g^{ij}$) imposed on the manifold. Specifically, a vector $\vec X=X^i\vec e_i$ is related to the $1$-form $\tilde X=X_i\tilde\omega^i$ by the equation $X^i=g^{ij}X_j$ under the standard coordinate bases $\vec e_i=\frac{\partial}{\partial q^i}$ and $\tilde\omega^i={\bf d} q^i$ for some set of coordinates $q^i$ if the usual $1$-form and metric requirements are imposed (viz. ${\bf d}q^i(\frac{\partial}{\partial q^j})=\delta^{i}_{\ \ j}$ and $g^{ik}g_{kj}=\delta^i_{\ \ j}$).

Using this relationship, ${\bf d}f=\frac{\partial f}{\partial q^i}{\bf d}q^i$ is related to $\vec\nabla f=(\partial^if)\frac{\partial}{\partial q^i}=(g^{ij}\frac{\partial f}{\partial q^j})\frac{\partial}{\partial q^i} $ through an obvious application of index juggling (which, remember, is the same as the musical isomorphism).

And you are correct that the raised index (which is hiding that inverse metric tensor) is important in other coordinate systems. Look at the definition of the gradient in other coordinates and you will notice that additional factors are present that resemble the elements of the inverse metric. If you look carefully, they appear (at least for cylindrical and spherical coordinates) to be the square root of those elements. The reason for this is that the definitions given on that Wikipedia page use normalized unit vectors instead of the standard basis of partial derivatives. The normalization factor is exactly the square root of those terms, so the unit vectors are actually hiding a portion of the inverse metric's contribution to the final answer.