I Gradient as vector vs differential one-form

[jbergman]

This is highly misleading, because it only works out using Cartesian coordinates for the vector space of an Euclidean affine space.

The natural general definition of a bare differentiable manifold is to define (alternating) differential forms as derivative operators on alternating tensor fields. Then the gradient of a scalar field naturally occurs as a one-form, i.e., a co-vector field,
$$\mathrm{d} \Phi=\mathrm{d}q^j \partial_j \Phi.$$

That's the exact same thing I wrote.

$$df = \frac{\partial f}{\partial x^i} dx^i$$

is a covector field.

 

[PeterDonis]

This is highly misleading

I'm as confused by this statement as @jbergman since what you wrote is the same as what he wrote (in the latter, coordinate independent part of his post). The only difference is that he also wrote an additional step where you use the metric to raise the index of the gradient covector field, which works if you have a metric on the manifold; you've already agreed that that's possible.

 

[vanhees71]

What is misleading is the first equation reading
$$\text{grad} f= \frac{\partial f}{\partial x^i} \vec{e}_i=(\partial_i f) \vec{e}_i,$$
which obviously is not an invariant vector, because you shouldn't sum over two lower indices. What's correct is
$$\text{grad} f= (\partial_i f)\vec{e}^i,$$
which clearly shows that it is a co-vector (1-form) not a vector.

Only for a Cartesian basis you can identify $\vec{e}_i$ with $\vec{e}^i$, because lowering and raising indices in this and only this case goes with the Kronecker $\delta_{ij}=\delta^{ij}$.

 

[PeterDonis]

What is misleading is the first equation reading
$$\text{grad} f= \frac{\partial f}{\partial x^i} \vec{e}_i=(\partial_i f) \vec{e}_i,$$
which obviously is not an invariant vector

And @jbergman explicitly said that this only works in $\mathbb{R}^n$ and does not work in the absence of coordinates.

Only for a Cartesian basis you can identify $\vec{e}_i$ with $\vec{e}^i$

And that part of @jbergman's post, as he explicitly said, was restricted to this case. He then went on, in the latter part of his post, to give a better formulation that is not dependent on any particular choice of coordinates. So he already gave the clarification that you are trying to give here. You're just repeating what he said in different words.

 

[jbergman]

What is misleading is the first equation reading
$$\text{grad} f= \frac{\partial f}{\partial x^i} \vec{e}_i=(\partial_i f) \vec{e}_i,$$
which obviously is not an invariant vector, because you shouldn't sum over two lower indices. What's correct is
$$\text{grad} f= (\partial_i f)\vec{e}^i,$$
which clearly shows that it is a co-vector (1-form) not a vector.

Only for a Cartesian basis you can identify $\vec{e}_i$ with $\vec{e}^i$, because lowering and raising indices in this and only this case goes with the Kronecker $\delta_{ij}=\delta^{ij}$.

That was the whole point of the post, that the definition that is often presented to students in multivariable calculus is not coordinate free.

 

[martinbn]

That was the whole point of the post, that the definition that is often presented to students in multivariable calculus is not coordinate free.

Just to emphasize the obvious, it is not coordinate free, but it is a perfectly valid and correct definition.

 

[vanhees71]

And @jbergman explicitly said that this only works in $\mathbb{R}^n$ and does not work in the absence of coordinates.And that part of @jbergman's post, as he explicitly said, was restricted to this case. He then went on, in the latter part of his post, to give a better formulation that is not dependent on any particular choice of coordinates. So he already gave the clarification that you are trying to give here. You're just repeating what he said in different words.

It works only with Cartesian bases, not for general bases, even not in $\mathbb{R}^n$.

 

[PeterDonis]

It works only with Cartesian bases

Yes, in other words, with a particular choice of coordinates. @jbergman even gave an example of how it does not work in polar coordinates. Did you actually read what he wrote?

 

[vanhees71]

Maybe I misunderstood the notation. I think we all agree finally.