Clarifying the difference between differential 1-form and covariant derivatives Upon trying to teach myself some basics of differential geometry I keep running into the notion of a differential 1-form, and the covariant derivative.


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*I understand intuitively what the covariant derivative ($\nabla_pu : TM \rightarrow TM$) aims to achieve. It provides a way to transport vectors in one tangent space to another over the manifold. If it is flat it has Christoffel symbol = 0.

*Now dual to the tangent space is the co-tangent space in which the differntial 1-form $dx : TM \rightarrow \mathbb{R}$ (for example) lives. 
(If any of the above definitions are incorrect please let me know)
In elementary school calculus we used "basic differentiation" to solve all issues in regards to differentiability. 
Moreover, to find the total derivative in university we used a formula which involves $\partial / \partial x$ notation in conjunction with $dx$. So they seemed to be doing the same job. The phrase everyone would throw around casually is:

"Use $dx$ when it's a function of one variable and $\partial / \partial x$ when there are multiple". But at the end of the day they both achieve to specify some sort of "rate of change". 

In differential geometry these are clearly different beasts! I can read a book and "get" each definition, but I am struggling to put everything together neatly like jigsaw.
Now my questions:


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*In elementary calculus when I "differentiated" what did I do? Did I implicitly use a 1-form or the covariant derivative over the Euclidean space in Cart co-ords. If they are the same, why are they same? Since in diff geo on the abstract level they clearly work on different objects. 

*So is integration ONLY concerned in working with 1-forms? (The $dx$ symbol is everywhere)

*When we solve DEs and PDEs there are derivatives involved. Do these stem from the covariant derivative, or from the differential 1-form? 

*On the abstract level when I say "I am going to find the derivative", how do people (versed in differential geometry) interpret that statement? Am I going to be working with 1-forms or with the covariant derivative?

*Are the ordinary derivatives we see in elementary school calculus (e.g. for performing gradient descent on a surface) covariant, or contra-variant? What implications does this idea have for the entire framework (i.e. I think I heard it is a contravariant object, but we have this thing called a "covariant derivative". So do we not use covariant derivative for gradient descent??)
Perhaps I have more questions to think of, but I feel if at least these are clarified it should be enough to help myself (and others) to continue with this self-learning journey!  :)
 A: There are many questions here. First, you should understand that on a manifold, without additional structure, the objects that you can naturally differentiate are only differential forms (functions are $0$-forms). If $\alpha$ is a $k$-differential form, $d\alpha$ is a $(k+1)$-differential form (and it doesn't depend on coordinates).
Of course, in euclidean space, you can identify everything with functions since you have a global chart. This is indeed what you do implictly in lower classes, where you identify $df = f'(x) dx$ with the "function" (which is coordinate dependant!) $x \mapsto f'(x)$. I think this answers 1). 
You can add another level of structure to a manifold by specifying a connexion. This, as you say, gives you a way to transport vectors on different tangent spaces, and enables you to differentiate different objects such as vector fields. But you don't need that structure for differentiating forms.


*Integration along curves is only coordinate-independant for 1-forms. Integration along $k$-dimensional submanifolds is coordinate independant for $k$-forms.

*Most PDEs require having chosen a connexion or a metric on the manifold. ODEs are equations of the form $\frac{d}{dt}\phi_t(x)=X(t,\phi_t(x))$, where $X$ is a vector field, and $\phi_0(x)=x$. For every $t \in \mathbb R$, the map $x \mapsto \phi_t(x)$ is a diffeomorphism of the manifold. No differential forms appear in that formulation.

*A gradient is a way to represent the differential of a scalar-valued function on your manifold. It is a (special kind of) a vector field. A covariant derivative is a way to make sense of the derivative of a vector field, using a connexion. Note that even when working in a finite dimensional vector space, talking about gradients require to have chosen an inner product.
A: I will start with 1. 
The answer here is neither. To get back to calculus we pick $M=\mathbb{R}^n$. Given a vector field $X$, the covariant derivative $\nabla_X$ maps another vector field $Y$ to a vector field $\nabla_XY$. A differential $1$-form $\omega$ maps a vector field to a function. Differentiating in elementary calculus takes a function and returns a function. That is the operation $f \mapsto X(f)$ in differential geometry language. If $M$ is just $\mathbb{R}^n$ you can pick for example $X=\partial/\partial_{x_1}$ the constant vector field in the direction of the first coordinate but you can also differentiate the function $f$ in different directions at different points.
