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In Lee's Introduction to Smooth manifolds, let $M$ be a smooth manifolds and given a smooth map $f \in C^\infty(M)$, we have two definitions of the "differential" at $f$:

(1): for each $p \in M$, define a map $$df_p: T_pM \to T_p\mathbb{R}$$ such that given $g: \mathbb{R} \to \mathbb{R}$, $$df_p(v) = v(g \circ f).$$

(2): $df$ is the covector field given at each point as $$df_p(v) = vf,$$ for $v \in T_pM$.

The first has $i^{th}$ coordinate representation $$\frac{\partial f}{\partial x^i}(p) \frac{d}{dx}|_p$$ via the bottom of page 61 here.

The second has coordinate representation (as a relation of covector fields)

$$df = \sum_{i}\frac{\partial f}{\partial x^i}(p) dx^i|_p.$$

I thus have two questions: How exactly are these the same object? And given they are the same object, what is the point of having covectors at all? Why not just use the first differential?

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Firsty, one defines the differential of a smooth function $f:M\to N$ as* the linear map $df_p:T_pM \to T_{f(p)}N$ such that for a vector $X \in T_p M$, the vector $df_p(X) $ is the derivation in $T_{f(p)} N$ such that for any $g:N\to \mathbb{R}$ smooth it holds that $df_p(X)(g)= X(g\circ f)$.

In the particular case where $N=\mathbb{R}$, since you can identify $T_{f(p)}\mathbb{R}$ with $\mathbb{R}$, the differential induces a smooth section of the cotangent bundle, called $df:M\to T^*M$. This diffential also coincides with the exterior derivative of $f$, considering $f$ as a $0$-form so there is no confusion.

If you have an arbitrary $N$ as codomain you cannot do this because the tangent bundle can be not so trivial as in the case of $\mathbb{R}$.

To answer your second question A differential form can also NOT be a differential of a function (otherwise we call it "exact"). For example consider in $\mathbb{R}^2$ the differential $1$-form $\omega = -x_2 dx_1 + x_1 dx_2$, since the exterior derivative $d\omega \neq 0$, then $\omega$ it is not $df$ for any $f$ since we know that $d d \eta=0$ for any $k$-form $\eta$.

*it is one possible way to define it

Reply to the comment.

One thing is the exterior derivative denoted as $d$, that takes a $k$-form as input (a $1$-form is a covector field) and gives a $k+1$-form as output. Another thing is the differential, also denoted as $d$, that takes a smooth function $f:M\to N$ and gives a map $df:TM\to TN$ between tangent bundles. In the particular case where $N=\mathbb{R}$, $f$ can be seen as a $0$-form, and its exerior derivative $df$ is a $1$-form that coincides with the 1-form induced by the differential $df$ of the function. In general you cannot identify an $N$ valued function with a $0$-form. Differential and exterior derivative are different things.

The second exterior derivative means that you are taking two times the exterior derivatives. If $f$ is a $0$-form $df$ is a $1$-form and $d(df)$ is a $2$-form, $d$ acts differently on $1$-forms than how it acts on $0$-forms or $k$-forms.

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  • $\begingroup$ Wait so if the second differential is just a special case of the first, then what is the reason for having covectors? $\endgroup$ – Anthony Peter Nov 13 '17 at 0:24
  • $\begingroup$ I have tried to reply but I still don't understand how you would bypass covectors. The exterior derivative is defined on $k$-forms fields that at a point $p$ are (sum) of wedge product of covectors $\omega_1\wedge \omega_2\wedge\dots \wedge \omega_k$. In the case $k=1$, you have a covector field. $\endgroup$ – Warlock of Firetop Mountain Nov 13 '17 at 9:35
  • $\begingroup$ Hello, is there something I can do to improve my answer? $\endgroup$ – Warlock of Firetop Mountain Nov 14 '17 at 21:20

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