# Confusion about Tangent vs Cotangent differentials

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?

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

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.
• 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. – Warlock of Firetop Mountain Nov 13 '17 at 9:35