Differential at a point and differential (Differential Geometry)

Question

Given $f\in C^\infty(U)$, $U$ open set of $\mathbb{R}^n$, we define the differential of $f$ at $p$ $$ (df)_p:T_p\mathbb{R}^n\to\mathbb{R}\\ (df)_p(v):=v(f) $$ and the differential of $f$ $$ df:U\to T^*U\\ p\mapsto (p,(df)_p) $$ where $T^*U$ is the cotagent bundle.

I can understand that $(df)_p$ is the map which associates for every point $p$ any derivative $v$ to $f$. While $df$ is just obtained by gluing together all these local differentials.

Now I'm asked to compute the differential of the i-th-coordinate map $x_i:\mathbb{R}^n\to\mathbb{R}$.

Here is my (unsuccesful) reasonement. I have to start by calculating, for every $p$, $(dx_i)_p$. By definition $(dx_i)_p\in (T_p\mathbb{R^n})^*$ and the latter set is spanned by $\{(dx_i)_p:1\le i\le n\}$. Now, I am a bit confused. My professor defined the basis of $(T_p\mathbb{R}^n)^*$ as the dual basis of the tangent plane and the symbols $(dx_i)_p$ are just formal symbols, nothing to do with the differential. (Right?) I would proceed by applying the differential of $x_i$ at $p$ to a general element of the tangent space $v=\sum_j v_j(\frac{\partial}{\partial x_j})_p$ $$ (dx_i)_p(v):=v_i $$ Then how shall I go on?

Answer 1

To remove notational confusion, let $\lambda^1,\ldots,\lambda^n$ be the dual basis of $\frac{\partial}{\partial x^1},\ldots,\frac{\partial}{\partial x^n}$. Then, your goal is to show that $$d(x^i)=\lambda^i.$$ i.e. $d(x^i)=dx^i$, where the later is the formal symbol usually used for $\lambda^i$. Note that this justifies the notation $dx^i$.

Now, compute: $$d(x^i)\left(\frac{\partial}{\partial x^j}\right)=\frac{\partial}{\partial x^j}x^i=\delta^i_j=\lambda^i\left(\frac{\partial}{\partial x^j}\right).$$ Hence, $d(x^i)$ agree with $\lambda^i$ on a basis, so they must be equal.

Answer 2

By whatever arguments, you yourself arrived at the correct answer: $dx_i$ computes at each point $p\in{\mathbb R}^n$ the $i$th coordinate of any tangent vector $v$ attached at $p$.

A simple way to see this is as follows: $x_i(\cdot)$ can be viewed as a real-valued function on ${\mathbb R}^n$. In order to compute the differential $dx_i(p)$ we have to look at increments $$\Delta x_i(v):=x_i(p+v)-x_i(p)$$ and to determine an approximation for $\Delta x_i(v)$ which is linear in $v$ when $v\to0$. Now in this special case an approximation is not needed at all, since $$\Delta x_i(v)=v_i$$ is already linear in $v$. This implies that the differential $dx_i(p)$ is given by $$dx_i(p).v= v_i\ ,$$ independently of the point $p$ under consideration.