Differentials as linear maps on $\Bbb R^n$

Question

Currently going through some multivariable analysis and I'm trying to understand why the differentials $\left(dx_{1}\right)_{p},\left(dx_{2}\right)_{p},\ldots ,\left(dx_{n}\right)_{p}$ at a point $p$ form a basis for the vector space of linear maps from $\mathbb {R}^{n}$ to $\mathbb{R}$ and I'm reading through the this article on Wikipedia.

Is there some simple example which I could use to understand this better? They refer to this trick on defining the coordinates $x_1, \dots, x_n$ in $\Bbb R^n$ such that for $p = (p_1, \dots, p_n)$ one has $x_j : \Bbb R^n \to \Bbb R$ for which $p \mapsto p_j$.

Is there any visual way to understand why the differentials $\left(dx_{1}\right)_{p},\left(dx_{2}\right)_{p},\ldots ,\left(dx_{n}\right)_{p}$ at a point $p$ form a basis for the dual space $(\Bbb R^n)^*$? I would be even interested in the cases when $n$ is small say $2$ or $3$.

It would be perhaps easier for me to first understand why the partial derivatives form a basis for the tangent space on $\Bbb R^n$ since we can use those to define the dual?

Answer 1

In short, you want the differential to be compatible with vector fields in the following sense. Remember that given a function $f$ and a vector field $X$, then $X(f)(p)$ is the directional derivative of $f$ at $p$ along $X(p)$. Since you still want $df$ to encode some sort of derivative information, the differential of a function $f$ is then defined as the object that maps vector fields to functions such that

$$X(f)(p) = df(X)(p).$$

The duality of these spaces then becomes clear as while vector fields consume functions to yield functions, differentials consume vector fields to yield the same functions. If you define the vector field basis in the standard $\partial/\partial x^i$, then the obvious basis for the differentials becomes $dx^i$ such that

$$dx^i\left(\frac{\partial}{\partial x^j}\right)=\delta^i_j$$

Answer 2

The differential of a map $f : U \to V$ between open subsets $U \subset \mathbb R^n$ and $V \subset \mathbb R^m$ at $p \in U$ is the linear map $$df\mid_p : \mathbb R^n \to \mathbb R^m $$ which is "the best linear approximation of $f$ at $p$."

In your question we consider the coordinate maps $x_i : \mathbb R^n \to \mathbb R$, i.e. the projections to the $i$-th coordinate of $p = (p_1,\ldots,p_n)$. These are linear maps and thus we have $(dx_i)\mid_p = x_i$ for all $p \in \mathbb R^n$.

Let us show that these maps form a basis for the dual space $(\mathbb R^n)^*$:

Given a linear $\phi : \mathbb R^n \to \mathbb R$, define $a_i = \phi(e_i)$, where $e_i$ is the $i$-th standard basis vector of $\mathbb R^n$. For $q = (q_1,\ldots,q_n) \in \mathbb R^n$ we have $q = \sum_{i=1}^nq_ie_i$ and therefore $$\phi(q) = \sum_{i=1}^n q_i\phi(e_i) = \sum_{i=1}^n q_ia_i = \sum_{i=1}^n a_ix_i(q) = \sum_{i=1}^n a_i(dx_i)\mid_p(q)$$ which means $$\phi = \sum_{i=1}^n a_i(dx_i)\mid_p .$$ This shows that the $(dx_i)\mid_p$ span $(\mathbb R^n)^*$. Since $\dim (\mathbb R^n)^* = n$, we are done.