Since a long time I was wondering how can we build proof that a basis applies to its dual give the Kronecker delta. I'm now following on-line lectures related to topology and find out that there is a proper way of doing this.
This proof, I think, will be useful to understand differential form and other somewhat related subjects.
But I'm stock at the end of the proof. Lets first start with some context.
Proof that: $\left(dx^a\right)_p \left(\left(\frac{\partial}{\partial x^b}\right)_p\right) = \delta^a_b$
Context : On a smooth manifold $(M,\mathscr{O}_M,\mathscr{A}_{C^\infty})$ let there be a point $p$, a curve $\gamma$, $\gamma: \mathbb{R} \to M$, where $\gamma(0)=p$ and a function $f: M\to \mathbb{R}$.
The velocity at the point $p$ is defined as :
$v_\gamma,_p : = (f\circ \gamma)^\prime(0)$
From a chart $(U,x)\in \mathscr{A}_{C^\infty}$ consider $x: U \to \mathbb{R}^d$ in particular $x^i: U \to \mathbb{R}$
$v_\gamma,_p = \left((f \circ x^{-1}) \circ (x \circ \gamma)\right)^\prime(0) = (x^i \circ \gamma)^\prime(0) \cdot \left(\partial_i(f \circ x^{-1})\right)(x(p))$
since $(x\circ\gamma)(0)= x(p)$.
Definition 1: By convention we right the notation:
$\left(\frac{\partial}{\partial x^i}\right)_p := \left(\partial_i(f \circ x^{-1})\right)(x(p))$
It is the vector basis $\left(\frac{\partial}{\partial x^i}\right)_p := e_i \in T_pM$
Definition 2: The dual basis is defined as the set of linear maps from $T_pM$ to $\mathbb{R}$ as fallow $T^*_pM := \{ \phi:T_pM \overset{\sim}{\to} \mathbb{R} \}$. In particular $(df)_p$ is defined:
$(df)_p :T_pM \overset{\sim}{\to} \mathbb{R}$
where
$X \mapsto (df)_p(X) :=Xf$
for a vector $X$ under the chart $(U,x)$ defines as $X:=X^i_{(x)}\left(\frac{\partial}{\partial x^i}\right)_p$
Proof : By applying the definition 2, $(df)_p(X) :=Xf$, we have that:
$\left(dx^a\right)_p \left(\left(\frac{\partial}{\partial x^b}\right)_p\right) = \left(\frac{\partial}{\partial x^b}\right)_p(x^a)= \left(\partial_b(x^a \circ x^{-1})\right)(x(p)) $
Now it is obvious that if we take the notation $\left(\frac{\partial}{\partial x^b}\right)_p(x^a)$ and apply undergraduate analysis we have $\delta_b^a$ but the question is what to do with the real thing: $\left(\partial_b(x^a \circ x^{-1})\right)(x(p)) $
Question
Why should $\left(\partial_b(x^a \circ x^{-1})\right)(x(p))$ equal $\delta_b^a$ ? If this is not the case, where is my mistake ?
Edit :
Differentiation of $x$ or $x^{-1}$ is not allowed since $x:U\to \mathbb{R}^d$ is not a $\mathbb{R}^n \to \mathbb{R}^m$ map and there is no differential operation define on the manifold the chart map is needed for that !
But $x^a \circ x^{-1} = \operatorname{id}^a$
so $\left(\partial_b(\operatorname{id}^a)\right)(x(p))$
Confusion
I'm confuse and I don't see how to get simply $\delta^a_b$ from these equations.
Thanks for your comment,
NLC
If $V$ is a vector space with basis $\mathcal{B}=\{v_1,\ldots,v_n\}$, then the dual set $\mathcal{B}^{*}=\{f_1,\ldots,f_n\}$ is defined by the equations: $$f_{i}:V\longrightarrow\mathbb{K}$$ $$f_{i}(v_j):=\delta_{i,j}=\begin{cases} 1 & \text{if} \ i=j\\ 0 & \text{if} \ i\neq j\end{cases}$$ this set turns out to be a basis of the dual space $V^{*}:=\mathrm{Hom}(V,\mathbb{K})$. In effect, the functionals $f_1,\ldots,f_n$ are linearly independent: $$\text{If} \quad \sum_{i=1}^{n}\lambda_{i}\cdot f_{i}=0_{V^{*}},\quad \text{then}$$ then we evaluate on each basis vector $v_{j}$ to get: $$\sum_{i=1}^{n}\lambda_{i}\cdot f_{i}(v_{j})=\left(\sum_{i=1}^{n}\lambda_{i}\cdot f_{i}\right)(v_{j})=0_{V^{*}}(v_{j})=0_{\mathbb{K}},\ \forall\, j$$ thus, $$0=\sum_{i=1}^{n}\lambda_{i}\cdot f_{i}(v_{j})=\sum_{i=1}^{n}\lambda_{i}\delta_{i,j}=\lambda_{j},\ \forall\, j$$. so $\lambda_1=\cdots=\lambda_{n}=0_{\mathbb{K}}$. Then, $\mathcal{B}^{*}$ is a basis of $V^{*}$ called the dual basis of $\mathcal{B}$.