Proof that $dx_i(\partial_{x_j})=\delta_{ij}$

56 Views Asked by At

Following O'neil "Elemenatry Differential Geometry", in pg.12 he defines the tangent vector $v_p \in T_p\Bbb R^n$ as an operator on the scalar field $f:\Bbb R^n\to \Bbb R$ s.t $v_p$ acting on $f$ is the directional derivative:

$$v_p[f]=\left. {d\over dt}f(p+vt)\right\rvert_{t=0} =\sum \left.{\partial f\over\partial x_i}\right\rvert_p v_i$$

and so you can write $v_p$ as

$$v_p=\sum v_i\left.{\partial \over\partial x_i}\right\rvert_p$$

because is a linear operator. Then in pg.24 he define $1-\text{form}\ df$ as a linear combination of the set $\{dx_1,dx_2,\dots,dx_n\}$ of covectors, and the acting of $v_p$ maps from $T_p\Bbb R^n$ to $\Bbb R$, which is an alternate definition for the directional derivative, and so:

$$df(v_p)=v_p[f]$$

I can understand the claim that $dx_i(v_p)=v_p[x_i]=v_i$, and then it must be that:

$$dx_i\left(\left.{\partial \over\partial x_j}\right\rvert_p\right)=\delta_{ij}$$

but for me, I'd expect the limit of something, like the definition of the derivative or the Riemann integral. Is it there a less abstract or some sort of algebraic manipulations to start from differentials as "infinitesimal displacement vector" to "the basis" of $(T_p\Bbb R^{n})^*$?