# [Physics] Intuition behind differential operators as the basis vectors of a manifold (space-time)

differential-geometrydifferentiationgeneral-relativityspacetimeVector Fields

I understand that in order to provide a basis for every point in space-time, the differential operators, $\partial_\mu$ (or partial derivative operator with respect to each one of the curvilinear coordinates in the manifold, i.e. $\left\{\frac{\partial}{\partial x^0}, \frac{\partial}{\partial x^1}, \frac{\partial}{\partial x^2},\frac{\partial}{\partial x^3}\right\}$) are typically chosen.

The actual function, $f$, is not explicitly stated, but it is understood that there would be a linear operator would coefficients $A^\mu$, such that at any point $p$

$$A^\mu\,\partial_\mu\; f\;\Big\vert_p$$

$\partial_\mu$ are the coordinate basis.

The basis of the dual space $V^*$ are the differential form $dx^\mu$, such that when the basis of $V$ are "fed" into the basis of the dual, $\frac{\partial}{\partial x^\mu}\,dx^\nu=\delta^\nu_\mu.$

I kind of get this idea, but I have problems visualizing $\partial_\mu$ as arrow vectors. Are there any visual aids?

Let $f$ be a smooth function on Euclidean space, and let $a$ be a vector. The point $P$ can be translated by $a$ into $P+a$. The translation acts on the function by an operator $\hat{T}_a$ as $$(\hat{T}_a[f])(P)=f(P+a).$$ If $\epsilon$ is an "infinitesimal", then $f(P+\epsilon a)$ is expanded as $$f(P+\epsilon a)=f(P)+\frac{\partial f}{\partial x^\mu}|_{x=P}a^\mu\epsilon+O(\epsilon^2),$$ but $f(P+\epsilon a)=(\hat T_{\epsilon a}[f])(P)$, so the operator that translates by $\epsilon a$ is given as $$\hat T_{\epsilon a}=1+\epsilon a^\mu\frac{\partial}{\partial x^\mu}.$$