Does $ \frac{\partial }{\partial x^l}|_p x^j =\frac{\partial x^j}{\partial x^l}|_p=\delta^j_l $ make sense in the general manifold case?

Question

If I have a manifold $M$ of $\dim M=m$ that has an atlas with one map $(M; \varphi)$. If we denote the components of $\varphi$ with $x^1,\dots, x^m$.

Does the following operation make sense?

$\frac{\partial}{\partial x^l}|_p$ is an element of $T_pM$ so it is aderivation that can act on smooth functions over the manifolds such as $x^l$

$ \frac{\partial }{\partial x^l}|_p x^j =\frac{\partial x^j}{\partial x^l}|_p=\delta^j_l $

Why or why not? Can I really see this a elementary derivation, or are there intermediate steps that justify this is true in the manifold case? I am a bit confused because $x^j$ is a function and a variable at the same time when I do that partial derivative and I am working as if M was $R^m$

Answer 1

Let $f :M \to \mathbb{R}$ a smooth function. If we denote the components of $\mathbb{R}^m$ with $r^1 \dots r^m$, by definition $$ \frac{\partial}{\partial x^j}\Bigg|_p(f) = \frac{\partial}{\partial r^j}\Bigg|_{\varphi(p)}(f \circ \varphi^{-1}) $$ where the RHS is a derivation with respect to Euclidean coordinates. So in your case you can compute $$ \frac{\partial}{\partial x^j}\Bigg|_p(x^i) = \frac{\partial}{\partial r^j}\Bigg|_{\varphi(p)}(x^i \circ \varphi^{-1} ) = \frac{\partial}{\partial r^j}\Bigg|_{\varphi(p)}(r^i \circ \varphi \circ \varphi^{-1} ) = \frac{\partial r^i}{\partial r^j}\Bigg|_{\varphi(p)} = \delta^i_j. $$

This reflects that manifolds are modeled as spaces which behave locally as the Euclidean space (at least when you make first order computations).

Answer 2

It makes sense. Have a look at https://www.physicsforums.com/insights/pantheon-derivatives-part-iv/#B-%E2%80%93-Left-Invariant-Vector-Fields-and-GL-n where the Lie groups $GL(n)$ and $SL(n)$ are the manifolds, and their tangent spaces at the identity matrix are computed. The calculation makes particular use of the equation, i.e. fact, you asked about.

The basic perspective is to consider a path $\gamma(t) $ through $p$ with $\gamma(0)=p$ and differentiate the function $f$ by $\left. \dfrac{d}{dt}\right|_{t=0}f(\gamma(t)).$ In the special case of your question, we have $f=x^j$ and $\gamma=x^l$.

Answer 3

The trick to learning this is to observe that if $\phi: U \rightarrow \mathbb{R}^n$ is a coordinate map, then we can write $$\phi=(\phi^1, \dots, \phi^n)$$ and observe that $\phi^i$ is equal to as $x^i$ viewed as a function.

For simplicity, let's do this for $\dim M = 2$. Given a point $p \in M$, let $\phi: U \rightarrow \mathbb{R}^2$ be a coordinate map such that $p \in U$ and $\phi(p) = (0,0)$. Given a function $f: M \rightarrow \mathbb{R}$, the partial derivative of $f$ at $p$ with respect to $x^1$ at $p$, $$ \frac{\partial f}{\partial x^1} (p)$$ is defined to be $$ \left.\frac{d}{dt}\right|_{t=0}f(c(t)), $$ where $$c(t) = \phi^{-1}(t,0).$$ The partial derivative of $f$ with respect to $x^2$ at $p$ is defined similarly. Therefore, \begin{align*} \frac{\partial x^1}{\partial x^1}(p) &= \frac{\partial \phi^1}{\partial x^1}(p)\\ &= \left.\frac{d}{dt}\right|_{t=0}\phi^1(\phi^{-1}(t,0))\\ &= \left.\frac{d}{dt}\right|_{t=0}t\\ &= 1. \end{align*} The calculations for the other partial derivatives of $\phi^1$ and $\phi^2$ are similar and therefore $$ \frac{\partial x^i}{\partial x^j}(p) = \delta^i_j. $$ Since this equation holds for any $p \in U$, we get $$ \frac{\partial x^i}{\partial x^j}(p) = \delta^i_j\text{, for all }p \in U. $$