# Partial differential with respect to local coordinates

Let $$M^n$$ be a smooth manifold and $$(U,\phi)$$ be a chart with local coordinates $$x^1,...,x^n$$, that is $$\phi(u)=(x^1(u),...,x^n(u))$$, with $$x^i:U\to\mathbb{R}$$. For $$f\in C^\infty(M)$$ we define $$\tfrac{\partial}{\partial x^i}f:U\to \mathbb{R},\quad p\mapsto\tfrac{\partial}{\partial x^i} f(p)=\partial_i(f\circ\phi^{-1})(\phi(p)),$$ where the partial derivative is w.r.t. the $$i$$-th coordinate is just partial derivative in $$\mathbb{R}^n$$. Now if we take a second chart $$(V,\psi)$$ with local coordinate functions $$y^1,...,y^n$$, why is it true that for $$p\in U\cap V$$ one has $$\tfrac{\partial}{\partial x^i}|_p=\tfrac{\partial}{\partial x^i}|_p(y^j)\tfrac{\partial}{\partial y^j}|_p=\sum_{j=1}^n \tfrac{\partial}{\partial x^i}|_p(y^j)\tfrac{\partial}{\partial y^j}|_p.$$ (In the identity the Einstein notation for sums was used). I think, I understand why it makes sense to define the partial derivative like this and for the case of $$M^n=\mathbb{R}^n$$, we can just take $$x^i$$ and $$y^i$$ to be the standard coordinates, in which case the identity would be trivial. But how to argue for general manifolds?

• Use the fact that $\frac{\partial}{\partial y^1}|_p, \ldots, \frac{\partial}{\partial y^n}|_p$ is a basis for the vector space $T_p M$. Apr 27 at 7:39

Since $$\frac{\partial}{\partial y^1}|_p, \ldots, \frac{\partial}{\partial y^n}|_p$$ is a basis for the vector space $$T_pM$$, we can express each vector $$\frac{\partial}{\partial x^i}|_p \in T_p M$$ as a linear combination $$\dfrac{\partial}{\partial x^i}\bigg|_p = a_1 \dfrac{\partial }{\partial y^1}\bigg|_p + \cdots + a_n \dfrac{\partial }{\partial y^n}\bigg|_p ,$$ for some $$a_1, \ldots, a_n \in \mathbb{R}$$. Now to find out the constants $$a_j$$, we apply both sides of the above identity on the functions $$y^j$$: $$\dfrac{\partial }{\partial x^i}\bigg|_p y^j= a_1 \dfrac{\partial }{\partial y^1}\bigg|_p y^j + \cdots + a_n \dfrac{\partial }{\partial y^n}\bigg|_p y^j.$$ Because $$\frac{\partial }{\partial y^i}|_p y^j= \delta^{i}_j$$, we get that $$\dfrac{\partial }{\partial x^i}\bigg|_p y^j= a_j.$$ Therefore substituting these values of $$a_j$$ on the first identity we get $$\dfrac{\partial}{\partial x^i}\bigg|_p = \dfrac{\partial y^1}{\partial x^i}(p)\dfrac{\partial}{\partial y^1}\bigg|_p + \cdots + \dfrac{\partial y^n}{\partial x^i}(p)\dfrac{\partial}{\partial y^n}\bigg|_p,$$ as required.