# Manifold and derivative: change of coordinate.

I'm stuck on the following problem.

Let $M$ a manifold of dimension $n$ and let $\varphi_1:U_1\longrightarrow W_2$ and $\varphi_2:U_2\longrightarrow W_2$ two charts at the neighborhood of $p\in U_1\cap U_2$. Let $$h=\varphi_2\circ \varphi^{-1}_1:\varphi_1(U_1\cap U_2)\longrightarrow \varphi_2(U_1\cap U_2).$$ Let $x^1,...,x^n$ the coordinates given by $\varphi_1$ and $y^1,...,y^n$ the coordinates given by $\varphi_2$ (therefore $y=h(x)$). Let $y^j=h^j(x_1,...,x_n)$. Let $u=u(y^1,...,y^n)\in\mathcal C^\infty (M)$. In what the fact that $$\frac{\partial }{\partial x^i}(u\circ h)=\sum_{\ell=1}^n\frac{\partial u}{\partial y^\ell}\frac{\partial y^\ell}{\partial x^i}$$ for all $u$ implies that $$\frac{\partial }{\partial x^i}=\sum_{\ell=1}^n\frac{\partial y^\ell}{\partial x^i}\frac{\partial }{\partial y^\ell}$$ I absolutely don't understand the manipulation here.

## 1 Answer

The first equation is essentially the chain rule. The chain rule says that $d(u \circ h) = du \circ dh$. First, $du$ is just the gradient vector $du = \left( \frac{\partial u}{\partial y^1},\dots,\frac{\partial u}{\partial y^n} \right)$. Second, $dh$ is the Jacobian matrix of partial derivatives of $h^i$ with respect to $x^j$. But since $h^i = y^i$, we have $dh = \left( \frac{\partial y^i}{\partial x^j} \right)_{i,j=1}^n$. Doing the matrix multiplication, $du \circ dh$ is a row vector, and $\frac{\partial}{\partial x^i}(u \circ h)$ is the $i^\mathrm{th}$ component of this vector, which is the dot product of $du$ with the $i^\mathrm{th}$ column of $dh$. This is where your first equation comes from:

$$\frac{\partial}{\partial x^i}(u \circ h) = \sum_{\ell = 1}^n \frac{\partial y^\ell}{\partial x^i} \frac{\partial u}{\partial y^\ell}$$

Notice that $u \circ h$ is just the same function $u$, but represented now in the $x$-coordinate patch. Your second equation comes from considering this observation, and also writing the right-hand side of the previous equation as a single differential operator acting on $u$:

$$\frac{\partial}{\partial x^i}(u \circ h) = \sum_{\ell = 1}^n \frac{\partial y^\ell}{\partial x^i} \frac{\partial u}{\partial y^\ell} = \left( \sum_{\ell=1}^n \frac{\partial y^\ell}{\partial x^i} \frac{\partial}{\partial y^\ell}\right)u$$

So $\frac{\partial}{\partial x^i}$ is the operator in parentheses, written in terms of the $\frac{\partial}{\partial y^i}$ basis.