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I'm reading through Spivak's Comprehensive Introduction to Differential Geometry, vol. 1. On page 35 I encountered the following:

Let $(x, U)$ be some coordinate system with $x(p) = (x^1(p), \ldots, x^n(p))$ and $f:M \to \mathbb{R}$ be a function. Define $$ \frac{\partial f}{\partial x^i}(p) = D_i(f \circ x^{-1})(x(p)) $$

Notice that $$ \frac{\partial x^i}{\partial x^j}(p) = \delta_j^i = \begin{cases} 1 & \mathrm{if}\;i=j \\ 0 & \mathrm{if}\; i\neq j \end{cases} $$

I have been trying to prove this assertion but I can't seem to get anywhere. Any help would be amazing!

EDIT:

I may have figured it out: $$ \begin{align*} \frac{\partial x^i}{\partial x^j}(p) &= D_i(x^j \circ x^{-1})(x(p)) \\& = D_i(x \circ x^{-1})^j(x(p)) \\& = D_i(I^j)(x(p)) = \begin{cases} 1 & \mathrm{if}\; i=j\\ 0 & \mathrm{if}\; i\neq j \end{cases} \end{align*} $$

Where $I$ is the identity map.

Tölvukisa
  • 53

1 Answers1

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As per my edit and peek-a-boo's confirmation, here is the proof:

$$ \begin{align*} \frac{\partial x^i}{\partial x^j}(p) &= D_i(x^j \circ x^{-1})(x(p)) \\& = D_i(x \circ x^{-1})^j(x(p)) \\& = D_i(I^j)(x(p)) = \begin{cases} 1 & \mathrm{if}\; i=j\\ 0 & \mathrm{if}\; i\neq j \end{cases} \end{align*} $$ Where $I$ is the identity map on $x(U)$.

Tölvukisa
  • 53