- #1
"Don't panic!"
- 601
- 8
So I know that this involves using the chain rule, but is the following attempt at a proof correct.
Let [itex]M[/itex] be an [itex]n[/itex]-dimensional manifold and let [itex](U,\phi)[/itex] and [itex](V,\psi)[/itex] be two overlapping coordinate charts (i.e. [itex]U\cap V\neq\emptyset[/itex]), with [itex]U,V\subset M[/itex], covering a neighbourhood of [itex]p\in M[/itex], such that [itex]p\in U\cap V[/itex]. Consider a function [itex]f:M\rightarrow\mathbb{R}[/itex], and let [itex]x=\phi(p)[/itex], [itex]y=\psi(p)[/itex]. It follows then that $$\frac{\partial f}{\partial x^{\mu}}(p)=\frac{\partial}{\partial x^{\mu}}\left((f\circ\phi^{-1})(\phi(p))\right)=\frac{\partial}{\partial x^{\mu}}\left[(f\circ\psi^{-1})\left((\psi\circ\phi^{-1})(\phi(p))\right)\right]\\ \qquad \quad\; =\frac{\partial}{\partial y^{\nu}}\left[(f\circ\psi^{-1})\left((\psi\circ\phi^{-1})(\phi(p))\right)\right]\frac{\partial}{\partial x^{\mu}}\left[\left((\psi\circ\phi^{-1})^{\nu}(\phi(p))\right)\right]\\ =\frac{\partial f}{\partial y^{\nu}}(p)\frac{\partial y^{\nu}}{\partial x^{\mu}}(p)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\,$$ where [itex]y(x)=(\psi\circ\phi^{-1})(\phi(p))[/itex].
Hence, as [itex]f[/itex] is an arbitrary differentiable function, we conclude that $$\frac{\partial }{\partial x^{\mu}}=\frac{\partial y^{\nu}}{\partial x^{\mu}}\frac{\partial }{\partial y^{\nu}}$$ From this, we note that as [itex]\lbrace\frac{\partial }{\partial x^{\mu}}\rbrace[/itex] and [itex]\lbrace\frac{\partial }{\partial y^{\nu}}\rbrace[/itex] are two coordinate bases for the tangent space [itex]T_{p}M[/itex] at the point [itex]p[/itex], the two bases must be related by the formula above.
Let [itex]M[/itex] be an [itex]n[/itex]-dimensional manifold and let [itex](U,\phi)[/itex] and [itex](V,\psi)[/itex] be two overlapping coordinate charts (i.e. [itex]U\cap V\neq\emptyset[/itex]), with [itex]U,V\subset M[/itex], covering a neighbourhood of [itex]p\in M[/itex], such that [itex]p\in U\cap V[/itex]. Consider a function [itex]f:M\rightarrow\mathbb{R}[/itex], and let [itex]x=\phi(p)[/itex], [itex]y=\psi(p)[/itex]. It follows then that $$\frac{\partial f}{\partial x^{\mu}}(p)=\frac{\partial}{\partial x^{\mu}}\left((f\circ\phi^{-1})(\phi(p))\right)=\frac{\partial}{\partial x^{\mu}}\left[(f\circ\psi^{-1})\left((\psi\circ\phi^{-1})(\phi(p))\right)\right]\\ \qquad \quad\; =\frac{\partial}{\partial y^{\nu}}\left[(f\circ\psi^{-1})\left((\psi\circ\phi^{-1})(\phi(p))\right)\right]\frac{\partial}{\partial x^{\mu}}\left[\left((\psi\circ\phi^{-1})^{\nu}(\phi(p))\right)\right]\\ =\frac{\partial f}{\partial y^{\nu}}(p)\frac{\partial y^{\nu}}{\partial x^{\mu}}(p)\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\,$$ where [itex]y(x)=(\psi\circ\phi^{-1})(\phi(p))[/itex].
Hence, as [itex]f[/itex] is an arbitrary differentiable function, we conclude that $$\frac{\partial }{\partial x^{\mu}}=\frac{\partial y^{\nu}}{\partial x^{\mu}}\frac{\partial }{\partial y^{\nu}}$$ From this, we note that as [itex]\lbrace\frac{\partial }{\partial x^{\mu}}\rbrace[/itex] and [itex]\lbrace\frac{\partial }{\partial y^{\nu}}\rbrace[/itex] are two coordinate bases for the tangent space [itex]T_{p}M[/itex] at the point [itex]p[/itex], the two bases must be related by the formula above.