Notes:Differential (manifolds)

Reason for page: I'm encountering expressions like:

• $$dF_p\left(\frac{\partial}{\partial x^i}\Big\vert_p\right)=dF_p\left(d(\varphi^{-1})_{\hat{p} }\left(\frac{\partial}{\partial x^i}\Big\vert_{\hat{p} }\right)\right)$$ $$=d(\psi^{-1})_{\hat{F}(\hat{p})}\left(d\hat{F}_{\hat{P} }\left(\frac{\partial}{\partial x^i}\Big\vert_{\hat{p} }\right)\right)$$ $$=d(\psi^{-1})_{\hat{F}(\hat{P})}\left(\frac{\partial \hat{F}^j}{\partial x^i}(\hat{p})\frac{\partial}{\partial}{y^j}\Big\vert_{\hat{F}(\hat{p})}\right)$$ $$=\frac{\partial\hat{F}^j}{\partial x^i}(\hat{P})\frac{\partial}{\partial y^j}\Big\vert_{F(p)}$$ and this is apparently a matrix! It's very easy to forget what the operations are, what their elements are, so forth. This notes page is a reminder for me. Example taken from page 62 of Books:Introduction to Smooth Manifolds - John M. Lee

Definitions

• Smoothness of a map (AKA: $C^\infty$ - a map, $f:U\subseteq\mathbb{R}^n\rightarrow V\subseteq\mathbb{R}^m$ is smooth if it has continuous partial derivatives of all orders.
• Smooth map - Given smooth manifolds, $M$ and $N$ and a map, $F:M\rightarrow N$. $F$ is a smooth map if:
  • $\forall p\in M\ \exists (U,\varphi)\in\mathcal{A}_M\ \exists(V,\psi)\in\mathcal{A}_N[p\in U\wedge F(p)\in V\wedge F(U)\subseteq V\implies \psi\circ F\circ \varphi^{-1}:\varphi(U)\rightarrow\psi(V)\text{ is smooth}]$
• Derivation - a map, $\omega:C^\infty(M)\rightarrow\mathbb{R}$ that is linear and satisfies the Leibniz rule:
  • $\forall f,g\in C^\infty(M)[w(fg)=f(a)w(g)+g(a)w(f)]$ (sometimes called the product rule)
• Tangent space to $M$ at $p$ $T_pM$ is a vector space called the tangent space to $M$ at $p$, it's the set of all derivations of $C^\infty(M)$
• Differential of $F$ at $p$. For smooth manifolds, $M$ and $N$ and a smooth map, $F:M\rightarrow N$ we define the differential of $F$ as $p\in M$ as:
  • $dF_p:T_pM\rightarrow T_{F(p)}M$ given by: $dF_p:v\mapsto\left\{\begin{array}{l}:C^\infty(N)\rightarrow \mathbb{R}\\:f\mapsto v(f\circ F)\end{array}\right.$

Moving about

• Changing coordinates - $$\frac{\partial}{\partial x_i}\Big\vert_p=\frac{\partial \bar{x}^j}{\partial x^i}(\varphi(p))\frac{\partial}{\partial \bar{x}^j}\Big\vert_p$$ - using Template:ESC
  • Note that this is actually a vector (as there's an implicit sum over $j$.