Notes:Differential (manifolds)

These notes are taken from Books:Introduction to Smooth Manifolds - John M. Lee unless otherwise noted

The skeleton of what is needed for manifolds is remarkably small. It just looks verbose because of the large amounts of discussion involved. This page is my attempt to "compress" the skeleton of what is needed for understanding the differential into one place. I have applied recursive decent to solve this, which means picking something, finding out about what it depends on, and picking them, until a foundation of known facts is reached.

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 $j$.

Todo

• Note that $(x^1,\ldots,x^n)=(x^1(x),\ldots,x^n(x)):=\varphi(x)$ for a chart $(U,\varphi)$
  • Then expressions like $$\frac{\partial \bar{x}^j}{\partial x^i}(\varphi(p))$$ make significantly more sense, as really it is just $$\frac{\partial \bar{x}^j}{\partial [\varphi(x)]^i}(\varphi(p))$$ where $[\cdot]^i$ is the $i$^th component of a vector in $\mathbb{R}^n$ (recall: $\varphi(U)\subseteq\mathbb{R}^n$)
    • Expand on this.
• There's another kind of derivation - although I suspect it is ALSO $K$-linear and satisfies that rule, however it slightly different.