Question about a proof in differential geometry and notation

Question

I'm currently studying about volume forms. The theorem I'm trying to understand is the following:

A k-manifold $M$ is orientable iff it admits a nowhere zero k-differential form $\omega$.

The proof I'm reading is from here:http://staff.ustc.edu.cn/~wangzuoq/Courses/18F-Manifolds/Notes/Lec23.pdf (theorem 1.1).

I have a couple of questions:

$1.$ What does the notation $\{U,x_1,...,x_n\}$ mean? I'm familiar with the notation $\{U,\phi\}$.

$2.$ Why is it that if we assume that $\mu$ is a never-vanishing differential form on $M$, then for every local chart $\{U,\phi\}$ there is a smooth function $f$ s.t $\mu=fdx_1\wedge...\wedge dx_n$?

$3.$ What does the notation $\mu(\partial_1,...,\partial_n)$ mean?

Thanks in advance

Answer 1

1. $(U,x^1,\dots,x^n)$ is the local chart written in components, i.e. $U\subseteq M$ is an open subset and $(x^1,\dots,x^n)\colon U\rightarrow\mathbb{R}^n$ is a homeomorphism onto an open subset of $\mathbb{R}^n$ from the given smooth atlas. The $x^1,\dots,x^n$ are known as the coordinates on $U$.

2. This holds true for any $n$-form. The point is that $\Lambda^nT^{\ast}M$ is a $1$-dimensional vector bundle over $M$ and if $(x^1,\dots,x^n)$ is a chart on $U$, $dx^1\wedge\dots\wedge dx^n$ is a frame for $\Lambda^nT^{\ast}M$ over $U$, i.e. for every $n$-form $\mu$ there is an $f\in C^{\infty}(U)$ such that $\mu\vert_U=fdx^1\wedge\dots\wedge dx^n$. The nowhere-vanishing of $\mu$ on $U$ then is equivalent to $f$ being nowhere zero on $U$.

3. $\partial_1,\dots,\partial_n$ are the vector fields induced by the chart $(x^1,\dots,x^n)$ that form a frame for $TM$ over $U$. You might be more familiar with the notation $\frac{\partial}{\partial x^1},\dots,\frac{\partial}{\partial x^n}$ for them. $\mu(\partial_1,\dots,\partial_n)$ simply denotes the evaluation of the $n$-form in these $n$ vectors fields, which is a smooth function on $U$. The fact that it is equal to $f$ follows since $\partial_1,\dots,\partial_n$ is the dual basis to $dx^1,\dots,dx^n$.

Answer 2

1. If $\phi:U\to\phi[U]\subset\Bbb{R}^n$ is a chart map, then you can consider the $n$ coordinate functions $x^i:= \text{pr}^i\circ \phi$, where $\text{pr}^i:\Bbb{R}^n\to\Bbb{R}$ is the mapping $(a^1,\dots, a^n)\mapsto a^i$. So, of course, specifying $\phi$ is the same as specifying all the maps $x^i$.

2. The fact that there exists $f$ such that $\mu=fdx^1\wedge\cdots dx^n$ is not special. This is just due to the fact $\mu$ is an $n$-form and because the space of $n$-forms is a $1$-dimensional space, for which $dx^1\wedge \cdots dx^n$ (pointwise) forms a basis. The crucial thing is that $\mu$ being nowhere vanishing and smooth means $f$ is smooth and nowhere vanishing (and hence constant sign on each connected component of $U$).

3. Given the chart $(U,\phi=(x^1,\dots, x^n))$ we have the coordinate induced vector fields $\frac{\partial}{\partial x^1},\dots, \frac{\partial}{\partial x^n}$. So, $\mu(\partial_1,\dots, \partial_n)$ is short hand for the function on $U$ given by \begin{align} p\mapsto\mu_p\left(\frac{\partial}{\partial x^1}(p),\dots, \frac{\partial}{\partial x^n}(p)\right). \end{align} i.e the evaluation of an $n$-form on $n$-tangent vectors to yield a real number.