# Question about a proof in differential geometry and notation

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?

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 $$dx^1,\dots,dx^n$$.
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.