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### Differential Forms in our sweet home, $\mathbb{R}^n$
> [!NOTE]
>
> I'm a bit deviated form the notation we used in the book, in the actual text, they use $\mathbb{R}^n_p$ to represent the tangent space of $\mathbb{R}^n$ at $p$. But to help you link those concepts as we see in smooth manifolds, $T_pM$, we will use $T_p\mathbb{R}^n$ to represent the tangent space of $\mathbb{R}^n$ at $p$.
Let $p$ be a point in $\mathbb{R}^n$. The tangent space of $\mathbb{R}^n$ at $p$ is denoted by $T_p\mathbb{R}^n$, is the set of all vectors in $\mathbb{R}^n$ that use $p$ as origin.
#### Definition of a vector field
A vector field is a map that associates to each point $p$ in $\mathbb{R}^n$ a vector $v(p)$ in $T_p\mathbb{R}^n$.
That is
@@ -40,3 +46,192 @@ $v$ is differentiable at $p$ if the function $a_i$ is differentiable at $p$.
This gives a vector field $v$ on $\mathbb{R}^n$.
#### Definition of dual space of tangent space
To each tangent space $T_p\mathbb{R}^n$ we can associate the dual space $(T_p\mathbb{R}^n)^*$, the set of all linear maps from $T_p\mathbb{R}^n$ to $\mathbb{R}$. ($\varphi:T_p\mathbb{R}^n\to \mathbb{R}$)
The basis for $(T_p\mathbb{R}^n)^*$ is obtained by taking $(dx_i)_p$ for $i=1,...,n$.
This is the dual basis for $\{(e_i)_p\}$ since.
$$
(dx_i)_p(e_j)=\frac{\partial x_i}{\partial x_j}=\begin{cases}0 \text{ if } i\neq j\\
1 \text{ if } i=j
\end{cases}
$$
#### Definition of a 1-form
A 1-form is a linear map from $(T_p\mathbb{R}^n)^*$ to $\mathbb{R}$.
$$
\omega(p)=a_1(p)(dx_1)_p+...+a_n(p)(dx_n)_p
$$
where $a_i(p)$ is a function that maps $\mathbb{R}^n$ to $\mathbb{R}$.
Generalization of 1-form is $k$-form defined as follows:
#### Definition of a $k$-form
We can define the set of linear map $\Lambda^2(\mathbb{R}^n_p)^*$ where $\varphi$ maps from $(T_p\mathbb{R}^n)^*\times ... \times (T_p\mathbb{R}^n)^*$ to $\mathbb{R}$, that are bilinear and alternate ($\varphi(v_1,v_2)=-\varphi(v_2,v_1$).
when $\varphi_1$ and $\varphi_2$ are linear maps from $(T_p\mathbb{R}^n)^*$ to $\mathbb{R}$, then $\varphi_1\wedge \varphi_2$ is a bilinear map from $(T_p\mathbb{R}^n)^*\times (T_p\mathbb{R}^n)^*$ to $\mathbb{R}$ by setting
$$
(\varphi_1\wedge \varphi_2)(v_1,v_2)=\varphi_1(v_1)\varphi_2(v_2)-\varphi_1(v_2)\varphi_2(v_1)=\det(\varphi_i(v_j))
$$
where $i,j=1,\ldots,k$, $k$ is the degree of the exterior form
More generally, $(\varphi_1\wedge \varphi_2\wedge\dots \wedge \varphi_k)(v_1,v_2,\dots,v_k)=\det(\varphi_i(v_j))$.
And $\{(dx_i\wedge dx_j)_p,i<j\}$ forms a basis for $\Lambda^2(\mathbb{R}^n_p)^*$.
- $(dx_i\wedge dx_j)_p=-(dx_j\wedge dx_i)_p$
- $(dx_i\wedge dx_i)_p=0$
An exterior fom of degree 2 in $\mathbb{R}^n$ is a correspondence $\omega$ that associates to each point $p$ in $\mathbb{R}^n$ an element $\omega(p)\in \Lambda^2(\mathbb{R}^n_p)^*$.
That is
$$
\omega(p)=a_{12}(p)(dx_1\wedge dx_2)_p+a_{13}(p)(dx_1\wedge dx_3)_p+a_{23}(p)(dx_2\wedge dx_3)_p
$$
In the case of $\mathbb{R}^3$.
<details>
<summary>Example for real space 4 product</summary>
0-forms: functino in $\mathbb{R}^4$
1-forms: $a_1(p)(dx_1)_p+a_2(p)(dx_2)_p+a_3(p)(dx_3)_p+a_4(p)(dx_4)_p$
2-forms: $a_{12}(p)(dx_1\wedge dx_2)_p+a_{13}(p)(dx_1\wedge dx_3)_p+a_{14}(p)(dx_1\wedge dx_4)_p+a_{23}(p)(dx_2\wedge dx_3)_p+a_{24}(p)(dx_2\wedge dx_4)_p+a_{34}(p)(dx_3\wedge dx_4)_p$
3-forms: $a_{123}(p)(dx_1\wedge dx_2\wedge dx_3)_p+a_{124}(p)(dx_1\wedge dx_2\wedge dx_4)_p+a_{134}(p)(dx_1\wedge dx_3\wedge dx_4)_p+a_{234}(p)(dx_2\wedge dx_3\wedge dx_4)_p$
4-forms: $a_{1234}(p)(dx_1\wedge dx_2\wedge dx_3\wedge dx_4)_p$
</details>
#### Exterior product of forms
Let $\omega=\sum a_{I}dx_I$ be a k form where $I=(i_1,i_2,\ldots,i_k)$ and $i_1<i_2<\cdots<i_k$.
$\varphi\wedge \omega=\sum b_jdx_j$ be a s form where $j=(j_1,j_2,\ldots,j_s)$ and $j_1<j_2<\cdots<j_s$.
The exterior product is defined as
$$
(\varphi\wedge \omega)(v_1,\ldots,v_k)=\sum_{IJ}a_I b_J dx_I\wedge dx_J
$$
<details>
<summary>Example for exterior product of forms</summary>
Let $\omega=x_1dx_1+x_2dx_2+x_3dx_3$ be a 1-form in $\mathbb{R}^3$ and $\varphi=x_1dx_1\wedge dx_1\wedge dx_3$ be a 2-form in $\mathbb{R}^3$.
Then
$$
\begin{aligned}
\omega\wedge \varphi&=x_2 dx_2\wedge dx_1\wedge dx_3+x_3x_1 dx_3\wedge dx_1\wedge dx_2\\
&=(x_1x_3-x_2)dx_1\wedge dx_2\wedge dx_3
\end{aligned}
$$
Note $dx_1\wedge dx_1=0$ therefore $dx_1\wedge dx_1\wedge dx_3=0$
</details>
#### Additional properties of exterior product
Let $\omega$ be a $k$ form, $\varphi$ be a $s$ form, and $\theta$ be an $r$ form.
- $(\omega\wedge\varphi)\wedge\theta=\omega\wedge(\varphi\wedge\theta)$
- $(\omega\wedge\varphi)=(-1)^{k+s}(\varphi\wedge\omega)$
- $\omega\wedge(\varphi+\theta)=\omega\wedge\varphi+\omega\wedge\theta$
#### Important implications with differential maps
Let $f:\mathbb{R}^n\to \mathbb{R}^m$ be a differentiable map. Then $f$ induces a map $f^*$ from k-forms in $\mathbb{R}^n$ to k-forms in $\mathbb{R}^m$.
That is
$$
(f^*\omega)(p)(v_1,\ldots,v_k)=\omega(f(p))(df(p)_1v_1,\ldots,df(p)_kv_k)
$$
Here $p\in \mathbb{R}^n$, $v_1,\ldots,v_k\in T_p\mathbb{R}^n$, and $df(p):T_p\mathbb{R}^n\to T_{f(p)}\mathbb{R}^m$.
If $g$ is a 0-form, we have
$f^*(g)=g\circ f$
#### Additional properties for differential maps
Let $f:\mathbb{R}^n\to \mathbb{R}^m$ be a differentiable map, $\omega,\varphi$ be k-forms on $\mathbb{R}^m$ and $g:\mathbb{R}^m\to \mathbb{R}$ be a 0-form on $\mathbb{R}^m$. Then:
- $f^*(\omega+\varphi)=f^*\omega+f^*\varphi$
- $f^*(g\omega)=f^*(g)f^*\omega$
- If $\varphi_1,\dots,\varphi_k$ are 1-forms in $\mathbb{R}^m$, $f^*(\varphi_1\wedge\cdots\wedge\varphi_k)=f^*\varphi_1\wedge\cdots\wedge f^*\varphi_k$
If $g:\mathbb{R}^p\to \mathbb{R}^n$ is a differential map and $\varphi,\omega$ are any two-forms in $\mathbb{R}^m$.
- $f^*(\omega\wedge\varphi)=f^*\omega\wedge f^*\varphi$
- $(f\circ g)^*omega=g^*(f^*\omega)$
#### Exterior Differential
Let $\omega=\sum a_{I}dx_I$ be a k form in $mathbb{R}^n$. The exterior differential $d\omega$ of $\omega$ is defined by
$$
d\omega=\sum da_{I}\wedge dx_I
$$
#### Additional properties of exterior differential
- $d(\omega_1+\omega_2)=d\omega_1+d\omega_2$ where $\omega_1,\omega_2$ are k-forms
- $d(\omega\wedge\varphi)=d\omega\wedge\varphi+(-1)^kw\wedge d\varphi$ where $\omega$ is a k-form and $\varphi$ is a s-form
- $d(d\omega)=d^2\omega=0$
- $d(f^*\omega)=f^*d\omega$ where $f$ is a differentiable map and $\omega$ is a k-form
## Differentiable manifolds
### A different flavor of differential manifolds
#### Definition of differentiable manifold
An $n$-dimensional differentiable manifold is a set $M$ together with a family of of injective maps $f_\alpha:U_\alpha\subseteq \mathbb{R}^n\to M$ of open sets $U_\alpha$ in $\mathbb{R}^n$ in to $M$ such that:
- $\bigcup_\alpha f_\alpha(U_\alpha)=M$
- For each pair $\alpha,\beta$, with $f_\alpha(U_\alpha)\cap f_\beta(U_\beta)=W\neq \emptyset$, the sets $f_\alpha^{-1}(W)$ and $f_\beta^{-1}(W)$ are open sets in $\mathbb{R}^n$ and the maps $f_\beta^{-1}\circ f_\alpha$ and $f_\alpha^{-1}\circ f_\beta$ are differentiable.
- The family $\{(U_\alpha,f_\alpha)\}$ is the maximal relative to the two properties above.
> This condition is weaker than smooth manifold, in smooth manifold, we require the function to be class of $C^\infty$ (continuous differentiable of all order), now we only needs it to be differentiable.
#### Definition of differentiable map between differentiable manifolds
Let $M_1^n$ and $M_2^M$ be differentiable manifolds. A map $\varphi:M_1\to M_2$ is a differentiable at a point $p\in M_1$ if given a parameterization $g:V\subset \mathbb{R}^m\to M_2$ around $\varphi(p)$, there exists a parameterization $f:U\subseteq \mathbb{R}^n\to M_1$ around $p$ such that:
$\varphi(f(U))\subset g(V)$ and the map
$$
g^{-1}\circ \varphi\circ f: U\subset \mathbb{R}^n\to \mathbb{R}^m
$$
is differentiable at $f^{-1}(p)$.
It is differentiable in an open set of $M_1$ if it is differentiable at all points in such set.
The map $g^{-1}\circ \varphi\circ f$ is the expression of parameterization of $f$ and $g$. (Since the change of parameterization is differentiable, the property that $f$ is differentiable does not depends on the choice of parameterization.)
#### Tangent vector over differentiable curve
Let $\alpha: I\to M$ be a differentiable curve on a differentiable manifold $M$, with $\alpha(0)=p\in M$, and let $D$ be the set of functions of $M$ which are differentiable at $p$. then tangent vector to the curve $\alpha$ at $p$ is the map $\alpha'(0):D\to \mathbb{R}$ given by
$$
\alpha'(0)\varphi=\frac{d}{dt}(\varphi\cdot \alpha(t))|_{t=0}
$$
A tangent vector at $p\in M$ is the