$M\subset \mathbb{R}^n$ is a $m$-dimensional smooth sub-manifold of $\mathbb{R}^n$ if $\forall \boldsymbol{p}\in M,\ \exists r > 0$ and $F: B_r(\boldsymbol{p}) \to \mathbb{R}^{n-m}$ such that

$\begin{aligned} M \cap B_r(\boldsymbol{p}) = \{\boldsymbol{x}\in B_r(\boldsymbol{p}) | F(\boldsymbol{x}) = 0\},\\ F\text{ is smooth,}\\ \forall \bar{\boldsymbol{x}} \in M \cap B_r(\boldsymbol{p}), \text{Rank}\left(\frac{\partial F}{\partial \boldsymbol{x}} \bigg\rvert_{\bar{\boldsymbol{x}}}\right) = n - m \end{aligned}$

By Definition 19, a manifold is essentially defined as the 0-level set of some smooth function $F$ and can be thought of as a surface embedded in a higher dimension.

The tangent space of a manifold $M$ at $\boldsymbol{p}\in M$ is given by

$T_{\boldsymbol{p}}M = \text{Null}\left(\frac{\partial F}{\partial \boldsymbol{x}}\bigg |_{\boldsymbol{p}}\right)$

The tangent space consists of all vectors tangent to the manifold at a particular point $\boldsymbol{p}$.

The Tangent Bundle of a manifold $M$ is the collection of all tangent spaces

$T_M = \bigcup_{\boldsymbol{p}\in M} T_{\boldsymbol{p}} M$

A vector field $f:M\to T_M$ on a manifold $M$ is an assignment of each point $\boldsymbol{p}\in M$ to a vector in the tangent space in that point $T_{\boldsymbol{p}}M$.

The Lie Derivative of a function $V$ with respect to a vector field $f$ is given by

$L_fV = (\nabla_{\boldsymbol{x}}V)^\top f(\boldsymbol{x}).$

Suppose that $f(\boldsymbol{x})$ and $g(\boldsymbol{x})$ are vector fields. The Lie Bracket of $f$ and $g$ is given by

The Lie Bracket is another vector field, and it essentially measures the difference between moving along vector field $f$ and vector field $g$ across some infinitesimal distance. Another way to think about the Lie Bracket is as a measure of the extent to which $f$ and $g$ commute with each other. The Lie Bracket is also sometimes denoted using the adjoint map

$[f,[f,[f,\cdots[f,g]]]] = \text{ad}_f^ig.$

For a function $h$ and vector fields $f$ and $g$,

$L_{[f,g]}h = L_fL_gh - L_gL_fh$

$L_gL_f^ih(\boldsymbol{x}) = 0 \Leftrightarrow L_{\text{ad}_f^ig}h(\boldsymbol{x}) = 0$

Suppose $f_1,f_2,\cdots,f_n$ are vector fields. A distribution $\Delta$ is the span of the vector fields at each point $\boldsymbol{x}$:

$\Delta(\boldsymbol{x}) = \text{span}\{f_1(\boldsymbol{x}), f_2(\boldsymbol{x}),\cdots,f_n(\boldsymbol{x})\}.$

At each point $\boldsymbol{x},\ \Delta(\boldsymbol{x})$ is a subspace of the tangent space at $\boldsymbol{x}$.

The dimension of a distribution at a point $\boldsymbol{x}$ is given by

$\text{Dim }\Delta(\boldsymbol{x}) = \text{Rank}\left(\begin{bmatrix} f_1(\boldsymbol{x}) & \bigg\lvert & f_2(\boldsymbol{x}) & \bigg\lvert & \cdots & \bigg\lvert & f_n(\boldsymbol{x}) \end{bmatrix}\right)$

A distribution $\Delta$is nonsingular, also known as regular, if its dimension is constant.

A distribution $\Delta$ is involutive if

$\forall f, g\in \Delta, \quad [f, g] \in \Delta$

A nonsingular $K$-dimensional distribution $\Delta(\boldsymbol{x}) = \text{span}\{f_1(\boldsymbol{x}), \cdots, f_k(\boldsymbol{x})\}$ is completely integrable if $\exists \phi_1,\cdots,\phi_{n-k}$ such that $\forall i,k,\ L_{f_k}\phi_i = 0$ and $abla_{\boldsymbol{x}}\phi_i$are linearly independent. \label{thm:involutive}

A nonsingular $\Delta$ is completely integrable if and only if $\Delta$is involutive.