# Differential Geometry

$M\subset \mathbb{R}^n$

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 consists of all vectors tangent to the manifold at a particular point

$\boldsymbol{p}$

.Therefore, a vector field can be thought of as a curve through the tangent bundle of a manifold.

A Lie Derivative is essentially a directional derivative, and it measures how a function changes along a vector field.

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

$\text{ad}_fg = [f, g].$

It is helpful when chaining Lie Brackets since we can denote

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

Since the Lie Bracket is a vector field, we can look at Lie Derivatives with respect to the Lie Bracket of two vector fields.

We can also use relate repeated Lie Derivatives to doing repeated Lie Brackets.

At each point

$\boldsymbol{x},\ \Delta(\boldsymbol{x})$

is a subspace of the tangent space at $\boldsymbol{x}$

.Distributions have different properties which are important to look at.

In involutive distributions, you can never leave the distribution by traveling along vectors inside the distribution.

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}It turns out that integrability and involutivity are equivalent to each other.

Last modified 1yr ago