# Sergei Yakovenko's blog: on Math and Teaching

## Friday, April 27, 2018

### Lecture 4 (April 15, 2018)

Filed under: Calculus on manifolds course — Sergei Yakovenko @ 4:09
Tags: , ,

## Smooth manifolds: howto and whatfor

Topological spaces with an extra smooth structure. Examples: subsets of the Euclidean spaces, Lie groups, quotient spaces. Examples.

How to work with manifolds. Charts, atlases.

How to simplify your thoughts once you know what they are about. Structural algebra of smooth functions, various definitions of the vector fields, functoriality.

### Lecture 3 (April 8, 2018)

Filed under: Calculus on manifolds course — Sergei Yakovenko @ 4:04
Tags: , ,

## Vector fields in open domains of $\mathbb R^n$

Ordinary differential equations, differential operators of the first order, local analysis and global consequences.

## Objects that live on manifolds: functions, curves, vector fields

We discussed how one may possibly define smooth functions on manifolds, smooth curves, tangent vectors, smooth vector fields. Next we discussed how these objects can be carried between manifolds if there exists a smooth map (or diffeomorphism) between these manifolds.

## Flow of vector field. Lie derivatives.

Every vector field $X$ on a compact smooth manifold $M$ defines a family of automorphisms $F^t_X$ (diffeomorphic self-maps) of $M$ which form a one-parametric group, called the flow. Any object living on $M$ can be carried by the flow by the operators $\bigl(F^t_X\bigr)^*$, $t\in\mathbb R$. The Lie derivative along $X$ is the velocity of this action at $t=0$, namely, $L_X=\frac{\mathrm d}{\mathrm dt}\big|_{t=0}\bigl(F^t_X\bigr)^*$.

We show that the Lie derivative of functions coincides with the action of the corresponding derivations, and the Lie derivation of another vector field is the Lie bracket $L_XY=[X,Y]$.

At the end of the day we establish the identities $[L_X,L_Y]=L_{[X,Y]}$ and the Leibniz rule for $L_X$ with respect to the Lie bracket, $L_X[Y,Z]=[Y,L_XZ]+[L_XY,Z]$. Both turn out to be equivalent to the Jacobi identity $[X,[Y,Z]]+[Y,[X,Z]]+[Z,[X,Y]]=0$ for the Lie bracket.

The lecture notes are available here.

* In fact, it is a slightly different Jacobi identity, not for the Lie bracket of vector fields, but for the vector product $\mathbb R^3\times\mathbb R^3\mapsto\mathbb R^3$, $u,v\mapsto [u,v]=u\times v$. But later we will see that this vector product is the commutator in the Lie algebra of vector fields on the group of orthogonal transformations of $\mathbb R^3$, thus the difference is purely technical.