Sergei Yakovenko's blog: on Math and Teaching

Friday, April 27, 2018

Lecture 3 (April 8, 2018)

Filed under: Calculus on manifolds course — Sergei Yakovenko @ 4:04
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Vector fields in open domains of $\mathbb R^n$

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

Differentiable maps

• Definition of differentiability at a point. Maps $f:U\to W$ between open subspaces of the Euclidean spaces $U\subseteq \mathbb R^n,\ W\subseteq\mathbb R^m$ smooth on their domain.
• Tangent spaces $T_a U$, tangent bundle $TU=\bigcup_{a\in U}T_a U\simeq U\times\mathbb R^n$.
• Differential of a smooth map: $\mathrm df:TU\to TW$.
• What is the derivative? (answer: exists only when $n=m=1$). Partial derivatives.
• How do we define functions “having more than one derivative”?

Algebraic formalism:

• Algebra $C^\infty(U)$ of functions infinitely smooth in a domain $U\subseteq\mathbb R^n$
• Pullback morphism of algebras $f^*:C^\infty(W)\to C^\infty(U)$.

Vector fields: smooth maps $v:U\to TU$, such that $v(a)\in T_a U$.
Lie (directional, flow) derivations $L_v:C^\infty(U)\to C^\infty(U)$. The Leibniz rule (algebra) and its meaning (“Any Leibniz linear map of $C^\infty(U)$ to itself is a Lie derivative along some vector field).
Commutator of two vector fields (to be discussed more in the future).
Push-forward of vector fields by smooth invertible maps.

Exterior derivation

The differential $\mathrm df$ of a smooth function $f$ is in a sense container which conceals all directional derivatives $L_Xf=\left\langle\mathrm df,X\right\rangle$ along all directions, and dependence on $X$ is linear.

If we consider the directional Lie derivative $L_X\omega$ for a form $\omega\in\Omega^k(M)$ of degree $k\ge 1$, then simple computations show that $L_{fX}\omega$ is no longer equal to $f\cdot L_X\omega$. However, one can “correct” the Lie derivative in such a way that the result will depend on $X$ linearly. For instance, if $\omega\in\Omega^1(M)$ and $X$ is a vector field, we can define the form $\eta_X\in\Omega^1(M)$ by the identity $\eta_X=L_X\omega-\mathrm d\left\langle\omega,X\right\rangle$ and show that the 2-form $\eta(X,Y)=\left\langle\eta_X,Y\right\rangle$ is indeed bilinear antisymmetric.

The 2-form $\eta$ is called the exterior derivative of $\omega$ and denoted $\mathrm d\omega\in\Omega^2(M)$. The correspondence $\mathrm d\colon\Omega^1(M)\to\Omega^2(M)$ is an $\mathbb R$-linear operator which satisfies the Leibniz rule $\mathrm d(f\omega)=f\,\mathrm d\omega+(\mathrm df)\land \omega$ and $\mathrm d^2 f=0$ for any function $f\in\Omega^0(M)$.

It turns out that this exterior derivation can be extended to all $k$-forms preserving the above properties and is a nice (algebraically) derivation of the graded exterior algebra $\Omega^\bullet(M)=\bigoplus_{k=0}^n\Omega^k(M)$.

The lecture notes are available here.

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