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On this page:

• Introduction
• Smooth Functions on Euclidean Spaces
• Tangent Vectors in $\RR^n$
  • Directional Derivative
  • Germs
  • Point Derivations
  • Vector Fields
• Exterior Algebra
• Worked Exercises

Introduction

This page is my notes is from my working through the first chapter of Loring Tu’s Introduction to Manifolds.

Smooth Functions on Euclidean Spaces

[definition] [deftitle]Definition %counter% (Smooth $U\subseteq \RR^n\to \RR$)[/deftitle]

Let $U\subseteq \RR^n$ be open. We say that a real-valued function $f:U \to \RR$ is $C^k$ at $p\in U$ provided that

of all order $0 \le j \le k$ exist and are continuous at $p$. We say that $f$ is $C^\infty$ or smooth at $p$ provided that it is $C^k$ for every $k \ge 0$. [/definition]

[definition] [deftitle]Definition %counter% (Smooth $U\subseteq \RR^n\to \RR^m$)[/deftitle]

Let $U\subseteq \RR^n$ be open. We say that $f:U \to \RR^m$ is $C^k$ at $p\in U$ provided that all of its component functions $f^1, \ldots, f^m$ are $C^k$ at $p$. Furthermore, it is $C^k$ on $U$ if it is $C^k$ at all points of $U$ and we define $C^\infty$ analogously. [/definition]

[example] [extitle]Example %counter% ($C^k$ maps)[/extitle]

Via the fundamental theorem of calculus, it is straightforward to construct $C^k$ functions that are not $C^{k+1}$. To do so, we just integrate a $C^{k-1}$ function $g$:

[/example]

[definition] [deftitle]Definition %counter% (Analytic)[/deftitle]

Let $f:U \to \RR$ is real-analytic at $p \in U$ if it can be represented by its Taylor series in some neighborhood of $p$. [/definition]

[example] [extitle]Example %counter% ($C^\infty$ does NOT mean analytic)[/extitle]

Let $f:\RR \to \RR$ be defined by

It is fairly clear that $f^{(k)}(x) = P_k(x^{-1})e^{-1/x^2}$ for $x > 0$. Furthermore,

This implies that the Taylor series of $f$ is just the zero map. [/example]

Tangent Vectors in $\RR^n$

Directional Derivative

[definition] [deftitle]Definition %counter% ($T_p\RR^n$)[/deftitle]

Let $p\in \RR$. We define the tangent space $T_p\RR^n$ of $\RR^n$ at $p$ as $T_\RR^n \coloneqq \RR^n$. The elements of $T_p\RR^n$ are called tangent vectors. [/definition]

The idea with the notation above is that we can visualize the tangent and translate it to the origin to get a linear space $\RR^n$.

[definition] [deftitle]Definition %counter% (Directional derivative)[/deftitle]

Let $p \in \RR^n$ and $v$ be a tangent vector at $p$. Furthermore, let $U$ be a neighborhood of $p$ and $f:U \to \RR$ be $C^k$ on $U$. The directional derivative of $f$ in the direction of $v$ at $p$ is defined as

[/definition]

By the chain rule, we know that

Later on, we will work with the operator

We will especially be interested in the mapping $v\mapsto D_v$.

Germs

Fix some $p\in \RR^n$. We define an relation on $C^\infty$ maps of the form $f:U \to \RR^n$ where $U$ is a neighborhood of $p$ as follows: Given another $C^\infty$ map $g:V \to \RR^n$ where $V$ is a neighborhood of $p$, we say that $f \sim g$ if and only if there is an open set $W \subseteq U \cap V$ containing $p$ such that $f|_W = g|_W$. This map is an equivalence relation:

• (Reflexivity) $f\sim f$ as $U \subseteq U \cap U$.
• (Symmetric) $f\sim g$ implies $g\sim f$ as $W \subseteq U \cap V = V \cap U$.
• (Transitivity) $f\sim g$ and $g \sim h$ implies $f\sim h$ as $f\sim g$ implies the existence of open $W_1 \subseteq U\cap V$ containing $p$ such that $f|_{W_1} = g|_{W_1}$ and $g\sim h$ implies the existence of open $W_2 \subseteq U \cap V$ containing $p$ such that $f|_{W_2} = g|_{W_2}$. Setting $W = W_1 \cap W_2$, it follows that $f|_W = h|_W$ and $W \subseteq U\cap V$.

[definition] [deftitle]Definition %counter% (Germs and $C_p^\infty$)[/deftitle]

The equivalence class of $f$ at $p$ is called the germ of $f$ at $p$. The set of all germs of $C^\infty$ functions on $\RR^n$ at $p$ is denoted $C_p^\infty$. [/definition]

[example] [extitle]Example %counter%[/extitle]

The maps $f:(\RR \setminus \{1\}) \to \RR$ and $g:(-1, 1) \to \RR$ given by

represent the same germ at any point $p$ in the open $(-1, 1)$. [/example]

[theorem] [thmtitle]Proposition %counter%[/thmtitle]

$C_p^\infty$ has a natural $\RR$-algebra structure. [/theorem]

[proof] (Addition) Let $[f:U \to \RR]_\sim, [g:V \to \RR]_\sim \in C_p^\infty$ and let $W \coloneqq U \cap V$ such that $f$ and $g$ agree on $W$. Then we define

We claim that this operation is well-defined. Suppose that $[f:U\to \RR]_\sim = [f_0:U_0\to\RR]_\sim$ and $[g:V \to \RR]_\sim = [g_0:V_0\to \RR]_\sim$. So there is $W_1 \subseteq U \cap U_0$ and $W_2 \subseteq V \cap V_0$ both containing $p$ such that

Now let $W_0 = W_1 \cap W_2$. Then

so it follows that

Thus,

(Multiplication) Let $[f:U \to \RR]_\sim, [g:V \to \RR]_\sim \in C_p^\infty$ and let $W\coloneqq U \cap V$. Then we define

The well-definedness is the same argument as above.

(Scalar multiplication) Let $[f:U \to \RR]_\sim$ and let $r \in \RR$. Then we define

These operations all define an $\RR$-algebra (not necessarily unital). [/proof]

Notice that the algebra $C_p^\infty$ is given by taking a direct limit over the neighborhoods of $p$.

Point Derivations

Let $v$ be a tangent vector at $p \in \RR^n$. Then the directional derivative at $p$ gives a mapping

It is easy to see that $D_v$ is $\RR$-linear as

Furthermore, we also see that $D_v$ satisfies the Leibniz rule

[definition] [deftitle]Definition %counter%[/deftitle]

A linear map $C_p^\infty \to \RR$ satisfying the Leibniz rule is a derivation at $p$. The set of all derivations at $p$ is written $\mathcal{D}_p(\RR^n)$. [/definition]

When we work on a manifold, there is typically no ambient Euclidean space to rely on. As a result, we cannot think of tangent vectors in the same way that we do in Euclidean space via affine transformations. The workaround is to notice that tangent vectors and derivations are in bijection.

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

The map

is an $\RR$-linear isomorphism. [/theorem]

[proof] (Linearity) Let $r\in \RR$ and $u,v\in T_p\RR^n$. Then

(Injectivity) Suppose that $D_v = 0$. Then applying $D_v$ to the coordinate function $x^j$ gives

So $v = 0$.

(Surjectivity) Let $D$ be a derivation at $p$. Let $f \in C_p^\infty$ be chosen so that $f$ has convex domain (can just pick the domain to be a very small ball). By Taylor’s theorem, it follows that there are funcitons $g_i$ in a neighborhood of $p$ such that

Thus,

Derivations of constants are 0 as

Thus,

So then set $v = (Dx^1, \ldots, Dx^n)$ and it follows that $D_v = D$. [/proof]

From here on out, we identity $T_p\RR^n$ with the space of derivations at $p$. In particular, we will write

Vector Fields

In multivariable calculus, we think of a vector field $X$ as a mapping from an open subset $U\subseteq \RR^n$ to $\RR^n$ that assigns each point $p$ a tangent vector. This is also how we think of it in the smooth manifold setting.

Since $T_p\RR^n$ has basis $\eval{\pdv{x^i}}_p$, it follows that

where $a_p^i \in \RR$. As a result, we very frequently will write

[definition] [deftitle]Definition %counter%[/deftitle]

The set of all $C^\infty$ vector fields on $U \subseteq \RR^n$ is denoted $\mathfrak{X}(U)$. [/definition]

Obviously $\mathfrak{X}(U)$ is an $\RR$-vector space. What is more interesting is that it is a $C^\infty(U)$-module (where $C^\infty (U)$ is the ring of $C^\infty$ functions on $U$).

[theorem] [thmtitle]Proposition %counter%[/thmtitle]

$\mathfrak{X}(U)$ is a $C^\infty(U)$-module where

[/theorem]

As result, we very frequently will write

Since vector fields map into the tangent space, it is possible to to think of them as a derivation (interpreted in an appropriate way). In particular, we let $X \in \mathfrak{X}(U)$ and $f\in C^\infty(U)$. Then we define $Xf:U \to \RR$ by

for every $p\in U$. Rewriting in terms of the standard basis of the tangent space, it follows that

In other words,

Since $X$ is $C^\infty$ on $U$, it follows that the $a^i$ must vary smoothly in $U$ so $Xf \in C^\infty(U)$.

[theorem] [thmtitle]Proposition %counter%[/thmtitle]

Let $U \subseteq \RR^n$ be open, $X \in \mathfrak{X}(U)$, and $f,g\in C^\infty(U)$. Then $X(fg)$ satisfies the Leibniz rule. [/theorem]

Let $p \in U$. Then

Since this holds for all $p$, it follows that

[definition] [deftitle]Definition %counter%[/deftitle]

Let $A$ be an algebra over the field $K$. We say that the $K$-linear map $D:A \to A$ is a derivation of $A$ given that

for every $a, b\in A$. [/definition]

It’s pretty immediate that $C^\infty(U)$ is an algebra over $\RR$. Additionally, note that there is a linear map

That is, the vector fields $\mathfrak{X}(U)$ can be associated with derivations of the algebra $C^\infty(U)$.

Exterior Algebra

[definition] [deftitle]Definition %counter%[/deftitle]

Let $V$ be a finite-dimensional $\RR$-vector space. We define the dual space $V^\vee$ of $V$ to be the vector space of all $\RR$-linear maps $V \to \RR$. The elements of $V^\vee$ are called covectors or $1$-covectors. [/definition]

[definition] [deftitle]Definition %counter%[/deftitle]

Let $e_1, \ldots, e_n$ be a basis for $V$. We define $\alpha^i:V \to \RR$ to be a linear map with the property that

where $\delta_{ij}$ is the Kronecker delta. We call $\alpha_1, \ldots, \alpha_n$ the dual basis of $e_1, \ldots, e_n$. [/definition]

[theorem] [thmtitle]Theorem %counter%[/thmtitle]

The dual basis is actually a basis. [/theorem]

[proof] (Spanning) Let $f \in V^\vee$ and let $v = \sum_i v^i e_i$. Then

Thus, $f = \sum_{i} \alpha^i f(e_i)$. This implies that the dual basis is spanning.

(Linear independence) Say \sum_{i} c_i \alpha^i = 0$ where each $c_i\in \RR$. Then

Thus, each $c_j = 0$ and so we obtain linear independence. [/proof]

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

If $V$ is finite-dimensional, then $\dim V^\vee = \dim V$. [/theorem]

[definition] [deftitle]Definition %counter%[/deftitle]

The map $f:V^k \to \RR$ is $k$-linear if it is linear in each of its $k$ arguments. A $k$-linear function whose codomain is $\RR$ is called a $k$-tensor on $V$ and $k$ is called the degree. We denote the space of all $k$-tensors by $L_k(V)$. [/definition]

Worked Exercises

[example] [extitle]Tu Problem 1.1[/extitle]

Let $g:\RR \to \RR$ be defined by $g(x) = \frac{3}{4}x^{4/3}$. Show that $h:\RR \to \RR$ given by $h(x) = \int_0^x g(t) \dd{t}$ is $C^2$ but not $C^3$ at $x = 0$. [/example]

[proof] By the fundamental theorem of calculus, $h’(x) = g(x)$. Thus, $h’$ is $C^1$ and so $h$ is $C^2$. [/proof]

[example] [extitle]Tu Problem 1.3[/extitle]

1. Show that the function $f:(-\pi/2, \pi/2) \to \RR$ given by $f(x) = \tan x$ is a diffeomorphism.
2. Let $a,b \in \RR$ with $a < b$. Find a linear function $h:(a, b) \to (-1, 1)$. In particular, this shows that all (bounded) open intervals are diffeomorphic.
3. The map $\exp:\RR \to (0, \infty)$ is a diffeomorphism. Use it to show that for any real numbers $a$ and $b$, the intervals $\RR$, $(a, \infty)$, and $(-\infty, b)$ are diffeomorphic. [/example]

[proof] (1) The function $\tan x$ is analytic, so $f$ is certainly $C^\infty$. Bijectivity follows from the fact that $\tan x$ is invertible on the given interval. Furthermore, $f^{-1}(x) = \arctan x$ and this is also analytic.

(2) The map $h:(a, b) \to (-1, 1)$ given by

So we have $h(a) = -1$ and $h(b) = 1$.

(3) Note that $a + e^x$ and $b - e^x$ give the desired diffeomorphisms. [/proof]

[example] [extitle]Tu Problem 1.4[/extitle]

Show that the map

is a diffeomorphism. [/example]

[proof] Each component function is analytic and have an inverse which consists of taking an inverse tangent in each component (also analytic). [/proof]

[example] [extitle]Tu Problem 2.1[/extitle]

Let $X:\RR^3 \to T_p\RR^3$ be the vector field

and $f:\RR^3 \to \RR$ be the function $x^2 + y^2 + z^2$. Compute $Xf$. [/example]

[solution] Recall that $Xf$ is given by

So then

\((Xf)(x, y) = 2x^3 + 2y^3.\) [/solution]