2. Tangent Vectors in Rn as Derivations

The Directional Derivative

One may visualize the tangent space $T_p({\mathbb{R}}^n)$ at $p\in {\mathbb{R}}^n$ as the vector space of arrows emanating from $p$ tangentially. We write vectors $v\in T_p({\mathbb{R}}^n)$ by $⟨v^1,\dots ,v^n⟩$. We denote the standard basis for ${\mathbb{R}}^n$ by $e_1,\dots ,e_n$. Hence, $v=\sum v^i{e}_i$. We call vectors $v\in T_p({\mathbb{R}}^n)$ tangent vectors.

If $f$ is $C^{\infty }$ in a neighborhood of $p\in {\mathbb{R}}^n$ and $v\in T_p({\mathbb{R}}^n)$, we define the directional derivative of $f$ in the direction of $v$ at $p$ by

$$D_{v}f=\underset{t\to 0}{\lim }\frac{f(p+tv)-f(p)}{t}=\frac{d}{dt}{|}_{t=0}f(p+tv).$$

Notice that

$$D_{v}f=v\cdot \text{grad}f(p)=\sum _{i=1}^n{v}^i\frac{\partial f}{\partial x^i}(p).$$

Note that $D_{v}f$ is a number, not a function. We write

$$D_v=\sum v^i\frac{\partial f}{\partial x^i}{|}_p$$

for the map that sends a function $f$ to $D_{v}f$.

Germs of Functions

Equivalence Relation

A relation on a set $S$ is a subset $R\subset S\times S$. We say $x\sim y$ for $x,y\in S$ if $(x,y)\in R$. The relation $R$ is an equivalence relation if for all $x,y,z\in S$

1. $x\sim x$
2. if $x\sim y$, then $y\sim x$
3. If $x\sim y$ and $y\sim z$, then $x\sim z$.

Let $(f,U)$ where $U$ is a neighborhood of $p$ and $f:U\to \mathbb{R}$ is $C^{\infty }$. We say that $(f,U)$ is equivalent to $(g,V)$ if there is an open set $W\subset U\cap V$ containing $p$ such that $f=g$ when restricted to $W$. The equivalence class of $(f,U)$ is called the germ of $f$ at $p$. We write $C_p^{\infty }({\mathbb{R}}^n)$ for the set of all germs of $C^{\infty }$ functions on ${\mathbb{R}}^n$ at $p$.

Example: The functions

$$f(x)=\frac{1}{1-x}$$

with domain $\mathbb{R}\setminus \{1\}$ and

$$g(x)=1+x+x^2+x^3+\dots$$

with domain $(-1,1)$ have the same germ at any point $p\in (-1,1)$ because $g$ is the geometric series

$$\sum _{n=1}^{\infty }{x}^{n-1}$$

which converges to $f(x)$ for all $|x|<1$.

Algebra

An algebra over a field $k$ is a vector space $A$ over $k$ with a multiplication map $\mu :A\times A\to A$ such that for all $a,b,c\in A$ and $r\in k$

1. $(a\cdot b)\cdot c=a\cdot (b\cdot c)$
2. $(a+b)\cdot c=a\cdot c+b\cdot c$ and $a\cdot (b+c)=a\cdot b+a\cdot c$
3. $r(a\cdot b)=(ra)\cdot b=a\cdot (rb)$.

Note that an algebra has three operations, addition, multiplication from the ring, and multiplication from the vector space.

Linear map

A map $L:V\to W$ between vector spaces over a field $k$ is called a linear map if for any $r\in k$ and $u,v\in V$,

1. $L(u+v)=L(u)+L(v)$
2. $L(rv)=rL(v)$

If $A$ and $A^{\prime }$ are algebras over a field $k$, then an algebra homomorphism is a linear map $L:A\to A^{\prime }$ such that $L(ab)=L(a)L(b)$ for all $a,b\in A$.

Prp

$C_p^{\infty }$ is an algebra over $\mathbb{R}$.

Proof: Let $(f,U),(g,V),(h,W)\in C_p^{\infty }$ and $s\in \mathbb{R}$. Define $(f,U)\cdot (g,V)$ to be the germ $(fg,U\cap V)$ and $(f,U)+(g,V)$ to be the germ $(f+g,U\cap V)$. Notice that

$$((f,U)\cdot (g,V))\cdot (h,W)=(fg,U\cap V)\cdot (h,W)=(fgh,U\cap V\cap W)$$

and

$$(f,U)\cdot ((g,V)\cdot (h,W))=(f,U)\cdot (gh,V\cap W)=(fgh,U\cap V\cap W).$$

Next, notice that

$$((f,U)+(g,V))\cdot (h,W)=(f+g,U\cap V)\cdot (h,W)=(fh+gh,U\cap V\cap W)=(f,U)\cdot (h,W)+(g,V)\cdot (h,W)$$

and

$$(f,U)\cdot ((g,V)+(h,W))=(f,U)\cdot (g+h,V\cap W)=(fg+fh,U\cap V\cap W)=(f,U)\cdot (g,V)+(f,U)\cdot (h,W).$$

Finally, we define $s(f,U)$ to be the germ $(sf,U)$. Then one may see that

$$s((f,U)\cdot (g,V))=(s(f,U))\cdot (g,V)=(f,U)\cdot (s(g,V)).$$

Therefore, $C_p^{\infty }$ is an algebra over $\mathbb{R}$.

Q.E.D.

Derivations at a Point

Notice that for each tangent vector $v\in T_p({\mathbb{R}}^n)$, the directional derivative at $p$ gives a map of real vector spaces

$$D_v:C_p^{\infty }\to \mathbb{R}$$

which maps germs $(f,U)$ to their directional derivative $D_{v}f$ at $p\in U$. One may show that $D_v$ is $\mathbb{R}$-linear and satisfies the Leibniz rule

$$D_v(fg)=(D_{v}f)g(p)+f(p)D_{v}g.$$

Indeed, notice that

$$\begin{array}{r}{D}_v(f+g)=\sum _{i=1}^n{v}^i\frac{\partial (f+g)}{\partial x^i}(p)=\sum _{i=1}^n{v}^i\frac{\partial f}{\partial x^i}(p)+\sum _{i=1}^n{v}^i\frac{\partial g}{\partial x^i}(p)=D_{v}f+D_{v}g\end{array}$$

and

$$D_v(rf)=\sum _{i=1}^n{v}^i\frac{\partial rf}{\partial x_i}(p)=r\sum _{i=1}^n{v}^i\frac{\partial f}{\partial x^i}(p)=r{D}_{v}f.$$

As for Leibniz’s rule,

$$D_v(fg)=\sum _{i=1}^n{v}^i\frac{\partial (fg)}{\partial x^i}(p)=\sum _{i=1}^n{v}^i\frac{\partial f}{\partial x^i}(p)g(p)+\sum _{i=1}^n{v}^i\frac{\partial g}{\partial x^i}(p)f(p)=(D_{v}f)g(p)+f(p)D_{v}g.$$

In general, any linear map $D:C_p^{\infty }\to \mathbb{R}$ satisfying Leibniz rule is called a derivation at $p$. We denote the set of all derivations at $p$ by ${\mathcal{D}}_p({\mathbb{R}}^n)$. It follows that all directional derivatives at a point $p$ are derivations at $p$ by the map

$$v\in T_p({\mathbb{R}}^n)\mapsto D_v=\sum v^i\frac{\partial }{\partial x^i}{|}_p.\text{\hspace{1em}(1)}$$

Lemma

If $D$ is a point-derivation of $C_p^{\infty }$, then $D(c)=0$ for any constant function $c$.

Proof: Notice that by $\mathbb{R}$-linearity, $D(c)=cD(1)$. So it suffices to show that $D(1)=0$. By making use of Leibniz’s rule,

$$D(1)=D(1\cdot 1)=D(1)1+1D(1)=2D(1)$$

which holds if and only if $D(1)=0$.

Q.E.D.

Thm

The linear map $\varphi :T_p({\mathbb{R}}^n)\to {\mathcal{D}}_p({\mathbb{R}}^n)$ defined in (1) is an isomorphism of vector spaces.

Proof: We need to show that $\varphi$ is a bijection and $\varphi$ is linear. Suppose $\varphi (v)=D_v=0$. Then it follows that

$$0=\sum v^i\frac{\partial x^j}{\partial x^i}(p)=\sum v^i{\delta }_i^j=v^j$$

where ${\delta }_i^j$ is the Kronecker delta. This holds for all $j$, hence $v=0$.

Now let $D\in D_p({\mathbb{R}}^n)$ be a point-derivation at $p$. Consider a germ $(f,V)$ in $C_p^{\infty }$ such that $V$ is star-shaped. By Taylor’s thoerem,

$$f(x)=f(p)+\sum _{i=1}^n(x^i-p^i)g_i(x),\text{ where }{g}_i(p)=\frac{\partial f}{\partial x^i}(p).$$

Notice that by applying $D$ to both sides

$$\begin{array}{rl}Df(x)& =D(f(p))+\sum D((x^i-p^i)g_i(x))\\ & =\sum D(x^i)g_i(p)+\sum (p^i-p^i)D{g}_i(x)\\ & =\sum D{x}^i\frac{\partial f}{\partial x^i}(p).\end{array}$$

Thus, $D=D_v$ where $v=⟨D{x}^1,\dots ,D{x}^n⟩$ for all $f\in C_p^{\infty }$. Hence $\varphi$ is surjective.

Linearity is trivial.

Q.E.D.

As a result, the standard basis $e_1,\dots ,e_n$ for $T_p({\mathbb{R}}^n)$ corresponds to the partial derivatives $\partial /\partial x^1{|}_p,\dots ,\partial /\partial x^n{|}_p$.

Vector Fields

Def

A vector field $X$ on a subset $U$ of ${\mathbb{R}}^n$ is a function that assigns to each point $p\in U$ a tangent vector $X_p\in T_p({\mathbb{R}}^n)$.

The vector $X_p\in T_p({\mathbb{R}}^n)$ can be expressed as a linear combination

$$X_p=\sum a^i(p)\frac{\partial }{\partial x^i}{|}_p.$$

Thus, one may view the vector space $X$ as the linear combination

$$X=\sum a^i\frac{\partial }{\partial x^i}$$

where $a^i:U\to \mathbb{R}$ are some functions. We say $X$ is $C^{\infty }$ if each coefficient function $a^i$ is $C^{\infty }$. One may identify vector fields on $U$ with column vectors:

$$X=\sum a^i\frac{\partial }{\partial x^i}⟷\left[\begin{array}{c}{a}^1\\ ⋮\\ a^n\end{array}\right].$$

Example: On ${\mathbb{R}}^2\setminus \{0\}$ let $p=(x,y)$. Then

$$X=\frac{-y}{\sqrt{x^2+y^2}}\frac{\partial }{\partial x}+\frac{x}{\sqrt{x^2+y^2}}\frac{\partial }{\partial y}$$

is the vector field below:

One may define multiplication of vector fields by functions on $U$ pointwise:

$$(fX{)}_p=f(p)X_p,p\in U.$$

Notice that if $X$ is a $C^{\infty }$ vector field, and $f$ is a $C^{\infty }$ function on $U$, then $fX$ is a $C^{\infty }$ vector field on $U$. Hence, the set of all $C^{\infty }$ vector fields on $U$, denoted $\mathfrak{X}(U)$, is a module over the ring $C^{\infty }(U$).

Def

If $R$ is a commutative ring with identity, then a (left) $R$-module is an abelian group $A$ with a scalar multiplication map $\mu :R\times A\to A$ such that for all $r,s\in R$ and $a,b\in A$

1. $(rs)a=r(sa)$
2. $1a=a$
3. $(r+s)a=ra+sa$, $r(a+b)=ra+rb$.

Def

Let $A$ and $A^{\prime }$ be $R$-modules. An $R$-module homomorphism from $A$ to $A^{\prime }$ is a map $f:A\to A^{\prime }$ such that

1. $f(a+b)=f(a)+f(b)$
2. $f(ra)=rf(a)$.

Vector Fields as Derivations

Let $X$ be a $C^{\infty }$ vector field on $U\subseteq {\mathbb{R}}^n$ and $f:U\to \mathbb{R}$ a $C^{\infty }$ function. Then we define a new function $Xf:U\to \mathbb{R}$ by

$$(Xf)(p)=X_{p}f=\sum a^i(p)\frac{\partial f}{\partial x^i}(p).$$

Thus, a $C^{\infty }$ vector field gives rise to an $\mathbb{R}$-linear map

$$\begin{array}{r}{C}^{\infty }(U)\to C^{\infty }(U)\\ f\mapsto Xf=\sum a^i\frac{\partial f}{\partial x^i}\end{array}$$

We claim this map is a derivation.

Prp

If $X$ is a $C^{\infty }$ vector field and $f$ and $g$ are $C^{\infty }$ functions on $U\subseteq {\mathbb{R}}^n$, then $X(fg)$ satisfies Leibniz rule

$$X(fg)=(Xf)g+fXg.$$

Proof: Notice that for $p\in U$

$$\begin{array}{rl}{X}_p(fg)& =\sum a^i(p)\frac{\partial (fg)}{\partial x^i}(p)\\ & =\sum a^i(p)\frac{\partial f}{\partial x^i}(p)g(p)+\sum a^i(p)f(p)\frac{\partial g}{\partial x^i}(p)\\ & =(X_{p}f)g(p)+f(p)X_{p}f.\end{array}$$

This holds for every point $p\in U$.

Q.E.D.

Problems

2.1. Vector fields

Let $X$ be the vector field $x\partial /\partial x+y\partial /\partial y$ and $f(x,y,z)=x^2+y^2+z^2$ on ${\mathbb{R}}^3$. Compute $Xf$.

Answer:

$$\begin{array}{rl}Xf& =x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}+0\frac{\partial f}{\partial z}=2x^2+2y^2\end{array}$$

2.3. Vector space structure on derivations at a point

Let $D$ and $D^{\prime }$ be derivations at $p$ in ${\mathbb{R}}^n$, and $c\in \mathbb{R}$. Prove that the sum $D+D^{\prime }$ and the scalar multiple $cD$ are derivations at $p$.

Answer: Clearly they are linear, so we simply need to show they satisfy Leibniz’s rule. Notice that

$$\begin{array}{rl}(D+D^{\prime })(fg)& =D(fg)+D^{\prime }(fg)\\ & =(Df)g+fDg+(D^{\prime }f)g+f{D}^{\prime }g\\ & =((D+D^{\prime })f)g+f(D+D^{\prime })g.\end{array}$$

Furthermore,

$$\begin{array}{rl}(cD)(fg)& =cD(fg)=c((Df)g+fDg)=((cD)f)g+f(cD)g.\end{array}$$