[Tu Differential Geometry] 4. Directional Dericatives in Euclidean Space

This article is one of Manifold, Differential Geometry, Fibre Bundle.

The directional derivative is one way of differentiating vector fields on $\mathbb{R}^n$ with respect to a tangent vector.

4.1 Directional Derivatives in Euclidean Space

Suppose $X_p=\sum a^i \partial /\partial x^i|_p$ is a tangent vector at a point $p=(p^1,\cdots,p^n)$ in $\mathbb{R}^n$ and $f(x^1,\cdots, x^n)$ is a $C^\infty$ function in a neighborhood of $p$ in $\mathbb{R}^n$. To compute the directional derivative of $f$ at $p$ in the direction $X_p$, we first write down a set  of parametric equations for the line through $p$ in the direction $X_p$: $$x^i=p^i+ta^i,\quad i=1,\cdots,n.$$ Let $a=(a^1,\cdots,a^n)$. Then the directional derivative $D_{X_p}f$ is $$\begin{align}D_{X_p}f=\lim_{t\rightarrow 0}\frac{f(p+ta)-f(p)}{t}=\left. \frac{d}{dt}\right|_{t=0}f(p+ta)\nonumber\\ = \sum \left. \frac{\partial f}{\partial x^i}\right|_p \cdot \left. \frac{dx^i}{dt}\right|_0=\sum \left. \frac{\partial f}{\partial x^i}\right|_p \cdot a^i=\left( \sum a_i\left. \frac{\partial}{\partial x^i}\right|_p\right) f=X_pf.\end{align}$$ In calculus, $X_p$ is required to be a unit vector, but we will allow $X_p$ to be an arbitrary vector at $p$.

As a shorthand, we write $\partial_i$ for $\partial/\partial x^i$. The directional derivative at $p$ of a $C^\infty$ vector field $Y=\sum b^i\partial_i=\sum b^i\partial/\partial x^i$ on $\mathbb{R}^n$ in the direction $X_p$ is defined to be $$\begin{align}D_{X_p}Y=\sum (X_p b^i)\left. \frac{\partial}{\partial x^i}\right|_p.\end{align}$$ This formula shows clearly that $D_{X_p}Y$ is $\mathbb{R}$-linear in $X_p$.

Although (1) computes the directional derivative using the values of $f$ along a line thorugh $p$, we can in fact use any curve $c(t)$ with initial point $p$ and initial vector $X_p$ (Figure 4.1), for $$D_{X_p}f=X_pf=c'(0)f=c_*\left( \left. \frac{d}{dt}\right|_{t=0}\right) f=\left. \frac{d}{dt} \right|_{t=0} f(c(t)).$$

When $X$ is a $C^\infty$ vector field on $\mathbb{R}^n$, not just a vector at $p$, we define the vector field $D_XY$ on $\mathbb{R}^n$ by $$(D_XY)_p=D_{X_p}Y\quad\mbox{for all }p\in \mathbb{R}^n.$$ Equation (2) shows that if $X$ and $Y$ are $C^\infty$ vector fields on $\mathbb{R}^n$, then so is $D_XY$. Let $\mathfrak{X}(\mathbb{R}^n)$ be the vector space of all $C^\infty$ vector fields on $\mathbb{R}^n$. The directional derivative in $\mathbb{R}^n$ gives a map $$D:\mathfrak{X}(\mathbb{R}^n)\times \mathfrak{X}(\mathbb{R}^n)\rightarrow \mathfrak{X}(\mathbb{R}^n),$$ which we write as $D_XY$ instead of $D(X,Y)$. Let $\mathcal{F}=C^\infty(\mathbb{R}^n)$ be the ring of $C^\infty$ functions on $\mathbb{R}^n$. Then $\mathfrak{X}(\mathbb{R}^n)$ is both a vector space over $\mathbb{R}$ and a module over $\mathcal{F}$.

Proposition 4.2 ({Condition of) Directional Derivative). For $X,Y\in \mathfrak{X}(\mathbb{R}^n)$, the directional derivcative $D_XY$ satisfies the following properties: 

(i) $D_XY$ is $\mathcal{F}$-linear in $X$ and $\mathbb{R}$-linear in $Y$;

(ii) (Leibniz rule) if $f$ is a $C^\infty$ function on $\mathbb{R}^n$, then $$D_X(fY)=(Xf)Y+fD_XY.$$

Proof. (i) Let $f$ be a $C^\infty$ function on $\mathbb{R}^n$ and $p$ an arbitrary point of $\mathbb{R}^n$. Then $$(D_{fX}Y)_p=D_{f(p)X_p}Y=f(p)D_{X_p}Y=(fD_XY))p.$$ For $Z\in \mathfrak{X}(\mathbb{R}^n)$, we have $$(D_{X+Z}Y)_p=D_{X_p+Z_p}Y=D_{X_p}Y+D_{Z_p}Y=(D_XY+D_ZY)_p.$$ This proves that $D_XY$ is $\mathcal{F}$-linear in $X$. The $\mathbb{R}$-linearity in $Y$ is clear from (2).

(ii) Suppose $Y=\sum b^i\partial_i$, where $b^i\in C^\infty(\mathbb{R}^n)$. Then $$(D_X(fY))_p=\sum X_p(fb^i)\partial_i|_p=\sum (X_pf)b^i(p)\partial_i|_p+\sum f(p)X_p b^i \partial_i|_p=(X_pf)Y_p+f(p)D_{X_p}Y=((Xf)Y+fD_XY)_p.$$

4.2 Other Properties of the Directional Derivative

Since the directional derivative $D$ in $\mathbb{R}^n$ is an $\mathbb{R}$-bilinear map $$D:\mathfrak{X}(\mathbb{R}^n)\times \mathfrak{X}(\mathbb{R}^n)\rightarrow \mathfrak{X}(\mathbb{R}^n),$$ the if $[X,Y]$ is the Lie bracket, then $$D_XY-D_YX=[X,Y].$$ The quantity $$T(X,Y)=D_XY-D_YX-[X,Y]$$ turns out to be fundamental in differential geometry and is called the torsion of the  directional derivative $D$.

For each smooth vector field $X\in \mathfrak{X}(\mathbb{R}^n)$, the directional derivative $D_X:\mathfrak{X}(\mathbb{R}^n)\rightarrow \mathfrak{X}(\mathbb{R}^n)$ is an $\mathbb{R}$-linear endomorphism. This gives rise to a map $$\begin{align}\mathfrak{X}(\mathbb{R}^n)\rightarrow \mbox{End}_\mathbb{R}(\mathfrak{X}(\mathbb{R}^n)),\nonumber\\ X\mapsto D_X.\end{align}$$ The vector space $\mathfrak{X}(\mathbb{R}^n)$ of $C^\infty$ vector fields on $\mathbb{R}^n$ is a Lie algebra under the Lie bracket of vector fields. For any vector space $V$, the endomorphism ring $\mbox{End}_\mathbb{R}(V)$ of endomorphisms of $V$ is also a Lie algebra, with Lie bracket $$[A,B]=A\circ B-B\circ A,\quad A,B\in \mbox{End}(V).$$ So the map in (3) is an $\mathbb{R}$-linear map of Lie algebras. It is natural to ask if it is a Lie algebra homomorphism, i.e., is $$[D_X,D_Y]=D_{[X,Y]}?$$ The answer is yes for the directional derivative in $\mathbb{R}^n$. A measure of the deviation of the linear map $X\mapsto D_X$ from being a Lie algebra homomorphism is given by the function $$R(X,Y)=[D_X,D_Y]-D_{[X,Y]}=D_XD_Y-D_YD_X-D_{[X,Y]}\in \mbox{End}_\mathbb{R}(\mathfrak{X}(M)),$$ called the curvature of $D$. !

Finally, one might ask if the product rule holds for the Euclidean inner product: $$D_Z\left\langle X,Y\right\rangle = \left\langle D_ZX,Y\right\rangle+\left\langle X,D_ZY\right\rangle.$$ The answer is yes. The following proposition summarizes the properties of the directional derivative in $\mathbb{R}^n$.

Proposition 4.3 (Property of Directional Derivative). Let $D$ be the directional derivative in $\mathbb{R}^n$ and $X,Y,Z\ C$ vector fields on $\mathbb{R}^n$. Then

(i) (zero torsion) $D_XY-D_YX-[X,Y]=0$,

(ii) (zero curvature) $D_XD_YZ-D_YD_XZ-D_{[X,Y]}Z=0,$

(iii) (compatibility with the metric) $X\left\langle Y,Z\right\rangle=\left\langle D_XY,Z\right\rangle+\left\langle Y,D_XZ\right\rangle.$

Proof. 

(i)

(ii) Let $Z=\sum c^i\partial_i\in \mathfrak{X}(\mathbb{R}^n)$. Then $$D_XD_YZ=D_X\left( \sum(Yc^i)\partial_i\right) = \sum(XYc^i)\partial_i.$$ By symmetry, $$D_YD_XZ=\sum (YXc^i)\partial_i.$$ So $$D_XD_YZ-D_YD_XZ=\sum (XY-YX)c^i\partial_i=D_{[X,Y]}Z.$$

(iii) Let $Y=\sum b^i\partial_i$ and $Z=\sum c^j\partial_j\in \mathfrak{X}(\mathbb{R}^n)$. Then $$X\left\langle Y,Z\right\rangle = X(\sum b^ic^i)=\sum (Xb^i)c^i+\sum b^u(Xc^i)=\left\langle D_XY,Z\right\rangle +\left\langle Y,D_XZ\right\rangle.$$

If $X$ and $Y$ are smooth vector fields on a manifold $M$, then the Lie derivative $\mathcal{L}_XY$ is another way of differentiating $Y$ with respect to $X$. While the directional derivative $D_XY$ in $\mathbb{R}^n$ is $\mathcal{F}$-linear in $X$, the Lie derivative $\mathcal{L}_XY$ is not, so the two concepts are not the same. Indeed, since $\mathcal{L}_XY=[X,Y]$, by Proposition 4.3(i), for vector fields on $\mathbb{R}^n$, $$\mathcal{L}_XY=D_XY-D_YX.$$

4.3 Vector Fields Along a Curve

Suppose $c:[a,b]\rightarrow M$ is a parametrized curve in a manifold $M$.

Definition 4.4. A vector field $V$ along $c:[a,b]\rightarrow M$ is the assignment of a tangent vector $V(t)\in T_{c(t)}M$ to each $t\in [a,b]$. Such a vector field is said to be $C^\infty$ if for every $C^\infty$ function $f$ on $M$, the function $V(t)f$ is $C^\infty$ as a function of $t$.

Now suppose $c:[a,b]\rightarrow \mathbb{R}^n$ is a curve in $\mathbb{R}^n$ and $V$ is a $C^\infty$ vector field along $c$. Then $V(t)$ can be written as a linear combination of the standard basis vectors: $$V(t)=\sum v^i(t)\partial_i|_{c(t)}.$$ So it makes sense to differentiate $V$ with respect to $t$: $$\frac{dV}{dt}(t)=\sum \frac{dv^i}{dt}(t)\partial_i|_{c(t)},$$ which is also a $C^\infty$ vector field along $c$.

For a smooth vector field $V$ along a curve $c$ in an arbitrary manifold $M$, the derivative $dV/dt$ is in general not defined. This is because an arbitrary manifold does not have a canonical frame of vector fields such as $\partial/\partial x^1,\cdots,\partial/\partial x^n$ on $\mathbb{R}^n$.

Proposition 4.7 (Compatibility of Directional Derivative). Let $c:[a,b]\rightarrow \mathbb{R}^n$ be a curve in $\mathbb{R}^n$ and let $V(t)$, $W(t)$ be smooth vector fields along $c$. Then $$\frac{d}{dt}\left\langle V(t),W(t)\right\rangle=\left\langle \frac{dV}{dt},W\right\rangle+\left\langle V,\frac{dW}{dt}\right\rangle.$$

Proof. Write $V(t)$ and $W(t)$ in terms of $\partial_i|_{c(t)}$ and differentiate.

4.4 Vector Fields Along a Submanifold

Let $M$ be a regular submanifold of a manifold of a manifold $\tilde{M}$. At a point $p$ in $M$, there are two kinds of tangent vectors to $\tilde{M}$, depending on whether they are tangent to $M$ or not. For example, if $M$ is a surface in $\mathbb{R}^3$, the vectors in a tangent vector field on $M$ are all tangent to $M$, but the vectors in a normal vector field on $M$ are not tangent to $M$ but to $\mathbb{R}^3$.

Definition 4.8. Let $M$ be a submanifold of a manifold $\tilde{M}$. A vector field $X$ on $M$ assigns to each $p\in M$ a tangent vector $X_p\in T_pM$. A vector field $X$ along $M$ in $\tilde{M}$ assigns to each $p\in M$ a tangent vector $X_p\in T_p\tilde{M}$. A vector field $X$ along $M$ is called $C^\infty$ if for every $C^\infty$ function $f$ on $\tilde{M}$, the function $Xf$ is $C^\infty$ on $M$.

The distinction between these two concepts if indicated by the prepositions 'on' and 'along'. While it may be dangerous for little prepositions to assume such grave duties, this appears to be common usage in the literature. In this terminology, a normal vector filed to a surface $M$ in $\mathbb{R}^3$ is a vector field along $M$ in $\mathbb{R}^3$, but not a vector field on $M$. Of course, a vector field on $M$ is a vector field along $M$.

The set of all $C^\infty$ vector fields on a manifold $M$ is denoted by $\mathfrak{X}(M)$. The set of all $C^\infty$ vector fields along a submanifold $M$ in a manifold $\tilde{M}$ will be denoted by $\Gamma(T\tilde{M}|_M)$. They are both modules over the ring $\mathcal{F}=C^\infty(M)$ of $C^\infty$ functions on $M$.

4.5 Directional Dericatives on a Submanifolds of $\mathbb{R}^n$

Suppose $M$ is a regular submanifold of $\mathbb{R}^n$. At any point $p\in M$, if $X_p\in T_pM$ is a tangent vector at $p$ and $Y=\sum b^i\partial_i$ is a vector field along $M$ in $\mathbb{R}^n$, then the directional derivative $D_{X_p}Y$ is defined. In fact, $D_{X_p}Y=\sum(X_pb^i)\partial_i|_p.$$

As $p$ varies over $M$, this allows us to associate to a $C^\infty$ vector field $X$ on $M$ and a $C^\infty$ vector field $Y$ along $M$ a $C^\infty$ vector field $D_XY$ along $M$ by $$D_XY=\sum(Xb^i)\partial_i\in \Gamma(T\mathbb{R}^n|_M).$$ For any $p\in M$, we have $(D_XY)_p=D_{X_p}Y$ by definition. It follows that there is an $\mathbb{R}$-bilinear map $$D:\mathfrak{X}(M)\times \Gamma(T\mathbb{R}^n|_M)\rightarrow \Gamma(T\mathbb{R}^n|_M)\\ D(X,Y)=D_XY.$$ We call $D$ the directional derivative on $M$. Because of the asymmetry between $X$ and $Y$-one is a vector field on $M$, the other a vector field along $M$, the torsion $T(X,Y):=D_XY-D_YX-[X,Y]$ no longer makes sense. Otherwise, $D$ satisfies the same properties as the directional derivative $D$ on $\mathbb{R}^n$.

Proposition 4.9 ((Condition of) D.D. on submanifold). Suppose $M$ is a regular submanifold of $\mathbb{R}^n$ and $D$ is the directional derivative on $M$. For $X\in \mathfrak{X}(M)$ and $Y\in \Gamma(T\mathbb{R}^n|_M)$,

(i) $D(X,Y)=D_XY$ is $\mathcal{F}$-linear in $X$ and $\mathbb{R}$-linear in $Y$;

(ii) (Leigniz rule) if $f\in C^\infty(M)$, then $$D(X,fY)=D_X(fY)=(Xf)Y+fD_X(Y).$$

Proposition 4.10 (Property of D.D. on submanifold). Suppose $M$ is a regualr submanifold of $\mathbb{R}^n$ and $D$ is the directional derivative on $M$.

(i) (zero curvature) If $X,Y\in \mathfrak{X}(M)$ and $Z\in \Gamma(T\mathbb{R}^n|_M)$, then $$D_XD_YZ-D_YD_XZ-D_{[X,Y]}Z=0.$$

(ii) (compatibility with the metric) If $X\in \mathfrak{X}(M)$ and $Y,Z\in \Gamma(T\mathbb{R}^n|_M)$, then $$X\left\langle Y,Z\right\rangle=\left\langle D_XY,Z\right\rangle+\left\langle Y,D_XZ\right\rangle.$$

Proof. By Remark 4.1, these are proven in exactly the same way as Proposition 4.3.

Proposition 4.11 (D.D. along Curve). Differentiation with respect to $t$ of a vector field along a curve is the directional derivative in the tangent direction: $$\frac{dV}{dt}=D_{c'(t)}\tilde{V}.$$

Proof. Let $$V(t)=\sum v^i(t)\partial_i|_{c(t)}\quad \mbox{and}\quad \tilde{V}_p=\sum \tilde{v}^i(p)\partial_i|_p\mbox{ for }p\in M.$$ Since $V(t)=\tilde{V}_{c(t)}$, $$v^i=\tilde{v}^i\circ c.$$ By the definition of $c'(t)$, $$c'(t)\tilde{v}^i=c_*\left( \frac{d}{dt}\right) \tilde{v}^i=\frac{d}{dt}(\tilde{v}^i\circ c)=\frac{d}{dt}v^i.$$ Therefore, $$D_{c'(t)}\tilde{V}=\sum (c'(t)\tilde{v}^i)\partial_i|_{c(t)} =\sum \frac{dv^i}{dt}\partial_i|_{c(t)}=\frac{dV}{dt},$$ by (2).