The relationship between the differential and the directional derivative of a function

I am currently studying differential manifolds (from John M. Lee 's book), and have a question concerning the difference between what is defined as the $$\textbf{differential of a function F}$$, and the $$\textbf{directional derivative of a function F}$$. Let $$M \subset \mathbb{R}^{m}$$, let $$N\subset \mathbb{R}$$, and suppose $$F:M \rightarrow N$$ is a smooth map. Then,

1. The differential of $$F$$ at $$p \in M$$ is a map

$$dF_{p}:T_{p}M \rightarrow T_{F(p)}N$$

defined as, for some $$v \in T_{p}M$$, $$dF_{p}(v)$$ is a derivation in $$T_{F(p)}N$$ defined as, for all $$f \in C^{\infty}(N)$$,

$$dF_{p}(v)(f)=v(f \circ F)$$.

Now, because $$M$$ and $$N$$ are Euclidean themselves, if $$v=(v_{1},...,v_{N})$$, this can be expressed as $$dF_{p}(v)(f)=v_{1} \cdot \frac{\partial f}{\partial F}\ \frac{\partial F}{\partial x_{1}}+...+v_{m} \cdot \frac{\partial f}{\partial F}\ \frac{\partial F}{\partial x_{m}}$$.

1. The directional derivative of $$F$$ at $$p$$ in direction $$v \in \mathbb{R}^{m}$$ is given by

$$D_{v}F(p)=v_{1} \cdot \frac{\partial F(p)}{\partial x_{1}}+...+v_{m}\cdot \frac{\partial F(p)}{\partial x_{m}}$$.

Now, it seems that if $$Id:\mathbb{R}\rightarrow \mathbb{R}$$ is the identity function on $$\mathbb{R}$$, we have that for $$v \in T_{p}M$$,

$$v(F)=dF_{p}(v)(Id)=D_{v}F(p)$$.

Am I reading this correctly? I am trying to weed through the abstraction of differentials between manifolds and ground it into something more familiar, the directional derivative. Are directional derivatives in the theory of manifolds expressed as the the differential evaluated at the identity function, which are equivalent to simply evaluating $$v(F)$$ itself?

• It is weird that you write $M \subset \mathbb{R}^{N}$, $N\subset \mathbb{R}$. You are taking $\Bbb R$ to the power of a subset of $\Bbb R$. Do you want $M \subset \mathbb{R}^{m}$, $N\subset \mathbb{R}^n$ instead?
– edm
Nov 7, 2018 at 14:59
• Ah yes, thanks. I will change that.
– Mark
Nov 7, 2018 at 15:07
• I don't get your question. Do you want to ask if your interpretation is correct? Or do you want to know the definition of directional derivative on manifolds? Or do you want to understand the motivation for the definition of differential? And are you asking for the difference or relation between differential of a function and directional derivative? And do you have several more questions in mind (as seen in your comments)?
– edm
Nov 7, 2018 at 18:04
• @edm My question is twofold: 1. First, if $T_{p}M$ is the tangent space, $v \in T_{p}M$ is a differential, then is $v(F)$ to be thought of as some directional derivative of $F$? 2. If 1. is true and $v(F)$ is the directional derivative, it seems that if we take the differential of $F$ at $p$, and the same $v \in T_{p}M$ as in question 1., then if $v=(v_{1},...,v_{m}) \in \mathbb{R}^{m}$ corresponds to $v \in T_{p}M$, then $v(F)=D_{v}F(p)$, where the RHS is the traditional directional derivative.
– Mark
Nov 7, 2018 at 21:07

If you work in open subsets of Euclidean spaces, you can write everything concretely. First I answer your interpretation in 1.

1. You do not take partial derivative with respect to a function, i.e. you should not write $$\frac{\partial f}{\partial F}$$. Instead, you write $$dF_p(v)=\sum_{i=1}^m v_i\frac{\partial F}{\partial x_i}$$ (don't act on $$f$$ yet). It is a number, because you require $$N\subset\Bbb R$$ and so $$\sum_{i=1}^m v_i\frac{\partial F}{\partial x_i}\in\Bbb R$$. This thing acts on a smooth function $$f\in C^\infty(N)$$ by $$dF_p(v)(f)=\left(\sum_{i=1}^m v_i\frac{\partial F}{\partial x_i}\right)f'(F(p)).$$ It looks similar to your formula; if you interpret $$\frac{\partial f}{\partial F}$$ as derivative of $$f$$, then you are correct.

This formula looks real boring. I am just multiplying a number to the usual derivative of $$f$$ at $$F(p)$$. You would see a more interesting formula if you assume $$N\subset\Bbb R^n$$, but you are not asking about that.

1. Your interpretation is correct. It is because $$dF_p(v)(Id):=v(Id\circ F)=v(F)$$ and $$v(F):=D_vF(p)$$.

Given a vector $$v$$ and a smooth real-valued function $$F$$ on a smooth manifold, the number $$v(F)$$ is to be directly interpreted as the directional derivative of $$F$$ towards the direction $$v$$. You can indeed say $$v(F)=dF_p(v)(Id)$$, by the above formula, but it complicates things.

And, regarding to your question in comments, you do not need to define partial derivatives on manifolds. Remember that partial derivatives are special cases of directional derivatives. Once you have defined all those directional derivatives, there is no need to define partial derivatives at all. What you might want to do is to choose a basis for a tangent space $$T_pM$$. You do this by using charts to transfer a basis from $$\Bbb R^n$$. Continue reading the book and you should see how to do this.

Edit: To correct your use of terminology in comment.

If you follow the terminology in Lee's book, an element $$v\in T_pM$$ is not called a differential, but is called a derivation or a tangent vector.

• Thank you. This answers all of my questions. I apologize for the confusion, as I am still getting all of the definitions/concepts straight in my head. This makes it much clearer for me.
– Mark
Nov 7, 2018 at 21:10
• @Mark Please see edit. I would like to correct your use of terminology.
– edm
Nov 8, 2018 at 4:33

The relationship between the differential and directional derivative is the same in differential manifolds as in Euclidean space. The derivative is a linear function. Linear functions take in vectors and output vectors. When the input vector is a unit vector, the output is called the directional derivative. That is because directions are defined by unit vectors.

Let's use an example.

Let $$f:\mathbb R^2\to\mathbb R^2$$ be defined by $$f(x,y)=(x^2-y^2,xy)$$

So $$f=(f_1,f_2)$$ where $$f_1(x,y)=x^2-y^2$$ and $$f_2(x)=xy$$

The derivative, or differential, is $$df= \begin{pmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \\ \end{pmatrix} = \begin{pmatrix} 2x & -2y \\ y & x \\ \end{pmatrix}$$

If we want the derivative of $$f$$ at say $$(1,2)$$ then we get $$df(1,2) = \begin{pmatrix} 2 & -4 \\ 2 & 1 \\ \end{pmatrix}$$

Imagine being at point $$(1,2)$$ in the plane. There are $$360^{\circ}$$ of directions. We represent each direction as a unit vector in that direction. Now we can ask questions like, what is the derivative of $$f$$ in the $$(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$$ direction at $$(1,2)$$?.

We evaluate the derivate to get $$\begin{pmatrix} 2 & -4 \\ 2 & 1 \\ \end{pmatrix} \begin{pmatrix} \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ \end{pmatrix}=\begin{pmatrix} \frac{-2}{\sqrt{2}} \\ \frac{-3}{\sqrt{2}} \\ \end{pmatrix}$$

• Thank you. I understand all of the calculations you have made above, but I am still unclear as to the connection to manifolds. I guess my more general question would be, how do we define partial derivatives on manifolds? Are we choosing the desired "movement way from $p \in M$", or $v \in T_{p}M$, and then evaluating the differential $dF_{p}(v)$ at the identity function?
– Mark
Nov 5, 2018 at 18:49
• @Mark Imagine that $M$ and $N$ are 2-manifolds. If $p\in M$ and $f(p)\in N$ then $T_p$ and $T_{f(p)}$ are both copies of $\mathbb R^2$ and the calculations are done in the tangent spaces. Nov 5, 2018 at 18:52
• @Mark differentiable manifolds are locally diffeomorphic to euclidean spaces, so we calculate derivatives in the coordinate system of these local euclidean spaces and after we can represent these derivatives using the coordinate system of the manifold. It can be shown that these derivatives are well-defined and independent of the choice of the diffeomorphism Nov 5, 2018 at 19:08
• @Masacroso Thank you. Can you tell me if I am wrong in my calculation that evaluating the differential of $F$ at $p$ at the identity, $F_{p}(v)(Id)$, gives me back the traditional directional derivative in direction $v$, given by $D_{v}F(p)$?
– Mark
Nov 5, 2018 at 19:12

The differential of a smooth real-valued function (with domain finite-dimensional) is a 1-form (also called Pfaff forms), that is a function from the domain of $$f$$ to it cotangential bundle, that is

$$d f: \Bbb R^n\to T^*\Bbb R^n,\quad p\mapsto d_p f$$

where $$T^*\Bbb R^n$$ is the cotangential bundle of $$\Bbb R^n$$, and $$d_pf:=\operatorname{pr_2}\circ T_pf=\partial f(p)$$ is the differential of $$f$$ at $$p$$, and $$\partial f(p)$$ is the Fréchet derivative of $$f$$ at $$p$$, that it can be represented by a vector (the gradient $$\nabla f(p)$$) due to the Riesz representation theorem.

And a directional derivative, for a Fréchet differentiable function, can be defined by $$D_v f(p)=\partial f(p)v$$. Then the relationship is $$(d_pf)(v)=D_v f(p)=(\nabla f(p)| v)$$.

I dont know the notation on the Lee's book so Im not sure how to understand your notation, or what is $$d_pF(v)(f)$$.

The cotangential bundle of $$X$$ is defined by $$T^*X=\bigcup_{p\in X}T^*_pX$$, where $$T^*_pX$$ is the dual space of $$T_pX$$. Indeed for finite-dimensional space $$X$$ the tangent bundle and the cotangential bundle are isomorphic.

• Thanks again. I am still starting out with the subject, and admit to not having studied the dual space or cotangential bundle, etc. I am only trying to figure out the relative notion of a derivative on a manifold. Maybe I can make my question clearer after some thought: If $T_{p}M$ is the tangent space, and $F:M \rightarrow \mathbb{R}$ is some differentiable function, is $v(F)$ considered to be a "directional derivative" of $F$ at $p \in M$ for some derivation $v \in T_{p}M$?
– Mark
Nov 6, 2018 at 19:16

The relation between differential of a function and directional derivative is essential.

Let $$f:\Bbb R^2\to \Bbb R$$ differentiable, $$a\in \Bbb R^2$$ and a vector $$h\in \Bbb R^2$$. Then set the real function $$g(t):=f(a+th), t\in \Bbb R$$ and we observe that $$g(0)=f(a)$$ and $$g(1)=f(a+h)$$, which means that for $$t\in[0,1]$$ we stand along the vector $$h\in T_a\Bbb R^3$$ (from $$a$$ to $$a+h$$). Thus we have

$$dg_0(1)=g'(0)\cdot 1=\frac {d}{dt}f(a+th)|_{t=0}=\nabla f(a+th)\cdot (a+th)'|_{t=0}=\nabla f(a)\cdot h=df_a(h).$$