# Use of partial derivatives as basis vector

I am trying to understand use of partial derivatives as basis functions from differential geometry

In tangent space $\mathbb{R^n}$ at point $p$, the basis vectors $e_1, e_2,...,e_n$ can be written as $\frac {\partial}{\partial x^1} \bigg|_p,\frac {\partial}{\partial x^2} \bigg|_p,...,\frac {\partial}{\partial x^n} \bigg|_p$

Let's say in 2 dimensional Euclidean space, a function $f : \mathbb {R^2}\rightarrow \mathbb {R^2}$ is

$x^2 + y^2=4$ , a circle with radius 2. Tangent at point $p$ (2,0) will be $0e_1 + e_2$. If I say $f =x^2 + y^2-4 =0$,

$\frac {\partial f}{\partial x} \bigg|_{p=(2,0)} = 4 \quad$ and $\quad \frac {\partial f}{\partial y} \bigg|_{p=(2,0)} = 4$

This does not make sense of the partial derivatives as basis vectors. Any comments?

• Hm, first of all, your $f$ is a funtion $\mathbb{R}^2\rightarrow \mathbb{R}$. Then, the $\frac{\partial}{\partial x^i}$ are said to be tangent vectors, but you are looking at $\frac{\partial f}{\partial x^i}$, which is an entirely different object. – Thomas Dec 25 '15 at 18:44
• Do you have any specific issues / confusions about these tangent space bases in general that you want addressed, or do you just want your example made coherent? – epimorphic Dec 27 '15 at 23:59
• @epimorphic, Wanted to know how, in general, partial derivatives can be used as basis functions to represent a vector. – 343_458 Dec 28 '15 at 23:36
• Sorry for the super late response. You might want to look at some existing answers on this topic first, such as math.stackexchange.com/a/509515. Let me know whether it helped or not. – epimorphic Jan 8 '16 at 3:44

## 4 Answers

Imagine a ruler. The ruler, when paired with an object, provides its length. The length is different from the ruler, of course. One could say that the ruler evaluates a length on a given object.

The ruler, here, is the tangent vector: $\frac{\partial}{\partial x}$. By doing $\frac{\partial f}{\partial x}$ you are evaluating your "ruler" on the object: the function. And that is what a tangent vector is (when interpretated as a derivation): it takes functions to real numbers. But the evaluation and the evaluator are two different objects altogether.

Cartesian coordinates constitute the forest that hides the trees, i.e. the intrinsic notion of a tangent vector. In $E={\Bbb R}^n$ or more generally in any normed (and complete) vector space the intuitive notion you have of a tangent vector works fine. It is probably something as follows: Let $c: ]-\epsilon,\epsilon[ \rightarrow E$ be a smooth curve. Then the tangent vector to the curve at $p=c(0)$ is given by: $$v = c'(0)= \lim_{h\rightarrow 0} \frac{1}{h} \left( c(h) - c(0) \right)$$ So in a vector space a tangent vector may simply be viewed as an element of the vector space itself. You may compose with functions, take derivatives etc. as in your example. No need for adapting a different point of view.

The statement you are asking about, may, however, be viewed as a preparation towards working with manifolds. Now, if $E$ were a manifold (whatever that is, I omit the details) then the situation is different. Although it is ok to talk of a smooth curve as before, the difference $c(h)-c(0)$ does not make sense in a general manifold, so how would you describe this $v$ that doesn't exist?

Well, it does make sense to look at a smooth function of the manifold into a vector space, say $A: E \rightarrow {\Bbb R}$. Then the composed function $A\circ c: ]-\epsilon,\epsilon[\rightarrow {\Bbb R}$ is a map between reals and this we know how to differentiate. So the wanted tangent vector "$v=c'(0)$" at $p=c(0)$ may be given an interpretation as a differential operator on $A$ acting as follows: $$L_v A = \frac{d}{dt}_{|t=0} A(c(t)) = (A\circ c)'(0)$$ Now, a coordinate system is a suitable collection of $n$ smooth maps $x_1,...,x_n: E \rightarrow {\Bbb R}$ so, in particular, we may act upon them to define a collection of real numbers: $$v_k = L_v x_k = (x_k \circ c)'(0), \; k=1,...,n$$ Also, when we have a coordinate system, we may express the function $A$ as some smooth function of the coordinates: $A(\xi) = a(x_1(\xi),...,x_n(\xi))$, $\xi\in E$. But then we are able to calculate another expression for the above derivative: $$L_v A = \frac{d}{dt}_{|t=0} a (x_1 \circ c(t)),...,x_n\circ c(t)) = v_1\frac{\partial}{\partial x_1} a (p) + \cdots + v_n \frac{\partial}{\partial x_n} a(p)$$ So in the given coordinate system we may express the above 'action' as: $$L_v = v_1 \left( \frac{\partial}{\partial x_1}\right)_{|p} + \cdots + v_n \left( \frac{\partial}{\partial x_n}\right )_{|p}$$ Our tangent vector has then coordinates $v_1$,...,$v_n$ ($n$ real numbers) in the basis consisting of partial derivatives $\partial_{x_1}$,...,$\partial_{x_n}$ (evaluated at the point $p$).

This point of view also allows you to describe how tangent vectors transform under change of coordinates and all this leads to the understanding of manifolds and how to do calculus on a manifold.

The space of the OP's circle is not a tangent space, that's why the simple example does not make any sense. One must stay inside the circle, so to speak. So let's start over again and define that circle with radius $2$ as: $$\begin{cases} x = 2\cos(\phi) \\ y = 2\sin(\phi) \end{cases} \quad \Longrightarrow \quad x^2+y^2=4$$ There are no partial derivatives because the manifold is one-dimensional. And there is only one basis (tangent) vector: $$\begin{bmatrix} dx/d\phi \\ dy/d\phi \end{bmatrix} = \begin{bmatrix} -2\sin(\phi) \\ 2\cos(\phi) \end{bmatrix}$$ Normed to a unit vector $\,\vec{t}\,$ if we divide by $2$ : $$\vec{t} = \begin{bmatrix} dx/d\phi \\ dy/d\phi \end{bmatrix} / 2 = \begin{bmatrix} -\sin(\phi) \\ \cos(\phi) \end{bmatrix} = -\sin(\phi)\,\vec{e_1} + \cos(\phi)\,\vec{e_2}$$ Now specify for $\,\phi=0\,$ and we're done, for this example at least.

The simplest way to approach this may be to fix $\vec p\in \mathbb R^n$,

let $f:\mathbb R^n\to \mathbb R$ and consider (assuming these exist) the directional derivatives

$\textbf D_\vec vf(\vec p)=\nabla f(\vec p)\cdot \vec v=\sum_{k=1}^{n}v_k\frac{\partial f}{\partial x_k}(\vec p)$.

Now notice that this motivates the following:

If we define for $1\leq k\leq n,\ \frac{\partial }{\partial x_k}:\mathcal C(\mathbb R^n,\mathbb R)\to \mathbb R$ in the obvious way by $f\mapsto \frac{\partial f}{\partial x_k}(\vec p)$ then $\left \{ \frac{\partial }{\partial x_k} \right \}_{1\leq k\leq n}$may be regarded as a basis for a vector space which we denote $T_{\vec p}(\mathbb R^n)$.

An arbitrary element of $T_{\vec p}(\mathbb R^n)$is then given by a linear combination of the basis elements, that is $\textbf v=\sum_{k=1}^{n}v_k\frac{\partial }{\partial x_k}$, whose effect on functions $f$ at the point $\vec p$ is simply $\textbf v(f)=\textbf D_\vec vf(\vec p)$.