# General vectors and tangent space

1. Dec 17, 2015

### sunrah

Given a scalar function g defined on a manifold and a curve f:λ -> xa, the change of the function along the curve is

$\frac{dg}{d\lambda} = \frac{dg}{dx^{a}}\frac{dx^{a}}{d\lambda} = T^{a}\frac{dg}{dx^{a}}$

where

$\frac{dx^{a}}{d\lambda} = T^{a}$ is the tangent to the curve.

The argument that I don't understand is that this

$T^{a}\frac{d}{dx^{a}}$

is a vector. To me it looks like the inner product of two vectors, $\vec{T} = (T_{x}, T_{y})$ and ∇x,y, so looks like a scalar to me.

Also who do the coefficients of the gradient necessarily form a basis for the tangent space?

2. Dec 17, 2015

### Fredrik

Staff Emeritus
Let M be an $n$-dimensional smooth manifold. Let F be the set of smooth real-valued functions with domain M. The set of all $v:F\to\mathbb R$ satisfies the definition of a vector space. So all its elements can be called "vectors". For each $p\in M$, the set of all linear $v:F\to\mathbb R$ that satisfy the condition $v(fg)=v(f)g(p)+f(p)v(g)$ is an $n$-dimensional subspace of that vector space. This subspace is denoted by $T_pM$ and is called the tangent space of M at p. The elements of $T_pM$ are called tangent vectors at p.

For each coordinate system (=chart) $x:U\to\mathbb R$ and each $i\in\{1,\dots,n\}$, the map $\frac{\partial}{\partial x^i}\big|_p:F\to\mathbb R$ defined by
$$\frac{\partial}{\partial x^i}\bigg|_p f=(f\circ x^{-1})_{,i}(x(p))$$ for all $f\in F$, is an element of $T_pM$. The comma denotes the kind of partial differentiation that you are familiar with from calculus. For example, if $g:\mathbb R^n\to\mathbb R$, I would denote the partial derivative of g with respect to its second variable as $g_{,2}$.

It's possible to show that the set
$$\bigg\{\frac{\partial}{\partial x^1}\bigg|_p,\dots,\frac{\partial}{\partial x^n}\bigg|_p\bigg\}$$ is a basis for $T_pM$. I've seen the proof in a couple of different books. I posted a version of it in post #14 in this thread.

So the thing you're asking about is a tangent vector at p because it's a linear combination of tangent vectors at p.

Last edited: Dec 17, 2015
3. Dec 17, 2015

### stevendaryl

Staff Emeritus
This is really just a convention, and if it bothers you, you don't have to adopt it, but you should understand it.

I think it's clear in terms of coordinates that if you have a parametrized path, $x^\mu(s)$, giving a path through space (or spacetime) as a function of a parameter $s$ (not necessarily time, nor even proper time, just a real-valued parameter that increases as you move down the path), then there is a corresponding vector $U$ whose components are given by $U^\mu = \dfrac{dx^\mu}{ds}$. This vector is called the tangent vector to the path $x^\mu(s)$. For each tangent vector $U$ with components $U^\mu$, there is a corresponding operator, the "directional derivative", that can be defined by: $(U \cdot \nabla) \phi = U^\mu \dfrac{d \phi}{dx^\mu}$

The above definitions are all in terms of components, which are specific to a coordinate system. Is there a coordinate-independent way to talk about directional derivatives, without mentioning components of the vector? Yes. Let $\mathcal{P}(s)$ be a smooth function from a real number $s$ to points in space (or spacetime). We define the directional derivative along path $\mathcal{P}$, $\frac{d \mathcal{P}}{ds}$, to be the operator $\hat{U}$ defined by:

For any scalar field $\phi$ (that is, function that assigns a real number to each point in space, or spacetime),

$\hat{U}(\phi) = \dfrac{d\phi(\mathcal{P}(s))}{ds}$

This definition of "directional derivative" doesn't mention coordinates or components. It only mentions scalar fields and parametrized paths and derivatives of real-valued functions.

At this point, we note that there is a one-to-one correspondence between directional derivatives (which are operators) and tangent vectors (which are...I don't know...abstract objects that can be represented by column matrices that transform in some particular way under coordinate transformations). We have a coordinate-free definition of a directional derivative, and every tangent vector corresponds to a directional derivative, so there is really no reason not to identify the two: A tangent vector simply IS a directional derivative.

This way of looking at it flips the idea of what is fundamental. Instead of defining a vector in terms of components, and using components to define a directional derivative, we view the directional derivative as fundamental, and components to be a derived concept:

Pick a coordinate system. Then since a scalar field is any function from points in space $\mathcal{P}$ to real numbers, then a coordinate system is equivalent to a collection of four scalar fields $X(\mathcal{P}), Y(\mathcal{P}), Z(\mathcal{P}), T(\mathcal{P})$, where $X(\mathcal{P})$ gives the value of the x-coordinate at point $\mathcal{P}$, etc. Then we can define the components of a directional derivative $\hat{U}$ via:

$U^x = \hat{U}(X)$
$U^y = \hat{U}(Y)$
$U^z = \hat{U}(Z)$
$U^t = \hat{U}(T)$

or more compactly: $U^\mu = \hat{U}(x^\mu)$ (where $x^\mu$ is understood as the scalar field corresponding to that coordinate).

Once we identify directional derivatives with tangent vectors, we can pick out a set of basis vectors corresponding to a coordinate system as follows:

$e_x = \frac{d}{dx}$
$e_y = \frac{d}{dy}$
$e_z = \frac{d}{dz}$
$e_t = \frac{d}{dt}$

Those are the directional derivatives in the x, y, z, and t-directions. An arbitrary vector $U$ can be written as a linear combination of these basis vectors:

$U = U^x e_x + U^y e_y + U^z e_z + U^t e_z = U^x \frac{d}{dx} + U^y \frac{d}{dy} + U^z \frac{d}{dz} + U^t \frac{d}{dt} = U^\mu \frac{d}{dx^\mu}$

So in the expression $U^\mu \frac{d}{dx^\mu}$, you're just creating a linear combination of basis vectors.

Last edited: Dec 17, 2015
4. Dec 17, 2015

### stevendaryl

Staff Emeritus
Just another word in clarification: when someone writes $U = U^\mu e_\mu$, the right-hand side looks like a scalar product of a vector with components $U^\mu$ and another vector with components $e_\mu$. But the index $\mu$ on $e_\mu$ doesn't indicate which component, it indicates which vector. $e_x, e_y, e_z, e_t$ are 4 different vectors, not 4 components of one vector.

5. Dec 17, 2015

### haushofer

That's why I like the convention in which the mu-components on e are written with brackets () to indicate they label vectors, not just components. It's confusing otherwise.

6. Dec 17, 2015

### pervect

Staff Emeritus
I usually use the following notation, which I think is helpful. $e_\mu$ is a vector, the components of that vector are $(e_\mu)^\nu$, where $\mu$ picks out "which vector" and $\nu$ picks out "which component".

7. Dec 17, 2015

### sunrah

Thanks everyone! You've been really helpful! So I still need to convince myself that $T^{a}\frac{d}{dx^{a}}$ is really a vector, but that just means going through the proof. I understand now that the $\frac{d}{dx^{a}}$ form a basis which spans the tangent space, but my problem was that it wasn't so obvious, having always viewed $\frac{d}{dx^{a}}$ as the coefficient of some vector-like operator, e.g. ∇.