### Contents

So, it turns out that dual spaces and dual vectors sneak into general relativity and, therefore, I decided to take a closer look at them. The videos from Dr Peyam on YouTube are also very informative and the examples I’m listing here are from his lectures. First, we will go through the math and then we will invoke some intuition to make sense of all these.

### Dual spaces

#### Definition

Given a vector space $$V$$, we define its dual space $$V^*$$ to be the set of all linear transformations $$\varphi: V \to \mathbb{F}$$. The $$\varphi$$ is called a linear functional. In other words, $$\varphi$$ is something that accepts a vector $$v \in V$$ as input and spits out an element of $$\mathbb{F}$$ (lets just assume that $$\mathbb{F} = \mathbb{R}$$, meaning that it spits out a real number). If you take all the possible (linear) ways that a $$\varphi$$ can eat such vectors and produce real numbers, you get $$V^*$$.

#### Examples of dual spaces

Here is a list of examples of dual spaces:

• Example 1: Let $$V = \mathbb{R}^3$$ and $$\varphi: \mathbb{R}^3 \to \mathbb{R}$$, then $$\varphi(x,y,z) = 2x+3y+4z$$ is a member of $$V^*$$.

• Example 2: Let $$V = P_n$$ (the set of polynomials with degreee $$n$$) and $$\varphi: P_n \to \mathbb{R}$$, then $$\varphi(p) = p(1)$$ is a member of $$V^*$$. Concretely, $$\varphi(1 + 2x + 3x^2) = 1 + 2\cdot 1 + 3\cdot 1^2 = 6$$.

• Example 3: Let $$V = M_{n\times n}$$ (the set of matrices with dimensions $$n\times n$$) and $$\varphi: M_{n\times n} \to \mathbb{R}$$, then $$\varphi(A) = \text{Trace}(A)$$ is a member of $$V^*$$. In specific,

$\varphi\left(\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}\right) = 1 + 5 + 9 = 15$
• Example 4: Let $$V = C([0,1])$$ (the set of all continuous function on the interval $$[0,1]$$) and $$\varphi: C[(0,1)] \to \mathbb{R}$$, then $$\varphi(g) = \int_0^1 g(x) \mathrm{d}x$$ is a member of $$V^*$$. For instance, $$\varphi(e^x) = \int_0^1 e^x \mathrm{d}x = e^1 - 1 = e -1$$.

As it turns out, the elements of $$V^*$$ satisfy the axioms of a vector space and therefore $$V^*$$ is indeed a vector space itself.

### The dual basis

If $$b = \{\mathbf{v_1}, \mathbf{v_2}, \ldots, \mathbf{v_n}\}$$ is a basis of vector space $$V$$, then $$b^* = \{ \varphi_1, \varphi_2, \ldots, \varphi_n\}$$ is a basis of $$V^*$$. If you define $$\varphi$$ via the following relations, then the basis you get is called the dual basis:

$\varphi_i \underbrace{(a_1 \mathbf{v_1} + \cdots + a_n \mathbf{v_n})}_{\text{A vector } \mathbf{v}\in V, a_i \in \mathbb{F}} = a_i, \quad i=1,\ldots,n$

It is as if the functional $$\varphi_i$$ acts on a vector $$\mathbf{v}\in V$$ and returns the $$i$$-th component $$a_i$$. Another way to write the above relations is if you set $$\varphi_i(\mathbf{v_j}) = \delta_{ij}$$.

Then any functional $$\varphi$$ can be written as a linear combination of the dual basis vectors, i.e.

$$\varphi = \varphi(\mathbf{v_1}) \varphi_1 + \varphi(\mathbf{v_2}) \varphi_2 + \ldots + \varphi(\mathbf{v_n}) \varphi_n$$

#### Example

Let’s see a concrete example. Assume $$V = \mathbb{R}^2$$ and a vector basis $$b = \{ (2,1), (3,1) \}$$, then what is the dual basis $$b^*$$?

By definition, it’s $$\varphi_i(\mathbf{v_j}) = \delta_{ij}$$, therefore:

\begin{align*} \varphi_1 (\mathbf{v_1}) &= \delta_{11} = 1 \Leftrightarrow \varphi_1 (2,1) = 1 \Leftrightarrow \varphi_1 \left[ 2(1,0) + 1(0,1) \right] = 1 \Leftrightarrow 2\varphi_1(1,0) + 1\varphi_1(0,1) = 1 \\ \varphi_1 (\mathbf{v_2}) &= \delta_{12} = 0 \Leftrightarrow \varphi_1 (3,1) = 0 \Leftrightarrow \varphi_1 \left[ 3(1,0) + 1(0,1) \right] = 0 \Leftrightarrow 3\varphi_1(1,0) + 1\varphi_1(0,1) = 0 \end{align*}

If you solve the system:

$\begin{bmatrix} 2 & 1 \\ 3 & 1 \end{bmatrix} \begin{bmatrix} \varphi_1(1,0)\\ \varphi_1(0,1) \end{bmatrix}= \begin{bmatrix} 1\\ 0 \end{bmatrix}$

You get $$\varphi_1(1,0) = -1, \quad \varphi_1(0,1) = 3$$. Therefore:

$\varphi_1(x, y) = x \varphi_1(1,0) + y\varphi_1(0,1) = -x + 3y$

Similarly one can prove that:

$\varphi_2(x, y) = x \varphi_2(1,0) + y\varphi_2(0,1) = x - 2y$

Therefore the dual basis $$b^*$$ is equal to $$\{ \varphi_1, \varphi_2 \} = \{ -x + 3y, x - 2y\}$$. Now here comes the magic. Suppose that you have a function $$\varphi = 8x - 7y$$ and you would like to write it as a linear combination of the dual basis. How would you do?

$\varphi = \varphi(\mathbf{v_1}) \varphi_1 + \varphi(\mathbf{v_2}) \varphi_2 + \ldots + \varphi(\mathbf{v_n}) \varphi_n$

Where $$\mathbf{v_1} = (2,1)$$ and $$\mathbf{v_2} = (3,1)$$. Let us do the math:

\begin{align*} 8x - 7y &= \overbrace{\varphi(2,1)}^{\varphi(\mathbf{v_1})} \cdot \underbrace{(-x + 3y)}_{\varphi_1} + \overbrace{\varphi(3,1)}^{\varphi(\mathbf{v_2})} \cdot \underbrace{(x-2y)}_{\varphi_2}\\ 8x - 7y &= (8 \cdot 2 - 7\cdot 1) \cdot (-x + 3y) + (8 \cdot 3 - 7\cdot 1) \cdot (x-2y)\\ 8x - 7y &= 9 (-x + 3y) + 17 (x-2y)\\ 8x - 7y &= 8x -7y \end{align*}

### The dual of a dual space

Recall that the dual of space is a vector space on its own right, since the linear functionals $$\varphi$$ satisfy the axioms of a vector space. But if $$V^*$$ is a vector space, then it is perfectly legitimate to think of its dual space, just like we do with any other vector space. This might feel too recursive, but hold on. The double dual space is $$(V^*)^* = V^{**}$$ and is the set of all linear transformations $$\varphi: V^* \to \mathbb{F}$$.

In other words, $$\varphi$$ is something that accepts a vector $$\hat{v} \in V^*$$ as input and spits out an element of $$\mathbb{F}$$ (again, we just assume that $$\mathbb{F} = \mathbb{R}$$, meaning that it spits out a real number). In plain English language, a double dual vector is a creature that eats “a creature that eats a vector and spits a real number” and spits a real number. Think about them as carnivore animals (double dual vectors) eating herbivore animals (dual vectors) eating plants (vectors), while at the same time they both produce feces (real numbers).

#### Isomorphisms

In several areas of mathematics isomorphism appears as a very general concept. The word derives from the Greek iso, meaning “equal”, and morphosis, meaning “to form” or “to shape.” Informally, an isomorphism is a map that preserves sets and relations among elements. When this map or this correspondence is established with no choices involved, it is called canonical isomorphism.

When we defined $$V^*$$ from $$V$$ we did so by picking a special basis (the dual basis), therefore the isomorphism from $$V$$ to $$V^*$$ is not canonical. It turns out that the isomorphism between the initial vetor space $$V$$ and its double dual, $$V^{**}$$, is canonical as we shall see right away. Let $$v \in V, \varphi \in V^*$$ and $$\hat{v} \in V^{**}$$. We can now define a linear map:

$\hat{u}(\varphi) = \varphi(u)$

This is a canonical isomorphism between $$V$$ and $$V^{**}$$. Mind that we are talking about finite dimensional vector spaces $$V$$, i.e. $$\text{dim}(V) < \infty$$.

#### The mind blowing intuition behind dual and double dual spaces

I’m basically copy/paste’ing the answer of Aloizio Macedo from here:

The dual is intuitively the space of “rulers” (or measurement-instruments) of our vector space. Its elements measure vectors. This is what makes the dual space and its relatives so important in Differential Geometry, for instance. This immediately motivates the study of the dual space.

This also happens to explain intuitively some facts. For instance, the fact that there is no canonical isomorphism between a vector space and its dual can then be seen as a consequence of the fact that rulers need scaling, and there is no canonical way to provide one scaling for space. However, if we were to measure the measure-instruments, how could we proceed? Is there a canonical way to do so? Well, if we want to measure our measures, why not measure them by how they act on what they are supposed to measure? We need no bases for that. This justifies intuitively why there is a natural embedding of the space on its bidual. (Note, however, that this fails to justify why it is an isomorphism in the finite-dimensional case).

This is the intuition behind the relation we wrote earlier: $$\hat{u}(\varphi) = \varphi(u)$$. Here $$\hat{u}$$ is the “ruler” by which we “measure” elements of the dual space $$V^*$$. And we “calilbrate” that ruler by taking into account how $$\varphi$$ (the “ruler” of $$V$$) “measures” elements $$u$$ of $$V$$.

### Connection to general relativity

Disclaimer: The following may be absolutely wrong.

Consider the space of partial derivative operators. How do we know that this is a vector space? Let’s check:

$\frac{\partial}{\partial x^k} \left(a f + b g\right) = \frac{\partial(a f + b g)}{\partial x^k} = a \frac{\partial f}{\partial x^k} + b \frac{\partial g}{\partial x^k} = a \frac{\partial }{\partial x^k}(f) + b \frac{\partial }{\partial x^k}(g)$

Then let us pick up a basis:

$b = \left\{\frac{\partial}{\partial x^1}, \frac{\partial}{\partial x^2}, \ldots, \frac{\partial}{\partial x^n}\right\}$

And then let us define its dual space $$b^* = \left\{ \mathrm{d} x^1, \mathrm{d} x^2,\ldots, \mathrm{d} x^n\right\}$$. By definition the functionals $$\mathrm{d} x^i$$ must fulfill the following relations:

$\mathrm{d}x^i \left( \frac{\partial}{\partial x^j} \right) = \delta_j^i$

So, $$\mathrm{d}x$$’s in reality are linear functionals that act on elements of the partial derivatives vector space. They are not be thought as scalars, i.e. as infinitesimal displacements along the coordinate axis.

Now suppose we have a scalar function $$f(x)$$ and we define its total differential as:

$\mathrm{d}f = \frac{\partial f}{\partial x^1} \mathrm{d} x^1 + \frac{\partial f}{\partial x^2} \mathrm{d} x^2 + \ldots + \frac{\partial f}{\partial x^n} \mathrm{d}x^n$

And we would like to calculate the directional derivative of $$f(x)$$ along the direction of some vector $$v \in V$$, that is calculate the rate of change of $$f(x)$$ along $$v$$.

$\mathrm{d}f(v) = \frac{\partial f}{\partial x^i} \mathrm{d}x^i \left(v^j \frac{\partial}{\partial x^j}\right)$

Recall now that the basis vectors $$\mathrm{d}x$$’s were chosen in such a way that when act upon $$\frac{\partial}{\partial x^j}$$ they “select” the $$i$$-th component. Therefore:

$\mathrm{d}f(u) = \frac{\partial f}{\partial x^i} v^i$