Tensor #3. Dual Spaces텐서 #3. 듀얼 스페이스

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 3 of the playlist, Dual Spaces.

Where should a map go?

In the previous post we dealt with a map $\map{\Lambda}{v}$ between two vector spaces. Once we decide the result of the map for each basis vector, linearity fixes the result for every vector.

But a map needs a destination. We made just one vector space $\V$, so do we have to make a second vector space whenever we want to use a map? We want to keep our universe as small as possible. Without creating anything new, there are two places a map can go.

1. $\V$ itself
2. The real numbers $\R$

The second is the one we care about. When we create a vector space, its definition already brings the scalar field $\R$ along, so there is nothing new to define. Moreover, $\R$ can be seen as a 1-dimensional vector space. It has just one basis vector, its addition and multiplication are those of the real numbers, and it draws its scalars from $\R$ itself. Implementing a 1-dimensional vector space with $\R$ in this way is called a realization.

The set of all maps

Making one map from $\V$ to $\R$ needs another choice, because we must decide which real number each basis vector goes to. If we do not even want to make this choice, there is one way out: think of the set of all possible maps. This set is the whole pool of candidates we would pick from when making a single map, so it is not something newly invented. It already exists, logically, the moment we create $\V$.

The set of all maps from $\V$ to $\R$ is called the dual space of $\V$ and written $\V^*$.

The dual space is a vector space too

To see why it deserves the name space, we check whether this set satisfies the conditions of a vector space. We need addition, scalar multiplication, and opposite elements. For maps $\Lambda, Z \in \V^*$, a vector $v \in \V$, and a scalar $a \in \R$, define them as follows.

• Addition
  $\map{\Lambda + Z}{v} = \map{\Lambda}{v} + \map{Z}{v}$
• Scalar Multiplication
  $\map{a\Lambda}{v} = a\map{\Lambda}{v}$
• Opposite
  $(-1)\Lambda$ is the opposite map, and adding the two gives the zero map, which sends every vector to $0$.

The $+$ and the multiplication on the right-hand sides all belong to $\R$, because $\map{\Lambda}{v}$ is a real number. So $\V^*$ is a vector space with the same scalar field as $\V$, and it has the same dimension too.

Q. Then what kind of vector is an element of $\V^*$, a dual vector?
A. It is a map that takes a vector and gives back one real number. What matters is that it is a vector inside a vector space and a map at the same time.

The dual basis

We wrote the basis vectors of $\V$ with a lower index $e_\mu$, so we write the basis vectors of the dual space with an upper index $e^\mu$. Choosing a dual basis is also up to us. Using the fact that the basis vector $e^\mu$ is a map, I want this rule.

That is, $e^0$ gives $1$ only when it meets $e_0$, and gives $0$ when it meets the other basis vectors $e_1, e_2, e_3$. $e^1, e^2, e^3$ likewise give $1$ only on their own partner.

There is something you must keep in mind here. This rule is only what I chose because it makes calculation convenient, and it does not say that one basis is more legitimate than another. Also, between $\V$ and $\V^*$ there is no map that naturally decides which vector corresponds to which dual vector. The dual space comes for free once we create $\V$. A way to connect the two does not exist unless we decide one separately.

Let us handle it by hand. In the plane below, drag $e_1, e_2$. The dual basis $e^1, e^2$ follows while keeping the rule $\map{e^{\mu}}{e_{\nu}} = \dl$, and you can see that the values in the box below always come out as $1$ and $0$. The number of stripes a vector crosses is $e^\mu(v)$.