Tensor #4. Cartesian Products텐서 #4. 카테시안 프로덕트

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 4 of the playlist, Cartesian Products.

What we have built so far

The moment we create a vector space $\V$, its dual space $\V^*$ comes along. Both are vector spaces over the real numbers $\R$ and they have the same dimension. We write the basis vectors of $\V$ with a lower index $e_\mu$ and those of $\V^*$ with an upper index $e^\mu$. The index tells us which space a basis vector came from.

The dual of the dual

$\V^*$ is a vector space too, so we can think of the collection of maps from it to $\R$ again. That is the double dual. Then do we get a dual, a dual of the dual, a dual of that, and so on forever?

No. The bracket notation we have been using for maps shows why.

This expression can be read in two ways. We can read it as a vector $v$ sitting in place and waiting for a map to come in, or as a map $\Lambda$ fixed in place and waiting for a vector. Read the first way, a vector is a map that takes an element of the dual space and gives back a real number. Read the second way, a dual vector is a map that takes a vector and gives back a real number. Neither side ranks above the other. So the double dual is treated as the same thing as the original $\V$ (for finite dimensions), and there are only two spaces, $\V$ and $\V^*$. Because we created $\V$ first, we call it the vector space and the other the dual space, and both are vector spaces.

Q. Then is a covector not a vector?
A. It is a vector. Every element of a vector space is a vector by definition. We call the elements of $\V^*$ covectors because we put $\V$ at the center, and the co means a map attached to $\V$. This distinction will later become the criterion that separates covariant from contravariant, so it is worth remembering.

How a map acts on a vector

An arbitrary map in the dual space can be written $b_\mu e^\mu$ and an arbitrary vector in the vector space $a^\nu e_\nu$. Choosing the coefficients $b_\mu$ freely gives an arbitrary map. By linearity the coefficients come out, and the dual basis rule $\map{e^{\mu}}{e_{\nu}} = \delta^{\mu}{}_{\nu}$ gives this.

Everything multiplied is a real number, so the result is a real number. I chose this rule because it makes calculation convenient. It does not mean that the vector $a^\mu e_\mu$ in $\V$ and the covector $a_\mu e^\mu$ in $\V^*$ are naturally partners. The only thing connecting the two spaces is the fact that each one sends the other to $\R$.

Q. $b_0 a^0 + b_1 a^1 + b_2 a^2 + b_3 a^3$ looks just like the dot product.
A. It resembles it, but it is a different operation. The dot product takes two vectors from the same vector space and gives a real number. Here we take one covector and one vector, elements of different spaces. The domains are completely different.

The Cartesian product

Now we build a new tool. The Cartesian product $A \times B$ is a binary operation that takes two sets and gives one set. A binary operation is one where you put one object on the left and one on the right, and a third object comes out. It is built like this.

Take one element from the first set and one from the second, form an ordered pair in every possible way, and collect them all in one set. The sets may be infinite and the recipe stays the same. All we need is for them to be sets, and they do not have to be vector spaces. Vector spaces are sets too, so we can multiply them. The products we will use in this series are exactly of that kind.

Let us build one by hand. Press elements to choose the sets, and switch between $A \times B$ and $B \times A$ to see how the cells change.

Q. Are $(a, 1)$ and $(1, a)$ the same element?
A. They are different. In an ordered pair the position carries meaning: the first slot comes from the first set and the second slot from the second set. So $A \times B$ and $B \times A$ are in general not the same set.

Why build this?

Because we want to write binary operations as maps too. To use a map we need a domain and a codomain, but a binary operation has two inputs. The collection of every way to choose one element from the first set $A$ and one from the second set $B$ is exactly $A \times B$. So we can write a binary operation $\circledast$ like this.

An operation that looked like it had two inputs becomes an ordinary map that takes a single ordered pair. Three or more inputs work the same way. If we form products of ordered triples or ordered quadruples, operations with three or four inputs can be written as maps as well.

In the next post we apply this tool to vectors and covectors to define a tensor.