Tensor #12. Contraction and Index Gymnastics텐서 #12. 컨트랙션과 인덱스 짐내스틱스

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 12 of the playlist, Contraction and Index Gymnastics. Continuing from the metric tensor of the previous post, I collect contraction, which removes two indices from a tensor, and index gymnastics, the knack of working with components alone.

Filling a map only partway

Say the tensor $A$ is an element of the space made from three copies of $\V$ and two copies of $\V^*$. An element of the space with p copies of $\V$ and q copies of $\V^*$ is called a tensor of type (p, q), so $A$ has type (3, 2). In components it is written like this.

An element of $\V$ can also be seen as a map that takes a covector and returns a real number (this is the double dual), so $A$ is a map that takes three covectors and two vectors and returns one real number. What happens if we do not fill all the slots? Put $\sigma$ in the first covector slot only and leave the rest empty, and we get a map waiting for two covectors and two vectors, that is, a tensor of type (2, 2). Putting it in the second slot also gives type (2, 2). So $A$ and $\sigma$ alone do not fix a single result, and we must also say which slot was filled. The point of this section is that a tensor can go not only to the real numbers but also to a smaller tensor product space.

Q. Maps between vector spaces were called linear. Why do we say multilinear here?
A. Because there are several slots. Move one slot and hold all the others fixed, and the result follows the input of that slot linearly. Multiply that input by $a$ and the result is multiplied by $a$. A map that behaves this way in every slot is a multilinear map, and tensor theory can be called the theory of these maps.

The universal property

Suppose there is a multilinear map $Z$ from the Cartesian product of three copies of $\V$ and two copies of $\V^*$ to some vector space $U$. The space $U$ may be $\R$. Then there is exactly one linear map $C$ from the tensor product space to $U$ that does the same job. This is called the universal property. For vectors $v_i$ and covectors $\sigma, \tau$ we write it like this.

The proof goes through a basis. Set the value of $C$ on each basis tensor to the value of $Z$, for example $C(e_0 \otimes e_0 \otimes e_1 \otimes e^3 \otimes e^2) = Z(e_0, e_0, e_1, e^3, e^2)$. Every tensor is a linear combination of basis tensors, so linearity fixes the rest. This is the only way to set it, so $C$ is unique. Checking one basis tensor at a time is enough because a basis tensor is a simple tensor, a chain of vectors joined by tensor products.

Contraction

Starting from $\V$, the tools we already have are the vector space, the dual space, and the map $\map{\sigma}{v}$ that connects them. With only these we build one multilinear map. It takes three vectors and three covectors, applies the second covector to the second vector to get a real number, and joins the rest with tensor products.

It is linear in every slot and its values lie in the (2, 2) tensor product space. The universal property then guarantees a linear map $C$ from the (3, 3) space to the (2, 2) space. This $C$ is the contraction.

Contraction matters because we decided nothing new for it. Creating $\V$ brought the dual space and the map between them for free, and the universal property carries that map over to the tensor product spaces. An object that is forced on us like this is called canonical. Each way of choosing one vector slot and one covector slot gives a contraction, so a (p, q) tensor has $p \times q$ of them, and all send it to type (p-1, q-1). Contracting several pairs is the same as doing one pair at a time, so one pair is enough to study.

Q. Can we contract the metric tensor $g_{\mu\nu}$?
A. Not directly. It has type (0, 2), so there is no $\V$ slot to pair with. Contraction needs both a $\V$ slot and a $\V^*$ slot, so types (0, q) and (p, 0) are out. Once we raise one index, as shown below, it becomes type (1, 1) and the contraction becomes possible.

Contraction in components

What happens on basis tensors is simple. Take only the basis elements of the two slots to be paired, $e_\beta$ and $e^\delta$, and replace them with $\map{e^\delta}{e_\beta} = \delta^\delta{}_\beta$. The delta makes the two indices equal, and a sum remains. For a (3, 2) tensor $T$, pairing the second $\V$ slot with the first $\V^*$ slot goes like this.

The coefficients $M^{\alpha\gamma}{}_{\epsilon} = T^{\alpha\beta\gamma}{}_{\beta\epsilon}$ are the components of a tensor of type (2, 1). In the language of components, writing one upper index and one lower index as the same letter and summing removes both indices. The operation is easy to compute but hard to see the importance of, and the reason is what we saw above: nothing new was decided.

Let us do it by hand with a two-dimensional tensor $T^{\alpha\beta}{}_{\gamma}$ of type (2, 1). With two upper indices there are two ways to choose the pair, and each pair leaves a different index and a different result.

Q. If we change the basis, does the result of a contraction still fit together?
A. In the toy, double the basis $e_0$. The component with upper index 0 is halved and the component with lower index 0 is doubled, and in the contracted pair these factors cancel. Only the factor of the remaining index is left, so the result changes just as the components of a vector do. (This explanation is based on a 2-dimensional example and was not covered in the lecture.)

Index gymnastics and raising and lowering indices

So far we carried the basis along, as in $A^{\mu\nu} e_\mu \otimes e_\nu$. Classical books drop the basis, call the components $A^{\mu\nu}$ themselves the tensor, and work by looking only at where the indices sit. What this does is follow the moves among $\V$, $\V^*$ and their tensor product spaces using components alone, so it works as a set of rules for checking that a move is valid even when no basis is in front of us.

The most famous moves are lowering and raising an index.

Here $g^{\mu\nu}$ is the inverse matrix of $g_{\mu\nu}$. (This explanation adds the standard definition and was not covered in the lecture.) Where the rule comes from shows up when we unfold it in a basis. Put the vector $A = A^\mu e_\mu$ into the second slot of the metric $g = g_{\alpha\beta}\, e^\alpha \otimes e^\beta$ and we get this.

The first slot is empty and waiting for a vector, so this is a covector, and its components are $A_\alpha = g_{\alpha\beta} A^\beta$. The metric is symmetric, so $g_{\mu\nu}A^\nu$ and $g_{\nu\mu}A^\nu$ have the same value, but it reads better to write the summed indices next to each other. Seen another way, $g_{\alpha\beta} A^\gamma$ is one tensor with three indices, and setting $\beta$ equal to $\gamma$ and summing is a contraction, so raising and lowering an index is a product followed by a contraction.

Slot order when multiplying

A product of components is also a tensor product once we unfold the basis.

Erasing the basis and keeping only $T^{\mu\nu} = A^\mu B^\nu$ is the classical notation. Components are real numbers, so the order of the factors does not matter, but the order of the slots in a tensor does.

The numbers in the two tables are the same, yet they are elements of different spaces. So we write the indices left to right in slot order, and when upper and lower indices mix we stagger them to keep the slots straight.

Q. If the numbers in the component table are the same, is it not the same tensor?
A. It is not the same tensor. In an element of $\V \otimes \V^*$ the first slot takes a covector and the second takes a vector, and in $\V^* \otimes \V$ the order is reversed. Even if we write the same table of numbers, what goes into which slot changes, so they are different maps.

Magnitude, and the symmetric and antisymmetric parts

The squared length of a vector $A$ comes from feeding $A$ to the metric twice.

After one contraction a single number remains. Finally, there is a notation that splits a tensor $F_{\alpha\beta}$ of type (0, 2) into a symmetric piece and an antisymmetric piece.

This does not make a tensor symmetric. It splits the tensor into two pieces, so holding only one piece throws away the information in the other. We need both pieces to recover the original tensor. Antisymmetric tensors get their own treatment later.

In the next post we turn to the realization of a vector space, which gives the basis vectors we have so far written only as symbols a concrete identity.