Tensor #6. Tensor Product Spaces텐서 #6. 텐서 프로덕트 스페이스

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 6 of the playlist, Tensor Product Spaces.

What we built last time

In the previous post we built one space. We took two dual vectors from the dual space $\V^*$, formed $\beta \otimes \alpha$, and put every possible combination into one set, $\V^* \otimes \V^*$. It has addition and multiplication by real numbers, and it also has opposite elements and a zero element, so this set is a vector space. Therefore a tensor is a vector. Tensors live in tensor product spaces, tensor product spaces are vector spaces, and every element of a vector space is a vector. (Later the word vector is also used in a narrower sense for the elements of $\V$ only, and a vector in that sense is one kind of tensor.)

Multiply more of them

The generalization is surprisingly easy. There is no reason to multiply $\V$ into the Cartesian product only twice. We define maps from a product of three or four copies to the real numbers in the same way.

The collection of these maps, $\V^* \otimes \V^* \otimes \V^*$, is a vector space exactly as before, and so it is for any number of factors. The basis is built the same way too: one dual basis vector per slot. With four factors it has the form $e^\mu \otimes e^\nu \otimes e^\lambda \otimes e^\sigma$, and in dimension 4 there are $4^4 = 256$ such tensors. With $k$ factors and dimension $n$ there are $n^k$ basis tensors. (The general formula $n^k$ is added for this explanation and was not covered in the lecture.) An arbitrary element is written like this.

All indices of the components $T_{\mu\nu\lambda\sigma}$ are down. The basis $e^\mu$ has an upper index, so for the Einstein summation convention to work the components must be lower. This sum holds both properties of a vector space. Summing is the addition, and the real-number component in front of each term is the scalar multiplication.

The vector side has the same structure

As we saw earlier, a vector $v$ is a map that takes a dual vector and gives a real number. So we can build maps from something like $\V^* \times \V^*$ to the real numbers, and the material for such maps is $\V \otimes \V$, which ties two vectors together. The logic is the same. It is the set of all possible tensor products, and this set is a tensor product space too.

The basis is $e_\mu \otimes e_\nu$, and this time the basis index is down, so the component indices go up.

The dual side has lower-index components and the vector side has upper-index components. The two sides are fully symmetric, and neither is less legitimate.

Mixing them

The Cartesian product multiplies any sets at all. The idea that we must multiply the same set again and again is only a habit; the definition used plain sets $A$ and $B$. So we may mix the vector space and the dual space. For example $\V \times \V^*$ holds ordered pairs of one vector and one dual vector, and the space of maps from it to the real numbers is $\V^* \otimes \V$. The basis tensor $e^\mu \otimes e_\nu$ takes a vector in the first slot and a dual vector in the second. Applied to a vector $a^\alpha e_\alpha$ and a dual vector $b_\beta e^\beta$ it gives this.

The same rule repeats in every slot. A slot holding a dual vector takes a vector, and a slot holding a vector takes a dual vector. The order and kinds of the factors can be mixed freely. Assemble one yourself in the toy below.

Q. Are $\V^* \otimes \V$ and $\V \otimes \V^*$ the same space?
A. No. The Cartesian products they take, $\V \times \V^*$ and $\V^* \times \V$, are different, because the order of the slots carries meaning in an ordered pair. It is the same reason we gave last time for $A \times B$ and $B \times A$ being different sets.

The order of the indices matters

When we write a basis tensor, the order of the slots is already visible. In a product like $e^\mu \otimes e_\nu \otimes e^\lambda$ we can see at a glance that the first is dual, the second is a vector, and the third is dual. So the indices of the component $T$ only have to say which basis they attach to, and even if the index order got mixed up, nothing would be confusing while the basis stands beside it.

But classical texts do not carry the basis around. General relativity and its relatives carry almost only the components and use separate rules for handling the indices. Everything the basis does is already contained in the components, so the basis begins to look redundant. For those rules to work, the components must keep the order of their indices by themselves. That is why the components of such a tensor are written with both the position and the order, as in $T_{\mu}{}^{\nu}{}_{\lambda}$.

Q. The basis already has an order, so why must the component indices keep it as well?
A. While we use the basis we do not need it, but this series is heading toward dropping the basis and working with components alone. By then the indices are the only thing that tells the order, so we have to write them in order from the start.

The spaces $\V$ and $\V^*$ are tensor product spaces too

The case of a single factor fits in here as well. A vector is a map that takes one dual vector and gives a real number, and a dual vector is a map that takes one vector and gives a real number. These come from Cartesian products with only one slot in each ordered pair, so they are the simplest tensor product spaces. The number of spaces multiplied is called the rank, and vectors and dual vectors are tensors of rank 1. The rank of a mixed tensor needs a little more explanation, so I leave it for the next post.

In the next post we define rank properly and compute how such tensors act on a pair of vectors or of dual vectors.