Tensor #7. Rank of Tensor Product Spaces텐서 #7. 텐서 프로덕트 스페이스의 랭크

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 7 of the playlist, Rank of Tensor Product Spaces.

Recap

In the previous post we grew the tensor product space to as many factors as we like. Products of dual spaces, products of vector spaces, and mixed products are all maps defined in the same way, and since all of them go to the real numbers, there is little reason to separate them. And the vector space $\V$ and the dual space $\V^*$ themselves are degenerate tensor product spaces with just one factor. An element of $\V$ is a map from $\V^*$ to the reals, and an element of $\V^*$ is a map from $\V$ to the reals.

Writing down bases and components

Let me write down the basis and components for a few products of four factors. The product of four dual spaces has the basis $e^\mu \otimes e^\nu \otimes e^\delta \otimes e^\lambda$ with components $T_{\mu\nu\delta\lambda}$, and the product of four vector spaces has the basis $e_\mu \otimes e_\nu \otimes e_\delta \otimes e_\lambda$ with components $S^{\mu\nu\delta\lambda}$. Mixing, with two vectors followed by two dual vectors, the components are $R^{\mu\nu}{}_{\delta\lambda}$. The indices of the components follow the order of the basis vectors, and to keep the order unambiguous we keep the left-to-right order as well as the up and down positions. With a basis present the order is already fixed, but soon we will write only components, so we keep the order in them too.

Rank

Now we define the rank of a tensor. The convention is this: vectors always come first and dual vectors always come last in the product. It is fine to have no vectors or no dual vectors at all, but when the two are mixed, vectors come first. Only products arranged in this order are called tensors, and the rank is written as two numbers.

A vector is $(1, 0)$ and a dual vector is $(0, 1)$. $\V \otimes \V \otimes \V^* \otimes \V^*$, two vectors followed by two dual vectors, is $(2, 2)$, a product of four dual vectors is $(0, 4)$, and a product of four vectors is $(4, 0)$. We read a rank with the first number as the number of vectors and the second as the number of dual vectors. That reading relies on the convention that vectors come first.

The set of all tensors of a given rank is named by that rank. In this post I will write it $\mathcal{T}^{(p,q)}$ (this notation is my own choice here). $\V \otimes \V^*$ is $\mathcal{T}^{(1,1)}$ and $\V \otimes \V^* \otimes \V^*$ is $\mathcal{T}^{(1,2)}$.

What about a product with a dual vector first, such as $\V^* \otimes \V$? It is a well-defined multilinear map and it acts in the same way we know. But it is not called a tensor, and we cannot write its rank in the form above. I think the reason is that a rank only makes sense if the order follows a rule. That everything here is a multilinear map stays the same.

Q. Isn’t the rank the sum of the two numbers?
A. In this lecture the rank is the pair $(p, q)$. Many books call the sum $p + q$ the rank, so when reading a book you have to check which meaning it uses. (The remark that many books use the sum is added here and was not covered in the lecture.)

How a tensor acts on its inputs

Let us compute directly how the basis tensor $e_\mu \otimes e^\nu \otimes e^\lambda$ of $\V \otimes \V^* \otimes \V^*$ acts on inputs. This tensor takes an element $(\beta, v, w)$ of $\V^* \times \V \times \V$. The first slot takes a dual vector and the others take vectors. We expand the inputs in components.

The first piece of the product acts on the first slot, the second piece on the second, and the third piece on the third. Everything is linear, so the coefficients come to the front.

By the definition of the dual basis, each pairing is a delta symbol. Since $\map{e^\eta}{e_\mu} = \delta^\eta{}_\mu$, this simplifies as follows.

The deltas removed the indices one by one and left only a product of three real numbers. The delta is used both as an index and as the delta function, which is confusing, but the context tells which is which. Put numbers in and watch it yourself.

Computing without the basis

Doing the computation once shows that this always happens the same way. Deltas appear, the deltas change the indices, and the basis vectors disappear. If we had put in an arbitrary tensor $T^{\mu}{}_{\nu\lambda}$ rather than a basis tensor, its components would stand in front and the result would be this.

So we can write this expression right from the start without the basis of the tensor product space and without the bases of the input vectors. We arrive at the same place just by the knack of matching indices, never meeting a delta. Most general relativity textbooks proceed this way, and so you never see what the map is really doing.

Q. There are no basis vectors here. Is this calculation really the same value?
A. It is the same value. The calculation above is the solution with the basis present, and since the deltas always give the same result, matching only the indices skips the basis. The two values at the bottom of the toy above always agreeing is the evidence.

Q. Why do textbooks teach the knack first?
A. Because skipping the basis makes calculations shorter. In return, phrases such as “a mixed tensor acts on one covariant vector and two contravariant vectors” appear at once with terms not yet defined. In this series we will also define covariant and contravariant step by step.

In the next post we set tensors aside for a moment and look at linear transformations, which change the basis within a single vector space.