Tensor #9. Transformations of Tensor Product Spaces텐서 #9. 텐서 프로덕트 스페이스의 변환

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 9 of the playlist, TPS transformations. It applies the linear transformations of the previous post to tensor product spaces.

Two bases

The 4-dimensional real vector space $\V$ has a basis $e_\mu$, and the dual space $\V^*$ has the dual basis $e^\mu$. They are tied together by the rule below.

We picked this rule because it makes calculation easy. The position of an index also announces something in advance. A lower index signals that the object will transform covariantly, and an upper index signals that it will transform contravariantly. Now we add a second basis. Last time we called it $f_\mu$, but this time we put a hat on it and write $\hat e_\mu$, and we fix its dual basis $\hat e^\mu$ by the same rule.

How the basis changes

The link between the hatted and unhatted bases is an invertible matrix $\Lambda$.

The dual basis does not change with the same matrix but with its inverse.

For the rule $\map{\hat e^\mu}{\hat e_\nu} = \delta^\mu{}_\nu$ to hold in the hatted basis as well, it has to be the inverse. Computing directly gives $\map{\hat e^\mu}{\hat e_\nu} = (\Lambda^{-1})^\mu{}_\alpha\, \Lambda^\beta{}_\nu\, \delta^\alpha{}_\beta = (\Lambda^{-1}\Lambda)^\mu{}_\nu = \delta^\mu{}_\nu$. (This check is my own supplement to a lecture that only states the result, and was not covered in the lecture.)

The meaning of the terms is fixed here. The first thing we built is the basis vectors of the vector space, and the matrix that changes that basis is our starting point. What changes with that matrix is called covariant, and what changes with its inverse is called contravariant. That is the whole origin of the names.

Q. Does contravariant mean going in the opposite direction?
A. The direction is the same. In both cases we are talking about the path from the hatted basis to the unhatted basis, and what differs is the tool used on that path. A covariant object uses $\Lambda$ and a contravariant object uses $\Lambda^{-1}$. We just use the right tool for each object.

Components of vectors and covectors

The vector $a^\mu e_\mu$ can also be written in the hatted basis. The vector itself is one object, and the basis and the components change together.

The basis changes with $\Lambda$, so the components must move the opposite way for the sum of products to stay the same. Solving both sides gives $a^\nu = \Lambda^\nu{}_\mu \hat a^\mu$, so $\hat a^\mu = (\Lambda^{-1})^\mu{}_\nu\, a^\nu$. An element $b_\mu e^\mu$ of the dual space has a basis that changes with $\Lambda^{-1}$, so its components change as $\hat b_\nu = \Lambda^\mu{}_\nu\, b_\mu$. (This solution is my own working-out of what follows from the rules above, and was not covered in the lecture.) There is one thing to remember. An object with an upper index changes with the opposite matrix to an object with a lower index.

Components in a tensor product space

Now the product spaces. Let us pick $\V \otimes \V^* \otimes \V^*$. It is a (1, 2) tensor product space, where 1 is the number of vector spaces placed first and 2 is the number of dual spaces placed after them. The basis and an arbitrary tensor are written like this.

Writing the indices of the basis and of the components in the same order may look redundant, but there is a reason. In general relativity we hardly ever write the basis and carry only the components around, because the positions and order of the indices hold all the information about the basis.

Let us change the basis. We have $e_\mu = (\Lambda^{-1})^\alpha{}_\mu\, \hat e_\alpha$ and $e^\nu = \Lambda^\nu{}_\beta\, \hat e^\beta$. Substitute these for every basis vector. The tensor product is multilinear, so all the matrix entries, which are constants, can be pulled to the front.

What stands in front of the hatted basis is the new component.

The components receive one $\Lambda^{-1}$ for each upper index and one $\Lambda$ for each lower index. The basis vectors change in exactly the opposite way, and the two cancel each other, so the tensor itself does not change. Let us check this by hand, reduced to 2 dimensions.

Q. Does the tensor itself change when we change the basis?
A. It does not. However you choose $\Lambda$ in the toy above, the real number computed in the old basis and the one computed in the new basis agree. Only the numbers used to write the tensor change. (This explanation was checked with a 2-dimensional example and was not covered in the lecture.)

Covariant, contravariant and mixed

A tensor is usually named after its components, because the components practically define it. But the indices of the basis vectors are exactly opposite to those of the components, so remember that the names are given by the positions on the component side.

The $A^\mu{}_{\nu\lambda}$ we just saw has both upper and lower indices in its components. It changes neither fully covariantly nor fully contravariantly, so it is a mixed tensor. Now consider the (0, 3) tensor product space, built only from dual spaces. Its basis is $e^\mu \otimes e^\nu \otimes e^\lambda$ and its components are $A_{\mu\nu\lambda}$, with every index down. Such a tensor is called fully covariant. This map takes three vectors and returns a real number. In the opposite case, a tensor of the (3, 0) space, built only from the vector space, has components $A^{\mu\nu\lambda}$ and is a fully contravariant tensor. Check once more that the basis vectors change opposite to the name.

Different spaces

The (0, 3) space and the (3, 0) space are completely different vector spaces. Just as we cannot add vectors from different vector spaces, there is no relation between the two. All they share is that they started from the same $\V$, have the same dimension, and sit over the real numbers. For now there is no way to pair a vector of $\V$ with a map in $\V^*$, and building that pairing is the whole point of this game. We will meet the method when we get to the inner product and the metric tensor.

Q. If I lower the indices of $A^{\mu\nu\lambda}$, don’t I get $A_{\mu\nu\lambda}$ right away?
A. Not for now. Raising and lowering indices means moving between two different spaces, and that needs a device that connects them. The device is the metric tensor. Before it exists, we have no ground to treat an element of (0, 3) and an element of (3, 0) as the same thing.

Classifying by transformation

As we saw here, the names we give objects all come from how they change when the basis changes. Covariant, contravariant and mixed are examples. This is a pattern that keeps coming back in physics. When we change the coordinate system or perform a gauge transformation, the way things change turns out to be a core property of the most fundamental objects. Every elementary particle is classified by how it changes under Lorentz and Poincaré transformations. In general relativity the criterion is how things change under general coordinate transformations.

In the next post we add one function to the vector space that takes two vectors and returns a real number, which turns it into a space with an inner product.