Tensor #31. Tensor Densities텐서 #31. 텐서 밀도

These are my notes on XylyXylyX’s What is Tensor lectures. This post covers lecture 31 of the playlist, Tensor Densities (Part 2 of Tensor Transformations). In the previous post we saw that the components of an $(n-1)$-form change like a vector with one extra factor of a determinant. This time we give objects with such a rule the name tensor density, define it, and follow where it lives.

The classical definition: the weight w

We first define it the classical way. In this approach a tensor is a collection of numbers whose components change by a fixed rule. If components $\mathcal{T}$ with $k$ upper indices and $q$ lower indices change like this for an integer $w$ when the basis is changed by $\Lambda$, we call it a $(k, q)$ tensor density of weight $w$.

This is the usual tensor rule, with upper indices changing by the inverse and lower indices by $\Lambda$, multiplied by the determinant of the transformation matrix to the power $w$. The determinant is a single number, so it does not touch the indices and multiplies the whole thing. For a $(1, 1)$ tensor density alone it reads as follows.

The weight can be a positive, zero or negative integer. If $w = 0$ the extra factor is 1 and the rule equals that of an ordinary tensor, so an ordinary tensor is a tensor density of weight 0. Some textbooks call the case $w < 0$ a tensor capacity, but the term is not widely standardized. The $V^k$ we built last time is the first example, a $(1, 0)$ tensor density with $w = 1$.

Q. If the weight is 0, is there a need to call it a density?
A. No. But the definition contains ordinary tensors as one case, so fixing just the weight lets us treat both in the same frame.

What the component-only definition leaves unsaid

In this series we have treated a tensor not as components alone but as a pair of a basis and components. When the components change by $\Lambda^{-1}$ the basis changes by $\Lambda$ and the product is 1, so the tensor itself does not change. The classical definition erases the basis and keeps only the rule for the components. But the rule for a density has a factor $(\dL)^{w}$ that does not multiply out to 1. So a question arises. Which space is an object whose components change this way an element of, and with what basis is it paired? If we only memorize the rule, the determinant looks stuck on at random, so the lecture answers by constructing the basis directly.

Multiplying in a top-degree wedge basis

Fix $n = 3$ and look at an object obtained by attaching, after the basis $\partial_\mu \otimes dx^\nu$ of a $(1, 1)$ tensor, one more top-degree wedge basis through a tensor product.

Since $n = 3$, the space of 3-forms is 1-dimensional, and its strict basis is the single element $dx^1 \wedge dx^2 \wedge dx^3$. This expression is an element of $V \otimes V^* \otimes \Lambda^3 V^*$. A tensor product can be made from any number of spaces, so this is a perfectly fine object. An element of a 1-dimensional space has a single number as its whole coefficient, so that coefficient can be multiplied into $K^\mu{}_\nu$ and absorbed. Attaching a few more wedge bases does not add indices to the components. The shape of the indices stays $(1, 1)$.

Q. Why does multiplying in a 1-dimensional space not change the rank?
A. That space has only one basis element, so no index is attached to the components. A single coefficient is just multiplied in front of the other components. But the basis is different, so this object does not belong to the space of ordinary $(1, 1)$ tensors.

Changing the basis brings out a determinant

The components and the $\partial_\mu \otimes dx^\nu$ part go as always. The only new thing is the last wedge basis. The covector basis changes as $dx^i = \Lambda^i{}_{j'} \, dx'^j$, so the top-degree wedge product becomes this.

The wedge product is antisymmetric, so only the terms with all different numbers survive, and the sum over the reordered terms is just the definition of the determinant. The lecture brings in the same fact as the rule that applying a linear transformation to the basis covectors one by one pulls the determinant to the front. The object $\mathcal{K}$ itself stays the same when the basis changes, so the components must take the $\dL$ that came from the basis. If we multiplied in $a$ wedge bases, we get this.

This has the same form as the definition in the first section, so $K$ is a tensor density of weight $a$. The lecture computes an example with the same wedge basis multiplied in twice, and shows that the factor becomes $(\dL)^2$ and the weight is 2. The odd factor in the definition is the determinant of the basis left over on the component side.

In $n = 2$, feeding the new basis vectors $\partial'_1, \partial'_2$ to $dx^1 \wedge dx^2$ gives $\dL$, which is the signed area of the parallelogram the two vectors make in the old coordinates. The new basis element $dx'^1 \wedge dx'^2$ gives 1 on the same input, so $dx^1 \wedge dx^2 = \dL \; dx'^1 \wedge dx'^2$. (This interpretation is added for understanding and was not covered in the lecture.) In the toy below, drag the two new basis vectors and check this factor yourself.

The weight is counted by number

Multiplying in a basis on the vector side brings out the opposite factor. The vector basis changes as $\partial_i = (\Lambda^{-1})^{j'}{}_i \, \partial'_j$, so the top-degree wedge product on the vector side is this.

So multiplying in the wedge basis of an $n$-vector gives the factor $(\dL)^{-1}$ and lowers the weight by 1. If we multiply in $a$ bases of $n$-forms and $b$ bases of $n$-vectors, the weight is as follows.

If we multiply in neither, $w = 0$ and it is an ordinary tensor. By adjusting how many we multiply in, we can make any integer weight. This works the same whether the dimension is 3 or 10. There is one condition, though. The rule that the determinant comes to the front holds only for the top-degree wedge product, and for a lower degree the minors of the previous post appear instead of the determinant.

Q. What happens if we multiply in one $n$-form basis and one $n$-vector basis together?
A. The weight is $1 - 1 = 0$ and it changes by the same rule as an ordinary tensor. The two factors $\dL$ and $(\dL)^{-1}$ cancel each other.

The strict basis makes the density

If we expand a wedge basis into the tensor product basis, it is a signed sum over all reorderings of the numbers. For $n = 3$ there are six terms.

Expanded this way, $\mathcal{K}$ is an ordinary tensor with $n$ more indices, and its completely antisymmetric components change by the ordinary tensor rule with no determinant. After the transformation, when we fold the antisymmetric terms back into a single strict basis element, the determinant finally appears in front of the coefficient $K$. So a density arises from using one strict basis element and looking at the coefficient in front of it alone.

The loose basis adds over all sequences of numbers and puts $1/n!$ in front, as in $\tfrac{1}{3!} \, \omega_{\alpha\beta\gamma} \, dx^\alpha \wedge dx^\beta \wedge dx^\gamma$. In that case the components change as an ordinary tensor with all its indices, and no density factor shows up separately. The lecture sums this difference up as the true nature of a tensor density.

The space of densities and the next post

The sum of two tensor densities of the same weight is a tensor density of that weight, and there are a zero density and additive inverses, so the densities of a fixed weight form a vector space. The lecture says there is no standard symbol yet for this space. Densities of different weights do not change by the same rule, so we do not add them. (This explanation follows directly from the transformation rules above and was not covered in the lecture.)

The lecture closes by previewing where this definition is used. The Levi-Civita symbol and the determinant of the metric are tensor densities, and they are central to the theory of integration on oriented manifolds. In the next post we check, as examples, how the Levi-Civita symbol and the determinant of the metric change as densities.