# Naive question about tensor densities

Well, a (contravariant) tensor is an element of tensor product:

$$V\otimes W =: \frac{F(V\times W)}{S}$$

as exposed here: https://textosdefisica.wordpress.com/textos-de-matematica/

A antisymmetric tensor is a element of de Wedge space:

$$\Lambda (V) =: \frac{T(V)}{I}$$

as exposed here: https://en.wikipedia.org/wiki/Exterior_algebra

Futhermore, consider then the following:

As exposed in Schrödinger's book called Space-time structure (pages 14-16), the definition of the mathematical object called tensor density comes inside a problem of integration of a scalar field $$A$$:

$$\int A dx^{4} = \int A \left\vert \frac{\partial x^{k}}{\partial x'^{i}} \right\vert dx'^{4} \neq \int A dx'^{4}$$

Where $$\Big \lvert \frac{\partial x^{k}}{\partial x'^{i}} \Big \rvert$$ is the determinant of jacobian transfomation matrix $$\frac{\partial x^{k}}{\partial x'^{i}}$$.

Now, to the integral holds as:

$$\int A dx^{4} = \int A' dx'^{4}$$

We must consider a new transformation for scalar fields:

$$A' = \left\vert \frac{\partial x^{k}}{\partial x'^{i}} \right\vert A \tag{1}$$

So $$(1)$$ is the definition of a scalar density, and then we generalize to tensors in a "natural" way:

$$D'^{ml} = \left\vert \frac{\partial x^{k}}{\partial x'^{i}} \right\vert\frac{\partial x'^{m}}{\partial x^{g}} \frac{\partial x'^{l}}{\partial x^{h}}D^{gh} \tag{2}$$

Then $$(2)$$ is a tensor density. My question is, how can I formalize this object with multilinear algebra? (consider only finite dimensional real vector spaces).

• Isn't the change of coordinates formula for a tensor field? A tensor itself is just an algebraic object? So I think you want information about a tensor density field. – Emil Jan 26 '19 at 7:40

If you know about tensor products and alternating (antisymmetric) products, as you seem to do, you don't need the concept of a tensor density. In physics texts, typically mention of quotient spaces is avoided, so tensors are defined as multilinear maps, (anti)symmetric tensors are defined as (anti)symmetric maps, etc. Since we are working over a field this works quite well (every subspace is a direct summand), but there are some places where the notation would get overly complex.

If we have an $$n$$-dimensional vector space, then its $$n$$-th exterior power is 1-dimensional. If this space is the tangent space of an $$n$$-dimensional manifold, elements of the $$n$$-th exterior power are the $$n$$-forms. If the manifold is orientable, there is a global nowhere zero $$n$$-form, which can be taken as a volume form. If the manifold has a metric, there is a canonical choice.

Now let a function (i.e. a scalar field) be given. Strictly speaking, we cannot integrate functions over a manifold, only $$n$$-forms. We can however multiply the $$n$$-form by the function, and integrate that. If we make a change of coordinates, we get a new function. If we integrate this in the same way, we get a different value. However, if instead of the function we transform the $$n$$-form that is the product of the function with the volume form, things work out.

Instead of saying that $$A$$ is a totally antisymmetric $$\binom0n$$-tensor, which would give it $$n$$ indices and a complicated looking transformation rule that works out to just being the determinant, a new quantity is introduced that is called a tensor density, and that manifestly only has a single component and a transformation rule that looks easy because we already know about determinants.

Your generalization is not really anything new either, if I am not mistaken that would be an ordinary tensor of type $$\binom 2n$$ that is antisymmetric in the lower indices.

• Could you please consider to give me an introductory text about integration,tensors and manifolds? – M.N.Raia Jan 26 '19 at 12:56
• This just sounds wrong. You seem to believe that a tensor density is a totally antisymmetric tensor with a certain valence, but that's not true. A tensor density is not a tensor. It has different transformation properties. – Ben Crowell Jan 26 '19 at 16:08
• @BenCrowell What I am saying is that a scalar density of is a totally antisymmetric tensor of valence $\binom 0n$, both sections of rank 1 vector bundles: the first one of the space of functions, the second of the top exterior power of the cotangent bundle, sometimes denoted $\Omega^n_X$ or $\omega_X$ if $X$ is the $n$-manifold and sometimes called the determinant bundle). – doetoe Jan 26 '19 at 17:35
• An ordinary scalar would be a $\binom00$ tensor. Likewise for $\binom kn$ tensor densities, which are products (sections of the tensor product bundle) of $\binom k0$ tensors and a section of the determinant bundle. Tensor densities of higher weight are sections of a bundle with more factors of $\omega_X$. – doetoe Jan 26 '19 at 17:35
• @BenCrowell If you don't believe me I would suggest you work out how $e_1\wedge\cdots\wedge e_n$ transforms under a matrix $A$ – doetoe Jan 26 '19 at 17:38