This answer elucidates the definition of a tensor density, which serves as a "generalization" of differential forms to extend the theory of integration to non-orientable manifolds. Just for the sake of completeness, let me quote (from the answer above) the definition of a tensor density here. A density of weight s is a map $\mu:V\times V\rightarrow\mathbb{R}$, where $V$ is a real vector space, such that for all linear operators $A:V\rightarrow V$, the following holds. $$\mu(AX_1,\dots,AX_n) = |\det A|^s\mu(X_1,\dots,X_n) \tag{1}\label{1}$$ with the obvious generalization to fields. On the other hand, there is another definition - more common in the Physics literature but also appearing on Ref. 1 - defining tensor densities based on their transformation law. For example, a scalar density $a$ of weight $s$ would transform, after a coordinate transformation with jacobian matrix $J$, as $$a=|\det J|^s a'\tag{2}\label{2}$$ or $$a=(\det J)^s a'\tag{3}\label{3}$$ the former if it's an even density and the latter if it's odd (or vice versa, the wiki article and Spivak seem to say different things, but that's irrelevant here). The primes denote the object in the new coordinates (\eqref{4} should make it clear) These are also called relative scalars and relative tensors. A notable example is the determinant of the metric tensor, which in two different coordinate charts is $$\det(g_{ij})=\det(\tilde{g}_{ij})\left[\det\frac{\partial\tilde{x}}{\partial x}\right]^2\tag{4}\label{4}$$

Now the question:

  1. Are these constructions different things sharing the same name or are they related?
  2. Given that $dx^1\wedge...\wedge dx^n=\det\left(\frac{\partial x}{\partial \tilde{x}}\right) d\tilde{x}^{1}\wedge...\wedge d\tilde{x}^n$. Define $d^n x:=dx^1\wedge...\wedge dx^n$ and $d^n\tilde{x}:=d\tilde{x}^{1}\wedge...\wedge d\tilde{x}^n$. Would $d^nx=\det\left(\frac{\partial x}{\partial \tilde{x}}\right) d^n\tilde{x}$ make it a density in the sense of \eqref{3}? Many Physics books (e.g. Carroll) assert $d^n x$ is a density for this reason$^1$.


$^1$ The idea behind would be that by $d^n x$ they mean an object that has this form in any coordinate frame (which for a differential form isn't the case for general coordinate transformations).


  1. A Comprehensive Introduction to Differential Geometry, Vol. 1, M. Spivak. Page 134.
  • $\begingroup$ I'd say they are related, with $\bar{x}^i=A^i_jx^j$ the relation is figuratively speaking $\bar{\mu}(\bar{x}^i)=\det (A^i_j)\mu(x^i)$. Since $A$ is any linear map we can use one in particular, $A^i_j=\frac{\partial \bar{x}^i}{\partial x^j }$. Finally with $J=\det (\frac{\partial\bar{x}^i }{\partial x^jl })$ $$\bar{\mu}(\bar{x}^i)=J\mu(x^j)$$ $\endgroup$ Commented Mar 30, 2023 at 1:31


You must log in to answer this question.

Browse other questions tagged .