Consider a tensor of arbitrary rank (2 for this case) $A_{ij}$, and dimension one. Granted there are two indices to specify a component, but since each index can only take one value, there is only one component in this entire tensor: $A_{11}$. So, are all one dimensional tensors scalars?
Further. transformation under coordinate transform for this case: $$(A')^{11}={\left (\frac{\partial x'}{\partial x}\right )}^2A^{11}$$ suggests that since in general $(A')^{11}$ is not equal to $A^{11}$, it is not a scalar.

So what exactly is this non-scalar one component object?


2 Answers 2


Perhaps an example is in order. Consider e.g. a 1D charge density $\rho$ in a 1D world. It transforms as a covariant (0,1) tensor $\rho^{\prime}=\frac{\partial x}{\partial x^{\prime}}\rho,$ so it is not a scalar.

  • $\begingroup$ I understand the transformation rule mathematically, and also showed it in the question, but the idea that charge density (or any other pure number) not being a scalar is odd. As suggested in the last line of my question is there any intuitive way to understand what that is? $\endgroup$ Feb 27, 2022 at 9:28
  • $\begingroup$ @GreasyToothBee If you change the scale of your coordinate, the density (defined with respect to that coordinate) will change. $\endgroup$
    – Javier
    Feb 27, 2022 at 13:32
  • $\begingroup$ @GreasyToothBee: When you fix a metric on your 1d manifold then antisymmetric covariant 1-tensors are identified with antisymmetric 0-tensors, these are just scalars. $\endgroup$ Mar 3, 2022 at 6:40
  • $\begingroup$ @Javier: But this is true of a scalar density! $\endgroup$ Mar 3, 2022 at 6:43
  • $\begingroup$ Given that 1d manifolds are always flat this means we can equip them with a metric and then antisymmetric covariant 1-tensors are canonically identified with antisymmetric covariant 0-tensors, aka scalar fields ... $\endgroup$ Mar 3, 2022 at 6:45

Tensors of rank $k$ over a vector space $V$ form the space $\otimes^k V$. When the vector space is 1d, then we may as well take it to be the ground field $\mathbb{K}$.

But $\otimes^k \mathbb{K} \simeq \mathbb{K}$.

A tensor is an element of the left side, so equivalently it is also an element of the right side and so it is just a scalar.

So, yes, it's true. Alternatively:

Fix a 1d connected manifold without boundary $M$. Now, there are only two such manifolds: the circle and the infinite line. They are both orientated and flat. That they are orientated means that they have a volume form (this is a nowhere vanishing top form) and that they are flat means that they are isometric to the standard circle or Euclidean line. So we may as well choose this metric.

Since the manifold $M$ is 1d, the forms of rank 1 are exactly the top forms. There are no higher rank forms, they vanish by antisymmetry. Now choosing a top form, say:

$\omega \in \Omega^1M$

This is not a scalar form, however because of the flatness of 1d manifolds, there is a standard metric and we can use the Hodge star to convert to a scalar field:

$*\omega \in \Omega^0 M \simeq C^{\infty}M$

In brief:

$\Omega^1 M \simeq \Omega^0 M \simeq C^{\infty} M$.

This resolves the paradox of a 1d charge density that is not a scalar field as referred by @QMechanic in another post. That is not the full picture because it is also correct that it is canonically a scalar field.

However, differential forms are not the only tensors on a 1d manifold, they are only the antisymmetric covariant tensors. However, the argument in the first paragraph shows that they are also isomorphic to scalar fields.

  • $\begingroup$ But a scalar should be invariant under coordinate transform, but the transformation equation given in my question suggests otherwise. $\endgroup$ Feb 27, 2022 at 12:20
  • $\begingroup$ Sure, scalars are invariant which suggests theres a mistake in your transformation equation. The definition I gave for a tensor is an invariant definition, and it gives the covariant definition common in physics books by taking coordinates. Anything that follows from an invariant definition is also invariant. $\endgroup$ Feb 27, 2022 at 12:30
  • $\begingroup$ The spaces are isomorphic (because they are all 1-dimensional) but the isomorphism is not natural, meaning that the space of vectors (to take a simple example) doesn't have a preferred basis. This means that looking at how things behave under a change of basis is important. $\endgroup$
    – Javier
    Feb 27, 2022 at 13:36
  • $\begingroup$ @Javier: Yeah, sure. But you can make the isomorphism natural by considering framed vector spaces. Nevertheless, the distinction isn't important here as the OP was simply asking whether tensors of arbitrary rank on a 1d space were scalars. What I've written above shows that. QED. $\endgroup$ Feb 27, 2022 at 15:48
  • $\begingroup$ @Javier: Alternatively, since a 1d manifold is always flat, it has a metric, then we have a Hodge star and so by its use: $\Omega^1 M \simeq \Omega^0 \simeq C^{\infty} M$. $\endgroup$ Mar 2, 2022 at 17:10

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