# Why are isotropic tensors not considered scalars?

In introductory textbooks (Griffiths, Shankar, Boas) a tensor is introduced as a mathematical objects which transform in a specific manner under changes of basis (i.e. changes of the coordinate system). Specifically, a 0-rank tensor (also called scalars) is independent of such changes, whereas any other $$n$$-rank tensor ($$n \neq 0$$) is not. Exceptions to this rule are the class of isotropic tensors (can however be of arbitrary rank), whose components are the same regardless of the choice/change in the coordinate system.

If isotropic tensors do not need to undergo any transformation when changing the coordinate system, why are they thereby not considered to be scalars (since the whole concept of a tensor of rank $$n \neq 0$$ is based on the concept of obeying certain tranformation rules)?

What am I missing here? (I know that by the quotient rule, the rank of these isotropic tensors can be confirmed to be non-zero.)

• I don’t see why a scalar isn’t an isotropic tensor of rank 0. – G. Smith Aug 20 '19 at 21:49
• I've never heard of this definition of an isotropic tensor. I suspect it's not widely recognized. By this definition, an isotropic tensor is not a tensor. – user4552 Aug 20 '19 at 22:44
• There are no tensors that will stay invariant under all coordinate transformations (except the trivial zero tensors). I suggest you misunderstood the definition you were given. – mmeent Aug 21 '19 at 6:59
• I think you mean they do not undergo any transformation when changing the coordinate system by an orthogonal transformation. But that's not the only kind of change of coordinates that's relevant. – leftaroundabout Aug 21 '19 at 9:49
• @leftaroundabout I think you should turn that comment into an answer, as it seems to be the root of the confusion. – mlk Aug 21 '19 at 10:12

From my point of view, you are conceptually equating “transform into themselves” with “don’t transform at all”. Isotropic tensors have the same transformation rule as non-isotropic tensors. You just don’t end up with anything different from what you started with.

• Isotropic tensors have the same transformation rule as non-isotropic tensors. This would contradict the definition given by the OP. – user4552 Aug 20 '19 at 22:46
• Perhaps he was taught a bad definition, or misunderstood what he was taught. – G. Smith Aug 20 '19 at 22:57

First, I would dispute the claim that the entire concept of tensors is centered around transformation rules. A tensor is a multilinear map on a vector space and it’s associated dual space (that is, a function which eats vectors and covectors); the transformation rules for its components simply reflect our demand that the action of a tensor on any arbitrary (co)vectors is independent of what basis we choose for the vector space.

In that sense, the fact that there exist (1,1)-tensors whose components are the same in any basis does not make those tensors scalars; they just look like scalars from a certain point of view.

More pragmatically, I don’t know how to add a tensor to a scalar. You could make up some rules like “a tensor plus a tensor is sometimes a tensor and sometimes a scalar depending on what the original two tensors were” but ... gross.

Ultimately, it’s possible to map all of the properties of invariant tensors (that is, tensors with invariant components) onto the properties of scalars and vice-versa. However, doing so would be both conceptually messy and utterly pointless.

Noting the similarities between certain tensors and scalars is fine, but merging the two concepts is not a good idea.

In the traditional definition of a tensor ("a tensor is something that transforms like a tensor"), you have to specify a set of transformation laws for the tensor. Frequently, we are most interested in rotations of coordinates, which have the nice property of preserving vector norms and inner products. In 3-D space, these transformations can be viewed as $$3 \times 3$$ matrices in the group $$SO(3)$$. Under this group of transformations, the Euclidean metric $$\delta_{ij}$$ is "a tensor that does not transform". In fact, it can't transform by definition: $$SO(3)$$ is defined to be the set of transformations which preserve inner products, and this turns out to be equivalent to saying that our transformation leaves $$\delta_{ij}$$ fixed.
But one could also consider more general coordinate changes that are not simply rotations of the axes: we could rescale our axes, for example, or even change to a "skewed" coordinate system where the coordinate axes aren't orthogonal. Such transformations are members of the group $$GL(3) \supset SO(3)$$, the group of all invertible $$3 \times 3$$ matrices. There are additional transformations in $$GL(3)$$ that do not leave the Euclidean metric $$\delta_{ij}$$ invariant, but the Euclidean metric still transforms "the right way" under the action of $$GL(3)$$ to be a rank-2 tensor.