In an arbitrary number of dimensions, one can naturally define two tensors, Kronecker delta and Levi-Civita epsilon tensor. However, why isn't it advantageous to define some totally symmetric tensor as well? Is there an intuitive reason for that or is it basically related to differential forms?

  • $\begingroup$ What do you expect from a 'totally symmetric tensor' that would not be an attribute of a plain old symmetric tensor? As user41431 says, the Kronecker delta seems to be what you want anyway. $\endgroup$ – Arthur Suvorov Jun 9 '14 at 5:30
  • $\begingroup$ Kronecker delta and Levi-Civita "epsilon tensor" are not invariant tensors, I mean that they take different components changing basis. Or, perhaps, are you considering Cartesian tensors? I.e. only orthogonal transformations between bases are allowed... $\endgroup$ – Valter Moretti Jun 9 '14 at 11:40

The Kronecker delta and the Levi-Civita tensors are very different kinds of tensors.

The Kronecker delta is, abstractly, a linear map: the identity map. That's all it is and all it can be.

The Levi-Civita, however, can be interpreted geometrically: as an oriented volumetric subspace (or four-volume or what-have-you).

What do I mean by "oriented"? Think of the right-hand rule in 3d. We say that a right-handed coordinate system is somehow oriented differently from a left-handed one. In fact, choice of orientation has nothing to do with the arrangement of the axes---you could use a "left-handed" orientation even with right-handed axes. This merely reflects an arbitrary choice that goes on top of the usual structure of the vector space.

An easier example might to think of a sheet of paper, a flat plane. Draw a spiral on said sheet of paper. You can turn the paper over, and the spiral runs the opposite direction now, even though the overall plane might not change. These are two different orientations of the same planar subspace.

A general term for tensors that represent subspaces is a blade. Vectors are blades: each vector is associated with a line-like subspace. Two vectors that aren't parallel can define a 2-blade, representing a planar subspace. Those vectors define an orientation by their ordering, and swapping the vectors reverses, or negates, the orientation. This is where the nature of blades being antisymmetric tensors comes into play.

It's important to note that linear combinations of 2-blades, called bivectors, are important geometric objects in relativity. These combinations may no longer be thought of as representing subspaces--for instance, how would you see the $xy$-plane and the $tz$-plane put together as a subspace?--but they are still of great physical significance. The EM tensor is a bivector. The Riemann tensor is a linear map from bivectors to bivectors (those bivectors are often broken down into plades, to talk about the planes in which covariant derivatives are considered, for whether they do or do not commute and in what planes they get distorted or changed to).

The highest dimensional blades in a space are often called pseudoscalars. You might have heard that the magnetic scalar potential is really a pseudoscalar. The Levi-Civita is really a pseudoscalar. The action of a linear map can be characterized by feeding it a pseudoscalar instead; geometrically, this is interpreted as the amount by which a volume is dilated or shrunk. The volume might also change orientation (change sign). This is the determinant of a linear map. Cool, huh?

$k$-blades and $k$-vectors are very important kinds of tensors to understand. They are antisymmetric to capture the idea that interchanging vectors changes the orientation of the underlying geometric object. There are no corresponding geometric analogues for symmetric tensors that are symmetric across many indices. Most of the geometry is covered with blades and multivectors.

  • $\begingroup$ Well, if consider the Kronecker delta as a symmetric tensor it is not the identity, since it transforms contravariant vectors to covariant ones. $\endgroup$ – Valter Moretti Jun 9 '14 at 9:55
  • $\begingroup$ @V.Moretti When both indices are down, yes. But in when there is a metric, we need not view vectors and covectors differently. Converting between contravariant and covariant components does not change the geometric quantity, so you see, the metric itself plays the role of the identity. $\endgroup$ – Muphrid Jun 9 '14 at 13:25
  • $\begingroup$ So, in other words, you are saying that $c\delta_{ij}$ is the metric in canonical form. It however does not work for metrics with different signatures, like Minkowski metric. $\endgroup$ – Valter Moretti Jun 9 '14 at 13:32
  • $\begingroup$ @V.Moretti Yes, I'm aware. My point is that the Kronecker delta, or the Minkowski metric, or the non-flat general (pseudo-)Riemannian metric should all be viewed as different aspects of the same basic concept: the identity map. Which one you use will of course depend on the signature of the space and the coordinates and so on, but I think of them all as conceptually related by this notion. $\endgroup$ – Muphrid Jun 9 '14 at 13:48

Isn't the Kronecker delta symmetric? Let's see:

$$\delta_{ij}=1 \, \, \, \text{if} \, \, \, i=j, \quad \delta_{ij}=0$$ otherwise. I'm pretty sure that is symmetric. The Levi-Cevita symbol is indeed anti-symmetric though.


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