Can we always rewrite a Tensor as a differential form?

[kay bei]

TL;DR

In the book Gravitation by Wheeler, they say any tensor can be completely antisymmetrized. Does this mean we can then convert any Tensor to differential form notation?

I read in the book Gravitation by Wheeler that "Any tensor can be completely symmetrized or antisymmetrized with an appropriate linear combination of itself and it's transpose (see page 83; also this is an exercise on page 86 Exercise 3.12).

And in Topology, Geometry and Physics by Michio Nakahara, on page 52 definition 5.4.1, it says a differential r-form (differential form of order r) is a totally antisymmetric tensor of type (0,r).

Putting these facts together, does this mean that we can always convert a tensor into a differential form by the following process; Tensor -> Antisymmetrize the Tensor -> Differential Form Notation.

Maybe I am reading this wrong and hopeful someone can clarify. I ask this question because many authors suggest that tensors are in fact more general than differential forms, but if you can always turn tensor into a differential form, I just see these as equivalent but different notations.

My follow-up question would be the following; given the Tensor form of the laws of physics, why would anyone want to rewrite them in the notation of differential forms anyway? I read that differential forms simplify higher dimensional calculations compared to tensors where we have to carry around all the indices, and they generalize many theorems such as stokes to higher dimensions, they unify and simplify some laws of physics such as Maxwell equations. Why then would we rather use differential forms as opposed to the Tensor case and would it just cause more headaches doing Tensor -> Antisymmetrize the Tensor -> Write in Differential Form notation.

I hope that someone with enough experience can clarify all of this. Very confusing. Thank you.

 

[anuttarasammyak]

Why then would we rather use differential forms as opposed to the Tensor case and would it just cause more headaches doing Tensor -> Antisymmetrize the Tensor -> Write in Differential Form notation.

I think the process of "Tensor -> Antisymmetrize the Tensor" is misleading. Any tensor is expressed as sum of symmetric and anti-symmetric tensors. Symmetric tensor component remains.

 

[Ibix]

Antisymmetrize the metric tensor and see if you can do anything useful with the result.

 

[martinbn]

The wording confuses you. When you antisymmetrize a tensor you obtain a new tensor that is antisymmetric. The two tensors are two different tensors unless the original one happened to be antisymmetric to begin with.

 

[Infrared]

Sure, you can anti-symmetrize, but I wouldn't really call the output a "rewriting" of the original tensor since you lose a lot of information by doing so. For example, if your original tensor is symmetric (like a metric), then you get zero.

Also, as your text notes, a differential form is an alternating tensor of type (0,n). You seem to be ignoring this last condition: You could have an alternating $(n,0)$ tensor for example, and it wouldn't be a differential form.

 

[zinq]

A tensor (algebraically) is a multilinear function from a product of vector spaces to scalars. It's not clear what "antisymmetrization" of a tensor T means unless the tensor takes just 2 inputs, i.e., is bilinear.

In that case, it can be written as the sum of its symmetrization and its antisymmetrization:

T = T_sym + T_antisym.

(Without additional information, neither T_sym nor T_antisym can be derived from the other.)

 

[Infrared]

It's not clear what "antisymmetrization" of a tensor T means unless the tensor takes just 2 inputs, i.e., is bilinear.

The generalization of "anti-symmetric" is alternating (meaning that if you swap any two inputs, you get a negative sign). As you probably know, the space of alternating $(0,n)$ tensors on $V$ is denoted by $\Lambda^n(V')$ (here $V'$ is the dual space of $V$).

There is a projection $\text{Alt}:(V')^{\otimes n}\to\Lambda^n(V')$ given by $\text{Alt}(T)(v_1,\ldots,v_n)=\frac{1}{n!}\sum_{\sigma\in S_n}(-1)^{\sigma}T(v_{\sigma(1)},\ldots,v_{\sigma(n)}),$ which generalizes the process of anti-symmetrization for arbitrary $(0,n)$ tensors.

 

[zinq]

Good point.

(But for more than 2 inputs, is there is a corresponding theorem about a tensor being equal to the sum of its symmetrization and its antisymmetrization?)

 

[Infrared]

Unfortunately not. By counting dimensions, one notices that the kernel of $\text{Alt}$ is larger than the space of symmetric tensors. For example, any tensor which is symmetric in just two arguments is in the kernel of $\text{Alt}$. I think it's a theorem in linear algebra that these tensors generate the kernel.