Differential Forms or Tensors for Theoretical Physics Today

[giulio_hep]

i don’t understand. Could you maybe provide an example.Maybe what the author was trying to say was a linear combination of a symmetric and antisymmetric form of itself?

An example from quantum physics: for the canonical symplectic 2-form in the cotangent bundle of the configuration space Q, the minus in the permutation sign for the anti-symmetry comes from the minus sign in the Hamilton equation.

 

[Ibix]

i don’t understand. Could you maybe provide an example.Maybe what the author was trying to say was a linear combination of a symmetric and antisymmetric form of itself?

Consider the Minkowski metric tensor, $\mathrm{diag}(1,-1,-1,-1)$. Antisymmetrize it (you can always do this for any tensor, as MTW say). Can the result possibly contain the same information as the original tensor? Antisymmetrize any other symmetric tensor. You should get the same result - so antisymmetrization is not invertible in general so, again, the result cannot contain the same information as the original.

 

[giulio_hep]

The differentiable structure on a manifold is not enough to define the so called covariant derivative on an arbitrary tensor. But on forms, this totally antisymmetric tensors, there is a notion of exterior derivative that can be defined just in terms of the differentiable structure of the manifold.

 

[weirdoguy]

since any tensor can be antisymmetrized

Why do you think that antisymmetrization of a tensor will give us exactly the same tensor? Why do you expect to do all those transformations and get exactly the same thing? What would be the point in doing them in the first place?

 

[robphy]

Gravitation by Wheeler, on page 83 they say "Any tensor can be symmetrized or antisymmetrized by constructing an appropriate linear combination of itself and it's transposes"

means that one can form the "totally-symmetric part" of a tensor and the "totally-antisymmetic part" of a tensor (not unlike finding the real-part of a complex number and the imaginary-part of a complex number).

For example,
$M_{ab} =\displaystyle \frac{(M_{ab} +M_{ba})}{2} + \frac{(M_{ab} -M_{ba})}{2} = M_{(ab)} + M_{[ab]}$,
where $M_{(ab)}$ is the symmetric-part and $M_{[ab]}$ is the antisymmetric-part.

If $M_{ab}=M_{(ab)}$ (or equivalently, $M_{[ab]}=0$), then $M_{ab}$ is said to be symmetric.

If $M_{ab}=M_{[ab]}$ (or equivalently, $M_{(ab)}=0$), then $M_{ab}$ is said to be antisymmetric.

But in general, since $M_{ab}=M_{(ab)} + M_{[ab]}$,
the general tensor $M_{ab}$ is neither symmetric nor antisymmetric.

 

[PrecPoint]

The differentiable structure on a manifold is not enough to define the so called covariant derivative on an arbitrary tensor. But on forms, this totally antisymmetric tensors, there is a notion of exterior derivative that can be defined just in terms of the differentiable structure of the manifold.

I don't understand, the exterior derivate still needs an affine connection?

 

[weirdoguy]

the exterior derivate still needs an affine connection?

No, it doesn't.

 

[PrecPoint]

On any differentiable manifold of positive dimension there are infinitely many affine connections. It should be possible to connect nearby tangent spaces, so it permits tangent vector fields to be differentiated as if they were functions on the manifold with values in a fixed vector space. A differentiable manifold Xn with endowed with a connection is an affinely connected space.

Can you please give an example? What kind of differentiable manifold does not allow this?

 

[weirdoguy]

what kind of manifold are you talking about?

Who and where?

 

[PrecPoint]

Your claim is that you can construct a differentiable manifold Xn that won't allow an affine connection (affinely connected space). Please give an example of such a differentiable manifold.

 

[weirdoguy]

Your claim is that you can construct a differentiable manifold Xn that won't allow an affine connection (affinely connected space).

Um, where did I claim such thing? You asked if exterior derivative needs an affine connection. The answer is no. That's all. I don't know what you are talking about in your last post.

 

[PrecPoint]

What do you mean when you say that the exterior derivative does not need an affine connection in the same way the ordinary covariant derivative does?

Lets be concrete. The covariant exterior derivative of the 2-form $\Pi^j= A_{hk}^j\mathrm{d}x^h\wedge \mathrm{d}x^k$ is (with the usual 1-form $\omega_h^j=\Gamma_{hl}^j\mathrm{d}x^l$ )

$$D\Pi^j=\mathrm{d}\Pi^j+\Pi^h\wedge\omega^j_h=\mathrm{d}\Pi^j+\omega_h^j\wedge\Pi^h$$

Please explain how you make $$\Gamma_{hj}^j$$ go away without actually using the exact same topological property of the differentiable manifold Xn in some other way (ie the affine connection).

 

[weirdoguy]

Lets be concrete.

Ok. Let's define exterior derivative:

Exterior derivative is a linear operator $d$, that takes $p$-forms to $p+1$-forms, fulfilling following conditions:
1. Acting on $0$-forms, ie functions, it gives us: $df\stackrel{df.}{=}\frac{\partial f}{\partial x^i}dx^i$.
2. For $k$-form $\alpha$ and $l$-form $\beta$ we have $d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^k\alpha\wedge d\beta$.
3. $d(d\alpha)=0$ for every form $\alpha$.

It can be shown that such operator exists and that this definition is unambiguous. Nowhere in this definition connection appears.

What do you mean when you say that the exterior derivative does not need an affine connection in the same way the ordinary covariant derivative does?

When I say that I mean that the very definition of exterior derivative does not use any additional structures on a manifold, including connection. And with the covariant derivative: it can be shown that existence of covariant derivative operator is equivalent to the existence of connection on a manifold (vector bundle in general). But again, both connection and covariant derivative can be defined without talking about exterior derivative. I have the definition of covariant derivative of a section of vector bundle at hand, but I'll omit it right now.

of the 2-form

It looks like a vector valued 2-form, not 2-form. If that is the definition of covariant derivative you know, then that's ok, but there are definitions that not make any use of exterior derivative.

 

[PrecPoint]

It looks like a vector valued 2-form, not 2-form.

It's a contravariant 2-form and the coefficients of $A_{hj}^k$ represents the components of a type (1,2) tensor field.

I have a hard time pinning down what parts of your answer actually addressed my question or how you used my concrete example.

Is your point that the covariant exterior derivative of the contravariant 2-form on a differentiable manifold endowed with a connection specified above is wrong?

 

[weirdoguy]

covariant exterior derivative

I guess that this might be the problem: covariant exterior derivative is something different than covariant derivative. We (me and @giulio_hep) were not talking about covariant exterior derivative, but covariant and exterior derivatives separately. Both of these can be defined without mentioning the other.

 

[PrecPoint]

Uh, ok, Let me be even more specific.

By "covariant derivative" on the differentiable manifold $X_n$ I assume giulio_hep mean stuff like the covariant derivative of a contravariant vector field $X^p(x^k)$ is the (1,1) tensor field

$$X^p_{|k}=\frac{\partial X^p}{\partial x^k}+\Gamma^p_{hk}X^h$$

We have not yet imposed a metric or anything like that. But we have used one additional assumption, namely that the differentiable manifold $X_n$ is endowed with an affine connection (notice the $\Gamma^p_{hk}$)

Now, the interesting question is. Can we calculate a covariant exterior derivative without using the same (or a topologically equivalent) assumption about $X_n$?

If so, please explain how using my example of a random contravariant 2-form above (so we don't get stuck in definitions).

 

[weirdoguy]

Can we calculate a covariant exterior derivative without using the same (or a topologically equivalent) assumption about ?

No. Citing wikipedia:

the exterior covariant derivative is an analog of an exterior derivative that takes into account the presence of a connection

But we can calculate exterior derivative without any connection, and that was what we were talking about in the very beginnig.

 

[giulio_hep]

Yes it is the exterior derivative of forms (= totally antisymmetric tensors ) that does not need any connection but can be defined just in terms of the differentiable structure of the manifold. This is what we are repeating in so many comments. The definition in terms of axioms has been given by @weirdoguy at #43 or you can find also the definition via the commutator, for example see this paragraph.

 

[PrecPoint]

Ok, assumed you were talking about the covariant exterior derivative. In practice, one is confronted with p-forms which are not scalars. It never occurred to me that you meant to compare $\mathrm{d}\omega$ to a covariant derivative.

Sorry for the confusion guys.

 

[robphy]

This section from Wald's General Relativity might help:

If we are given a derivative operator, $\nabla_a$, we could define a map from smooth p-form fields to (p + 1)-form fields by
$$\omega_{a_1\cdots a_p}\rightarrow(p+1)\nabla_{[b}\omega_{a_1\cdots a_p]}\qquad \rm (B.1.4)$$
If instead we were given another derivative operator $\widetilde\nabla_a$, we would obtain the map
$$\omega_{a_1\cdots a_p}\rightarrow(p+1)\widetilde\nabla_{[b}\omega_{a_1\cdots a_p]}\qquad \rm (B.1.5)$$
However, according to equation (3.1.14) [which characterizes the difference between two derivative operators applied to a tensor field by $C^c{}_{ab}$], we have
$$\nabla_{[b}\omega_{a_1\cdots a_p]}<br /> -\widetilde\nabla_{[b}\omega_{a_1\cdots a_p]}<br /> =\sum_{j=1}^{p} C^d{}_{[ba_j}\omega_{a_1\cdots|d|\cdots a_p]}=0<br /> \qquad \rm (B.1.6)$$
since $C^c{}_{ab}$ is symmetric in a and b.
Thus the map defined by equation (B.1.4) is independent of derivative operator, i .e., it is well defined without the presence of a preferred derivative operator on M.
We denote this map by $d$.
In particular, we may use the ordinary derivative, $\partial_a$, associated with any coordinate system to calculate $d$.

 

[giulio_hep]

By the way, since @PrecPoint has mentioned the covariant exterior derivative, now it is maybe interesting to notice that, via the Chern-Weil theory, the coefficients of the characteristic polynomial of the curvature form (= covariant exterior derivative) do not depend on the choice of connection, but they can be seen as an obstruction to find global sections and are topological invariants, appearing also in many applications in physics, for example the geometric quantization. A similar treatment can be done in Higgs bundles... As an aside, since @lavinia added a comment about the relationship between the two disciplines and a metaphor, you can hear Nigel Hitchin quoting J W von Goethe in the beginning of his Higgs bundles talk:

Mathematicians are like Frenchmen: whatever you say to them they translate into their own language and right away it is something entirely different.

 

[kay bei]

means that one can form the "totally-symmetric part" of a tensor and the "totally-antisymmetic part" of a tensor (not unlike finding the real-part of a complex number and the imaginary-part of a complex number).

For example,
$M_{ab} =\displaystyle \frac{(M_{ab} +M_{ba})}{2} + \frac{(M_{ab} -M_{ba})}{2} = M_{(ab)} + M_{[ab]}$,
where $M_{(ab)}$ is the symmetric-part and $M_{[ab]}$ is the antisymmetric-part.

If $M_{ab}=M_{(ab)}$ (or equivalently, $M_{[ab]}=0$), then $M_{ab}$ is said to be symmetric.

If $M_{ab}=M_{[ab]}$ (or equivalently, $M_{(ab)}=0$), then $M_{ab}$ is said to be antisymmetric.

But in general, since $M_{ab}=M_{(ab)} + M_{[ab]}$,
the general tensor $M_{ab}$ is neither symmetric nor antisymmetric.

Robphy, that is a great analogy with complex numbers. So then from what I understand, a Tensor with a non-zero symmetric part cannot be written in differential forms (because forms are antisymmetric). So from my understanding, you can rewrite the antisymmetric type (0,n) tensor in a differential form notation. And that is all it is, a convenient notation with properties which can often help intuition when working in higher dimensions and often times saving space in calculations.

I am also aware that the generalization of antisymmetric tensor to higher dimensions is called alternating tensor. So my question here is, does the definition of differential form as antisymmetric (0,n) tensor generalize to , a differential form is a alternating type (0,n) tensor?

 

[weirdoguy]

I am also aware that the generalization of antisymmetric tensor to higher dimensions

There is no generalization, tensors (including antisymmetric ones) are defined on vector spaces of any arbitrary dimension. So (in this context) they are as general as they can be. You can use different names, but that does not change the object you are talking about. In most cases, alternating tensor is a different name for an (totally) antisymmetric tensor.

 

[giulio_hep]

So then from what I understand, a Tensor with a non-zero symmetric part cannot be written in differential forms (because forms are antisymmetric).

Notice that multilinear forms are covariant tensors.
Differential (forms) are totally antisymmetric (covariant tensors).

And that is all it is, a convenient notation...

It is much more than notation convenience: it is the natural language to describe the notions of volume and orientation.

 

[kay bei]

Notice that multilinear forms are covariant tensors.
Differential (forms) are totally antisymmetric (covariant tensors).
It is much more than notation convenience: it is the natural language to describe the notions of volume and orientation.

Oh okay, so differential forms are a subset of multilinear forms. And more specifically differential forms are totally antisymmetric multilinear forms. So can we write any Tensor as multilinear form? Why are differential forms much more important to differential geometry than multilinear forms? Is it differential forms or is it multilinear forms that provide the natural language for volume and orientation? Can you suggest a good textbook or set of lecture notes for physicists or engineers on differential geometry which provides a good explanation of all of this. It would be good to have a problem based book and informal style of mathematics as used by many physicists.

 

[giulio_hep]

Is it differential forms or is it multilinear forms that provide the natural language for volume and orientation?

Absolutely differential forms. If it is permitted here, I'd simply suggest you this succinct but effective answer.

And Terence Tao wrote a short introduction to explain why:

The integration on differential forms concept is of fundamental importance in differential topology, geometry, and physics, and also yields one of the most important examples of cohomology.

 

[afi1188]

To answer the initial question, forms are a subset of tensors...So obviously not all formulas using tensors can be converted to differential forms equivalent.

If i may ask a question as i am not a specialist: Is there any known computational/logic process (by this i mean kind of an algorithm...that can be implemented using a computer for example) that allows to convert the syntax of an equation using diff forms to equivalent equation using tensors (and the reverse, of course with the requirement that the equation only uses tensors within the forms subset) ? ...of course i know that for a given equation a mathematician would be able to do it...

thanks

see below extract form Fortney A visual introduction to diff forms...
https://i.stack.imgur.com/aJU2u.jpg

 

[WWGD]

To answer the initial question, forms are a subset of tensors...So obviously not all formulas using tensors can be converted to differential forms equivalent.

If i may ask a question as i am not a specialist: Is there any known computational/logic process (by this i mean kind of an algorithm...that can be implemented using a computer for example) that allows to convert the syntax of an equation using diff forms to equivalent equation using tensors (and the reverse, of course with the requirement that the equation only uses tensors within the forms subset) ? ...of course i know that for a given equation a mathematician would be able to do it...

thanks

see below extract form Fortney A visual introduction to diff forms...
https://i.stack.imgur.com/aJU2u.jpg

Not clear on your question: Every Differential Form is a tensor ( but obviously not vice versa). In particular, they are alternating antisymmetric tensors.

 

[afi1188]

i know that…my question was about the existence of an algorithm/method to convert a formula using forms onto a formula using tensors….mainly is this doable using a computer for example?
the reverse should also be possible in the specific cases where the tensors involved are all aternating and antisymétric
i understand that forms are a tool used for easy computational results in specific cases/conditions,
tensors being the reference in more general cases : required for example when a metric tensor is involved like einstein field equations.

 

[weirdoguy]

method to convert a formula using forms onto a formula using tensors

Forms are tensors, so every formula involving forms is, by definition of a form, also a formula involving tensors.