5
$\begingroup$

During a course I took on General Relativity, the professor started with an introduction on differential geometry. Vectors were properly motivated: he said that since the differential manifold doesn't have distances it didn't make sense to define vectors as the displacement between two points; we had to use something infinitesimal instead. Then, he gave several advantages for using derivatives as vectors instead of the classic euclidean two-point arrows.

However, when he starting explaining 1-forms... he just said what they were and move on. I don't really understand why we need 1-forms. Also, I don't know if 1-forms are also a generalization of an euclidean concept (such as vectors).

I've read many questions about 1-forms but none of them asked about clear motivation for introducing them in a General Relativity course So that's the question: What is the motivation for using 1-forms in General Relativity? what do they are useful for? Can't we just use vectors and then introduce a metric to have a direct product?

Note: I have checked many books looking for a proper motivation but I just find the definition followed by the usual interpretation of 1-forms as perpendicular planes in space. I've read Gravitation, Carrol's book and Schutz' both books.

To be clear, I don't need a physical explanation, what I want is motivation for using 1-forms when we can just use the metric and two vectors if we want a inner product.

$\endgroup$
9
  • $\begingroup$ Have you tried looking into a book about GR? $\endgroup$ Jan 28, 2018 at 18:28
  • $\begingroup$ Yes! I should have cleared that out in the question. Let me edit it. $\endgroup$ Jan 28, 2018 at 18:30
  • 2
    $\begingroup$ I just did a search on this Physics.SE forum of the string "general relativity vectors one forms" and found lots and lots of pretty good articles that should answer your question. Or, at least put you into a position of asking a more specific question. Note that one-forms and tangent vectors are dual spaces of each other and using the Metric you can turn one into the other and visa-versa. But, some things are more natural as one-forms then vectors such as a basis for a vector space, or a gradient of a scalar function, and other examples exist. Is that motivation enough? $\endgroup$
    – K7PEH
    Jan 28, 2018 at 18:49
  • $\begingroup$ Also, the big MTW Gravitation book has some very good explanatory content with diagrams on one forms and how they are different from vectors. $\endgroup$
    – K7PEH
    Jan 28, 2018 at 18:51
  • 1
    $\begingroup$ This post (v2) seems like a list question. One-forms aka. co-vectors are used all over the place in GR. $\endgroup$
    – Qmechanic
    Jan 28, 2018 at 19:44

4 Answers 4

14
$\begingroup$

The notion of differential forms depends on several structures:

  1. the wedge product
  2. the dual space
  3. the tangent bundle
  4. sections

So its not suprising that your lecturer wasn't as easily able to motivate them as vectors! Lets take these step by step:

1. The wedge product

In a 3d vector space we have the additional structure of an inner product and the cross product. These have geometric interpretations. However, when we generalise to a vector space of any dimension it's easy to see that the inner product generalises in an obvious way. Not so the cross product. In fact, this is only available in 3d.

Recall, that the scalar triple product $u.(v \times w)$ gives the volume of the parallelopid formed by the sides $u,v,w$. It is this property that generalises.

Given a parallelepid in a n-dimensional vector space $V$ (this is the generalisation of a parallelogram in the plane) whose sides are $v_1,..., v_n$. Then the wedge product $v_1\wedge ... v_n$ gives us the signed volume. It turns out that this is a vector, but they don't lie in the same vector space as $v$. We call them $k$-vectors and say that $v_1\wedge ... v_k$ lies in $\wedge^k V$.

2. The dual space

The dual space of a vector space $V$ is usually written $V^*$. It consists of all linear functions to the real line, $f:V\rightarrow R$. What does this mean? Each function is linear, so we can think of it as a kind of measurement or metric on the vector space. It tells us how to measure a vector. Thus $V^*$ is the space of all the ways we can measure vectors in $V$.

3. The Tangent Space

Given a manifold $M$, we can construct its tangent bundle $TM$. The easiest example to visualise is when the manifold is a curve or surface. Lets take the curve first: at every point of the curve $C$ we can draw the tangent line to it, this line extends to infinity and is a 1d vector space. We bundle them up all together into the bundle $TC$, and the tangent line at the point $p$ on the curve is $T_pC$. Similarly for a surface $S$, at each point $p$ of the surface we can draw the tangent plane to it, we write this as $T_pS$ and we bundle them all together into the bundle $TS$.

Now any bundle $E$ over a manifold $M$, has a projection map $\pi:E\rightarrow M$ and this is how they are usually referred to. It tells us where the 'fibres' are attached to. If we take the first example, $TC$, the tangent bundle of the curve; let $v$ be a vector in one of the tangent spaces, say $T_pC$ - this means that $v$ is in the tangent line (rather vector space) - that is defined (or attached) to the point $p$ of the curve. The projection map $\pi$ simply maps $v$ to the point $p$. So we can see that the image of the entire space $T_pC$ is just the point $p$.

4. Sections

Given a bundle $\pi:E\rightarrow M$ then we can take its space of sections $CE$. This is the space of all maps $s:M\rightarrow E$ such that $pi\circ s =Id_M $. For example, suppose $E$ was a bundle of vector spaces over the manifold $M$, then a section is a choice of a vector in each fibre. It is a vector field.

Construction of differential forms

Finally we put all these structures together: We construct the bundles $\wedge^k T^*M$. That is we take the manifold $M$, we construct the tangent space $TM$ over it, and then take it's dual space $T^*M$ and the finally we take the $k^{th}$ wedge $\wedge^k T^*M$. The sections of this bundle is $C(\wedge^k T^*M)$ and this is the space of all $k$-differential forms and is usually written (at least by mathematicians and sometimes others) as $\Omega^kM$.

Uses

It turns out that we have a map $d^k:\Omega^kM \rightarrow \Omega^{k+1}M$ called the exterior derivative (another name for the wedge product is the exterior product) and this generalises the $grad$ operator in vector analysis. That is $d^0=grad$. The other vector analysis operators - $div$ & $curl$ - are variants of this.

It also turns out that when we integrate a form $\omega$ over a manifold $M$ we get a generalisation of Stokes theorem: $\int_M d\omega=\int_{dM} \omega$, where the symbol $dM$ is the boundary of the manifold.

conclusion

Thus we see that differential forms allow us to generalise the vector analysis that we're already familiar with in 3d Euclidean space to the context of manifolds of any dimensions. This is important given the importance of vector analysis in physics. But they have many other uses, for example de Rham cohomology. They also bring in many other notions that are important, for example vector, fibre and principal bundles.

There is a formulation of General Relativity that uses a connection on the frame bundle of the tangent bundle and this a principal bundle with structure group the Lorentz group. This connects to the way the other forces are described, for example electromagnetism, the electroweak and the strong force are described as principal bundles with structure group $U(1),SU(2)$ and $SU(3)$ respectively in the Standard Model.

$\endgroup$
7
  • $\begingroup$ Great answer. I would maybe talk about the Riesz theorem in section 2 as the theoretical grounding of index raising and lowering, through the bijection between tangent and cotangent spaces that it guarantees (usually of course this bijection is chosen in GR to be through the metric). $\endgroup$ Jan 29, 2018 at 7:56
  • 1
    $\begingroup$ I think 90% of this answer is not what the OP wanted or needed. They didn’t ask for general differential forms and it’s overkill to toss in bundles too. Besides the fact that you could start with one-form fields (instead of sections of a bundle) the OP didn’t even ask for fields in the first place, i.e. this answer has three superfluous levels of complexity. $\endgroup$
    – knzhou
    Jan 29, 2018 at 9:29
  • 1
    $\begingroup$ In the section 4 shouldn't it be $\pi \circ s =\pi $.? And a bit earlier maybe it is "...simply maps $v$ to the point $p$ "? Can't propose an edit cause it would be too short $\endgroup$ Jan 29, 2018 at 10:24
  • 1
    $\begingroup$ @knzhou: I addressed the main focus of the question in the first two points. Without differential forms I couldn't explain how they generalise vector analysis and that's an important part of the formalism - and the OP was asking for motivation. Personally, I think bundles are simple enough, in principle, that everyone ought to know about them ... they certainly make it much simpler to define differential forms amongst many other things. $\endgroup$ Jan 29, 2018 at 15:57
  • 1
    $\begingroup$ The point of introducing all these structures is to obtain conceptual clarification by placing them all in perspective and this can be done without getting into the technical niceties which can be intimidating to the uninitiated...! $\endgroup$ Jan 29, 2018 at 22:48
6
$\begingroup$

The idea is to see what object really are without using the metric (with which you can convert freely between one forms and vectors). It may happen that you have been using 1 forms all the time without noticing. The gradient of a function is a one form. If you think it is a vector, you silently used the isomorphism between vectors and 1-forms.

To address your last paragraph: it looks convenient to map everything to a vector. But later you want to vary term with respect to the metric (variational principle of the action), which you cannot do if you dont know where and where not the metric is used.

$\endgroup$
4
$\begingroup$

Adding to some of the other answers:

One of the main geometrical interpretations of 1-forms are that they are quantities that can be integrated over a 1-dimensional curve. (This is essentially the dual of the interpretation of vector (field)s as derivatives.) More generally, an n-form can be integrated over an n-dimensional surface. This integration does not rely on any additional structures such as a metric.

$\endgroup$
0
$\begingroup$

What is the motivation for using 1-forms in General Relativity? what do they are useful for?

In my view, it's very simple : 1-forms (especially the tetrad's vierbeins and spin-connection) and 2-forms (curvature) are very often much more easier to manipulate and make calculations easier (faster), than using the Christoffel symbols.

Take the Schwarzschild metric as an example. Without the differential forms, you'll have to calculate laboriously all the Christoffel symbols just to get the components of the Riemann tensor, and then contract them to get the Einstein tensor. Doing the same with the tetrad 1-forms is much more easier.

You could check what Weinberg says about forms, in his classical book on General Relativity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.