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Home , Exterior derivative

- Problem set 4: solutions Spring 2021

Here are some solutions to selected exercises from the semester. The idea is to give you a sense of what a ‘model’ answer should be (and also to elaborate on some discussions we have in the help sessions). They are supposed to be complimentary to the help sessions and the problems which I grade - i.e. not in lieu of actually doing the problem sets, working together, and asking questions!

If you spot any typos, mistakes, or have questions in general, don’t hesitate to let me know (my email: [email protected]). Special thanks to Jeriek Van den Abeele and Magdalena Kersting.

Question 13. for a 2-torus T 2 is the direct product S1 × S1 of two . An atlas for those two will then be one for the torus. The circle S1 needs two charts, for example (this is by no means a unique choice) covering the circle with two overlapping domains (0, 2π) and (−π, π) and then mapping the points in those domains to their corresponding real-valued angle. We can extend this to the torus, of which every point can be identified by two angular coordinates, one on the interior circle and one on the exterior one:

T 2 = {(θ, φ) | 0 ≤ θ < 2π, 0 ≤ φ < 2π} .

The torus is covered by the following four domains:

U1 = {(θ, φ) | 0 < θ < 2π, 0 < φ < 2π}

U2 = {(θ, φ) | − π < θ < π, 0 < φ < 2π}

U3 = {(θ, φ) | 0 < θ < 2π, −π < φ < π}

U4 = {(θ, φ) | − π < θ < π, −π < φ < π} ,

and the atlas is completed by a set of corresponding charts (i = 1, 2, 3, 4)

2 φi : Ui → φi(Ui) ⊂ R :(θ, φ) 7→ (x = θ, y = φ).

Just so we don’t miss the forest for the trees here: all we’re saying that we need two charts to cover a circle and simply do this twice, for two different angles, and as such will cover the torus.

Question 14. The Lie bracket

• Vector field on manifolds are examples of what’s known as derivations on C∞(M). By definition, derivations obey Leibniz’s rule. Let X,Y ∈ X(M), and let f, g ∈ C∞(M), then

[X,Y ](fg) = X(Y (fg)) − Y (X(fg)) = X {gY (f) + fY (g)} − Y {gX(f) + fX(g)} = g {XY (f) − YX(f)} + f {XY (g) − YX(g)} = g[X,Y ](f) + f[X,Y ](g).

Hence [X,Y ] is a derivation C∞(M) → C∞(M), which makes it a vector field. Note that in line 2, Y (fg)) = gY (f) + fY (g) (and the same for X) since we already know X,Y are individually vector fields and so are already known to obey the .

1 FYS4160 – General Relativity Spring 2021

Note that this a very abstract means of proof. There are other ways to show this: for example, by using the bundle definition of vector fields, and demonstrating those properties - and product rule; another way is to express the Lie bracket in coordinate form and show it transforms as a vector field does.

• The Lie bracket is independent of the choice of coordinates on the M, and so we are free  i ∂ to write it in terms of a specific of our choosing, x . We write ∂i = ∂xµ for the µ associated local of the tangent bundle, so that general vector fields can be witten X = X ∂µ µ and Y = Y ∂µ. Then the Lie bracket can be computed as:

µ µ ν µ ν µ [X,Y ] := (X (Y ) − Y (X )) ∂µ = (X ∂ν Y − Y ∂ν X ) ∂µ

Since the basis chosen is ∂µ, the coordinates of the Lie bracket are thus

µ ν µ ν µ [X,Y ] := X ∂ν Y − Y ∂ν X

• Follows immediately from the fact that partial commute.

Question 14. The exterior In , the question of notation is important. For this problem, you have various choices of notation: , differential forms, or some impure, bastardised, mixed form 1. You may wonder which is best. The answer is that both are useful for different purposes, and in different contexts. An important skill to develop is the ability to recognise when one of these two notations is more useful and to develop the ability to pass easily and effortlessly from one notation to the other.

That said, for my purposes, the choice is obvious: I will 100% be using differential forms, because there are only so many hours in a day and I will not be wasting all of them LateX-ing the bazillion indices.

(a) My laziness aside, differential forms also illuminate the conceptual similarity of the exterior derivative to other notions of taking derivatives.

We use a condensed shorthand for multi-index notation such that an ordered differential p-form has labels I = (i1, ..., ip). We additionally make use of the fact that we can write an arbitrary p-form as a linear combination of wedge products of the coordinate one-forms dx(i).

Now, with all the preamble stuff out of the way, let’s calculate d(ω ∧ η):

d(ω ∧ η) = d((fdxI ) ∧ (gdxJ )) = d(fgdxI ∧ dxJ ) = (gdf + fdg) ∧ dxI ∧ dxJ = (df ∧ dxI ) ∧ (gdxJ ) + (−1)p(fdxI ) ∧ (dg ∧ dxJ ) = dω ∧ η + (−1)pω ∧ dη

1I joke, of course - it’s actually a transparent way to do this calculation on paper.

Jake Gordin, University of Oslo 2 FYS4160 – General Relativity Spring 2021

In line 1 I used the fact that a p-form can be expressed as ω = fdxI ; in line 2, note that f, g are merely components and can be moved in front of the wedge products. Line 3 is an application of the and the fact that dω = d(fdxI ) = df ∧ dxI , and line 4 uses the antisymmetry property of the ordering of forms, i.e. dg ∧ dxI = (−1)pdxI ∧ dg.

(b) Let’s take the “double exterior derivative" of a 0-form f first, since that’s easy to do –

 ∂f  d(df) = d dxj ∂xj ∂2f = dxi ∧ dxj ∂xi∂xj  ∂2f ∂2f  = − ∂xi∂xj ∂xj∂xi = 0

where in line 3 we used the antisymmetry of differential form ordering and to get zero in line 4 we note that partial derivatives commute.

J This result then straightforwardly generalises, using a p-form ωj1...jp = ωJ = ωdx :

! X J d(dω) = d d(ωJ ) ∧ dx J p X J X X p j1 ji jp = d(dωJ ) ∧ dx + (−1) dωJ ∧ dx ∧ ... ∧ dx ∧ ∧dx J J i=1 = 0

Again we have made use of differential form ordering (note that the partial derivatives form dxji had been moved to the end) and the fact that partial derivatives commute.

Jake Gordin, University of Oslo 3