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Home , Multiple integral, Volume form

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3.3. Integration. 3.3.1. Boundaries. In order to present Stokes’ theorem, we first need to generalize the concept of a slightly. Recall that an n-dimensional manifold is locally homeomorphic to Rn; that is, any point has some neighborhood which is homeo- morphic to an open set in Rn. An n-dimensional manifold with boundary is locally n−1 homeomorphic to R− R , where R− := {x R | x 0}. The terminology is unfortunate,× because this∈ is more≤ general (not more specific) than manifold. A manifold with boundary might not have a boundary, in which case it is just a manifold. If M is a manifold with boundary, then the boundary of M is denoted ∂M. This n−1 n−1 is the set of points that correspond to points of {0} R R− R in the local homeomorphisms (coordinate systems. The orientation× ⊂ of ∂M×is such that if (x1,x2, . . . , xn) (with x1 0) is an oriented coordinate system on M, then (x2, . . . , xn) is an oriented coordinate≤ system on ∂M. The interior is the rest of M, i.e., M˚ := M ! ∂M. A compact manifold without boundary is also called a closed manifold. The boundary of a compact manifold with boundary is automatically closed. So, ∂∂M = or “The boundary of a boundary is zero.” (the mantra of John Archibald Wheeler). ∅ with boundary will be important when we discuss the structure of infinity. The conformal compactification of a spacetime M is a compact manifold with boundary whose interior is M and whose boundary is “infinity”. 3.3.2. Computing . A p-form on M can be integrated over a p-dimensional submanifold with boundary of M, or more generally over the image of a smooth map from a manifold with boundary to M. We can compute integrals of differential forms by using a few simple properties. (1) If U Rn is an open set (and therefore a manifold) with the canonical ⊆ orientation of the coordinate system (x1, . . . , xn), then the of an n- form over U is exactly what the notation suggests:

f dx1 ∧ ∧ dxn = f dx1 . . . dxn ··· ··· !U ! !U can be computed by iterated integration. The notation of differential forms is inspired by the notation of iterated integrals. The exterior product en- codes the correct rules for a in a multiple integral. (2) Integration respects pull-backs. If ϕ : N M is a smooth map, then

ω = ϕ∗ω. → !ϕ(N) !N If we have a submanifold of M that is covered by a coordinate chart, then this tells us how to integrate a form over that submanifold. (3) If Σ,Γ M do not intersect, then ⊂ ω = ω + ω. Σ Γ Σ Γ ! ∪ ! ! (4) If the closure of Σ M is also a submanifold with boundary, then ⊂ ω = ω. !Σ !Σ 18

So, the general strategy to integrate Σ ω is to find a set of coordinate charts which do not intersect, but which almost cover Σ. Each coordinate system gives a # homeomorphism from an open subset of Rdim Σ to a coordinate patch in Σ. We can pull σ back to Rdim Σ by this map and integrate it there. The sum of these integrals is Σ ω. Example.# Let S2 = {(x, y, z) R3 | x2 + y2 + z2 = 1} be the unit in , oriented as the boundary∈ of the unit . Let ω = x dy ∧ dz. We shall compute the integral S2 ω. It is impossible to cover S2 with a single coordinate chart, but we can almost # cover it with one chart. Let C := {(x, y, z) S2 | y = 0, x 0}; this is a semicircle ∈ ≤ connecting the north and south poles. The closure of the subset V := S2 ! C is V = S2, and V is a coordinate patch for the standard . The coordinates give a homeomorphism from V to an open subset of R2, namely, U = {(θ,φ) R2 | 0<θ<π, −π<φ<π }. We are more interested in the inverse ∈ homeomorphism ϕ : U V R3. This is given by ⊂ 1 ϕ∗x = ϕ (θ,φ)=sin θ cos φ → 2 ϕ∗y = ϕ (θ,φ)=sin θ sin φ 3 ϕ∗z = ϕ (θ,φ)=cos θ

Now, compute the pull-back ϕ∗ω. First,

ϕ∗(dy)=cos θ sin φdθ + sin θ cos φdφ

ϕ∗(dz)=− sin θdθ So,

ϕ∗ω = sin θ cos φ sin θ cos φdφ ∧ (− sin θdθ) · = sin3 θ cos2 φdθ ∧ dφ. Now, integrate this over U,

3 2 ω = ω = ϕ∗ω = sin θ cos φdθ ∧ dφ 2 !S !V !U !U π π 4π = sin3 θdθ cos2 φdφ = 3 !0 !−π Note that it was important to write dθ ∧ dφ in that order to conform with the orientation and get the correct sign from the iterated integral. Exercise 3.9. Let Σ by the piece of the graph of z = x + y2 where 0 x 1 and 0 y 1, with the orientation given by the coordinates (x, y). Integrate≤ ≤ ≤ ≤

(y dx + z dy) ∧ dz. !Σ 19

3.3.3. Flux integrals. A particularly interesting case is when dim Σ = dim M − 1. Suppose that M is an N-dimensional, oriented, ; then we can use the form * Ω2(M) to write any N − 1-form as v ! * in terms of a vector field v. ∈ Let i : Σ ! M be the inclusion map. Let n be the unit normal 1-form to Σ, extended continuously over a neighborhood of Σ. In a neighborhood of Σ, the volume form→ factorizes as * = n ∧ *Σ. The form *Σ is slightly arbitrary, but its pull-back i∗*Σ is just the Riemannian volume form on Σ. Now,

v ! * = v ! (n ∧ *Σ)= n, v *Σ − n ∧ (v ! *Σ). ' ( To compute Σ v ! *, we need the pull-back of v ! *, but because n is normal to Σ, we have i∗n = 0. So, #

v ! * = n, v *Σ. ' ( !Σ !Σ You may have seen this before, denoted as

v n dS = v dS. · · !Σ !Σ 3.3.4. Stokes’ theorem. Let ω Ωp(M) and Σ M be a compact p + 1-dimensional submanifold with boundary.∈ Stokes’ theorem⊂ states that

dω = ω. !Σ !∂Σ Example. If p = 0, then Σ is a curve, and this is the fundamental theorem of cal- culus for line integrals, provided that we understand 0-dimensional integration as follows. Let a and b be the beginning and end of the curve Σ. The boundary is ∂Σ = {a, b}, but with a “oriented negatively”. An integral over ∂Σ is

f = f(b)− f(a). !∂Σ Example. In R2, any 1-form can be written as α = f dx + g dy. The exterior deriva- tive is ∂g ∂f dα = − dx ∧ dy. ∂x ∂y ! " So, Stokes’ theorem (for p = 1) gives Green’s theorem ∂g ∂f f dx + g dy = − dx ∧ dy. ∂x ∂y !∂Σ !Σ ! " Example. Suppose that M is Euclidean R3. As before, we can identify the exterior of a 1-form as the of a vector field. Since the integral of a 2-form is identified with the flux integral, Stokes’ theorem for p = 1 is just the classical Stokes’ theorem,

α ds = α = dα = ( α) dS. · ∇× · !∂Σ !∂Σ !Σ !Σ 20

Example. For p = dim M−1, Stokes’ theorem becomes Gauss’ theorem (generalized to any dimension). For V M a compact region, ⊂ v ! * = ( v)*. ∇· !∂V !V Example. The boundary of a boundary is always trivial. This is intimately con- nected to the fact that the of an exterior derivative is always 0. ω = dω = ddω !∂∂Σ !∂Σ !Σ The first expression is 0 because ∂∂Σ = . The last expression is 0 because d2 = 0. ∅ Example. Suppose that Σ ,Σ M are two p-dimensional submanifolds with bound- 0 1 ⊂ ary such that we can deform continuously from Σ0 to Σ1 while leaving the bound- ary ∂Σ = ∂Σ fixed. This implies that there is a p + 1-dimensional Γ M between 0 1 ⊂ them such that the difference between Σ0 and Σ1 is ∂Γ. Now, let ω be a closed p-form. ω − ω = ω = dω = 0 !Σ1 !Σ0 !∂Γ !Γ So, the integral Σ ω of a closed form does not change if we deform Σ in this way. Example. Roughly# speaking, the homology of M is the set of closed submanifolds of M, but with two submanifolds considered equivalent if they differ by a bound- ary. The deRham cohomology of M is the set of closed differential forms on M, but with two forms considered equivalent if the difference is an exact form. Stokes’ theorem shows that integration gives a well defined pairing between homology and deRham cohomology. That is, if Σ is a closed submanifold and ω is a closed form, then Σ ω only depends upon the homology class of Σ and the cohomology class of ω. # 3.3.5. Currents. Suppose that some “stuff” is moving around space. This is de- scribed by two fields. The density ρ C (M) describes where the stuff is; the ∈ integral V ρ* is the amount of stuff in V ∞ M. The current j X(M) describes how the stuff is moving; the flux ⊂ ∈ # j ! * !Σ is the rate at which stuff is crossing a Σ. The amount of stuff in V is decreasing at the rate that it is flowing out across ∂V. So, d ρ* = ρ*˙ =− j ! * =− ( j)*. dt ∇· !V !V !∂V !V Since this should be true for any V M, it implies the conservation equation ⊂ 0 = ρ˙ + j. ∇· We will see later that this is even simpler from the perspective of spacetime.