1. THEVOLUME (OR SURFACE) ELEMENTONASUBMANIFOLD. Let M be a submanifold of Rn of dimension k, p a point in M. We can think of M in three different ways: 1. M is (locally) given by a parametrization ψ : D → Rn where D is a domain in Rk and ψ is a map whose differential is injective at all points in D. 2. M is (locally) given by an equation, n {x in R ; g(x) = 0} where g is a map from a domain U in Rn to Rn−k with surjective differential at all points in U. 3. M is given (locally) as −1 n G {x in R ; xk+1 = ...xn = 0} where G is a diffeomorphism. You should go through the implications connecting each of these definitions to see that they are all equivalent! Let µ be a continuous nonvanishing k-form on M defining an orientation of M. The tangent plane Tp(M) of M at p is now the image of the differential of ψ, Dψ|0, if ψ(0) = M p. (This is the same thing as the kernel of Dg|p.) The volume elemnt on M at p , ω |p is the k form on Tp satisfying M ω (e1, ...ek) = 1

is e1, ...ek is an orthonormal basis for Tp which is oriented, i e satisfies µ(e1, ...ek) > 0. The volume of an open subset V of M is Z ωM . V Let now η be a k-form on a neighbourhood of p in Rn, and denote also by η the pullback of η to M under the inclusion map from M to Rn. This just means that we restrict the form η that is defined on all k-tuples of vectors in Rn to the tangent plane of M. Since all forms of maximal M degree k are multiples of each other, η|p = cω |p. Theorem 1.1. Let M be a hypersurface, i e suppose k = n − 1. At p in M let let η be a form of degree n − 1 in Rn. Let M be defined by an equation g = 0, where dg 6= 0. Then, if

(dg/|dg|) ∧ η = fdx1 ∧ ...dxn, at p η = fωM at p. Proof. After a linear change of coordinates we may assume that the tangent space of M at p is Tp(M) = {x, x1 = 0}. Then, by definition M ω = ±dxˆ1, 1 2

depending on which orientation of M that we have chosen. On the other hand dg/|dg| = ±dx1 P since the kernel of dg is T(M) and dg/|dg| has length 1. Then, if η = ηjdxˆj, the restriction of M η to Tp(M) equals η1dxˆ1. So, at p, η = ±η1ω . On the other hand (dg/|dg| ∧ η = ±dx1 ∧ η = ±η1dxˆ1. Hence f = ±η1, so the claim follows.  Another way to state this is P Corollary 1.2. Let η = ηjdxˆj. Then X M η = ± njηjω

on M, where nj = gj/|dg| is a unit normal to M. Proof. X (dg/|dg|) ∧ η = njηjdx1 ∧ ...dxn.  Let us look at a few examples of this. (i) Let F be a smooth function from (an open set in) Rn−1 to R and let M be the graph of 0 F . Then M is defined by the equation g = 0 if g = xn − F (x ). Hence dg = dxn − dF and p 2 |dg| = 1 + |dF | . Let η = dx1 ∧ ...dxn−1 Then

p 2 −1 (dg/|dg|) ∧ η = ±( 1 + |dF | ) dx1 ∧ ...dxn. Hence M p 2 ω = 1 + |dF | dx1 ∧ ...dxn, and the volume of M is Z p 2 1 + |dF | dx1...dxn−1.

2 p 2 (ii) If M is the semicircle in R defined by |x| = 1 and x2 > 0 we can choose F = 1 − x1 and get a formula for the volume (length) element on the circle

M 1 ω = dx1. p 2 1 − x1 On the other hand we may also use the formula in the corollary and get M ω = x1dx2 − x2dx1. From this we see that there are many forms on Rn defining the volume form when restricted to M. The last choice gives the easiest way to compute the length, via the parametrisation x = (cos θ, sin θ). (iii) Let now M be the unit sphere in R3. First we claim that

ω1 = (1/x3)dx1 ∧ dx2 3 is a volume element on M (up to sign). To check this we take g = |x|2 − 1. Then dg/|dg| = P xjdxj, so (dg/|dg|) ∧ ω1 = dx1 ∧ dx2 ∧ dx3 so our claim follows from the theorem. In a similar way ω2 = −(1/x2)dx1 ∧ dx3 would work. Finally, one also checks that 2 2 ω = dx1 ∧ (x2dx3 − x3dx2)/(x2 + x3) is a volume form. But 2 2 (x2dx3 − x3dx2)/(x2 + x3) = dθ if θ is the angle defined by (x2, x3) in the (x2, x3)-plane. (To check this, write θ = arctan x3/x2+ C.) Hence ω = dx1 ∧ dθ is a volume form. Integrating we get that the volume of the unit 2-sphere is Z 1 Z 2π dx1 dθ = 4π. −1 0 Notice that using the volume form ω we also see that the surface area of the slice of the sphere where −1 < a < x1 < a + h < 1 equals Z a+h Z 2π dx1 dθ = 2πh, a 0 and is thus independent of a. This was already shown by Archimedes and used by him to compute the area of the sphere.

2. THE COAREAFORMULA As an application we will prove the following Theorem 2.1. Let ρ be a smooth function such that {x; ρ(x) < 1} is a bounded domain. Assume dρ 6= 0 for all x such that ρ(x) = r. Then Z Z (d/dt)|t=r fdx = fdS/|dρ| ρ