satisfying the following condition: If T : V V is any linear map, then µ(x 1,,X n )= det T µ(x 1,,X n ).

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1 ensities Although differential forms are natural objects to integrate on manifolds, and are essential for use in Stoke s theorem, they have the disadvantage of requiring oriented manifolds in order for their integrals to be defined. There is a way to define integration on nonorientable manifolds as well, which we describe below. The reason an orientation is needed for integrals of differential forms to make sense has to do with transformation law under change of coordinates. The transformation law for an n-form on an n-manifold under a change of coordinates involves the jacobian determinant of the transition map, while the transformation law for integrals involves the absolute value of the determinant. We will define below objects whose transformation law for integrals involves the absolute value of the determinant. We begin, as always, in the linear-algebraic setting. efinition. Let V be an n-dimensional vector space. A density of V is a function µ : V V R }{{} n copies satisfying the following condition: If T : V V is any linear map, then µ(x 1,,X n )= det T µ(x 1,,X n ). Observe that a density is not a tensor, because it is not linear over R in any of its arguments. Let Ω(V ) denote the set of all densities on V. Proposition 1 (Properties of ensities). Let V be a vector space of dim V = n 1. (a) Ω(V ) is a vector space under the obvious vector operations: (c 2 µ 1 + c 2 µ 2 )(X 1,,X n )=c 1 µ 1 (X 1,,X n )+c 2 µ 2 (X 1,,X n ). (b) If µ 1, µ 2 Ω(V ) and µ 1 (E 1,,E n )=µ 2 (E 1,,E n ) for some basis (E i ) of V, then µ 1 = µ 2. (c) If ω Λ n (V ), the map ω : V V R defined by ω (X 1,,X n )= ω(x 1,,X n ) is a density (d) Ω(V ) is a 1-dimensional spanned by ω for any nonzero ω Λ n (V ). Proof. (b) Suppose µ 1 and µ 2 give the sams value when applied to (E 1,,E n ). Typeset by AS-TEX 1

2 2 If X 1,,X n are arbitrary vectors in V, let T : V V be the unique linear map that takes E i to X i, for i =1,,n. It follows that (c) We have µ 1 (X 1,,X n )=µ 1 (TE 1,,TE n ) = det T µ 1 (E 1,,E n ) = det T µ 2 (E 1,,E n ) =µ 2 (TE 1,,TE n ) =µ 2 (X 1,,X n ). ω (TX 1,,TX n )= ω(tx 1,,TX n ) = (det T )ω(x 1,,X n ) = det T ω (X 1,,X n ). (d) Suppose ω is any nonzero element of Λ n (V ). It suffices to claim: µ Ω(V ), c R such that µ = c ω. Let (E i ) be a basis for V, and define a, b R by Since ω 0, we have a 0. Thus a = ω (E 1,,E n )= ω(e 1,,E n ), b =µ(e 1,,E n ). µ(e 1,,E n )=(b/a) ω (E 1,,E n ). Hence µ =(b/a) ω by part (b). efinition. (1) A positive density on V is a density µ with µ(x 1,,X n ) > 0 for all linearly independent (X 1,,X n ). (2) A negative density is defined similarly. Every density on V is either positive, negative or zero. The set of positive densities is a convex subset of Ω(V ), namely, a half-line. Now let be a smooth manifold. efinition. The set Ω = Ω(T p ) p is called the density bundle of. Let π :Ω be the projection map taking each element of Ω(T p ) to p.

3 Lemma 2. If is a smooth manifold, its density bundle is a smooth line bundle over. Proof. We will construct local trivilizations. Let (U, (x i )) be any smooth coordinate chart on, and let ω = dx 1 dx n. Proposition 1 shows that ω p is a basis for Ω(T p ) at each point p U. Therefore, the map Φ : π 1 (U) U R given by Φ(c ω p )=(p, c) is a bijecton. Suppose (Ũ,( xj )) is another smooth chart with U Ũ. Let ω = d x 1 d x n, and define Φ 1 :π (Ũ) Ũ R corresponding;y: Φ(c ω p )=(p, c). Then Φ Φ 1 (p, c) =Φ(c ω p )=Φ(c det( xj x i ) ωp ) ( ) x j ). =(p, c det x i ) Thus the transition functions are equal to det ( x j / x i. efinition. A section Ω is called a density on. A density on is said to be positive or negative if its value at each point has that property. Any nonvanishing n-form ω determines a positive density ω, defined by ω p = ω p for each p. If ω is a nonvanishing n-form on an open set U, then any density µ on U can be written µ = f ω for some real-valued function f. One important fact about densities is that every manifold admits a global smooth positive density, without any orientability assumptions. Lemma 3. If is a smooth manifold, there exists a smooth positive density on. Proof. Because the set pf positive elements of Ω is an open subset whose intersection with each fiber is convex, the usual partition of unity argument allows us to piece togeter local densities to obtain a global smooth positive density. Remark. This lemma works because positivity of a density is a well-defined property, independent of any choices of coordinates or orientations. There is no corresponding existence result for orientation forms because without a choice of orientations, there is no way to decide whether n-forms are positive. 3

4 4 Under smooth maps, densities pull back in the same way as differential forms. efinition. If F : N is a smooth map between n-manifolds and µ is a density on N, we define a density F µ on by (F µ) p (X 1,,X n )=µ F (p) (F X 1,,F X n ). Lemma 4. Let G : P N and F : N be smooth maps between n-manifolds, and let µ be a density on N. (a) For any f C (N), F (fµ)=(f F )F µ. (b) If ω is an n-form on N, then F ω = F ω. (c) If µ is smooth, then F µ is a smooth density on. (d) (F G) µ = G (F µ). The next result shows how to compute the pullback of a density in coordinates. Proposition 5. Suppose F : N is a smooth map between n-manifolds. If (x i ) and (y j ) are smooth coordinates on open sets U and V N, respectively, and u is a smooth real-valued function on V, then the following holds on U F 1 (V ): (1) F (u dy 1 dy n )=(u F ) det F dx 1 dx n, where F represents the matrix of partial derivatives of F in these coordinates. Proof. We have F (u dy 1 dy n )=(u F )F ( dy 1 dy n ) =(u F ) F (dy 1 dy n ), by Lemma 4(b), =(u F ) (det F)dx 1 dx n =(u F ) det F dx 1 dx n.

5 Integration of ensities If R n is a compact domain of integration and µ is a density on, we can write µ = f dx 1 dx n for some uniquely determined continuous function f : R. efinition. efine the integral of µ over by µ = fdv, or more suggestively, f dx 1 dx n = fdx 1 dx n. Similarly, if U is an open subset of R n and µ is compactly supported in U, we define µ = µ, U where U is any compact domain of integration containing the support of µ. The key fact is that this is diffeomorphism-invariant. Proposition 6. (1) If and E are compact domains of integration in R n, and G : E is a smooth map that restricts to a diffeomorphism from Int to Int E, then µ = G µ E for any density µ on E. (2) Similarly, if U, V R n are open sets and G : U V is a diffeomorphism, then µ = G µ for any compactly supported density µ on V. Proof. Use (1) in Proposition 5. V Now let be a smooth n-manifold. If µ is a density on whose support is contained in the domain of a single smooth chart (U, ϕ), the integral of µ over is defined as µ = (ϕ) 1 µ. This is extended to arbitrary densities µ by setting µ = ψ i µ, i where {ψ i } is a smooth partition of unity subordinate to an open cover of by smooth charts. This is independent of the choices of coordinates or partition of unity. U ϕ(u) 5

6 6 Proposition 7 (Properties of Integrals of ensities). Suppose and N are smooth manifolds with or without boundaries, and µ, ν are compactly supported densities on. (a) Linearity: If a, b R, then aµ + bη = a µ + b η. (b) Positivity: If µ is a positive density, then µ>0. (c) iffeomorphism Invariance: If F : N is a diffeomorphism, then µ = F µ. N The Riemannian ensity Lemma 8 (The Riemannian ensity). Let (,g) be a Riemannian manifold with or without boundary. There is a unique smooth positive density µ on, called the Riemannian density, with the property that (14) µ(e 1,,E n )=1 for any local orthonormal frame (E i ). Proof. (i) Uniqueness is obvious, because any two densities that agree on the elements of a basis must be equal. (ii) Existence: Given any point p, let U be a connected smooth coordinate neighborhood of p. Since U is diffeomorphic to an open subset of Euclidean space, it is orientable. Any choice of orientation of U uniquely determines a Riemannian volume form dv g with the property that dv g (E 1,,E n )=1 for any oriented orthonormal frame. If we put µ = dv g, it follows easily that µ is a smooth positive density on U satisfying (14). If U and V are two overlapping smooth coordinate neighborhoods, the two definitions of µ agree where they overlap by uniqueness, so this defines µ globally. Proposition. Let (,g) be a Riemannian manifold with or without boundary and let dv g be its Riemannian volume form. (a) The Riemannian density of ie equal to dv g. (b) For every compactly supported continuous function f : R, f dv g = fdv g. Because of part (b), it is customary to denote the Riemannian density simply by dv g, and to specify when necessary whether the notation refers to a density or a form.