A COMPREHENSIBLE PROOF OF THE CHANGE OF VARIABLES THEOREM USING HOMOTOPY THEORY

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1 A COMPREHENSIBLE PROOF OF THE CHANGE OF VARIABLES THEOREM USING HOMOTOPY THEORY BEN KORDESH AND WILLIAM RICHTER Abstract. We give a comprehensible proof of the CVT using a homotopy theory result close to π n 1 (R n {0}) 0. The proof is suggested by Lax s amazing proof of a version of the CVT [Lax99], which Taylor improved on [Tay02]. 1. Introduction We assume knowledge of Lang s [Lan83, XX 1 3]. We will give a comprehensible proof of [Lan83, Thm. XX-4.5]: Theorem 1.1. Let C R n be a cube contained in some open set U and φ: U V be a C 1 diffeomorphism, for an open set V R n. Then Vol(φ(C)) = det dφ. C We use this, our primer on the Riemann integral [KR17a], and [Lan83, Cor. XX-4.6] to prove the CVT needed to show integration on smooth manifolds is well-defined, Theorem 3.2 below. Taylor [Tay02] writes controversially that The typical proofs of [the CVT] that one finds in advanced calculus texts tend to be grim affairs, involving careful estimates of volumes of images of small cubes under the [diffeomorphism] and numerous pesky details. We applaud Taylor s criticism, but say instead that most proofs of the CVT (including Lax s [Lax01, Thm. 2]) are too long or unmotivated. We assert that motivation is found after adding the homotopy theory which Lax and Taylor s paper [Lax99, Tay02] suggest a simple proof of, which we proved in [KR17b, Thm. 4.1]: Theorem 1.2. Take y in the interior of Cube. There is no continuous function φ: Cube R n {y} which is the identity on Cube. Here Cube = [ 1, 1] n R n. There are two proofs in the literature of the CVT we re quite happy with [LS68, Lan83] except for their nonuse of Theorem 1.2. Our proof of Theorem 1.1 generalizes the following n = 1 proof of the CVT, which requires only the IVT and the MVT: 1

2 2 BEN KORDESH AND WILLIAM RICHTER Let C = [a, b]. We may assume φ > 0 on U. Then we must prove (1) φ(b) φ(a) = length(φ([a, b])) = since by the IVT, φ([a, b]) = [φ(a), φ(b)]. φ is uniformly continuous on [a, b], so for every ɛ > 0 there exists δ > 0 such that for c, t [a, b] with c t < δ, we have φ (c) φ (t) < ɛ. Given a partition a = x 0 < < x n = b of [a, b] with x i < δ, for all i {1,..., n}, there exists by the MVT c i (x i 1, x i ) such that φ(x i ) φ(x i 1 ) = φ (c i )(x i x i 1 ). The absolute value of the difference between the sides of (1) is n xi n xi φ(x i ) φ(x i 1 ) φ (c i ) φ (t) dt (b a)ɛ. x i 1 i=1 x i 1 φ i=1 As this is true for every ɛ > 0, we conclude that (1) holds. We contend this proof was not a grim affair of pesky details. Our proof of Theorem 1.1 is much longer, but we believe that Taylor would not consider it to be grim affair of pesky details. Lang s next result [Lan83, Cor. XX-4.6] has a beautiful proof we can add nothing to, and we do not believe that Taylor could object to it. However, Lang s most general CVT [Lan83, Cor. XX-4.7] is exactly the sort of grim affair that we think Taylor would object to. We believe this general CVT grim affair is standard in among Advanced Calculus texts. The solution is that Lang s general CVT is not needed to show integration on manifolds is well-defined. For that we need only a mild generalization of [Lan83, Cor. XX-4.6], as explained in [Lee00, Lem & Cor ]. Thanks to Jared Wunsch for pointing out Lax s and Taylor s papers and for much valuable encouragement. Thanks to David Groisser for pointing out the value of Lee s text. 2. The Cube CVT We follow [Lan83, Thm. XX-4.5] except by using Theorem 1.2, in place of his ingenious but unmotivated use of the Shrinking Lemma. Note Cube = B 1 (0), and Cube = {x R n : x = 1}. We heavily use our primer on the Riemann integral, where we proved [KR17a, Thm. 5.1]: Theorem 2.1. Take any invertible linear function λ: R n R n and any rectangle R R n. Then Vol(λ(R)) = det λ v(r). Let C R n be a cube contained in some open set U and φ: U V be a C 1 diffeomorphism, for an open set V R n. Take ɛ > 0. There b a φ,

3 A COMPREHENSIBLE PROOF OF THE CVT 3 exists d > 0 and p R n such that C = B d (p). Since C is compact, dφ is uniformly continuous on C. Let M = max x C ( (dφ x ) 1 ). Then there exists δ > 0 such that dφ x dφ y < ɛ/m, for all x, y C with x y < δ. Let P be a cubical partition {S j } N j=1 of C, where S j = B r (a j ), for a j C and 0 < r < δ. As Lang argues, since the images φ(s j ) have only negligible sets in common, N (2) Vol(φ(C)) = Vol(φ(S j )) j=1 We will prove that Vol(φ(S j )) is sufficiently close to Vol(dφ aj (S j )). Recall the MVT we will need. The two results [Lan83, Cor. XVII-4.4] and [LS68, Cor. Ch. 3, 7] simplify to prove Lemma 2.2. For all a C and all x B r (a) C, φ(x) φ(a) dφ a (x a) (ɛ/m) x a. Let r = (1 + ɛ)r and r = (1 ɛ)r. Write λ j = dφ aj. We will prove (3) λ j (B r (0)) φ(s j ) φ(a j ) λ j (B r (0)). Define θ j : B p (0) R n by θ j (x) = φ(x + a j ) φ(a j ). Then θ j (0) = 0 and d(θ j ) 0 = λ j. Define η j = λ 1 j θ j : B p (0) R n. By the uniform norm property, Lemma 2.2 says that for all j N and x B r (0), (4) η j (x) x ɛ x. By the triangle inequality and (4), for all j N and x B r (0), (5) η j (x) (1 + ɛ) x and η j (x) (1 ɛ) x. Our statement (3) is now (6) B r (0) η j (B r (0)) B r (0), for all j N. The righthand subset statement follows immediately from (5). The lefthand subset statement needs our homotopy theory result Theorem 1.2: Lemma 2.3. B r (0) η j (B r (0)), for all j N. Proof. Immediately from (5), we have η j (x) r, for all x B r (0). Define the straight line homotopy h: B r (0) [0, 1] R n by h(x, t) = x + t(η j (x) x). By (4) and the triangle inequality, h(x, t) x t η j (x) x (1 tɛ) x r, for all x B r (0) and t [0, 1]. Using the well-known homotopy extension property, we define g : B r (0) R n by { η j (2x) if x r/2, g(x) = h((r/ x )x, 2 (2/r) x ) if x r/2.

4 4 BEN KORDESH AND WILLIAM RICHTER By Theorem 1.2, B r (0) Im g, since g is continuous and the identity on B r (0). But η j (B r (0)) contains the open ball Br (0), since Im g = η j (B r (0)) Im h and h r. Since the continuous image of a compact set is compact, and a compact subset of a Hausdorff space is closed, η j (B r (0)) is closed, and thus contains B r (0), the closure of Br (0). Proof of Theorem 1.1. Define φ = det dφ: U R. We have now established (6) and hence (3). Taking Vol of (3) and using Theorem 2.1, (7) det λ j Vol(B r (0)) Vol(φ(S j )) det λ j Vol(B r (0)), for all j N. Note that Vol(B r (0)) = (1 ɛ) n Vol(S j ) and Vol(B r (0)) = (1 + ɛ) n Vol(S j ). Let D 1 = 2 n 1. We may assume ɛ (0, 1/D 1 ). Then (1+ɛ) n 1+ɛD 1 and (1 ɛ) n 1 ɛd 1. Now we use Lang s argument. Sum the inequalities (7) over all j N, use (2), and estimate φ by an upper and lower bound to obtain (8) (1 ɛd 1 )L(P, φ ) Vol(φ(C)) (1 + ɛd 1 )U(P, φ ). Since φ is continuous and thus admissible, by [Lan83, Thm. XX-1.3], there exists η > 0 such that if 2r < η, then U(P, φ ) L(P, φ ) < ɛ. We may assume that our cubical partition P chosen above satisfies r < η/2. Let M = max x C φ (x). Now C φ satisfies the bounds of (8), and thus Vol(φ(C)) C φ < ɛ + ɛd1 2MVol(C). 3. The integration on manifolds CVT Lang [Lan83, Cor. XX-4.6] easily deduced the following from our Theorem 1.1: Theorem 3.1 (Lang). Let φ: U V be a C 1 diffeomorphism between open subsets of R n. Take a cube C U. Let f : φ(c) R be an admissible function. Then f = (f φ) det dφ. φ(c) C Lee [Lee00, Lem ] gives a nice explanation of why the following CVT show that integration on smooth manifolds is well-defined. Theorem 3.2. Let φ: U V be a C 1 diffeomorphism between open subsets of R n, and f : V R be a compactly supported admissible function. Then f = (f φ) det dφ. V U

5 A COMPREHENSIBLE PROOF OF THE CVT 5 Proof. The function g = (f φ) det dφ : U R is compactly supported and admissible, because φ: supp(f φ) supp(f) is a homeomorphism and det dφ is continuous and always nonzero. Let K = supp(g) U, which is disjoint from the closed C = R n U. Take a cube T = B r (a) containing K. Take ɛ > 0. By [KR17a, Lem. 4.3], there is a minimum distance d > 0 between K and C. Take a partition P k of T where 2r/k < d, using the even partition P k given by subdividing the cube T into k n subcubes of radius r/k. Then all of the subcubes of P k which intersect K are subsets of U. Let ζ = {S P k : S K }. Since g U = g on any S ζ and g U = 0 on any S ζ, we know (9) g = g U = g U = g = f U T S P S k S ζ S S ζ φ(s) by [KR17a, Lem. 3.6, INT 1] and Theorem 3.1. Let A = S ζ S. We know A is admissible and that K A U. We use the proof of [KR17a, Lem. 3.6]. Let α = S ζ S. Since φ is Lipschitz and α is negligible, φ(α) is negligible, and therefore admissible, by [Lan83, Rem. after Prop. XX-2.2]. Similarly, φ( S) is negligible and admissible for all S ζ. By [Lan83, Prop. XX-2.3], φ(int S), φ(s), and φ(a) are admissible. Since φ is one-to-one, φ(a) is the disjoint union of φ(α) and all φ(int S), for S ζ. Furthermore, each φ(s) is the disjoint union of φ(int S) and φ( S). By [KR17a, Lems. 3.4, 3.5] and (9), g = f = f = f, U S ζ φ(s) φ(a) V because supp f φ(a) V and [KR17a, Lem. 4.2]. References [KR17a] B. Kordesh and W. Richter, A Primer on Riemann Integration in R n, 7 pages, available at RiemannPrimer.pdf, [KR17b], A Rectangular Version of Lax s IVT, 6 pages, available at http: // [Lan83] S. Lang, Undergraduate Analysis, Undergraduate Texts in Math., Springer, New York, [Lax99] P. Lax, Change of Variables in Multiple Integrals, Amer. Math. Monthly 106 (1999), , available from [Lax01], Change of Variables in Multiple Integrals II, Amer. Math. Monthly 108 (2001), , available from stable/ [Lee00] J. Lee, Introduction to Smooth Manifolds, Springer, New York, 2000.

6 6 BEN KORDESH AND WILLIAM RICHTER [LS68] [Tay02] L. Loomis and S. Sternberg, Advanced Calculus, Addison-Wesley, Readng, Mass., M. Taylor, Differential Forms and the Change of Variable, Journal of Math. Analysis and Applications 268 (2002),