1. Tangent Vectors to R n ; Vector Fields and One Forms All the vector spaces in this note are all real vector spaces.

Transcription

1 1. Tangent Vectors to R n ; Vector Fields and One Forms All the vector spaces in this note are all real vector spaces. The set of n-tuples of real numbers is denoted by R n. Suppose that a is a real number and x = (x 1,, x n ) and y = (y 1,, y n ) are elements of R n. We define the sum x + y and the scalar product a x by x + y = (x 1 + y 1,, x n + y n ), a x = (ax 1,, ax n ). It is well known that the set R n together with the addition + and the scalar multiplication forms a vector space (R n, +, ). In fact, the set R n can be equipped with other vector space structures. Let p be an element of R n. For each element x of R n, we represent x as a sum p + v where v = x p. We denote p + v by v p or (v) p. If x = (v) p and y = (w) p are elements of R n and a R is a real number, we define x + p y and a p x by x + p y = (v + w) p, a p x = (av) p. Lemma 1.1. The set R n together with the addition + p and the scalar multiplication p forms a vector space. Proof. This is left to the reader as an exercise. The vector space (R n, + p, p) is denoted by T p R n and called the tangent space to R n at p. We remark that T R n is the usual R n we have learned before. One can also see that the vector space T p R n is isomorphic to R n for each p R n. This can be proved by constructing a linear isomorphism from T p R n onto R n by sending v p to v. If {e 1,, e n } is the standard basis for R n, {(e 1 ) p,, (e n ) p } forms a basis for T p R n. We can also identify T p R n with the subset {(p, v) R n R n : v R n } of R n R n by sending v p to (p, v). Instead of using the notation (R n, + p, p) for T p R n, we may use the identification T p R n = {(p, v) R n R n : v R n }. On T p R n, we define the addition and the scalar multiplication by (p, v) + p (p, w) = (p, v + w), Furthermore, we introduce an inner product on T p R n by (p, v), (p, w) TpR n = v, w R n a p (p, v) = (p, av). where the right hand side of the equation is the standard Euclidean inner product of v, w on R n. Definition 1.1. Let U be an open subset of R n. The set T U = p U T pr n is called the tangent bundle over U. A vector field on U is a function V : U T U such that V (p) T p R n for any p U. By definition, T U = U R n as a set. Let U be an open subset of R n and f : U R be a C 1 -function. Let p U and v p T p R n. We define df p (v p ) to be the directional derivative of f at p along v, i.e. df p (v p ) = d dt f(p + tv). t= By chain rule, we know that df p (v p ) = Df(p)(v). 1

2 2 Here Df(p) : R n R is the derivative of f at p. For each v p, w p T p R n and a, b R, by linearity of Df(p), we obtain that df p (av p + bw p ) = df p ((av + bw) p ) This shows that df p : T p R n R is a linear map. = Df(p)(av + bw) = adf(p)(v) + bdf(p)(w) = adf p (v p ) + bdf p (w p ). Definition 1.2. Let V be a finite dimensional vector space. A linear map ψ : V R is called a linear functional on V. The set of all linear functionals denoted by V is called the dual space to V. By linear algebra, V forms a vector space. The dual space to T p R n is denoted by T p R n. For each 1 i n, let x i : R n R be the function defined by x i (p) = p i where p = (p 1,, p n ). Then {x i : 1 i n} are smooth functions on R n and called the rectangular coordinate functions on R n. Example 1.1. Let p = (p 1, p 2, p 3 ) be an element of R 3 and {e 1, e 2, e 3 } be the standard basis for R 3. Then {(e 1 ) p, (e 2 ) p, (e 3 ) p } forms a basis for T p R 3. Let v p = (v 1, v 2, v 3 ) p be any vector in T p R 3. For t R, p + tv = (p 1 + tv 1, p 2 + tv 2, p 3 + tv 3 ) and hence By definition, x 1 (p + tv) = (p 1 + tv 1, p 2 + tv 2, p 3 + tv 3 ) = p 1 + tv 1. (dx 1 ) p (v p ) = d dt x 1(p + tv) = d t= dt (p 1 + tv 1 ) = v 1. t= In general, one can show that (dx i ) p (v p ) = v i for 1 i 3. One can easily see that (dx 1 ) p ((e 1 ) p ) = 1, (dx 1 ) p ((e 2 ) p ) =, (dx 1 ) p ((e 3 ) p ) =, (dx 2 ) p ((e 1 ) p ) =, (dx 2 ) p ((e 2 ) p ) = 1, (dx 2 ) p ((e 3 ) p ) =, (dx 3 ) p ((e 1 ) p ) =, (dx 3 ) p ((e 2 ) p ) =, (dx 3 ) p ((e 3 ) p ) = 1. This example motivates the following definition: Definition 1.3. Let V be an n-dimensional real vector space and {v 1,, v n } be a basis for V. A set of linear functionals {ϕ 1,, ϕ n } on V is called the dual basis to {v 1,, v n } providing that ϕ i (v j ) = δ ij, 1 i, j n. Lemma 1.2. Let V be an n-dimensional real vector space and {v 1,, v n } be a basis for V. Suppose that {ϕ 1,, ϕ n } is the dual basis to {v 1,, v n }. Then {ϕ 1,, ϕ n } forms a basis for V. Proof. Let us first prove that the set is linearly independent. Suppose a 1 ϕ 1 + +a n ϕ n =. By ϕ i (v j ) = δ ij, one has for 1 j n. = (a 1 ϕ a n ϕ n )(v j ) = a 1 ϕ 1 (v j ) + + a n ϕ n (v j ) = a j.

3 Let ϕ : V R be a linear functional on V. Define ψ = n ϕ(v j)ϕ j V. Then ψ(v i ) = ϕ(v i ) for all 1 i n. Hence ψ(v) = ϕ(v) for any v V. We show that ψ = ϕ. Thus ϕ is a linear combination of {ϕ 1,, ϕ n }. We find that {ϕ 1,, ϕ n } spans V. The above lemma also implies that for any basis {v 1,, v n } for V and its dual basis {ϕ 1,, ϕ n }, any linear functional ϕ : V R on V has the following expression: (1.1) ϕ = ϕ(v i )ϕ i. It follows from the definition that the set {(dx 1 ) p, (dx 2 ) p, (dx 3 ) p } is the dual basis to {(e 1 ) p, (e 2 ) p, (e 3 ) p } and hence is a basis for Tp R 3. Let U be an open subset of R n and f : U R be a smooth function. Let p U. Since df p : T p R n R is a linear functional, by equation (1.1) and Corollary 1.1, we have df p = df p ((e i ) p )(dx i ) p. Let us now compute df p ((e i ) p ). By definition, df p ((e i ) p ) = d dt f(p + te i ) = f (p). t= x i From here, we obtain that Lemma 1.2 implies the following results: (dx i ) p ((e j ) p ) = x i x j (p) = δ ij, 1 i, j n. Corollary 1.1. The set {(dx 1 ) p,, (dx n ) p } forms a basis to T p R n ; it is the dual basis to {(e 1 ) p,, (e n ) p }. Since {(dx 1 ) p,, (dx n ) p } is the dual basis to {(e 1 ) p,, (e n ) p }, by Lemma 1.2 and its remark (1.1), we conclude that f (1.2) df p = (p)(dx i ) p. x i Example 1.2. Let f : (a, b) R be a smooth function and p (a, b). Then df p = f (p)dx p. Theorem 1.1. (Riesz representation theorem) Let V be a finite dimensional inner product space. For any linear functional ψ : V R, there exists a unique ξ V such that ϕ(v) = v, ξ for all v V. Proof. Let us prove the uniqueness first. Suppose ξ and η are two vectors in V such that ϕ(v) = v, ξ = v, η for all v V. Let α = ξ η. Then v, α = for all v V. We choose v = α. Then α, α =. By the property of inner product, α = and hence ξ = η. Let us prove the existence. Assume that dim V = n. Choose an orthonormal basis {e 1,, e n } for V. We set ξ = n ϕ(e i)e i and define ψ : V R by ψ(v) = v, ξ. Then ψ is a linear functional on V. Furthermore, for 1 j n, ψ(e j ) = e j, ξ = ϕ(e i ) e j, e i = ϕ(e j ). This implies that ψ(v) = ϕ(v) for all v V and thus ψ = ϕ. 3

4 4 Remark. Let V be an n-dimensional inner product space and {e i : 1 i n} be an orthonormal basis for V. If ϕ : V R is a linear functional on V, the unique vector ξ so that ϕ(v) = v, ξ has the following representation: ξ = ϕ(e i )e i. Now we are ready to introduce the notion of gradient of a smooth function at a point p. Since df p : T p R n R is a linear functional and T p R n is a finite dimensional inner product space, there is a unique vector in T p R n, denoted by f(p), such that df p (v) = v, f(p) p. The vector f(p) in T p R n is called the gradient vector of f at p. Definition 1.4. If f : U R n R is smooth, f defines a function f : U T U, p f(p). This vector field is called the gradient vector field of f on U. By Theorem 1.1, and its remark, we see that f(p) = df p ((e i ) p )(e i ) p. Since df p ((e i ) p ) = f xi (p), we find that ( ) f f f(p) = (p)(e i ) p = (p)e i x i x i Example 1.3. Let f : R 2 R be a smooth function and p R 2. Then f(p) = (f x (p), f y (p)) p. If f : U R n R is smooth, df p T p R n for any p U. We define the notion of cotangent bundle over an open subset U of R n. Definition 1.5. The cotangent bundle T U of an open subset U of R n is the union p U T p R n. An one-form on U is a function ω : U T U. Let ω be an one-form on U. For each p U, ω(p) Tp R n. By equation 1.1, if we set a i (p) = ω(p)((e i ) p ), then ω(p) = a 1 (p)(dx 1 ) p + + a n (p)(dx n ) p. We obtain functions a 1,, a n : U R. We write ω = a 1 (x)dx a n (x)dx n. We say that ω is a continuous/differentiable/c k /smooth one-form if a 1,, a n : U R are continuous /differentiable/c k /smooth functions on U. For example, let f : U R n R be a C k function. Then df defines a one-form df : U T U by sending p to df p. Since df = f x1 dx f xn dx n and f xi are C k 1 functions on U, df is a C k 1 one-form. Remark. The tangent bundle T U = U R n over an open subset U of R n and the cotangent bundle T U = U (R n ) over U are open subsets of T R n = R n R n and T R n = R n (R n ). We can define the continuities/differentiability/smoothness of a vector field V : U T U over U or an one form ω : U T U in the usual sense. p

5 Example 1.4. Let f(x, y) = e x cos y for (x, y) R 2. Then df = f x dx + f y dy = e x cos ydx e x sin ydy. Example 1.5. Let y = sin x for x R. Then dy = f (x)dx = cos xdx. Example 1.6. Let f(x, y) = x 3 y + 2x 2 y for (x, y) R 2. Then df = f x dx + f y dy = (3x 2 y + 2y 2 )dx + (x 3 + 4xy)dy. Example 1.7. Let U = R 2 \ {(, )}. Then U is an open subset of R 2. Define ydx + xdy ω = x 2 + y 2 for (x, y) R 2. Then ω is a smooth one-form on U. A one form on an open subset U of R n is of the form ω = a 1 (x)dx a n (x)dx n. It is natural to ask when ω = df for some f C (U)? This is equivalent to the following family of differential equations a i (x) = f x i (x), 1 i n. We will begin with the case when n = 2. Consider a smooth one-form on an open set U R 2 of the form ω = M(x, y)dx + N(x, y)dy and solve this problem when U is an open ball. If ω = df holds, then M = f x and N = f y. By smoothness of f, M y = f xy = f yx = N x. We find that if ω = df, then M y = N x. In fact, we have the following result: Proposition 1.1. Let B be an open ball in R 2 and ω = M(x, y)dx+n(x, y)dy be a smooth one form on U. Then ω = df for some smooth function f : B R if and only if M y = N x. Proof. We have proved one direction. Let us prove the reverse direction. We may assume that B the open unit ball B(, 1) centered at of radius 1. We define a function f : B R by f(x, y) = c + x M(tx, ty)dt + y N(tx, ty)dt. Here c is any real number. By taking the partial derivative of f with respect to x, we obtain f x (x, y) = M(tx, ty)dt + x By the relation M y = N x, we see that Therefore By chain rule, f x (x, y) = N x (tx, ty)tdt = M(tx, ty)dt + M x (tx, ty)tdt + y x M y (tx, ty)tdt. N x (tx, ty)tdt. t(m x (tx, ty)x + M y (tx, ty)y)dt. d dt M(tx, ty) = M x(tx, ty)x + M y (tx, ty)y. 5

6 6 Therefore f x can be rewritten as f x (x, y) = = = M(x, y). M(tx, ty)dt + t ( ) d M(tx, ty) dt dt M(tx, ty)dt + tm(tx, ty) 1 t= M(tx, ty)dt (use the integration by parts) Similarly, one can show that f y (x, y) = N(x, y). This proves that ω = df if M y = N x. In calculus, we have learned the notion of antiderivative of a C 1 -function on a closed interval [a, b]. Let f : [a, b] R be a continuous function. A C 1 -function F : [a, b] R is said to be an antiderivative of f provided that F (x) = f(x). If we denote ω = f(x)dx, then the condition F (x) = f(x) is equivalent to the condition ω = df. We now can define the notion of antiderivative of a smooth one form. Definition 1.6. Let ω be a smooth one form on an open subset U of R n. If there exists a smooth function f : U R such that ω = df, then ω is said to have an antiderivative on U. In this case, f is said to be an antiderivative of f. Proposition 1.1 says that a smooth one form ω = M(x, y)dx + N(x, y)dy on an open ball B in R 2 has an antiderivative on B if and only if M y = N x. It is natural for us to ask when a smooth one form on an open set possess an antiderivative on that open set? What are the necessary and sufficient conditions for a smooth one form to have an antiderivative? Example 1.8. Let ω = f(x)dx be a smooth one form on R. We define F (x) = x f(t)dt, x R. By fundamental theorem of calculus, F (x) = f(x) for any x R and hence ω = df. In other words, any smooth one form on R always possess an antiderivative on R by the fundamental Theorem of calculus. Now we will express the condition M y = N x in terms of the derivative of differential forms. Let us introduce the notion of r-forms for any r. 2. Differential r-forms Suppose v p = (a, b) p and w p = (c, d) p are two tangent vectors to R 2 at p. The area of the parallelogram spanned by {v p, w p } is the absolute value of the determinant a c b d. Since dx p (v p ) = a and dy p (v p ) = b and dx p (w p ) = c and dy p (w p ) = d, the above determinant can be expressed as : dx p(v p ) dx p (w p ) dy p (v p ) dy p (w p ). This motivates the following definition. Definition 2.1. Let V be a finite dimensional vector space and ϕ, ψ be two linear functionals on V, i.e. ϕ, ψ V. We define the wedge product ϕ ψ of ϕ and ψ as follows. We define ϕ ψ : V V R, (v, w) ϕ(v) ϕ(w) ψ(v) ψ(w).

7 Lemma 2.1. Let ϕ, ψ be linear functionals on a finite dimensional vector space V. Their wedge product ϕ ψ : V V R is a skew symmetric bilinear form on V, i.e. (1) ϕ ψ : V V R is bilinear (2) (ϕ ψ)(w, v) = (ϕ ψ)(v, w). Especially, when ϕ = ψ, ϕ ϕ =. Proof. The proof is left to the reader as an exercise. Definition 2.2. Let ψ 1,, ψ r : V R be linear functionals on a finite dimensional vector space V. We define the wedge product ψ 1 ψ r : V r R by Here (v 1,, v r ) V r. (ψ 1 ψ r )(v 1,, v r ) = det[ψ i (v j )] r i,j=1. Example 2.1. Let ψ 1, ψ 2, ψ 3 : V R be linear functions on V. For v 1, v 2, v 3 V, one has ψ 1 (v 1 ) ψ 1 (v 2 ) ψ 1 (v 3 ) (ψ 1 ψ 2 ψ 3 )(v 1, v 2, v 3 ) = ψ 2 (v 1 ) ψ 2 (v 2 ) ψ 2 (v 3 ) ψ 3 (v 1 ) ψ 3 (v 2 ) ψ 3 (v 3 ) Lemma 2.2. Let ψ 1,, ψ r : V R and ψ 1 ψ r : V r R be as above. Then ψ 1 ψ r : V r R is a alternating r-linear form on V. Proof. This is left to the reader. Example 2.2. Let u p = (a 1, b 1, c 1 ) p and v p = (a 2, b 2, c 2 ) p, and w p = (a 3, b 3, c 3 ) p be vectors in T p R 3 for p R 3. In high school, we have learned that the volume of the parallelepiped spanned by u p, v p, w p is the absolute value of the determinant (dx p dy p dz p )(u p, v p, w p ) = a 1 b 1 c 1 a 2 b 2 c 2 a 3 b 3 c 3 Definition 2.3. Let V be a finite dimensional vector space. We set T (V ) = R and T 1 (V ) = V and T r (V ) be the space of all r-linear forms f on V, i.e. f : V r R is r-linear. Elements of T r (V ) are called r-tensors on V. Let T 1, T 2, T : V r R be r-tensors and a R. We define the sum T 1 + T 2 and the scalar product at by for (v 1,, v r ) V r. (T 1 + T 2 )(v 1,, v r ) = T 1 (v 1,, v r ) + T 2 (v 1,, v r ), (at )(v 1,, v r ) = at (v 1,, v r ), Lemma 2.3. The set T r (V ) forms a vector space over R. (In fact, it is a vector subspace of the space of functions F(V r, R) from V r into R.) For any linear functionals ψ 1,, ψ r : V R, we define the tensor product ψ 1 ψ r : V r R by (ψ 1 ψ r )(v 1,, v r ) = ψ 1 (v 1 )ψ 2 (v 2 ) ψ r (v r ) for any (v 1,, v r ) V. We leave it to the reader to verify that see quiz 11. ϕ ψ = ϕ ψ ψ ϕ, 7

8 8 Proposition 2.1. Let β = {v 1,, v n } be a basis for a n dimensional vector space V and {ϕ 1,, ϕ n } be the dual basis to β. Then {ϕ i1 ϕ ir : 1 i 1,, i r n} forms a basis for T r (V ); the dimension of T r (V ) is n r. Proof. See quiz 11. A r-tensor T T r (V ) is alternating if T (v σ1,, v σr ) = (sgn σ)t (v 1,, v r ), (v 1,, v r ) V r for any permutation σ : {1,, r} {1,, r}, 1 where { 1 if σ is an even permutation; sgn σ = 1 if σ is an odd permutation. Lemma 2.4. Let Λ r (V ) the subset of T r (V ) consisting of all alternating r-tensors on V. Then Λ r (V ) forms a vector subspace of T r (V ). Proof. This is left to the reader as an exercise. Proposition 2.2. Let β = {v 1,, v n } be a basis for a n dimensional vector space V and {ϕ 1,, ϕ n } be the dual basis to β. Then {ϕ i1 ( ) ϕ ir : 1 i 1 < < i r n} forms a n basis for Λ r (V ). Hence the dimension of Λ r (V ) is. r Proof. We will prove the case when r = 2. For general case, see quiz 11. Definition 2.4. Let U be an open subset of R n. We denote Λ r (T U) = p U Λr (T p R n ). A r-form on U is a function η : U Λ r (T U) such that η(p) Λ r (T p R n ) for any p U. Since {(dx i ) p : 1 i n} forms a basis of Tp R n (dual to the standard basis {(e i ) p : 1 i n} in T p R n,) the set {(dx i1 ) p (dx ir ) p : 1 i 1 < < i r n} forms a basis for the vector space Λ r (Tp R n ) by Lemma 2.2. An element of Λ r (Tp R n ) is of the form a i1 i r (dx i1 ) p (dx ir ) p. 1 i 1 < <i r n If U is an open subset of R n, a smooth r form on U is of the form η = η i1 i r (x)dx i1 dx ir, 1 i 1 < <i r n where η i1 i r : U R are smooth functions on U. Example 2.3. A two form on an open subset U of R 2 is of the form Here f : U R is a function. η = f(x, y)dx dy. Example 2.4. A two form on an open subset U of R 3 is of the form η = P (x, y, z)dx dy + Q(x, y, z)dx dz + R(x, y, z)dy dz; a three form on U is of the form Here P, Q, R, g : U R are functions. α = g(x, y, z)dx dy dz. 1 A permutation on a nonempty set X is a bijection on X.

9 Let I r,n be the set of all r-tuple of nonnegative integers (i 1,, i r ) such that 1 i 1 < < i r n. Let η be a r form on R n. Then we denote η by η = I I r,n η I (x)dx I where dx I = dx i1 dx ir for any I = (i 1,, i r ) I r,n. The set of all smooth r-forms on an open subset U of R n is denoted by Ω r (U). Let η = I η Idx I and ω = I ω Idx I be r-forms on U and f C (U). We define η + ω and fη by η + ω = (ω I (x) + η I (x))dx I, fη = f(x)η I (x)dx I. I I Proposition 2.3. The set Ω r (U) forms a vector space over R such that for f, g C (U) and ω, η Ω r (U), (1) f(η + ω) = fη + fω; (2) (f + g)η = fη + gη; (3) (fg)η = f(gη); (4) 1η = η Proof. This is left to the reader as an exercise. Remark. In algebra, the above properties tell us that that Ω r (U) is a C (U)-module. Let η = I η I(x)dx I be a r-form and ω = J ω Jdx J be a s-form. We define their wedge product by η ω = η I (x)ω J (x)dx I dx J. I,J One can easily check that η ω = ( 1) rs ω η. Definition 2.5. Let η = I η I(x)dx I be a smooth r forms on U R n. We define the derivative d r η of η by d r η = dη I dx I. I Here df denotes the differential of a smooth function f on U. Example 2.5. Let ω = M(x, y)dx + N(x, y)dy be a smooth one form on an open set U R 2. The derivative of ω is given by d 1 ω = (N x M x )dx dy. We see that the condition N y = M x is equivalent to the condition d 1 ω =. Example 2.6. Let η = P dy dz Qdx dz + Rdx dy be a smooth two form on an open subset of R 3. Then d 2 η = (P x + Q y + R z )dx dy dz. Definition 2.6. Let η be an r-form on an open subset U of R n. If d r η =, η is called a closed r-form. The set of all closed smooth r-forms on U is denoted by Z r dr (U). Lemma 2.5. The derivative d r of r-forms on an open subset of R n defines a linear map d r : Ω r (U) Ω r+1 (U). Hence Z r dr (U) = ker dr is a vector subspace of Ω r (U). Proof. The proof follows from the definition. 9

10 1 We will set Ω (U) = C (U); i.e. a smooth zero form on U is a smooth function on U. We set d = d. Example 2.7. Let f : U R be a smooth function on an open subset U of R 2. Then df = f x dx + f y dy. The previous example tells us that d 1 (d f) = (f yx f xy )dx dy. Since f is smooth, f yx = f xy on U. Hence d 1 (d f) =. Since f is arbitrary, we find d 1 d =. In fact, we have a general result: Lemma 2.6. The linear map d r : Ω r (U) Ω r+1 (U) satisfies the following properties: (1) d r+1 d r =. (2) (graded Leibniz rule) d r+s (η ω) = d r η ω + ( 1) r η d s ω for any r form η and for any s form ω. Let us go back to define the notion of antiderivatives of differential forms. Definition 2.7. Let ω be a smooth r form on an open set U R n. We say that ω has an antiderivative on U if there exists a smooth r 1 form η on U such that ω = d r 1 η. If ω has an antiderivative on U, we say that ω is an exact r-form. The set of all exact r-forms on U is denoted by B r dr (U). The Proposition 1.1 can be reformulated as follows. Proposition 2.4. Let B be an open ball in R 2 and ω be a smooth one form on B. Then ω is an exact r-form on B if and only if it is a closed one form on U. Since B r dr (U) consists of r form of the form dr 1 η for η Ω r 1 (U), by definition, B r dr (U) = Im dr 1. Since d r 1 is linear, B r dr (U) is a vector subspace of Ωr (U). Furthermore, since d r d r 1 =, d r (d r 1 η) = for any η Ω r 1 (U). This implies that Im d r 1 ker d r, i.e. B r dr (U) is a vector subspace of Zr dr (U). Now let us state the fundamental theorem of calculus for smooth differential forms on open sets in an Euclidean space. Definition 2.8. Let U be an open subset of R n. We say that the fundamental theorem of calculus holds for r forms on U if Z r dr (U) = Br dr (U) Proposition 2.7 can be reformulated as follows: Proposition 2.5. The fundamental theorem of calculus for one forms holds on any open ball in R 2. To know whether the fundamental theorem of calculus holds or not, we need to study the difference between ZdR r (U) and Br dr (U). Thus we consider the quotient space HdR r (U) = Zr dr (U)/Br dr (U). The quotient space HdR r (U) is called the r-th de Rham cohomology of U. Thus we can reformulate the fundamental theorem of calculus in terms of the quotient space HdR r (U) : Proposition 2.6. Let r 1. The fundamental Theorem of calculus holds for smooth r-forms on U R n if and only if HdR r (U) = {}. The Proposition 2.7 can be rewritten as: Proposition 2.7. Let r 1. For any open ball B in R 2, HdR 1 (B) = {}.