Notation We will use the following abbreviations: if Rn has coordinates xi , we will use ∂i = ∂xi = ∂ ∂xi to denote the differentiation with respect to xi . In addition, we will use Einstein’s summation notation, in which the sum symbol is not written if two indices appear, one as a lower index and one as an upper index, e.g. the directional derivative is DAϕ = ∂iϕ · Ai . In such a situation, the summation is implicit (and thus, it is necessary to state explicitly if the summation is not intended). We believe that this highly improves readability of formulas. We will also use a notation ϕ: V // W that means that ϕ is defined on an open subset of V , denoted dom ϕ. 1. Analysis in a vector space A derivative of a mapping ϕ: V // W at a point x is the best linear approximation of ϕ at x and as such is a linear map V → W (more precisely, it is the linear part of the best affine approximation). We will use this geometric object in general considerations and, only in computations, we will use the matrix (∂jϕi ) representing this linear map in coordinates on V and W. Definition 1.1. Given a mapping ϕ: V // W, its derivative (or differential) at x is a linear map D|xϕ: V → W that satisfies ϕ(x + v) = ϕ(x) + D|xϕ(v) + o(v). Remark. A function h(v) lies in o(v) if limv→0 |h(v)| |v| = 0. Similarly, h(v) lies in O(v) if |h(v)| |v| is bounded in some neighbourhood of 0. Since any pair of norms is comparable, the notions of O(v) and o(v) do not depend on a particular choice of a norm. Applying to the |−|∞ norm on W, h(v) ∈ o(v) if and only if each component hi (v) ∈ o(v). Example 1.2. Functions of one variable, i.e. V = R, W = R. Then D|tϕ is a (1 × 1)-matrix with the sole entry ϕ (t), the usual derivative, i.e. D|tϕ = (ϕ (t)). We will usually identify these two objects. Example 1.3. Functions of several variables, i.e. V = Rn , W = R. Then D|xϕ is a (1 × n)matrix, or a linear form. It associates to each vector the rate of growth of ϕ along this vector. The gradient of ϕ at x is then the direction of the largest growth (geometrically, it is perpendicular to the level set ϕ−1 (ϕ(x)) and of some particular magnitude) and as such depends on a scalar product: D|xϕ(v) = gradx ϕ, v . In orthonormal coordinates, it corresponds to transposing the linear form D|xϕ. For these reasons, we will use mostly the more geometric D|xϕ. Example 1.4. Paths, i.e. V = R, W = Rn . Then D|tϕ is an (n × 1)-matrix. There is a natural identification between matrices of this type and vectors (since Hom(R, Rn ) ∼= Rn , given by evaluation at 1). Denoting ϕ (t) = D|tϕ(1), we thus obtain D|tϕ(τ) = τ · ϕ (t). This is a generalization of the first example. The geometric meaning of ϕ (t) is that of a tangent vector to ϕ at time t. It is thus natural to think of it as a vector based at ϕ(t) – we will continue to develop such formalism later and reserve the notation ϕ (t) for the based version. Lemma 1.5 (chain rule). Given two mappings ϕ: V // W, ψ: W // X, we have D|x(ψ ◦ ϕ) = D|ϕ(x)ψ ◦ D|xϕ. It will be useful to consider the derivatives at different points and organize them into a mapping Dϕ: V // Hom(V, W), x → D|xϕ. We also denote D|xϕ(v) = Dv|xϕ = Dvϕ(x) (the directional derivative of ϕ along v at x) and, thus, obtain a mapping Dvϕ: V // W. It is called the directional derivative of ϕ along v. Since the derivative of a linear map ψ is the very same map ψ, the chain rule gives: Corollary 1.6. For any linear map ψ, we have Dv(ψ ◦ ϕ) = ψ ◦ Dvϕ, i.e. linear maps commute with directional derivatives. 1 2 Lemma 1.7 (Leibniz rule). Given two mappings ϕ: V // W, g: V // X, we have Dv(ϕ ⊗ ψ) = Dvϕ ⊗ ψ + ϕ ⊗ Dvψ. The last two results together prove that for any bilinear map Φ: W × X → Y , we get Leibniz rule: DvΦ(ϕ, ψ) = Φ(Dvϕ, ψ) + Φ(ϕ, Dvψ) (since Φ(ϕ, ψ) = F(ϕ⊗ψ) with F : W ⊗X → Y the linear map corresponding to Φ). Special cases are the multiplication R × R → R (this gives the usual Lebiniz rule), multiplication by scalars V × R → V , inner product V × V → R, matrix multiplication Matk×n × Matn×m → Matk×m etc. Lemma 1.8 (symmetry of second derivatives). Assuming that ϕ is C2 in a neighbourhood of x, we have DuDvϕ(x) = DvDuϕ(x). The second derivative is a mapping D2 ϕ: V // Hom(V, Hom(V, W)) ∼= Hom(V ⊗ V, W). As a reformulation, the previous lemma says that D2 ϕ: V // Hom(S2 V, W). Changing coordinates. We have seen (or we know) that, with respect to any given bases on V and W, (Dϕ)i j = ∂jϕi . We will now study the effect of changing coordinates (non-linearly!), i.e. we compose ϕ with maps (changes of coordinates) on both sides: writing x = χ(y) and z = ϕ(x), we have a chain rule D|y(ψ ◦ ϕ ◦ χ) = D|zψ ◦ D|xϕ ◦ D|yχ. In words, the linear map D|xϕ gets replaced by an equivalent one, i.e. one modified by linear coordinate changes of both domain and codomain. The second derivative is more complicated and it is useful to rewrite the above formula first in coordinates: ∂j(ψ ◦ ϕ ◦ χ)i (y) = ∂lψi (ϕχ(y)) · ∂mϕl (χ(y)) · ∂jχm (y). Now we can differentiate once more to get ∂k∂j(ψ ◦ ϕ ◦ χ)i (y) = ∂l ∂lψi (ϕχ(y)) · ∂m ϕl (χ(y)) · ∂kχm (y) · ∂mϕl (χ(y)) · ∂jχm (y) + ∂lψi (ϕχ(y)) · ∂n∂mϕl (χ(y)) · ∂kχn (y) · ∂jχm (y) + ∂lψi (ϕχ(y)) · ∂mϕl (χ(y)) · ∂k∂jχm (y). The second term is the standard way of transforming a bilinear map, the existence of the first and the third term however makes it clear that the second derivative depends on coordinates – it is relatively simple to come up with an example where D2 |xϕ is zero and D2 |y(ψ ◦ ϕ ◦ χ) is non-zero. The invariance under coordinate changes holds in two special cases: ψ and χ linear (so that the second derivatives vanish) and D|xϕ = 0 (more generally, the first non-vanishing derivative is invariant). To get an invariant object, it is necessary to take the Taylor polynomial, i.e. to include also the first derivative (and the value), but this results in an object very different from a symmetric bilinear form. 2. Bump functions Lemma 2.1. Let ε > 0. There exists a smooth function ρ: Rn → [0, 1] such that ρ(0) = 1 and ρ(x) = 0 for |x| ≥ ε. Proof. In the case n = 1, using the smooth function λ of the next lemma, we set ρ(x) = λ(ε + x)λ(ε − x)/λ(ε)2 = e − 2x2 ε(ε2−x2) on (−ε, ε) and zero otherwise – the equality is a straightforward calculation, giving ρ(x) ∈ [0, 1]. For general n, we use the constructed function for n = 1 (possibly for smaller ε) in the following way: ρ(x1 , . . . , xn ) = ρ(x1 ) · · · ρ(xn ). 3 Lemma 2.2. The function λ(x) = e−1/x x > 0 0 x ≤ 0 is smooth. Proof. For any positive integer n, it is known that ex > xn for x 0. Thus, e1/x > (1/x)n for x > 0 small or, in other words, e−1/x < xn for x > 0 small. It is easy to see that, in such a case, the right-sided derivatives of e−1/x at zero vanish up to order n − 1. 3. Derivations Let U ⊆ V be an open subset of a vector space and x0 ∈ U. We say that a mapping (an “operator”) A: C∞ U → R is a derivation at x0 if it is R-linear and satisfies the following “Leibniz rule at x0” A(f · g) = Af · g(x0) + f(x0) · Ag. An example is the differentiation along any vector at x0, i.e. A = Dv|x0 . We will now show that this is the only example. Theorem 3.1. Let A: C∞ U → R be a derivation at x0. Then there exists a unique vector v ∈ V such that Af = Dv|x0 f. In other words, there is an isomorphism between V and the space of derivations at x0. Proof. To make our formulas simpler, we assume x0 = 0. First we make a simple observation. From A(1 · 1) = A1 · 1 + 1 · A1 we see that A1 = 0. Consequently, Ac = c · A1 = 0 for any constant function c. In the first part, we prove the statement for U = V or, more generally, for functions that extend smoothly to V . Let f ∈ C∞ V and write f(x) − f(0) = [f(t · x)]1 t=0 = 1 0 ∂t(f(t · x)) dt = 1 0 ∂if(t · x) · xi dt = 1 0 ∂if(t · x) dt · xi . Denoting gi(x) = 1 0 ∂if(t · x) dt, a smooth function on V with gi(0) = ∂if(0), we get f = f(0) + gi · xi . Now we apply A to obtain Af = A(f(0)) 0 + Agi · xi (0) 0 + gi(0) ∂if(0) · Axi = ∂if(0) · Axi = Dv|x0 f, the derivative of f at x0 = 0 along v = (Ax1 , . . . , Axn ). Now we proceed to general open subset U ⊆ V . Since the first part applies to functions extendable to V , it will be our goal now to show that, in some sense, every function f : U → R extends to V . Let λ: V → [0, 1] be such that λ(x0) = 1 and such that supp λ ⊆ U. Then λ · f is clearly a smooth function that extends to V (by declaring it zero on V supp λ). According to the first part, we get Aλ = Dv|x0 λ = 0, since x0 is a maximum point of λ, and therefore A(λ · f) = Aλ · f(x0) + λ(x0) · Af = Af. As a special case, we get (this can be also seen directly) Dv|x0 (λ · f) = Dv|x0 f. Since the left hand sides of the two equalities agree by the first part (again, λ · f extends to V ), so do the right hand sides, i.e. Af = Dv|x0 f. We say that a mapping (an “operator”) X : C∞ U → C∞ U is a derivation if it is R-linear and satisfies the following “Leibniz rule” X(f · g) = Xf · g + f · Xg. Theorem 3.2. Let X : C∞ U → C∞ U be a derivation. Then there exists a unique smooth map v: U → V such that Xf(x) = Dv(x)|xf. 4 The map v should be thought of as a smooth vector field on U. We will be more precise on this matter later. Proof. For any x ∈ M, the composition of X with the evaluation map at x (i.e. evx(f) = f(x)), X|x : C∞ U X −→ C∞ U evx −−→ R, is a derivation at x and, thus, there is a unique vector v(x) such that Xf(x) = X|xf = Dv(x)|xf. It remains to show that x → v(x) is a smooth vector field. However, we have seen that v(x) = (X|xx1 , . . . , X|xxn ) = (Xx1 (x), . . . , Xxn (x)), i.e. v = (Xx1 , . . . , Xxn ), and as such is smooth. 4. Tangent map We have seen that vectors are in bijection with derivations at x. It will be useful to think of the corresponding vector v as “based at x” and, for emphasis, we will denote it as a pair A = (x, v) and use this based vector A interchangeably as a derivation, i.e. Af = Dv|xf. We denote by TxV = {x} × V the set of all vectors based at x, clearly a vector space isomorphic to V . It is called the tangent space of V at x and its elements, i.e. vectors based at x, the tangent vectors. We will now rephrase the last theorem in terms of based vectors. We denote by TU = U × V the set of all vectors based at all points of U and call it the tangent bundle of U. Then a vector field is a map X : U → TU with values X|x ∈ TxU, i.e. X|x = (x, v(x)). We will use Xf to denote the function Xf(x) = X|xf = Dv(x)|xf. An important special case is the vector field ∂i with ∂i|x = (x, ei) that gives the partial differentiation. Thus, any tangent vector A = (x, v) ∈ TxU with coordinates v = Ai · ei can be written as A = Ai · ∂i|x and any vector field can be written as X = Xi · ∂i, where Xi are now smooth functions. In addition to being simply the coordinate expression, these formulas also suggest how to take a derivative along A: Af = (Ai · ∂i|x)f = Ai · ∂i|xf, i.e. one multiplies the row of partial derivatives by the column of coordinates of A (or coordinate functions of a vector field X). It is easy to see that, for a smooth map ϕ: V // W and a tangent vector A ∈ TxV , the association f → A(f ◦ ϕ) is a derivation at ϕ(x) and is thus given by a tangent vector from Tϕ(x)W. Let us derive a formula for this vector. Writing A = (x, v), we have A(f ◦ ϕ) = D|x(f ◦ ϕ)(v) = D|ϕ(x)f ◦ D|xϕ(v) = DD|xϕ(v)|ϕ(x)f. Denoting ϕ∗A = (ϕ(x), D|xϕ(v)), we thus arrive at a formula A(f ◦ ϕ) = (ϕ∗A)f. Note that A is based at x and its image ϕ∗A is based at ϕ(x); this should be in correspondence with one’s geometric intuition. There results a map ϕ∗ : TU → TW or, again, ϕ∗ : TV // TW, called the tangent map of ϕ. The chain rule for this kind of “derivative” is particularly simple: (ψ ◦ ϕ)∗ = ψ∗ ◦ ϕ∗ (one can prove this either directly from the usual chain rule or formally from the above formula). Proposition 4.1. For a smooth map ϕ: V // W, the tangent map ϕ∗ : TV // TW is also smooth. Proof. This is clear from the definition, since ϕ∗(x, v) = (ϕ(x), D|xϕ · v) and ϕ: V // W, Dϕ: V // Hom(V, W) and the evaluation Hom(V, W) × V → W (the matrix multiplication) are smooth maps. 5 5. Implicit function theorem and its applications Theorem 5.1 (Implicit function theorem). Let F : Rn × Rk // Rk be a smooth map such that F(a, b) = 0 and such that D|(a,b)F ∈ Matk×(n+k) has the right (k × k)-block invertible. Then there exist neighbourhoods U a, V b such that for each x ∈ U there is a unique y = ϕ(x) ∈ V with F(x, y) = 0. Moreover, the resulting map ϕ: U → V is smooth. Now we will be concerned with applications of the theorem. We start with an inverse function theorem. Theorem 5.2 (Inverse function theorem). Let ϕ: Rn // Rn be a smooth map such that D|aϕ is regular. Then there exist neighbourhoods U a, V ϕ(a) such that the restriction ϕ: U → V is invertible with a smooth inverse. In particular, ϕ(a) lies in the interior of the image. Proof. We set F(x, y) = ϕ(y) − x. This satisfies the assumptions of the previous theorem and thus has a unique solution y ∈ V for each x ∈ U. Clearly, this y equals ϕ−1 (x) and, in particular, ϕ: V ∩ ϕ−1 (U) −→ U is a smooth bijection with an inverse smooth by the previous theorem. Now, we generalize the previous theorem in two ways to maps between spaces of different dimensions. Theorem 5.3 (Submersion theorem). Let ϕ: Rn // Rk be a smooth map such that D|aϕ is surjective. Then there exist neighbourhoods U a, V ϕ(a) and a diffeomorphism ψ such that Rn ϕ // Rk Rn ψ OO pr == where pr(x1 , . . . , xn ) = (x1 , . . . , xk ). In other words, in the coordinates given by ψ, the map ϕ is the projection map. Proof. We assume for simplicity that D|aϕ = A B has the left (k × k)-block A invertible. Let χ: Rn → Rn be the map χ    x1 ... xn    =           ϕ1 (x1 , . . . , xn ) ... ϕk (x1 , . . . , xn ) xk+1 ... xn           . Clearly, D|aχ = A B 0 E and is thus invertible. We set ψ = χ−1 according to the inverse function theorem. Since the diagram Rn ϕ // χ  Rk Rn pr == clearly commutes, the same is true for the diagram from the statement. Theorem 5.4 (Immersion theorem). Let ϕ: Rk // Rn be a smooth map such that D|bϕ is injective. Then there exist neighbourhoods V b, U ϕ(a) and a diffeomorphism ψ such that Rk ϕ // in !! Rn ψ  Rn 6 where in(x1 , . . . , xk ) = (x1 , . . . , xk , 0, . . . , 0). In other words, in the coordinates given by ψ, the map ϕ is the inclusion map. Proof. We assume for simplicity that D|bϕ = A B has the top (k × k)-block A invertible. Let χ: Rn → Rn be the map χ    x1 ... xn    =           ϕ1 (x1 , . . . , xk ) ... ϕk (x1 , . . . , xk ) xk+1 + ϕk+1 (x1 , . . . , xk ) ... xn + ϕn (x1 , . . . , xk )           . Clearly, D|aχ = A 0 B E and is thus invertible. We set ψ = χ−1 according to the inverse function theorem. Since the diagram Rk ϕ // in !! Rn Rn χ OO clearly commutes, the same is true for the diagram from the statement. 6. Submanifolds of a vector space Definition 6.1. A subset M ⊆ Rn is a smooth submanifold of dimension m if, for every x ∈ M, there is a diffeomorphism ψ: Rn // Rn defined near x such that ψ(M ∩ dom ψ) = Rm ∩ im ψ, where we understand Rm ⊆ Rn . Clearly, we may replace Rm in the definition by any affine subspace of Rn , since there is always an (affine) diffeomorphism that maps this subspace to Rm and we may compose the original diffeomorphism with the affine one and get a diffeomorphism from the definition. Theorem 6.2. Let F : Rn → Rk be a smooth map, b ∈ Rk and denote M = F−1 (b). If F is a submersion at every point of M then M is a smooth submanifold of dimension n − k. Proof. Under the assumptions of the theorem, near every x ∈ M, there is a diffeomorphism ϕ: Rn // Rn such that F = pr ◦ϕ. Therefore, F−1 (b) = ϕ−1 (pr−1 (b)) and clearly pr−1 (b) = {(b1 , . . . , bk , xk+1 , . . . , xn ) | xk+1 , . . . , xn ∈ R} is an affine subspace of dimension n − k. By definition, ψ restricts to a homeomorphism ϕ: M // Rm and we think of this map as introducing local coordinates on M and thus call it local coordinates or a local chart on M. Its inverse ϕ−1 : Rm // M is called a local parametrization of M. Let ϕ1, ϕ2 be two local charts, i.e. restrictions of diffeomorphisms ψ1, ψ2 from the definition of a submanifold. Then ϕ12 = ϕ2 ◦ ϕ−1 1 is called the transition map for the coordinates. It is clearly a diffeomorphism as a restriction of the diffeomorphism ψ2 ◦ ψ−1 1 . The above serves as a motivation for the definition of an abstract manifold (i.e. not a submanifold of some vector space). 7 7. Smooth manifolds Definition 7.1. A topological manifold is a topological space M that • is Hausdorff, • has a countable basis for topology, and • is locally euclidean, i.e. each point x ∈ M has an open neighbourhood U x that is homeomorphic to some open subset V ⊆ Rm , i.e. there exists ϕ: U ∼= −−→ V . We say that M has dimension m. We remind the reader that M is Hausdorff if any pair of distinct points x = y admits a pair of disjoint neighbourhoods U x, V y, U ∩ V = ∅. Equivalently, any sequence1 has at most one limit point (“⇒”: if a sequence converged to both x and y, it would have to lie eventually in U and V , thus eventually in U ∩ V = ∅, a contradiction; “⇐”: if B1/n(x) ∩ B1/n(y) was non-empty, containing a point zn, the sequence zn would converge to both x and y, a contradiction). Further, we remind that a basis for topology is a collection U of open sets such that any open set W is a union of some elements from this collection, W = U∈U,U⊆W U. Any open subset V ⊆ Rn is second countable, generated by all open balls Bε(x1 , . . . , xn ) ⊆ V with all x1 , . . . , xn and ε rational. For a pair of local charts ϕ1, ϕ2 on M, we again form the transition map ϕ12 = ϕ2 ◦ ϕ−1 1 . We say that the charts are compatible if the transition map is a diffeomorphism. An atlas is a collection of charts A = {ϕi | i ∈ I} whose domains cover M, i.e. i∈I dom ϕi = M, and such that any pair is compatible. Lemma 7.2. Any pair of charts ψ, χ compatible with an atlas A is itself compatible. Proof. We consider M ψ || ϕi  χ "" Rm ϕi◦ψ−1 // Rm χ◦ϕ−1 i // Rm with both horizontal maps smooth by assumption. Their composition χ ◦ ψ−1 is then also smooth at points where this composition is defined. But for any x ∈ dom ψ ∩ dom χ, we may choose ϕi with x ∈ dom ϕi and then the horizontal composition is defined at ψ(x). Corollary 7.3. Let A be an atlas. Then there exists a unique maximal atlas containing A, consisting of all charts compatible with A. Definition 7.4. A smooth manifold is a topological manifold M equipped with a maximal atlas. When speaking of charts (or coordinates) on a smooth manifold, we will always mean a chart from the given maximal atlas. Definition 7.5. A continuous map F : M → N between smooth manifolds is said to be smooth if, for every chart ϕ on M and every chart ψ on N, the composition ψ ◦ F ◦ ϕ−1 : Rm → Rn is smooth. M F // ϕ  N ψ  Rm ψ◦F ◦ϕ−1 // Rn Similarly to the previous lemma (in fact, the lemma is a special case for F = id), it is enough to check smoothness for some atlas on M and a collection of charts covering im F. Example 7.6. A smooth submanifold M ⊆ Rn is a smooth manifold and the inclusion ι is smooth. 1Since M is locally euclidean, every point has a countable basis of neighbourhoods and thus sequences suffice to capture topology. In general, one would replace them by nets or filters. 8 Example 7.7. A local chart is precisely a diffeomorphism M // Rm . Construct two local charts on Sm ⊆ Rm+1 and study the transition map (it should be a disc inversion). Remark. The charts are (x0, x) → 1 1±x0 · x with inverses x → 1 1+|x|2 (±(1 − |x|2), 2 · x). 8. Tangent bundle Definition 8.1. We say that A: C∞ M → R is a derivation at x0 if A is R-linear and satisfies the Leibniz rule at x0, A(f · g) = Af · g(x0) + f(x0) · Ag. Definition 8.2. We define the tangent space TxM of a smooth manifold M at a point x ∈ M to be TxM = {A: C∞ M → R | A is derivation at x}, the set of all derivations at x. It is clear that derivations at x are closed under addition and multiplication by real scalars and thus TxM is a vector space over R. Our main aim will now be to show that this vector space can be computed in coordinates and is thus isomorphic to Rm (in particular, it is finite dimensional, something that is not obvious from the definition). In order to compare the tangent spaces of various manifolds, we define the tangent map ϕ∗ of a smooth map ϕ: M → N in two steps. First, the precomposition with ϕ defines a map ϕ∗ : C∞ N → C∞ M given by ϕ∗ f = f ◦ ϕ and it is clearly a homomorphism of algebras, e.g. ϕ∗ (f · g) = ϕ∗ f · ϕ∗ g. Next comes: Lemma 8.3. The algebra homomorphism ϕ∗ determines a map ϕ∗ : Derx(C∞ M, R) → Derϕ(x)(C∞ N, R), given by ϕ∗A = A ◦ ϕ∗ , i.e. by (ϕ∗A)f = A(ϕ∗ f) = A(f ◦ ϕ). Proof. We need to show that ϕ∗A is indeed a derivation at ϕ(x). The linearity of ϕ∗A is clear and the Leibniz rule is (ϕ∗A)(f · g) = A(ϕ∗ (f · g)) = A(ϕ∗ f · ϕ∗ g) = A(ϕ∗ f) · ϕ∗ g(x) + ϕ∗ f(x) · A(ϕ∗ g) = (ϕ∗A)f · g(ϕ(x)) + f(ϕ(x)) · (ϕ∗A)g. By our definition of tangent spaces, ϕ∗ is thus a map ϕ∗ : TxM → Tϕ(x)N. We define TM = x∈M TxM and call it the tangent bundle of M. The various maps ϕ∗ : TxM → Tϕ(x)N define together a map ϕ∗ : TM → TN, called again the tangent map of ϕ. We use the notation ϕ∗x : TxM → Tϕ(x)N if we want to stress that it is defined only on the tangent space at x (see e.g. the next theorem). Theorem 8.4. If ϕ: M // N is a local diffeomorphism at x then ϕ∗x is an isomorphism. 9 Germs. In fact, we have not defined a tangent map to a partially defined map ϕ so far. This and the proof of the theorem will be achieved by passing to germs of functions at a point x. We consider smooth functions f : M // R defined on an open neighbourhood dom f x and an equivalence relation on such functions: f ∼ g if and only if f = g in some neighbourhood of x. The equivalence class of f is called the germ of f at x and denoted germx f. The set of all germs at x will be denoted C∞ x M. In particular, we obtain a map germx : C∞ M → C∞ x M associating to each (globally defined) function its germ at x. Lemma 8.5. The above map germx is surjective. Proof. Let f : U → R be a smooth function. We choose a function λ: M → [0, 1] such that λ = 1 in a neighbourhood of x and such that supp λ ⊆ U. Then germx f = germx(λ·f), since λ = 1 near x, and the product λ · f can be extended by zero to a smooth function of M, giving a preimage of germx f. It is easy to define linear combinations and products of germs in terms of their representatives (more abstractly, one may use the above lemma to view C∞ x M as the quotient algebra of C∞ M). Thus, it makes sense to say that a map C∞ x M → R is a derivation at x (on the other hand, it makes no sense to speak about derivations at other points, since the value of a germ at a point y different from x is ill-defined). Lemma 8.6. Every derivation A: C∞ M → R at x factors uniquely C∞ M A // germx  R C∞ x M ∃! << through a derivation C∞ x M → R, i.e. the precomposition with germx gives an isomorphism Derx(C∞ x M, R) ∼= −−→ Derx(C∞ M, R) = TxM. In words, the above lemma says that it is possible to define a unique derivation of germs if we impose that the derivation of a germ of a globally defined function f ∈ C∞ M is Af. Proof. The factorization, if exists, is unique by the surjectivity of germx. On the other hand, a factorization exists if any only if germx f = 0 ⇒ Af = 0. The condition germx f = 0 means that f = 0 in some neighbourhood U x. Let λ: M → [0, 1] be such that λ = 1 in a neighbourhood of x and such that supp λ ⊆ U. Set ρ = 1 − λ, so that ρ = 0 in a neighbourhood of x and ρ = 1 on M U. Thus, f = ρ · f and Af = A(ρ · f) = Aρ · f(x) 0 + ρ(x) 0 · Af = 0. Now we are ready to prove the theorem. Proof of Theorem 8.4. Clearly, any smooth map ϕ: M // N defines a map ϕ∗ : C∞ ϕ(x)N → C∞ x M. In the case that ϕ is a local diffeomorphism at x, the map ϕ∗ has an inverse (ϕ−1 )∗ and is therefore an isomorphism. In particular, it induces an isomorphism ϕ∗ : TxM = Derx(C∞ x M, R) ∼= −−→ Derϕ(x)(C∞ ϕ(x)N, R) = Tϕ(x)N. 10 Tangent bundle as a smooth manifold. We start with a simple observation. If U is a covering of a topological space X, then a subset V ⊆ X is open if and only if U ∩V is open for every U ∈ U. Thus, given an atlas on M, a subset V ⊆ M is open if and only if for every chart ϕ: M // Rm the image ϕ(V ∩ dom ϕ) ⊆ Rm is open. Consequently, an atlas also determines the toplogy of M. Now, we would like to equip TM with a structure of a smooth manifold. Its atlas will consist of the tangent maps to charts on M, i.e. for ϕ: M // Rm we consider A = {ϕ∗ : TM // TRm ∼= R2m }. Theorem 8.7. There is a structure of a smooth manifold of dimension 2m on TM, given by the above atlas. State the main ingredient formally. Proof. A smooth structure is given by a countable collection of maps M // Rm (charts), bijections from the domain onto the image, such that the transition maps are smooth maps between open subsets of Rm and such that any pair of points lies in a domain of a chart. Now the above property for M implies the same for TM, since the transition maps ϕij∗ are smooth. 9. Vector fields There is a canonical projection p: TM → M, associating to each A ∈ TxM its base p(A) = x. Definition 9.1. A smooth vector field is a smooth map X : M → TM such that X|x ∈ TxM, i.e. TM p  M id // X << M The definition of a smooth structure on TM implies that a vector field is smooth if and only if its expression in coordinates is smooth, i.e. if X = Xi ∂i with coordinate functions Xi smooth. Thus, every smooth vector field X induces a derivation X : C∞ M → C∞ M (denoted by the same symbol), given by Xf(x) = X|xf; locally Xf = Xi · ∂if and is clearly smooth. We have the following converse. Theorem 9.2. For a derivation X : C∞ M → C∞ M there exists a unique smooth vector field inducing it. Proof. The composition evx ◦X with the evaluation map evx : C∞ M → R gives a derivation at x and is thus given by a unique vector X|x ∈ TxM, Xf(x) = X|xf. It remains to show that X is smooth. In local coordinates, Xi = Xxi which almost gives smoothness except xi /∈ C∞ M so that Xxi does not make sense. This is corrected by passing, at each point, to germs and the extension of X|x to germs at x. We choose, in a coordinate neighbourhood U x0, a smooth function λ with λ = 1 in a neighbourhood V x0 and with supp λ ⊆ U; then for x ∈ V Xi (x) = X|xxi = X|x(λxi ) = X(λxi )(x) since germx xi = germx λxi . As λxi is a smooth function on M, so is its image X(λxi ) and Xi is smooth on V . Let γ : R // M be a path, i.e. we assume that dom γ is an interval. We define the tangent vector to γ at time t0 to be γ |t0 = γ∗(∂t|t0 ). The chain rule then easily gives ϕ∗(γ |t) = (ϕ ◦ γ) |t, i.e. the image of a tangent vector to a path γ under the tangent map ϕ∗ is the tangent vector to the image of the path under ϕ. 11 Definition 9.3. We say that γ is an integral curve of a vector field X if γ |t = X|γ(t) for each t ∈ dom γ. In local coordinates this reads ∂tγi (t) = Xi (γ1 (t), . . . , γm (t)), i.e. the tuple of coordinate functions γi forms a solution of a system of ordinary differential equations. Since the involved functions Xi are smooth, there exists a solution γx with any given initial value γx(0) = x. Together these form a map FlX : R × M // M, FlX (t, x) = γx(t), defined in a neighbourhood of {0} × M and smooth. This map is called the flow of X. Theorem 9.4. FlX (t, FlX (s, x)) = FlX (t + s, x). Proof. We have to show that the right hand side γ(t) = FlX (t + s, x) is an integral curve of X. However, since the translation by s clearly takes ∂t|t0 to ∂t|t0+s, we get γ |t0 = (FlX (−, x)) |t0+s = X|FlX (t0+s,x) = X|γ(t0). Thus, it is indeed an integral curve; the initial value is also correct, γ(0) = FlX (s, x). We say that X is complete if the flow is defined on R × M. We define a support of a vector field X, denoted supp X, to be the closure of the set {x ∈ M | X|x = 0}. Theorem 9.5. A compactly supported vector field is complete. In particular, any vector field on a compact manifold is complete. Proof. For every x ∈ supp X there is a neighbourhood Ux and εx > 0 such that FlX is defined on (−εx, εx) × Ux. Since the support is compact, we have supp X ⊆ Ux1 ∪ · · · ∪ Uxk . Taking ε = min{εx1 , . . . , εxk }, the flow of X is defined on (−ε, ε) × M (at points not in the support, the integral curve through that point is constant and thus defined on R). By the previous theorem, we may write, for any t ∈ R and any x ∈ M, FlX (t, x) = FlX (t/T, · · · FlX (t/T, x) · · · ) and for T 0 we have t/T ∈ (−ε, ε) so that the right hand side is defined. Definition 9.6. Let ϕ: M → N be a smooth map, X a vector field on M and Y a vector field on N. We say that X and Y are ϕ-related if ϕ∗X|x = Y |ϕ(x), i.e. if the following diagram commutes: TM ϕ∗ // TN M ϕ // X OO N Y OO We will occasionally denote this by X ∼ϕ Y . A special case is that of an inclusion of a submanifold, that we will denote in: M ⊆ N. Since each in∗x is injective, we may and will think of it as an inclusion of a subspace; then for each vector field Y ∈ XN, a vector field X ∈ XM with X ∼ϕ Y exists if and only if for each x ∈ M, the value Y |x lies in TxM ⊆ TxN. We say that Y is tangent to N. We have the following two characterizations using the induced derivation of functions and using the flows. Lemma 9.7. Vector fields X and Y are ϕ-related if and only if (Y f) ◦ ϕ = X(f ◦ ϕ). 12 Proof. This is clear upon unfolding the value of the right hand side at x: X(f ◦ ϕ)(x) = X|x(f ◦ ϕ) = (ϕ∗X|x)f while the value of the left hand side is simply Y |ϕ(x)f. Lemma 9.8. Vector fields X and Y are ϕ-related if and only if ϕ(FlX (t, x)) = FlY (t, ϕ(x)) In other words ϕ maps the flow lines of X to the flow lines of Y . We will use this property quite often. Proof. Taking the tangent vectors to the paths in the above equality, we get ϕ∗X|x = (ϕ(FlX (−, x))) |0 = (FlY (−, ϕ(x))) |0 = Y |ϕ(x), which is precisely the definition of ϕ-relatedness. In the opposite direction, given that X and Y are ϕ-related, we wish to prove the equality from the statement, i.e. we want to prove that γ(t) = ϕ(FlX (t, x)) is an integral curve of Y through ϕ(x). Since the initial value γ(0) = ϕ(x) is correct, we need only check that it satisfies the differential equation of an integral curve: γ (t0) = (ϕ(FlX (−, x))) |t0 = ϕ∗X|FlX (t0,x) = Y |ϕ(FlX (t0,x)) = Y |γ(t0) (the third equality uses the ϕ-relatedness). 10. Lie bracket We define the Lie bracket through derivations. Definition 10.1. Let X and Y be two vector fields on M. Then it is easy to see that f → XY f − Y Xf is a derivation and the corresponding vector field is denoted [X, Y ] and called the Lie bracket of the vector fields X and Y . We will now derive a coordinate formula XY f − Y Xf = Xj · ∂j(Y i · ∂if) − Y j · ∂j(Xi · ∂if) = Xj · (∂jY i · ∂if + Y i · ∂j∂if) − Y j · (∂jXi · ∂if + Xi · ∂j∂if) = (Xj · ∂jY i − Y j · ∂jXi ) · ∂if so that [X, Y ] = (Xj · ∂jY i − Y j · ∂jXi ) · ∂i. Proposition 10.2. The Lie bracket has the following properties: • [X, Y1 + Y2] = [X, Y1] + [X, Y2], • [X, f · Y ] = Xf · Y + f · [X, Y ], • [X, Y ] = −[Y, X], • [X, [Y, Z]] = [[X, Y ], Z] + [Y, [X, Z]]. • X ∼ϕ Z, Y ∼ϕ W ⇒ [X, Y ] ∼ϕ [Z, W]. Proof. All points are rather straightforward, we explain the most interesting one – the second: [X, f · Y ]g = X(f · Y g) − f · Y (Xg) = Xf · Y g + f · X(Y g) − f · Y (Xg) = Xf · Y g + f · ([X, Y ]g) = (Xf · Y + f · [X, Y ])g. The last point is also interesting: ([Z, W]f) ◦ ϕ = Z(Wf) ◦ ϕ − W(Zf) ◦ ϕ = X(Wf ◦ ϕ) − Y (Zf ◦ ϕ) = X(Y (f ◦ ϕ)) − Y (X(f ◦ ϕ)) = [X, Y ](f ◦ ϕ). 13 Denoting LXf = Xf and LXY = [X, Y ] (the Lie derivatives of f and Y along X), the second point becomes LX(f · Y ) = LXf · Y + f · LXY and the fourth becomes LX[Y, Z] = [LXY, Z] + [Y, LXZ], i.e. both are some forms of the Leibniz rule. Corollary 10.3. Let M ⊆ N be a submanifold. If X, Y ∈ XN are tangent to M, so is [X, Y ]. Proof. This is just the last point applied to the inclusion ϕ = in. Definition 10.4. Let X, Y be two vector fields on a manifold M. Then we denote (FlX t )∗ Y (x) = (FlX −t)∗Y (FlX t (x)) ∈ TxM the pullback of Y along the flow FlX t of X. For each x ∈ M it is defined for t small. (In other words, the pullback (FlX t )∗ Y is the unique vector field that is FlX t -related to Y .) The Lie derivative of Y along X is LXY (x) = ∂t|0((FlX t )∗ Y (x)). Theorem 10.5. Let X be a vector field and x ∈ M a point. If X|x = 0 then, in a neighbourhood of x, there exists a coordinate chart in which X = ∂1. Proof. We set X1 = X and choose vector fields X2, . . . Xm so that X1, . . . , Xm form a basis in a neighbourhood of x. We define a map ϕ: Rm // M, (t1 , . . . , tm ) → FlX1 t1 · · · FlXm tm (x) (it is defined in a neighbourhood of 0). The image of the coordinate vector field ∂i|0 = ∂t|0(0, . . . , t, . . . , 0) at the origin then equals ϕ∗∂i|0 = ∂t|0 FlXi t (x) = Xi|x and ϕ is a local diffeomorphism at 0. We may thus use its inverse ϕ−1 as a coordinate chart on M. Now, for i = 1, we get more generally ϕ∗∂1|t0 = ∂t|t1 0 FlX1 t FlX2 t2 0 · · · FlXm tm 0 (x) = X1(FlX1 t1 0 FlX2 t2 0 · · · FlXm tm 0 (x)) = X1|ϕ(t0) and this shows that, in the coordinates given by ϕ−1 , we have ∂1 = X1. Proposition 10.6. The following holds LXY |x = [X, Y ]|x. More generally, ∂t|t0 (FlX t )∗ Y |x = (FlX t0 )∗ [X, Y ]|x. Proof. First assume that t0 = 0. Let x ∈ M be such that X|x = 0. Then, by Theorem 10.5, there is a coordinate chart in which X = ∂1 near x. Then FlX t (x) = x + t · ∂1|x is the translation by the t-multiple of the coordinate vector ∂1|x, its derivative is then the identity. Consequently, (FlX −t)∗∂i|x+t·∂1 = ∂i|x, so that, for Y = Y i · ∂i, we get (FlX t )∗ Y |x = (FlX −t)∗Y (FlX t (x)) = Y i (x + t · ∂1) · ∂i|x and finally LXY |x = ∂t|0(FlX t )∗ Y |x = ∂1Y i (x) · ∂i. This equals [X, Y ]|x by the coordinate formula for the Lie bracket, proving the claim in this case. By continuity, the same holds for points in the closure of the set {x ∈ M | X|x = 0}, i.e. on the support of X. On the other hand, if x /∈ supp X, then X = 0 in a neighbourhood of x and then both sides equal zero. For a general t0, we have (FlX t )∗ Y |x = (FlX t0 )∗ (FlX t−t0 )∗ Y |x. Since (FlX t0 )∗ is a linear map we can interchange with ∂t. 14 Remark. We will need a useful property for the proof of the next proposition. It is based on an observation that for a function ϕ(s, t) of two variables, with values in a vector space, we have ∂t|t0 ϕ(t, t) = ∂t|t0 ϕ(t, t0) + ∂t|t0 ϕ(t0, t). Now let X be a time-dependent vector field, i.e. a map X : R × M → TM such that X(t, x) ∈ TxM (not true in the second application of (∗), but also not necessary there). Let f : R × M → R be a “time-dependent” function on M (just a function on R × M). Write Xt and ft for the vector field and function obtained by plugging in a specific value of t. Then we may form the directional derivative Xtft and ∂t|t0 Xtft = ∂t|t0 (Xtft0 ) + ∂t|t0 (Xt0 ft). (∗) (Locally, we have Xtft(x) = X(t, x)i · ∂if(t, pX(t, x)) and we apply the previous observation.) Proof. First assume that t0 = 0 and let f : M → R be a smooth function. We differentiate f in the direction of the left hand side: ∂t|0(FlX t )∗ Y (x) f (∗) = ∂t|0 (FlX t )∗ Y (x)f = ∂t|0 (FlX −t)∗Y (FlX t (x))f = ∂t|0 Y (FlX t (x))(f ◦ FlX −t) (∗) = ∂t|0 Y (x)(f ◦ FlX −t) + ∂t|0 Y (FlX t (x))(f) (∗) = Y (x) ∂t|0(f ◦ FlX −t) + ∂t|0 (Y f)(FlX t (x)) = Y (x)(−Xf) + X(x)(Y f) = −(Y Xf)(x) + (XY f)(x) = [X, Y ](x) f (the steps labeled by (∗) involve the observation made before the proposition, of which the first in the opposite direction). For a general t0, we have (FlX t )∗Y (x) = (FlX t0 )∗(FlX t−t0 )∗Y (x). Since (FlX t0 )∗ is a linear map we can interchange with ∂t. Corollary 10.7. The following conditions are equivalent: • [X, Y ] = 0, • (FlX t )∗ Y = Y , i.e. Y is FlX t -related with itself for all t, • FlX t FlY s (x) = FlY s FlX t (x), i.e. the flow lines commute. In general we have FlY −s FlX −t FlY s FlX t (x) = x + st[X, Y ](x) + o(s, t)2 . Proof. The equivalence of the three conditions follows immediately from the previous proposition – the second condition states that (FlX t )∗ Y is a constant function of t, i.e. that the derivative is zero and this is clearly equivalent to the first condition. At the same time, the second condition is equivalent to Y being FlX t -related to itself and this is equivalent to FlX t preserving the integral curves of Y , which is precisely the third condition. Differentiating the commutator of the flows twice, we get ∂t|0∂s|0 FlY −s FlX −t FlY s FlX t (x) = ∂s|0 −Y (x) + (FlX t )∗ Y (x) = [X, Y ](x) The remaining derivatives of order at most two are clearly zero. We will need the following generalization of Theorem 10.5. Theorem 10.8. If vector fields X1, . . . , Xk are linearly independent and satisfy [Xi, Xj] = 0 then, in a neighbourhood of any point x, there exists a coordinate chart in which Xi = ∂i. Proof. We choose vector fields Xk+1, . . . Xm so that X1, . . . , Xm form a basis in a neighbourhood of x. We define a map ϕ: Rm // M, (t1 , . . . , tm ) → FlX1 t1 · · · FlXm tm (x) (it is defined in a neighbourhood of 0). The image of the coordinate vector field ∂i|0 = ∂t|0(0, . . . , t, . . . , 0) at the origin then equals ϕ∗∂i|0 = ∂t|0 FlXi t (x) = Xi|x 15 and ϕ is a local diffeomorphism at 0. We may thus use its inverse ϕ−1 as a coordinate chart on M. Now, for i ≤ k, we study the image of the coordinate vector field ∂i|t0 at a general point t0. Since we may interchange the flows (since [Xi, Xj] = 0), we get ϕ∗∂i|t0 = ∂t|ti 0 FlXi t FlX1 t1 0 · · · FlXi ti 0 · · · FlXm tm 0 (x) = Xi(FlXi ti 0 FlX1 t1 0 · · · FlXi ti 0 · · · FlXm tm 0 (x)) = Xi|ϕ(t0) and this shows that, in the coordinates given by ϕ−1 , we have ∂i = Xi, for i ≤ k. 11. Distributions Definition 11.1. A (non-smooth) distribution S of dimension k is a mapping x → S(x) that associates to each point x ∈ M a k-dimensional vector subspace S(x) ⊆ TxM. A distribution S is smooth if, for each point x0 ∈ M, there exist a neighbourhood U and local vector fields X1, . . . , Xk ∈ XU such that X1|x, . . . , Xk|x form a basis of S(x) for x ∈ U. From now on, all our distributions will be smooth. Definition 11.2. An submanifold N ⊆ M is said to be an integral manifold of a distribution S if, for each x ∈ N, one hase S(x) = TxN. A distribution S on M is called integrable if, for each x ∈ M, there exists an integral manifold passing through x. We say that a vector field X lies in S if, for each x, we have X|x ∈ S(x). Integrable distributions have a special property of being involutive. In fact, the converse also holds, as we will see shortly. Definition 11.3. A distribution S on M is called involutive if, for every pair of vector fields X, Y lying in S, their bracket [X, Y ] also lies in S. Theorem 11.4. Every integrable distribution is involutive. Proof. Let X and Y be vector fields lying in S, let x ∈ M be an arbitrary point and let N be an integral manifold passing through x. Then both X and Y are tangent to N and, thus, so is their Lie bracket [X, Y ]. In particular, [X, Y ]|x ∈ S(x). Since x was arbitrary, [X, Y ] indeed lies in S. Now we are ready to prove the converse, in fact it proves a stronger version of integrability, since it also describes how the integral manifolds vary locally – they form a so-called foliation. Theorem 11.5 (Frobenius theorem). If S is involutive then for every x0 ∈ M there exists a local coordinate system in a neighbourhood U x0 such that the vector fields ∂1, . . . , ∂k form a basis of the distribution S on U. In particular, S is integrable. Proof. Let X1, . . . , Xk be vector fields defined in a neighbourhood of x0 that form a basis of S. By composing the local chart M // Rm with a suitable projection map Rm → Rk , we get a map pr: M // Rk for which pr∗x : S(x) ⊆ TxRm → Tpr(x)Rk is an isomorphism for x = x0 and consequently also for x in a neighbourhood of x0. Since ∂1, . . . , ∂k form a basis, we may write pr∗(X1|x, . . . , Xk|x) = (pr∗ X1|x, . . . , pr∗ Xk|x) = (∂1|pr(x), . . . , ∂k|pr(x)) · A(x) for an invertible matrix A(x) ∈ GL(k) that depends smoothly on x. Since the map GL(k) → GL(k), M → M−1 is smooth (it is given by a rational map using the determinant of the matrix and its minors), the matrix A−1 (x) also depends smoothly on x. Defining (Y1|x, . . . , Yk|x) = (X1|x, . . . , Xk|x) · A−1 (x), clearly vector fields lying in S and also giving a basis of S in a neighbourhood of x0, we easily get pr∗(Y1|x, . . . , Yk|x) = pr∗(X1|x, . . . , Xk|x) · A−1 (x) = (∂1|pr(x), . . . , ∂k|pr(x)) so that the Yi are pr-related to the ∂i. But then [Yi, Yj] is pr-related to [∂i, ∂j] = 0. Since [Yi, Yj] lies in S by involutivity and pr∗x is an isomorphism on S(x), we get [Yi, Yj] = 0. Thus, Theorem 10.8 applies. 16 Proof. Let X1, . . . , Xk be local vector fields which span the distribution S near x and choose vector fields Xk+1, . . . , Xn so that (X1, . . . , Xn) form a basis near x. We then define a map ϕ : Rn ⊇ U −→ M (t1 , . . . , tn ) −→ FlX1 t1 · · · FlXn tn (x) The partial derivatives at the origin clearly consist of the vectors Xi(x) and thus ϕ is a local diffeomorphism – its inverse will form our coordinate system. Let us compute the partial derivative with respect to ti for i ≤ k at a general point. ∂iϕ(t1 , . . . , tn ) = FlX1 t1 ∗ · · · Fl Xi−1 ti−1 ∗ Xi Fl Xi ti · · · FlXn tn (x) To conclude the proof it is therefore enough to show that for any Y belonging to S the pullbacks (FlY t )∗Xi also belong to S (then the same will be true for pullbacks (FlY t )∗X with X ∈ S by linearity, and we apply the claim to Xi, Fl Xi−1 ti−1 ∗ Xi, etc.) Denote this pullback by Yi(t) = (FlY t )∗ Xi(x) ∈ TxM and write [Y, Xi] = aj i Xj. By Lemma 10.6 the paths Yi(t) satisfy the following system of differential equations d dt Yi(t) = (FlY t )∗ [Y, Xi] = aj i (FlY t (x))Yj(t) We have Yi(0) = Xi(x) ∈ S(x) and, since the system is linear, we must have Yi(t) ∈ S(x) for all t (namely, there exists a solution of the system d dt Zi(t) = aj i (FlY t (x))Zj(t) with Zi ∈ S(x) and with Zi(0) = Xi(x). By uniqueness, we must have Yi(t) = Zi(t) and, thus, Yi(t) ∈ S(x).) Theorem 11.6 (Frobenius theorem through 1-forms). Let ω: TM → V be a smooth map that is linear on each TxM (we say that ω is a V -valued 1-form) and surjective. Then ker ω is a distribution. It is integrable if and only if ω(X) = 0, ω(Y ) = 0 ⇒ dω(X, Y ) = 0. This uses the exterior differential of the next section. Proof. In local coordinates on M and in a basis of V , the 1-form ω is given by a matrix of maximal rank. We may assume that the left most square block is regular in a neighbourhood of a given point and use the Gauss elimination to make this matrix (E | A). Then ker ω is given by (ξn−k+1 , . . . , ξn )T = −A(ξ1 , . . . , ξn−k ), proving that it is a (smooth) distribution. Now dω(X, Y ) = Xω(Y ) − Y ω(X) − ω([X, Y ]) gives easily the result. 12. Cotangent bundle Every smooth function f ∈ C∞ M defines a mapping df : TM → R, called the differential of f at x, given by df(A) = Af i.e. its values are the various directional derivatives of f along various tangent vectors. The restriction to each tangent space TxM is then a linear form df|x ∈ (TxM)∗ , i.e. an element of the cotangent space T∗ x M = (TxM)∗ . We denote T∗ M = x∈M T∗ x M and call it the cotangent bundle of M. Its elements are called cotangent vectors. We will soon equip the cotangent bundle T∗ M with a structure of a smooth manifold. At this point, we want to rephrase the differential: it can be viewed as a mapping df : M → T∗ M, x → df|x ∈ T∗ x M. Again, it is a field in the sense that it maps each point to an object of the appropriate vector space. In coordinates, for A = Ai ∂i|x, we get df|x(A) = Af = (Ai ∂i|x)f = Ai · ∂i|xf, (the row of partial derivatives of f, as expected). As a special case, we may apply this to the coordinate functions xi to get dxi |x(A) = Aj · ∂j|xxi = Ai so that we may rewrite the formula for df as df|x(A) = dxi |x(A) · ∂i|xf, or simply df = ∂if · dxi . 17 This formula is well known from calculus, but now it has an exact meaning, the differentials are sections of cotangent bundles, partial derivatives are functions and both the product and the (implicit) sum make sense. We will now study how cotangent vectors transform along a smooth map ϕ: M → N. Namely, for x ∈ M, the tangent map is ϕ∗x : TxM → Tϕ(x)N and thus induces a dual linear map (ϕ∗x)∗ : T∗ ϕ(x)N → T∗ x M. When ϕ is a local diffeomorphism its inverse is then a linear map ϕ∗x : T∗ x M → T∗ ϕ(x)N. For varying x ∈ M, these form a map ϕ∗ : T∗ M → T∗ N. We will now show that this map is smooth for M and N open subsets of Rm . First, we have (ϕ∗x)∗ (ϕ(x), η) = (x, η ◦ D|xϕ) and thus ϕ∗(x, θ) = (ϕ(x), η ◦ (D|xϕ)−1 ). Since the composition (matrix multiplication) is smooth and so is the map GL(m) → GL(m), A → A−1 , the result follows. Since the transition maps are smooth, we may use the induced maps ϕ∗ : T∗ M // T∗ Rm to give a smooth atlas for T∗ M. Definition 12.1. A 1-form on M is a smooth map ω: M → T∗ M such that ω|x ∈ T∗ x M. The set of all 1-forms on M will be denoted Ω1 M. In local coordinates, we have ω = ωi·dxi and the smoothness is equivalent to the functions ωi being smooth, since in the charts ϕ and ϕ∗ the coordinate expression is ω(x) = (x, (ω1(x), · · · , ωm(x))). In particular, for every smooth function f, its differential df is a 1-form. An important feature of forms is that they pull back along smooth maps: for a 1-form ω on N and a smooth map ϕ: M → N, we get a 1-form ϕ∗ ω on M given by (ϕ∗ ω)|x = (ϕ∗x)∗ (ω|ϕ(x)) = ω|ϕ(x) ◦ ϕ∗x. It will be useful to give a local formula for the pull back: if ω = ωj · dyj , we obtain from the following lemma that ϕ∗ ω = ϕ∗ ωj · d(ϕ∗ yj ϕj ) = ωj ◦ ϕ · ∂iϕj · dxi . (writing yj = ϕj the essential part of the formula becomes very intuitive: dyj = ∂yj /∂xi · dxi .) Lemma 12.2. ϕ∗ df = d(ϕ∗ f) and ϕ∗ (f · ω) = ϕ∗ f · ϕ∗ ω. Proof. The first equality is just the definition of the push forward, (ϕ∗A)f = A(f ◦ ϕ), rewritten in terms of the differential – the left hand side is df(ϕ∗A) = (ϕ∗ df)(A) and the right hand side is d(f ◦ ϕ)(A) = d(ϕ∗ f)(A). The second equality is straightforward using the linearity of the dual map: (ϕ∗ (f · ω))|x = (ϕ∗x)∗ (f(ϕ(x)) · ω|ϕ(x)) = f(ϕ(x)) · (ϕ∗x)∗ (ω|ϕ(x)) = (ϕ∗ f · ϕ∗ ω)|x Remark. We have the following relation of df(A) = Af to f∗A ∈ Tf(x)R Af = A(id ◦f) = (f∗A) id where, clearly, (t, τ) id = τ, i.e. taking the derivative of the identity makes from a based vector the corresponding free vector. 18 13. Tensor fields A common generalization of vector fields and 1-forms are the so called tensor fields. They are associations x → ω|x ∈ r TxM ⊗ s T∗ x M that are smooth in a sense similar to that of vector fields and 1-forms. This is again achieved by defining a smooth manifold r TM ⊗ s T∗ M = x∈M r TxM ⊗ s T∗ x M called the tensor bundle. Again, this is done via coordinate charts ϕ: M // Rm on M by using their induced maps on tensor bundles ϕ∗x = r ϕ∗x ⊗ s ϕ∗x (the two maps ϕ∗x are different – the first is for the tangent bundle and the second for the cotangent bundle). Again, it is easy to see that these are smooth for open subsets U ⊆ Rm where r TU ⊗ s T∗ U = U × r Rm ⊗ s (Rm )∗ . The corresponding field is then called a tensor field of type (r, s). Of special importance are tensor fields of type (0, k), i.e. those with values in the tensor power of the cotangent bundle, since they again pull back along smooth maps: if ϕ: M → N is a smooth map and ω: N → k T∗ N a tensor field then (ϕ∗ ω)|x = ω|ϕ(x) ◦ (ϕ∗x)⊗k Since (T∗ x M)⊗k is naturally isomorphic to the vector space Link(TxM, . . . , TxM; R) of k-linear forms on TxM, every tensor field ω of type (0, k) can be also seen as a collection of k-linear forms ω|x on TxM and we will use ω(A1, . . . , Ak) to denote the values on a k-tuple of tangent vectors, necessarily in a single tangent space TxM. We may then rewrite the above definition of the pull back of ω as (ϕ∗ ω)(A1, . . . , Ak) = ω(ϕ∗A1, . . . , ϕ∗Ak). We will now explain in more detail how we view a tensor product of 1-forms as a k-linear form: ω1 ⊗ · · · ⊗ ωk (A1, . . . , Ak) = ω1 (A1) · · · ωk (Ak). In particular, we may write every tensor field of type (0, k) locally as ω = ωi1···ik · dxi1 ⊗ · · · ⊗ dxik and then we get ω(A1, . . . Ak) = ωi1···ik (A1)i1 · · · (Ak)ik . We clearly have ϕ∗ (ω ⊗ θ) = ϕ∗ ω ⊗ ϕ∗ θ and thus the pull back ϕ∗ ω has the form ϕ∗ ω = (ωi1···ik ◦ ϕ) · ∂j1 ϕi1 · · · ∂jk ϕik · dyj1 ⊗ · · · ⊗ dyjk . An important special case are the antisymmetric tensor fields of type (0, k), also called (exterior) k-forms. We adopt the convention for ωi ∈ Ωki M: ω1 ∧ · · · ∧ ωr = (k1+···+kr)! k1!···kr! · Alt(ω1 ⊗ · · · ⊗ ωr ). It is not difficult to check that this wedge product is associative. The advantage of the factor stems from the fact that it eliminates the appearance of such factors in many subsequent formulas (there are also some more subtle advantages that we will not discuss here). Also, for ki = 1, i.e. for 1-forms ωi , the formula becomes ω1 ∧ · · · ∧ ωr = σ∈Σr sign(σ) · ωσ(1) ⊗ · · · ⊗ ωσ(r) . The evaluation at a k-tuple of vectors is thus ω1 ∧ · · · ∧ ωk (A1, . . . , Ak) = σ∈Σk sign(σ) · ω1 (Aσ(1)) · · · ωk (Aσ(k)). 19 Applying these formulas to the 1-forms dxi results, for ω = ωi1···ik · dxi1 ∧ · · · ∧ dxik , in the following ω(A1, . . . , Ak) = σ∈Σk sign(σ) · ωi1···ik · (Aσ(1))i1 · · · (Aσ(k))ik . 14. Exterior differential 14.1. Differentiation of tensor fields on a vector space. A tensor field of type (r, s) on an open subset of a vector space V = Rm may be interpreted as a map ω: V // r V ⊗ s V ∗ and as such may be differentiated along any vector field X, giving another tensor field DXω of type (r, s). The total derivative Dω is then a map Dω: V // Hom(V, r V ⊗ s V ∗ ) ∼= r V ⊗ s+1 V ∗ , i.e. a tensor field of type (r, s + 1). In coordinates, this consists simply of differentiating the coordinate functions of the tensor field, i.e. for ω = ωj1···jr i1···is · ∂j1 ⊗ · · · ⊗ ∂jr ⊗ dxi1 ⊗ · · · ⊗ dxis we obtain (DXω)j1···jr i1···is = ∂i0 ωj1···jr i1···is · Xi0 , so that we obtain the final formula (Dω)j1···jr i0i1···is = ∂i0 ωj1···jr i1···is . 14.2. Differentiating vector fields along paths. We concentrate in this paragraph on the case of vector fields Y , since this is the only case needed later. We observe that DXY |x depends only on the value X|x and on the values of Y “in the direction of X|x”. More concretely, if X|x = γ (t0) and if we denote Z = Y ◦ γ, then DXY depends only on Z: Dγ (t0)Y = Dγ (t0)Y j · ∂j = ∂t|t0 (Y j ◦ γ) · ∂j = ∂t|t0 Zj · ∂j = DtZ|t0 or more concisely Dγ Y = DtZ. More generally, a vector field along γ is a smooth map Z : R // TM such that Z|t ∈ Tγ(t)M. Then the formula DtZ|t0 = ∂t|t0 Zj · ∂j as above defines, for an arbitrary vector field Z along γ, another vector field DtZ along γ. In particular, the tangent vectors γ = ∂tγi · ∂i form a vector field along γ and we get Dtγ = γ = ∂2 ttγi · ∂i. 14.3. Non-invariance of D. Let us study the invariance of D under the change of coordinates, i.e. let ω be a tensor field of type (0, k) with components ωi1···ik , ω = ωi1···ik · dxi1 ⊗ · · · ⊗ dxik , and recall that (Dω)i0i1···ik = ∂i0 ωi1···ik . Now we apply the change of coordinates ϕ to get (ϕ∗ (Dω))j0j1···jk = (∂i0 ωi1···ik ◦ ϕ) · ∂j0 ϕi0 · ∂j1 ϕi1 · · · · · ∂jk ϕik . On the other hand D(ϕ∗ ω) equals (D(ϕ∗ ω))j0j1···jk = ∂j0 ((ωi1···ik ◦ ϕ) · ∂j1 ϕi1 · · · · · ∂jk ϕik ) = ∂j0 ((ωi1···ik ◦ ϕ)) · ∂j1 ϕi1 · · · · · ∂jk ϕik + r (ωi1···ik ◦ ϕ) · ∂j1 ϕi1 · · · · · ∂2 j0jr ϕir · · · · · ∂jk ϕik The first term equals ϕ∗ (Dω)j0j1···jk by the chain rule. Now the point is that in order to get ϕ∗ (Dω) = D(ϕ∗ ω), i.e. to get a differential that does not depend on coordinates as we will see 20 shortly, we have to get rid of the second term involving the second derivative ∂2 j0jr ϕir . It clearly disappears after antisymmetrization. Some representation theory would be required to get that no “other part” is invariant and we will not attempt to do this here. Thus, we get an invariant differentiation operator – the exterior differential – on antisymmetric forms by antisymmetrizing DX0 ω(X1, . . . , Xk); for technical reasons, we multiply the antisymmetrization by (k+1)! 1!k! , since the form already was antisymmetric in the variables X1, . . . , Xk and obtain dω(X0, . . . , Xk) = i (−1)i DXi ω(X0, . . . , Xi, . . . , Xk) Starting from the formula for D, Dω = ∂i0 ωi1···ik dxi0 ⊗ dxi1 ∧ · · · ∧ dxik , we obtain a formula for d by antisymmetrization, i.e. dω = ∂i0 ωi1···ik dxi0 ∧ dxi1 ∧ · · · ∧ dxik . We will now explain a useful formalism. We denote, for a k-tuple I = (i1, . . . , ik), dxI = dxi1 ∧ · · · ∧ dxik and also ωI = ωi1···ik so that we may write ω = ωI · dxI where the implicit summation occurs over all ordered k-tuples I. We assume ωI antisymmetric in I, as usual. We then get a very simple formula for all operations: (ωI · dxI ) ∧ (θJ · dxJ ) = ωIθJ · dxIJ and most importantly dω = ∂kωI · dxk dωI ∧dxI = dωI ∧ dxI . Proposition 14.1. The exterior differential satisfies ϕ∗ (dω) = d(ϕ∗ ω) and, therefore, it induces an operator d: Ωk M → Ωk+1 M on any smooth manifold. Proof. The invariance was proved earlier. This enables to define dω in the domain of a chart ϕi by the formula dω = (ϕi)∗ d((ϕ−1 i )∗ ω), i.e. ω is translated to the chart, differentiated there and then translated back to M. For any other chart ϕj, we have ϕj = ϕij ◦ ϕi and then (ϕj)∗ d((ϕ−1 j )∗ ω) = (ϕi)∗ (ϕij)∗ d((ϕ−1 ij )∗ d (ϕ−1 i )∗ ω) = (ϕi)∗ d((ϕ−1 i )∗ ω) by the invariance with respect to the transition map ϕij, wherever both sides are defined. Theorem 14.2. The exterior differential satisfies the following properties: (1) the exterior differential of a 0-form, i.e. a function f, is the usual differential, (2) d(ω + θ) = dω + dθ, (3) d(ω ∧ θ) = dω ∧ θ + (−1)|ω| · ω ∧ dθ, (4) d(dω) = 0. It is the unique invariant operator satisfying these properties. Proof. The first point is clear from the definition and so is the second. For the third point, write ω = ωI · dxI , θ = θJ · dxJ and thus we get d(ω ∧ θ) = d(ωIθJ · dxI ∧ dxJ ) = (∂kωI · θJ + ωI · ∂kθJ ) · dxk ∧ dxI ∧ dxJ = (∂kωI · dxk ∧ dxI ) ∧ (θJ · dxJ ) + (−1)|ω| · (ωI · dxI ) ∧ (∂kθJ · dxk ∧ dxJ ), where the sign comes from interchanging dxk with dxI . First observe d(df) = d(∂if · dxi ) = ∂j∂if · dxj ∧ dxi 21 and, for each i and j, the term ∂jif · dxji cancels with the term ∂ijf · dxij since the partial derivatives are symmetric, while dxij = −dxji . By the Leibniz rule, we then get d(dxI ) = d(dxi1 ∧ · · · ∧ dxik ) = (−1)r−1 · dxi1 ∧ · · · ∧ d(dxir ) 0 ∧ · · · ∧ dxik = 0. Consequently, for general ω = ωI · dxI , we get d(dω) = d(dωI · dxI ) = d(dωI) ∧ dxI − dωI ∧ d(dxI ) = 0 − 0 = 0. 14.4. Coordinate-free formula for the exterior differential. The differentiation operator DX0 satisfies the Leibniz rule DX0 (ωi1···ik (X1)i1 · · · (Xk)ik ) = DX0 ωi1···ik · (X1)i1 · · · (Xk)ik + ωi1···ik (X1)i1 · · · DX0 (Xj)ij · · · (Xk)ik which (after subtracting the sum from the right hand side) translates to (DX0 ω)(X1, . . . , Xk) = DX0 (ω(X1, . . . , Xk)) − ω(X1, . . . , Xj−1, DX0 Xj, Xj+1, . . . , Xk). Here the first DX0 on the right is the directional derivative of the function ω(X1, . . . , Xk). The second appearance is, however, very different and we have [X, Y ] = DXY − DY X. The exterior differential then equals dω(X0, . . . , Xk) = i (−1)i (DXi ω)(X0, . . . , Xi, . . . , Xk) = i (−1)i Xiω(X0, . . . , Xi, . . . , Xk) − ij (−1)i ω(X0, . . . , Xj−1, DXi Xj, Xj+1, . . . , Xi, . . . , Xk) where we split the antisymmetrization of the second term according to whether i < j or i > j. Next we move the term DXi Xj onto the first spot (here the sign differs for the two possibilities): = i (−1)i Xiω(X0, . . . , Xi, . . . , Xk) + ij (−1)i+j ω(DXi Xj, X0, . . . , Xj, . . . , Xi, . . . , Xk) and finally swap the indices i, j in the last sum and subtract, using DXi Xj − DXj Xi = [Xi, Xj], to obtain the final formula = i (−1)i Xiω(X0, . . . , Xi, . . . , Xk) + i 0 and supp λx ⊆ Ux. Clearly the open sets Vx = {y ∈ M | λx(y) > 0} cover M (since Vx x) and thus, by compactness, there is a finite subcover M = Vx1 ∪ · · · ∪ Vxk . This means that λ = λx1 + · · · + λxk > 0 on M. Take λU = ( Uxi =U λxi )/λ. Corollary 15.3. There exists a Riemannian metric on every manifold. Proof. We choose a Riemannian metric gU arbitrarily in every coordinate chart U. Using a partition of unity λU (assuming that M is compact or using the non-compact version of the previous theorem), we set g = U∈U λU gU . 15.2. Orientation. An orientation of a manifold is an orientation of each tangent space TxM that is “smooth” in the following sense: for every m-tuple α = (X1, . . . , Xm) of local vector fields that form a basis of TxM where defined (the so-called local frame), the function sign α is smooth. Since it takes values in {±1}, it must in fact be locally constant. Let ϕ: M → N be a local diffeomorphism. Then, for each x ∈ M, the tangent map ϕ∗x : TxM → Tϕ(x) is an isomorphism and we define sign ϕ∗x to be +1 if ϕ∗x preserves orientation and −1 if it reverses orientation. In formula, sign(ϕ∗xα|x) = sign ϕ∗x · sign α|x For each (local) frame β on N, there is a (local) frame α = ϕ∗ β on M (consisting of the pullbacks (ϕ∗ Y )|x = (ϕ∗x)−1 Y |ϕ(x)) so that the two frames are ϕ-related. Thus, sign β|ϕ(x) = sign(ϕ∗xα|x) = sign ϕ∗x · sign α|x and since both sign α|x and sign β|ϕ(x) are locally constant, so is sign ϕ∗x. In particular, if M is connected, then ϕ either preserves orientation at every point or it reverses orientation at every point. In particular, every connected chart either preserves orientation or its composition with a reflection Rm → Rm preserves orientation. Consequently, the collection of all orientation preserving charts forms an atlas – the maximal oriented atlas on M. For this atlas, the transition maps also preserve orientation. (There is a special case m = 0, where no reflection exists and, thus, it is not always possible to get an oriented atlas.) 15.3. Integral. Let ω be an m-form on an open subset V ⊆ Rm and we assume for simplicity that it has compact support. Writing ω = a · dx1 ∧ · · · ∧ dxm , i.e. a = ω(∂1, . . . , ∂m), we define V ω = · · · V a(x) dx1 · · · dxm . Clearly, the above defined integral is additive in ω. For a smooth map ϕ: Rm // Rm , we get ϕ∗ ω(y) = a ◦ ϕ(y) · ϕ∗ ydx1 ∧ · · · ∧ ϕ∗ ydxm = a ◦ ϕ(y) · det ϕ∗y · dy1 ∧ · · · ∧ dym . with det ϕ∗y the Jacobian; thus, for a diffeomorphism ϕ: W → V with positive Jacobian, we get W ϕ∗ ω = V ω. 23 Let M be an oriented manifold of dimension m. We assume for simplicity that M is compact. Consider the maximal oriented atlas on M and choose a partition of unity λi so that supp λi is a subset of a domain Ui of a chart ϕi : Ui → Vi. Let ω be an m-form on M. Then we define M ω = i Vi (ϕ−1 i )∗ (λiω) We note that λiω has a compact support inside Ui and, thus, the pullback (ϕ−1 i )∗ (λiω) has a compact support inside Vi so that the integral exists and is finite. It remains to show that it does not depend on the choice of the partition λi. Thus, let µi be another partition. Then we get i Vi (ϕ−1 i )∗ (λiω) = i,j Vi (ϕ−1 i )∗ (λiµjω) Denoting θ = λiµjω, a compactly supported m-form inside Ui ∩ Uj, and further Vij = ϕi(Ui ∩ Uj) and Vji = ϕj(Ui ∩ Uj), we have Vi (ϕ−1 i )∗ θ = Vij (ϕ−1 i )∗ θ = Vij ϕ∗ ij((ϕ−1 j )∗ θ) = Vji (ϕ−1 j )∗ θ = Vj (ϕ−1 j )∗ θ by the invariance of the integral with respect to the diffeomorphism ϕij = ϕj ◦ ϕ−1 i : Vij → Vji. By the additivity of the integral, we may reformulate the procedure: write the m-form ω as a finite sum ω = ωi of m-forms with each ωi concentrated in an oriented coordinate patch. Then ω = ωi and each of the integrals is computed in coordinates, ωi = · · · ωi (∂1, . . . , ∂m) dx1 · · · dxm . 15.4. Manifolds with boundary. The main idea here is: in exactly the same manner in which manifolds are built from the Euclidean space Rm , manifolds with boundary are built from the Euclidean halfspace Hm+1 = {(x0 , x1 , . . . , xm ) ∈ Rm+1 | x0 ≤ 0}. It is however important that we allow tangent vectors at the boundary hyperplane to be all vectors from Rm+1 , i.e. THm+1 = Hm+1 × Rm+1 . Thus, the geometric definition using paths is inappropriate. Derivations work well if we interpret ∂0f(x) for a boundary point x to be the one-sided partial derivative. Formally, a map between open subsets of the half-spaces is said to be smooth, if all partial derivatives exist (one-sided where needed) and are continuous. A diffeomorphism between open subsets of Hm+1 preserves the boundary points, since at an interior point, any (local) diffeomorphism has a local inverse and as such maps to an interior point. With this notion, we define a (smooth) manifold with boundary W as a topological space, Hausdorff and with countable basis of topology, equipped with a maximal atlas consisting of homeomorphisms ϕ: U → V with V an open subset of Hm+1 and with all change of coordinate maps smooth in the above sense. We define the boundary of W to be the set ∂W of points that correspond to the boundary points in a chart (equivalently, in all charts). The standard bases (e0, e1, . . . , em) of Hm+1 and (e1, . . . , em) of Rn are considered positive. We say that ∂Hm+1 is oriented via its outward normal: The outward normal is by definition u0 = e0 (or any combination u0 = x0 e0 + x1 e1 + · · · xm em with x0 > 0) and a basis (u1, . . . , um) is then a positive basis of ∂Hm+1 according to this principle if and only if (u0, u1, . . . , um) is a positive basis of Hm+1 . This gives a way of orienting the boundary ∂W of any oriented manifold with boundary W. We will always consider ∂W with this induced orientation. 15.5. Stokes’ theorem. Theorem 15.4. For a compact manifold with boundary W of dimension m + 1 and an m-form ω on W, we have ∂W ω = W dω. 24 (The left hand side is really the integral of the pullback j∗ ω along the inclusion j : ∂W → W.) Proof. We may write ω as a sum of m-forms supported in a coordinate chart and thus reduce to a local situation, i.e. we may assume that W = Hm+1 . Since ω is an m-form on Hm+1 , we may write ω = i ai · dx0 ∧ · · · ∧ dxi ∧ · · · ∧ dxm . Since j(x1 , . . . , xm ) = (0, x1 , . . . , xm ), we get j∗ dx0 = 0 and j∗ dxi = dxi , for i > 0. Thus, j∗ ω = a0 ◦ j · dx1 ∧ · · · ∧ dxm and the integral on the left is ∂Hm+1 j∗ ω = · · · Rm a0(0, x1 , . . . , xm ) dx1 · · · dxm . Now we simplify the integral on the right, i.e. we compute dω = i ∂iai · dxi ∧ dx0 ∧ · · · ∧ dxi ∧ · · · ∧ dxm = i (−1)i ∂iai · dx0 ∧ · · · ∧ dxm . Now the integral simplifies to Hm+1 dω = i (−1)i · · · Hm+1 ∂iai dx0 · · · dxm = i (−1)i · · · Hm+1 ∂iai dxi dx0 · · · dxi · · · dxm For i > 0, we get ∞ −∞ ∂iai · dxi = ai|xi=∞ − ai|xi=−∞ = 0 − 0 = 0, since ω is assumed compactly supproted, while 0 −∞ ∂iai · dx0 = a0|x0=0 − a0|x0=−∞ = a0(0, x1 , . . . , xm ). Thus, the integral also equals Hm+1 dω = · · · Rm a0(0, x1 , . . . , xm ) dx1 · · · dxm . Remark. It is interesting to see what we would get if we integrated over a cube instead. Then the “boundary conditions” ai|xi=±∞ = 0 would be replaced by the non-zero restrictions to the faces of the cube and the resulting formula would be Im+1 dω = i (−1)i ∂+ i Im+1 ω − i (−1)i ∂− i Im+1 ω where the ∂ε i Im+1 denotes the subset {(x0 , . . . , xm ) ∈ Im+1 | xi = ε}. The signs reflect the orientations of these faces so that the right hand side actually equals ∂Im+1 ω when ∂Im+1 is interpreted correctly. 15.6. An interpretation of Stoke’s theorem. First we prove that a k-form ω is uniquely determined by integrals Dk ι∗ ω for arbitrary embeddings ι: Dk → M. Here, Dk is a k-dimensional unit ball. Alternatively, the same holds for cubes. To prove this claim, observe that Dk ι∗ ω = · · · Dk ω(ι∗∂1, . . . , ι∗∂k) dt1 · · · dtk where the integrand is the function obtained by evaluating ω at the images of the canonical vector fields on Dk ⊆ Rk . Clearly, this equals roughly Dk ι∗ ω ≈ Vol(Dk ) · ω(ι∗0∂1, . . . , ι∗0∂k) Restricting to an ε-ball εDk , we obtain an equality in limit lim ε→0 1 εk εDk ι∗ ω = Vol(Dk ) · ω(ι∗0∂1, . . . , ι∗0∂k) 25 Thus, if ω integrates to 0 over any embedded k-dimensional ball, then ω = 0. As an immediate consequence, we see that ω = 0 if and only if ω integrates to 0 over any k-dimensional ∂-submanifold W ⊆ M. Now ω is closed if and only if 0 = W dω = ∂W ω, i.e. if and only if ω integrates to 0 over any k-dimensional boundary. Similarly, if ω is exact, say ω = dθ, then for any submanifold N ⊆ M (without boundary!) N ω = N dθ = ∂N θ = 0, since ∂N = ∅. Thus, when N ω = 0 for ω closed and N submanifold without boundary, we conclude that ω is not exact and that N is not a boundary (of a compact ∂-submanifold). In particular, Hk M = 0. (This, in general, is far from an equivalence.) 15.7. Cohomology in top dimension. In order to distinguish compact manifolds without boundary from those with boundary, we call them closed. Theorem 15.5. For any closed oriented Riemannian manifold M of dimension m, Hm (M) = 0. Proof. Since every m-form on M is closed, it is enough to find one that is not exact. We know that oriented Euclidean spaces admit a canonical volume form specified by the requirement Vol(e1, . . . , em) = 1 for any positive orthonormal basis (e1, . . . , em). In this way, we obtain a volume form Vol ∈ Ωm M. In any chart compatible with the orientation, Vol = a · dx1 ∧ · · · ∧ dxm with a = Vol(∂1, . . . , ∂m) > 0. Thus, M Vol is an integral of a positive function and as such must also be positive. Thus, Vol cannot be exact. 15.8. Homotopy invariance. We would like to show that Hk Rn = 0 for k > 0. This will follow from the following “homotopy invariance” property. Theorem 15.6. Let h: [−1, 1] × M → N be a smooth map and denote ht = h(t, −). Then for any closed k-form ω, we get [h∗ −1ω] = [h∗ 1ω] ∈ Hk M. Proof. Employing, for X ∈ XM and ω ∈ Ωk+1 M, the notation X ω(A1, . . . , Ak) = ω(X, A1, . . . , Ak), we define a homotopy operator K : Ωk+1 ([−1, 1] × M) → Ωk M via Kω = 1 −1 j∗ t (∂t ω) dt, where jt : M → [−1, 1] × M is the map x → (t, x). Writing in local coordinates ω = ωi1···ik dt ∧ dxi1 ∧ · · · ∧ dxik + ωi0···ik dxi0 ∧ · · · ∧ dxik , we get a formula Kω = 1 −1 j∗ t ωi1···ik dt · dxi1 ∧ · · · ∧ dxik . (the coordinate functions corresponding to terms that involve dt get integrated along t, the remaining terms disappear). Now we compute dKω = 1 −1 j∗ t ∂i0 ωi1···ik dt · dxi0 ∧ dxi1 ∧ · · · ∧ dxik while dω = ∂i0 ωi1···ik dxi0 ∧ dt ∧ dxi1 ∧ · · · ∧ dxik −dt∧dxi0 ∧dxi1 ∧···∧dxik + ∂tωi0···ik dt ∧ dxi0 ∧ · · · ∧ dxik + terms not involving dt and consequently Kdω = −dKω + 1 −1 j∗ t ∂tω dt · dxi0 ∧ dxi1 ∧ · · · ∧ dxik = −dKω + j∗ 1 ω − j∗ −1ω. 26 This implies rather easily the result, since, for ω closed, the term on the left vanishes and, thus, the difference j∗ 1 ω −j∗ −1ω = dKω is exact, i.e. the two terms represent the same cohomology class; now h∗ εω = h∗ j∗ ε ω. Proof. The idea of the proof is simple. Any k-form is determined by its integrals along k-dimensional cubes embedded in M. This is so because any embedding [−1, 1]k → M that maps ∂i(0) to Ai ∈ TxM restricts to the cube [−ε, ε]k to an embedding iε such that (iε)∗ω ∼ (2ε)kω(A1, . . . , Ak) (equality holds in limε→0). Now for an embedding i: [−1, 1]k → M, we get an associated embedding id ×i: [−1, 1]k+1 → [−1, 1] × M. Denote by jt : [−1, 1]k → [−1, 1]k+1 the embedding given by jt(t1, . . . , tk) = (t, t1, . . . , tk). Then h∗ t ω = j∗ t h∗ω and both j±1 are embeddings as part of the boundary. Thus, the Stokes’ theorem relates [−1,1]×[−1,1]k d(h∗ ω) = ∂([−1,1]×[−1,1]k) h∗ ω = [−1,1]k h∗ 1ω − [−1,1]k h∗ −1ω − [−1,1]×∂[−1,1]k h∗ ω (∗) (the first two terms correspond to ∂[−1, 1] × [−1, 1]k). Writing d(h∗ ω) = a · dt ∧ dt1 ∧ · · · ∧ dtk , the integral on the left can be computed using Fubini’s theorem as [−1,1]×[−1,1]k d(h∗ ω) = [−1,1]k [−1,1] a(t, t1 , . . . , tk ) dt dt1 · · · dtk . This can be rephrased in terms of an operator K : Ωk+1([−1, 1] × M) → ΩkM, given by the integral K(η)x(A1, . . . , Ak) = [−1,1] θ(t,x)(A1, . . . , Ak) dt η = dt ∧ θ 0 η(∂t, −, . . . , −) = 0 (formally, there is an isomorphism Λk(V ⊕ W) ∼= i+j=k ΛiV ⊗ ΛjW; apply this to the canonical decomposition T(t,x)([−1, 1]×M) ∼= Tt[−1, 1]⊕TxM; then K is defined by projecting to Λ1T∗ t [−1, 1]⊗ΛkT∗ x M, writing the image uniquely as dt ⊗ θ and then integrating θ as above) as [−1,1]×[−1,1]k d(h∗ ω) = [−1,1]k Kd(h∗ ω) = [−1,1]k K(h∗ dω). The remaining boundary term in (∗) is then [−1,1]×∂[−1,1]k h∗ ω = ∂[−1,1]k K(h∗ ω) = [−1,1]k dK(h∗ ω), again by the Stokes’ theorem. Thus, we have finally obtained [−1,1]k h∗ 1ω − [−1,1]k h∗ −1ω = [−1,1]k (dK(h∗ ω) + K(h∗ dω))) or, in other words, h∗ 1ω − h∗ −1ω = dK(h∗ω) + K(h∗dω). This implies rather easily the result, since, for ω closed, the first term on the right vanishes and, thus, the difference on the left is exact, i.e. the two terms represent the same cohomology class. In the situation from the above proof, we say that two chain maps (maps that commute with differentials, such as pullback maps j∗ ε ) are chain homotopic if there exists a collection of maps η such that g − f = dη + ηd. Then, f and g induce the same map in cohomology. Corollary 15.7. Hk Rm = 0 for k > 0. Proof. There is a homotopy id ∼ 0 between the identity and the constant map onto the zero. Then for any closed k-form ω we have [ω] = [id∗ ω] = [0∗ ω] = [0]. Remark. The case k = 1 gives the following: a 1-form ω = gi · dxi is a differential of a function f, i.e. we have gi = ∂if, if and only if dω = i