8.1 Examples, definitions, and basic properties


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1 8 De Rham cohomology Last updated: May 21, Examples, definitions, and basic properties A kform ω Ω k (M) is closed if dω =. It is exact if there is a (k 1)form σ Ω k 1 (M) such that dσ = ω. A form σ such that dσ = ω is called a primitive for ω. (Of course, when it exists, it is not unique.) The condition dω = is local, in the sense that a form on M is closed if it is closed near every point of M (in other words, if all the restrictions ω U are closed as forms on U M). To check that dω =, we simply check that in local coordinates. Compared to that, the property of being exact is not a local one. It is necessary that a form σ such that dσ = ω is a form on the whole M, i.e., σ should be everywhere defined. We know that every exact form is closed, since d(dω) =. On the other hand, there are closed but not exact forms. Example 8.1. Consider the 1form ω = dθ on R 2 \ {} where θ is the polar angle. In standard Cartesian coordinates, as one can check, ω = xdy ydx x 2 + y 2. This form is not exact, because for a cycle going around the origin once, say, counterclockwise, ω = 2π. C (The exact number is not important; that it is nonzero, is important.) Remark 8.1. In the above example, dθ appear as exact (the differential of the function θ), but it is not, because the function θ is bad. It cannot be made everywhere defined, smooth and singlevalued function on the whole of R 2 \ {} at the same time. It can be made a genuine function only locally, in a sufficiently small domain, in particular where it is not possible to have a cycle going around the origin. The case of dθ is typical. Every closed form on a manifold can be regarded as exact if we surrender some of the properties: 1
2 allow discontinuities or multivaluedness, or consider the form only locally. (The last claim is the content of the socalled Poincaré lemma, that every closed form is locally exact, which will be discussed shortly.) Therefore, the difference between closed forms and exact forms is a global feature of the manifold in question and measures its topological complexity. Definition 8.1. A closed kform is cohomologous to zero if it is exact. Closed kforms ω and ω on a manifold M are cohomologous if their difference is exact: ω ω = dτ for some (k 1)form τ. (The form τ should be a welldefined smooth form on the whole M.) This is an equivalence relation. Equivalence classes w.r.t. this relation are called cohomology classes. The set of all cohomology classes of degree k (sometimes people also say: in dimension k), is denoted H k (M). Notation for cohomology classes: [ω] H k (M), if ω Ω k (M) is a closed form representing an element [ω]. (Check that it is indeed an equivalence relation!) Theorem 8.1. For each k =, 1,, 2,..., the set of cohomology classes H k (M) is a vector space over R w.r.t. addition and multiplication by constants defined on representatives. Proof. Consider the sum of two closed forms, ω + σ. It is closed, by the linearity of d. If we replace one of the summands by a cohomologous form, the sum will remain in the same cohomology class: ω + σ = ω + (σ + dτ) = (ω +σ)+dτ. Thus addition of cohomology classes is welldefined and H k (M) inherits the structure of an Abelian group. In a similar way we can show that multiplication by real numbers is welldefined on cohomology classes. Therefore H k (M) is a vector space with respect to the operations [ω 1 ] + [ω 2 ] := [ω 1 + ω 2 ], k[ω] := [kω] defined by the usual addition and multiplication by constants of closed forms representing cohomology classes. Hence, for each k, there is the cohomology space H k (M). In particular, it is an Abelian group and the traditional name for the vector space H k (M) is the kth cohomology group. Remark 8.2. Using the notion of the quotient space V/W of a vector space V w.r.t. a subspace W V, we can say, shortly, that H k (M) = Z k (M)/B k (M) 2
3 where Z k (M) := Ker(d: Ω k (M) Ω k+1 (M)) stands for the space of all closed kform and B k (M) := d ( Ω k 1 (M) ) stands for the space of all exact kforms. Remark 8.3. In the definition of cohomology groups we have used forms with real coefficients. Sometimes it is convenient to consider complexvalued functions and forms. The definition of cohomology applies to them as well and to distinguish the two cases, we may use the notation such as H k (M, R) and H k (M, C), which are vector spaces over R and C respectively. The cohomology group H k (M) defined using differential forms is also called the de Rham cohomology group, after the Swiss mathematician George de Rham. The name suggests that there are other cohomology theories, and this is indeed true 1, but for manifolds they all give the same objects. This is exactly the statement of the fundamental de Rham theorem conjectured by Élie Cartan and proved by G. de Rham. Sometimes the notation HDR k (M) is used for de Rham cohomology. Definition 8.2. The dimension dim H k (M) is denoted b k (M) and called, the kth Betti number of M. (One can see that we can use both realvalued or complexvalued forms and this makes no difference for the dimension of cohomology: b k (M) = dim R H k (M, R) = dim C H k (M, C).) As we know, forms can be multiplied and they form a graded algebra Ω (M) = ( Ω k (M) ), which is associative and gradedcommutative. Theorem 8.2. The exterior multiplication of forms induces a multiplication of cohomology classes making the vectors spaces H k (M) a gradedcommutative algebra. It is denoted H (M) and is called the (de Rham) cohomology algebra of a manifold M. Proof. Consider the product of closed forms: ω σ. The Leibniz rule implies that it is closed: d(ω σ) = dω σ ± ω dσ = + =. If one of the factors 1 We may mention simplicial cohomology defined using a triangulation, singular cohomology based on the notion of singular simplex or, as a variant, singular cube, and Čech cohomology defined using open covers. A feature distinguishing other cohomology constructions from the Cartan de Rham theory is the possibility to use the coefficients more general than the real numbers. For example, using theories such as simplicial or Čech it is possible to introduce certain cohomology groups H k (M, Z) or H k (M, Z 2 ), which may carry more information than the cohomology with real and complex coefficients. 3
4 is replaced by a different representative of the same cohomology class, then the cohomology class of the product will not change: ω σ = (ω + dτ) σ = ω σ+dτ σ = ω σ+d(τ σ), by the same Leibniz rule, since dσ =. Hence, the exterior multiplication of forms induces a welldefined multiplication on cohomology classes: [ω] [σ] := [ω σ]. We arrive at the structure of a graded algebra on the graded vector space H (M) = ( H k (M) ), called the cohomology algebra of M. The cohomology algebra H (M) is associative and gradedcommutative, inheriting these properties from the algebra of forms Ω (M). Consider a smooth map F : M N. It induces the pullback of forms: F : Ω k (N) Ω k (M). Let [ω] H k (N) be a cohomology class represented by a closed form ω Ω k (N). Definition 8.3. The pullback (or induced map) on cohomology classes is defined by the formula: F ([ω]) := [F (ω)], where at the r.h.s. there is the pullback of a differential form representing a given cohomology class [ω]. Theorem 8.3. The pullback of cohomology classes is welldefined. It is a linear map F : H k (N) H k (M), for each k =, 1,.... For the composition of smooth maps, we have (G F ) = F G (1) on cohomology. The pullback F is an algebra homomorphism of cohomology algebras. Proof. Suppose a form ω Ω k (N) is closed and represents a cohomology class [ω] H k (N). We first need to check that the form F (ω) Ω k (M) is also closed and therefore gives a cohomology class in H k (M). Indeed, df (ω) = F (dω) =. (The key fact here is that the exterior differential commutes with pullbacks of forms.) Next, suppose a representative ω is replaced by ω = ω+dσ. Then F (ω ) = F (ω+dσ) = F (ω)+f (dσ) = F (ω)+df (σ), 4
5 so [F (ω )] = F (ω). Therefore the linear map F : H k (N) H k (M) is welldefined. Equation (1) follows from the same property for the pullbacks of forms. To see that F is an algebra homomorphism on cohomology, we write F ( [ω][σ] ) = F ( [ω σ] ) = [F ( ω σ ) ] = [F ω F σ] = so it is as claimed. [F ω][f σ] = F [ω] F [σ]. Corollary 8.1. If manifolds M and N are diffeomorphic, then the cohomology groups H k (M) and H k (N) are isomorphic (for all k). Proof. Suppose F : M N and G: N M are mutually inverse smooth maps giving a diffeomorphism M = N. Consider the vector spaces H k (M) and H k (N) for some given k. There are linear maps F : H k (N) H k (M) and G : H k (M) H k (N). The identities F G = id and G F = id imply (F G) = G F = id and (G F ) = F G = id. Hence the linear maps F and G are mutually inverse, so the vector spaces H k (M) and H k (N) are isomorphic (in particular, have the same dimension). Remark 8.4. Moreover, if M and N are diffeomorphic, the cohomology algebras H (M) and H (N) are isomorphic. Indeed, we have isomorphism of vector spaces (in each degree) that also preserves the multiplication. Corollary 8.1 means that the de Rham cohomology is a diffeomorphism invariant of a smooth manifold. We shall see shortly that de Rham cohomology is, in fact, an invariant under a much coarser equivalence relation. Namely, it is a homotopy invariant. Proposition 8.1. For any manifold M, the dimension of the space H (M) is the number of connected components of M (if it is finite). 5
6 Proof. The space of forms is the space of smooth functions C (M). A function f Ω (M) = C (M) is closed if df =. What does it mean? Near each point f must be a constant (indeed, we may introduce coordinates and write df = in coordinates). Hence f is a local constant. It need not be a constant on the whole M (a global constant), which is demonstrated by the example of a manifold consisting of two connected components such as two disjoint copies of R n. The function may be zero on one component and 1, on another. Notice that there no exact forms (because there are no 1forms). Therefore H (M) is the space of all local constants on M. It is intuitively clear and can be proved by a simple topological argument that on a connected topological space any local constant is a constant. Therefore, if a topological space, in particular, a manifold M, is the disjoint union of its connected components (maximal connected subspaces), then any local constant is defined by its values of the components (constants); thus it is a function on the set of components. If the number of components is finite, the space of functions on this set is finitedimensional. As a basis one take functions that are identically 1 on one component and identically on all other components. Hence the dimension is the number of components. In most of our examples, the manifolds in question are connected. Then, automatically, H (M) = R for them. Consider examples of cohomology classes in degrees higher than zero. Example 8.2. We saw above that [dθ] is a nonzero class in H 1 (R 2 \ {}). Example 8.3. If we consider θ as a parameter on the circle S 1 (defined up to 2π), the 1form dθ is welldefined and gives a nonzero class [dθ] in H 1 (S 1 ). We shall shortly see that all other cohomology classes of the circle in degree 1 are proportional to [dθ], i.e., dim H 1 (S 1 ) = 1 and [dθ] can be taken as a basis (consisting of a single element). 8.2 Homotopy invariance and the Poincaré Lemma Definition 8.4. Two smooth maps f, f 1 : M N are homotopic if there is a smooth map F : M [, 1] N such that F (x, ) = f (x) and F (x, 1) = f 1 (x) for all x M. Notation: f f 1. (In other words, there is a family f t = F (, t) giving smooth interpolation between f and f 1.) The map F is known as a homotopy between f and f 1. 6
7 Of course, homotopy also makes sense just for topological spaces, not manifolds. In such a case instead of smoothness we impose continuity. What we have defined above is known as smooth homotopy, but we shall skip the adjective. Theorem 8.4 (Homotopy invariance). Homotopic maps of smooth manifolds induce the same linear map of their de Rham cohomology. A proof is given below. An immediate application of this property is as follows. Definition 8.5. Two manifolds (more generally, two topological spaces) are homotopy equivalent if there are maps in the opposite directions such that their compositions are homotopic to identities. That is, for X and Y, there are f : X Y and g : Y X such that f g id Y and g f id X. Notation: X Y. A manifold (more generally, topological space) homotopy equivalent to a point is called contractible. Example 8.4 (Contractibility of R n ). Let us show that R n { } (a single point). We can identify the point with R n. Consider the obvious maps i: {} R n (inclusion) and p: R n {} (projection). We have p i = id, but i p is not the identity, it is the map that send every vector to zero. However, it is homotopic to the identity, the family f t : x tx, where t [, 1], giving the desired homotopy. Example 8.5. Open and closed cylinders, S 1 R and S 1 [ 1, 1], are both homotopy equivalent to the circle S 1. (The same is true if we replace S 1 by any space X.) (Construct the maps and homotopies following the previous example.) The previous example shows that homotopy equivalence is much coarser relation than diffeomorphism or homeomorphism: it does not have to preserve dimension (the cylinders are 2dimensional and S 1 is 1dimensional) and a noncompact space (such as the open cylinder) may be homotopy equivalent to a compact space (such as the closed cylinder). Corollary 8.2 (from Theorem 8.4). Homotopy equivalent manifolds M and N have isomorphic cohomology groups H k (M) and H k (N) for all k. Proof. Let f : M N and g : N M be such maps that f g id N and g f id M. Then by Theorem 8.4, for each k, we have linear maps of vector 7
8 spaces f : H k (N) H k (M) and g : H k (M) H k (N) with the property f g = id on H k (M) and g f = id on H k (N) ( homotopic to identity becomes equal to identity for the maps of cohomology). That mean that these vector spaces are isomorphic. Example 8.6. All generalized cylinders over a manifold M such as M R, M ( 1, 1), and M [ 1, 1] have the same cohomology as M. For example, H 1 (S 1 ( 1, 1)) = R and H 2 (S 1 ( 1, 1)) =. Applying Corollary 8.2 to the special case of R n domain in R n, see below) we obtain the following. (or any starshaped Theorem 8.5 ( The Poincaré Lemma ). On R n every closed kform, k 1, is exact. Or: { H k (R n ) R for k = = for k >. Proof. Indeed, R n is contractible. Therefore, by Corollary 8.2, H k (R n ) = H k ({ }) for all k =, 1, 2,.... The claim follows. Although the Poincaré lemma is the statement about vanishing of certain cohomology, namely, H k (R n ) = for k, it plays a key role in calculating the cohomology for manifolds when it is possibly nonzero. Indeed, since in a neighborhood of a point every manifold M n is undistinguishable from R n or an open ball in R n, the Poincaré lemma immediately implies that there is no local obstruction for a closed form on M n to be exact, but only locally. See the discussion at the beginning of this section. So if there is an open cover of a manifold by contractible domains U α (for example, diffeomorphic to R n ), it is possible to find a local primitive for each closed kform: i.e., write ω = dσ α for σ α Ω k 1 (U α ). Then the task of finding a global primitive for ω (or an obstruction for that) is reduced to considering the differences σ α σ β on the intersections and whether or not, by a choice of each σ α, these differences can be made zero. We shall see an application of this idea in the next subsection 2. Let us now give a proof of the homotopy invariance of de Rham cohomology. 2 Moreover, although this remains beyond the scope of this course, provided all the pairwise and higher intersections such as U α U β, U α U β U γ, etc., are also contractible, it is possible to reduce the description of the cohomology classes of M n to a purely combinatorial data in terms of the open cover U = (U α ), encoding in this case the topology of M n. 8
9 Proof of Theorem 8.4. Suppose f and f 1 are two homotopic maps M N. Let F : M [, 1] N be a homotopy. We wish to show that the linear maps f : H k (N) H k (M) and f 1 : H k (N) H k (M) coincide for all k. That means that if we take a cohomology class [ω] H k (N) represented by a closed kform ω Ω k (N), the pullbacks f (ω) and f 1 (ω) not necessarily coincide, but they must differ only by an exact form. To this end, we shall introduce a linear transformation K : Ω k (N) Ω k 1 (M), decreasing degrees, such that for any form ω Ω k (N), f 1 ω f ω = (d K + K d)ω. (2) Thus for closed forms there will be f1 ω f ω = d(kω). We define K, using the homotopy F. Consider F ω Ω k (M [, 1]). Each kform σ on M [, 1] can be uniquely written as σ = σ () t + dt σ (1) t where σ () t and σ (1) t are forms on M depending on the parameter t. (Use some local coordinates on M to see how it works.) The degree of σ () t is k, while the degree of σ (1) t is k 1. Indeed, the term with σ (1) t contains an extra factor of dt, which raises the degree by 1. We define the operation K by Kω = ( K F )ω, where K sends any σ = σ () t + dt σ (1) t on M [, 1] to 1 dt(σ(1) t ) (integration over t as a parameter). The result is a (k 1)form. Let us check that K has the desired property (2). Firstly, d F = F d, so we need to calculate the commutator of d and K. On M [, 1], d = d x + dt t where d x can be considered as the differential on M. In local coordinates, d x = dx i. If we apply it to σ Ω k (M [, 1]) written as above, we x i arrive at ( dσ = d x + dt ) ( ) ( ) σ () t + dt σ (1) t = d x σ () t + dt d x σ (1) t + σ() t. t t Hence Kdσ = 1 dt ( d x σ (1) t ) + σ() t = t 1 ( ) dt d x σ (1) t + σ () 1 σ (). 9
10 On the other hand, clearly d Kσ = d 1 dt(σ (1) t ) = 1 ( ) dt d x σ (1) t. Therefore ( d K + K ) d σ = σ () 1 σ (). (3) Now the desired relation (2) follows from Equation (3) applied to σ = F ω. We have, for K = K F, ( (d K + K d) ω = d K F + K ) F d ω = (d K F + K ) ( d F ω = d K + K ) d (F ω) = (F ω) () 1 (F ω) () = f ω f 1 ω. Indeed, when we view F as the family f t = F (, t), we see that ft ω is obtained from F ω by setting dt to zero, i.e., by passing to (F ω) () t in the above notation. Hence f ω and f1 ω are obtained by additionally setting t to and to 1, respectively, which gives the desired formula. Remark 8.5. The method used in the proof above is a prototype of a general construction called algebraic homotopy. More precisely, the algebraic homotopy in the proof is the operator K : Ω k (N) Ω k 1 (M) satisfying the property f1 f = [d, K] (the commutator understood in the sense of graded algebras). We deduced the existence of such an operator K from the existence of a homotopy of smooth maps F = (f t ). From the existence of K immediately follows that the maps given by f and f1 coincide on the level of cohomology. These constructions can be axiomatized and generalized as it is done in the area of mathematics called homological algebra. The Poincaré Lemma is often given a direct proof independent from the homotopy invariance of de Rham cohomology. The underlying idea is very simple and is related with integration of forms. Example 8.7. On the real line R, any 1form ω = f(x)dx (which is automatically closed), is exact. Indeed, consider the function F (x) = x f(y)dy. It is defined on the whole line R and we have df = ω by the Newton Leibniz theorem (the main theorem of calculus ). 1
11 This can be seen as a prototype of the general statement concerning R n or the socalled starshaped domains in R n. Let us define them Definition 8.6. An open domain U R n is called starshaped if there is a point x U such that for each x U, all points of the segment [x x] are in U. (The segment [x x] consists of all points of the form (1 t)x + tx where t [, 1].) The point x is often referred to as the center of the starshaped domain U. For example, any convex domain (i.e., such that for any two points the segment joining them is also contained in the domain) is starshaped. However, a starshaped domain does not have to be convex. Example 8.8. An open ball in R n is convex, therefore starshaped. Example 8.9. The interior of any star in R 2 is starshaped. (Notice that a star is not convex.) A version of the Poincaré Lemma is as follows. Theorem 8.6 (Poincaré Lemma). On R n or any starshaped domain in R n, every closed kform is exact, if k 1. We can give an independent direct proof for 1forms. (It is in fact generalizes to kforms as well, but we shall not go into that here.) Direct proof of the Poincaré Lemma for 1forms. Let A = A i (x)dx i be a closed 1 form on a starshaped domain U R n. That means that i A j j A i =. Without loss of generality we may assume that the center x of U is the origin R n. For any point x U, the segment [x] consists of points tx where t [, 1]. Define a function f by the formula f(x) = 1 x i A i (tx)dt for all x U. It makes sense because tx U. We have df = dx j 1 1 x j x i A i (tx)dt = dx j ( ) x j x i A i (tx) dt 1 ( ) = dx j δja i i (tx) + x i t j A i (tx) dt } {{ } = i A j (tx) 1 )dt = dx j = dx j 1 ( A j (tx) + tx i i A j (tx) therefore A = df is exact, as claimed. 11 d ( ) ta j (tx) dt dt = dx j ta j (tx) = A, t= t=1
12 Remark 8.6. In the proof for 1forms above, the function f is nothing but the integral [x x] A over the straight line segment [x x]. A similar proof for k > 1, which we do not give, also uses integration, but in a more sophisticated way (since the integral of a kform over a ksurface would give a number, but we need to obtain a (k 1)form). 8.3 Further examples of calculation Consider some examples of calculating cohomology. We know that for any connected manifold M, the space H (M) is onedimensional. Consider the opposite extreme. What can be said of the nth cohomology of a manifold M n, where dim M n = n? All nforms on M n are automatically closed. Suppose M n is compact and orientable. It is not difficult to see that in such case on a manifold M n there exists an nform ω such that for all charts of an oriented atlas, ω = n1! ω 12...n dx 1... dx n where ω 12...n >. (For surfaces in R 3, one can take the area element; similarly, a volume element can be constructed for higher dimensional manifolds, for example, using an embedding into R N.) For the form ω, M n ω > due to the positivity of ω 12...n. Suppose M n is also without boundary. Then, by the Stokes theorem, the integral of any exact form dσ vanishes: dσ = σ = σ =. M n M Therefore the form ω cannot be exact, and we conclude that the space H n (M n ) for an arbitrary closed manifold is nonzero. (Recall that a manifold is called closed if it is compact, orientable and without boundary 3.) In fact, more can be said: Theorem 8.7. For a connected closed manifold M n, the top cohomology space is onedimensional: H n (M n ) = R. 3 Hopefully the terminology in the context of de Rham cohomology will not confuse the reader. Yes, there are manifolds that are closed and there are differential forms that are closed. What is in common, is that M = for a closed manifold and dω = for a closed form, so the two notions are not so far apart. 12
13 A proof of that is beyond the scope of this course. We can only note that a map H n (M n ) R giving the desired isomorphism is nothing but the integration [ω] for ω Ω n (M n ). M n ω Remark 8.7. The condition for M n to be connected is important. If M n has several connected components, then one can consider integrals over each of them, arriving therefore at several real numbers instead of one. Theorem 8.7 generalizes to the statement that for an arbitrary closed manifold, the vector space H n (M n ) is isomorphic to the dual of the vector space H (M n ) and therefore there is the equality of the Betti numbers: b n (M n ) = b (M n ). More generally, for a closed kform ω and a closed (n k)form σ on a closed ndimensional manifold M n, the map (ω, σ) ω σ R M n defines a linear transformation H k (M n ) ( H n k (M n ) ). (Check that it is welldefined!) A generalization of Theorem 8.7 tells that this linear transformation is an isomorphism, so for an arbitrary closed manifold M n, and therefore H k (M n ) = ( H n k (M n ) ) b k (M n ) = b n k (M n ), for all k =, 1,..., n. This is known as the Poincaré Duality Theorem. Again, its proof is outside our course. Let us now calculate the de Rham cohomology in some simple examples where it can be done by hands (without developing special technique). 13
14 Example 8.1 (Cohomology of the circle S 1 ). Consider S 1 as the unit circle with center at the origin in R 2. We already know that [dθ], where θ is the polar angle, is a nonzero class in H 1 (S 1 ). This is because S 1 dθ. Claim: dim H 1 (S 1 ) = 1 and one can take [dθ] as a basis. To prove it, note that arbitrary 1forms on the circle can be written as f(θ)θ where f(θ+2π) = f(θ), i.e., as periodic 1forms on the real line. Any such form is the differential of a function g on R, which is possibly not periodic: f(θ)dθ = dg where g(θ) = θ f(s)ds. It turns out that the condition for g(θ + 2π) = g(θ) is exactly the vanishing of the integral 2π f(s)ds = f(θ)θ. S 1 Therefore a 1form on S 1 is exact if and only if it has zero integral over S 1. Any 1form ω Ω 1 (S 1 ) (automatically closed because all 2forms are zero) is cohomologous to c [dθ] where c = 1 2π S 1 ω. Thus the cohomology classes in H 1 (S 1 ) are in 11correspondence with numbers c R and H 1 (S 1 ) = R. Conclusion: R for k = H k (S 1 ) = R for k = 1 for k > 1. (It is also zero for k <, but we shall always omit such obvious equality.) Example 8.11 (Cohomology of the cylinder S 1 R). A similar direct argument can be applied to the cylinder S 1 R (or a finite open cylinder S 1 ( 1, 1), diffeomorphic to it) to show that H 1 (S 1 R) = R (a representative of a nonzero class is again [dθ] where θ is the angular coordinate on S 1 ) and that H 2 (S 1 R) =. It is also possible to apply a homotopy argument, that S 1 R is homotopy equivalent to S 1, therefore H k (S 1 R) = H k (S 1 ) for all k. 14
15 Example 8.12 (Cohomology of the sphere S 2 ). To find the cohomology of S 2 we may argue in the following way. Consider an open cover of S 2 by two open sets, open neighborhoods of the north and south poles respectively, such that each of them is diffeomorphic to a disk and the intersection is diffeomorphic to a cylinder S 1 ( 1, 1). For example, one can take the coordinate domains for the stereographic atlas, where U = S 2 \ {N} and U 1 = S 2 \ {S}. Then it is possible to compute the cohomology of S 2 by reducing it to the known cohomology of U, U 1 and U 1 = U U 1. Suppose ω Ω 1 (S 2 ) is closed. Consider ω U and ω U1. By the Poincaré lemma, ω U = df and ω U1 = df 1 where f C (U ) and f 1 C (U 1 ), respectively. On the intersection U 1, d(f f 1 ) =, so f f 1 is a constant (since U 1 is connected), say, f f 1 = c R. Consider f 1 := f 1 + c on U 1 ; for it, ω = df 1 still holds, and f and f 1 agree on U 1. Therefore there is a globally defined function f such that f U = f and f U1 = f 1, and ω = df. We can conclude that every closed 1form on S 2 is exact, i.e., H 1 (S 2 ) =. Consider now ω Ω 2 (S 2 ). It is automatically closed. We know that there are 2forms whose cohomology classes are nonzero. For example, the area element ds = sin θ dθ dϕ on S 2. We shall show that H 2 (S 2 ) is onedimensional. In a similar way to the above, ω U = dσ and ω U1 = dσ 1 where σ Ω 1 (U ) and σ 1 Ω 1 (U 1 ), respectively. On the intersection, d(σ σ 1 ) =, hence σ 1 := σ σ 1 represents a cohomology class in H 1 (U 1 ). As U 1 is diffeomorphic to the cylinder S 1 ( 1, 1), we know that H 1 (U 1 ) = R. If we choose σ or σ 1 differently, say, take σ = σ +df instead of σ, the cohomology class of σ 1 = σ σ 1 on U 1 will not change. If we replace the original ω by a cohomologous form on S 2, say, take, ω = ω + dσ for some global σ Ω 1 (S 2 ), then for the new form it is possible to take σ = σ + σ and σ 1 = σ 1 + σ, so σ 1 = σ 1. Hence we arrive at a welldefined map H 2 (S 2 ) H 1 (U 1 ). It is linear (check!). It is an isomorphism. Indeed, consider a closed 1form σ 1 on U 1. Using a partition of unity on S 2 subordinate to the cover (U, U 1 ), g +g 1 = 1, Supp g U and Supp g 1 U 1, we can define σ := g 1 σ 1 as a 1form on U (by extending by zero from U 1 ) and, similarly, σ 1 := g σ 1 as a 1form on U 1. Note the minus sign. Then σ σ 1 = g σ 1 ( g 1 σ 1 ) = (g + g 1 )σ 1 = σ 1 on U 1. Applying d, we see that dσ dσ 1 = on U 1. Therefore there is a welldefined 2form ω on S 2 = U U 1 such that ω U = dσ and ω U = dσ respectively. In this way we have constructed the inverse map for H 2 (S 2 ) H 1 (U 1 ). So it is an 15
16 isomorphism and we conclude that H 2 (S 2 ) = H 1 (U 1 ) = R. To summarize: R for k = H k (S 2 ) = for k = 1 R for k = 2. (It is of course zero also for k < or k > 2.) Example 8.13 (Cohomology of S n for n > 2). The cohomology of S n for n > 2 can be calculated by the same method as for S 2. It is clear that H (S n ) = R because the sphere is connected. By arguing as above, one can see that H 1 (S n ) =. In the same way one can establish an isomorphism H k (S n ) H k 1 (S n 1 ) for all k > 1. Using induction on n, one can arrive at the following answer: R for k = H k (S n ) = for < k < n R for k = n. (Question: what is the cohomology of S?) We shall give more examples without proofs, which would require developing some technical tools. They are not particularly difficult, but we do not have the time. Example 8.14 (Real projective space). The cohomology of RP n is as follows. The answer depends on the parity of n. For the evendimensional real projective spaces, { H k (RP 2m ) R for k = = for k >. For the odddimensional real projective spaces, R for k = H k (RP 2m+1 ) = for < k < 2m + 1 R for k = 2m
17 Here m =, 1, 2,.... (For example, RP 1 = S 1 and H (RP 1 ) = H (S 1 ).) The difference of the two cases (and the whole answer) can be explained by the fact that RP n is obtained from S n by the identification of the antipodal points. So the kforms on RP n can be naturally considered as the kforms on the sphere S n invariant under the antipodal map x x. With some more care, one can see that for all k the vector space H k (RP n ) can be identified with a subspace in H k (S n ). The question is therefore about the cohomology in degree n. Here the difference between n = 2m and n = 2m + 1 arises: in the first case the manifold is nonorientable and there is no volume form giving a nontrivial class in H n (RP n ), while in the second case the volume form on S n is invariant, thus giving a volume form for RP n and a nontrivial class in H n (RP n ). Example 8.15 (Complex projective space). The answer is as follows: R for k = R for k = 2 H k (CP n )... = R for k = 2p... R for k = 1 for k = 2p + 1 for k = 2n (recall that dim CP n = 2n). For example, when n = 1, we know that CP 1 = S 2 and the answer above agrees with the results for the sphere. We have a onedimensional space in all even degrees 2p 2n and zero otherwise. In the previous examples there was no need to consider the multiplicative structure in cohomology (since by dimensional considerations, the multiplication was always zero). Here the algebra structure of H (CP n ) can help to describe the answer better. The following is true. Take a nonzero cohomology class x H 2 (CP n ). (It spans the onedimensional vector space H 2 (CP n ).) Then all its powers: x, x 2,..., x n (up to the nth power), are nonzero and therefore span the corresponding cohomology spaces H 2p (CP n ). The cohomology algebra of CP n can be described as the truncated polynomial algebra H (CP n ) = R[x]/(x n+1 ), 17
18 i.e., the factoralgebra of the algebra of polynomials R[x] w.r.t. the relation x n+1 =. There is a nice choice of a basis class x H 2 (CP n ) given by a certain closed 2form ω Ω 2 (CP n ). (We skip the formula for ω in coordinates.) In particular, this form satisfies ω = 1 (S 2, i) for a natural embedding i of S 2 = CP 1 into CP n. (When n grows, the relation x n+1 = in the algebra H (CP n ) goes away to infinity. The limit CP makes sense, which is called infinite complex projective space. It is not a finitedimensional manifold, but cohomology still makes sense. For it, the cohomology algebra is just the algebra of polynomials: H (CP ) = R[x].) Example 8.16 (the ntorus). By definition, T n is the product S 1... S 1. Denote the angular coordinate on each circle as θ i, where i = 1... n. The 1 forms dθ i are obviously closed and define the nontrivial cohomology classes y i = [dθ i ] H 1 (T n ). (Indeed, the forms dθ i give nonzero integrals over the corresponding embedded circles.) They can be multiplied and we have y i y j = y j y i (as represented by the wedge product of 1forms, dθ i dθ j = dθ j dθ i ). The statement is that there are no other relations between the products of the classes y i and that the products y i1... y ik with i 1 <... < i k span the cohomology spaces H k (T n ). So the Betti numbers are given by the binomial coefficients: b k (T n ) = Cn k n! = k!(n k)!, and the cohomology algebra H (T n ) can be described as the exterior algebra on n generators y 1,..., y n. In particular, for the 2torus we have R for k = H k (T 2 ) = R 2 for k = 1 R for k = 2, with [dθ 1 ], [dθ 2 ] giving a basis of H 1 (T 2 ), and [dθ 1 dθ 2 ], the basis of H 2 (T 2 ). 18
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