MAT 531 (Spring 1996) Topology/Geometry II

Week 9
1. Elements of de Rham cohomology. There are two results which allow our elementary calculations for the point and the circle to be extended to a large collection of manifolds.
Homotopy Theorem. If f_0 and f_1: M --> N are smoothly homotopic, then f*_0 = f*_1: H*N --> H*M.
Mayer-Vietoris Theorem. Suppose a manifold M can be written as a union of two open sets U and V. Then there exist a set of linear d*: H^p(M)--> H^(p+1)(U intersect V) which fit, together with the maps induced by the inclusions i_1: U --> M, i_2: V --> M, j_1: (U intersect V) --> U, j_2: (U intersect V) --> V, into an exact sequence (the Mayer-Vietoris sequence)

    d*      (i*_1,i*_2)       j*_1 - j*_2             d*

... --> H^k(M) --> H^k(U) + H^k(V) --> H^k(U inter V) --> H^(k+1)(M) --> ... .
Example Calculation of H*S^2. We will need the following easy consequence of the Homotopy Theorem: if i: N --> M as smooth deformation retract, then i*: H*M --> H*N is an isomorphism. Let U and V be the complements of the North and South poles respectively; then U and V are open discs and (U inter V) is an open annulus. The discs have points as deformation retracts, and the annulus has a circle, so using the homotopy theorem we know H*U, H*V and H*(U inter V) in the Mayer-Vietoris sequence. We need the additional elementary piece of information: H^0(S^2) = R. Then the beginning of the sequence is, writing S^2 = M,
0 --> H^0(M) --> H^0(U) + H^0(V) --> H^0(U inter V) --> H^1(M)


0 -->   R    -->      R + R      -->        R       --> H^1(M) 
Exactness implies that the map R + R --> R is onto and therefore that the map R --> H^1(M) is the zero map. The next map in the sequence
H^1(M) --> H^1(U) + H^1(V) 


H^1(M) -->  0  
must therefore be injective, so H^1(S^2)=0. In the next part of the sequence
H^1(U) + H^1(V) --> H^1(U inter V) --> H^2(M)  --> H^2(U) + H^2(V)


      0        -->         R       --> H^2(M)  -->    0
exactness implies H^2(S^2) = Z.
Class exercise: Use the same method to calculate H*S^3.

The homotopy theorem.
Proposition A: Let i_0, i_1 : M --> R x M be the inclusions at levels 0 and 1. So i_0(x) = (0,x), i_1(x) = (1,x). Then i*_0 = i*_1: H*(R x M) --> H*M.
Note: this proposition implies the homotopy theorem, if we take as definition of smooth homotopy between f_0 and f_1 the existence of F: R x M --> N with F(0,x)=f_0 and F(1,x)=f_1. Because then f_0 = F o i_0 and f_1 = F o i_1, so that f*_0 = i*_0 o F* = i*_1 o F* = f*_1.
Proof of Proposition A. A local coordinate system (u_1,...,u_n) on M gives a local coordinate system (t,u_1,...,u_n) on R x M. In terms of these coordinates, any p-form on R x M may be written as omega =Sum_I a_I(t,x)du_I + Sum_J b_J(t,x)dt^du_J, where the multi-index I ranges over all p-tuples i_1 < ... < i_p, and the multi-index J ranges over all (p-1)-tuples j_1 < ... < j_(p-1). For for such an I, du_I means du_(i_1)^...^du_(i_p), and similarly for J.
Let the linear map Q: Omega_p(R x M) --> Omega_(p-1)(M) be defined by Q(omega)(x) = Sum_J(Int_0^1 b_J(t,x)dt) du_J, with omega as above. First we check that this definition is independent of the choice of (u_1,...,u_n). In another system (v_1,...,v_n) suppose omega =Sum_K A_K(t,x)dv_K + Sum_L B_L(t,x)dt^dv_L. The change of coordinates from the (t,u) system to the (t,v) system has the form
dt = dt
dv_i = Sum_j X_ij du_j
i.e. the t's and the other coordinates transform independently. Consequently B_L(t,x) = Sum_J X_LJ b_J(t,x) where the X_LJ are appropriate (p-1)x(p-1) minors of the matrix X_ij; and similarly for A_K(t,x); so calculating Q(omega)(x) in the v-cordinates gives
Sum_L(Int_0^1 B_L(t,x)dt) dv_L
= Sum_L(Int_0^1 Sum_J X_LJ b_J(t,x) dt) dv_L
= Sum_L Sum_J X_LJ (Int_0^1 b_J(t,x) dt) dv_L,
this last step because the X_LJ are constant in t,
=Sum_J Sum_L X_LJ (Int_0^1 b_J(t,x) dt) dv_L
=Sum_J (Int_0^1 b_J(t,x) dt) Sum_L X_LJ dv_L
=Sum_J (Int_0^1 b_J(t,x) dt) du_J,
the same as the calculation in the u-coordinates.

Next we verify the formula (*) dQ+Qd = i*_1 - i*_0. In fact, d(omega)=Sum_I da_I(t,x)^du_I + Sum_J db_J(t,x)^dt^du_J
= Sum_I Da_I(t,x)/Dt dt^du_I + Sum_J Sum_i Db_J(t,x)/Du_i du_i^dt^du_J + terms with no dt
= Sum_I Da_I(t,x)/Dt dt^du_I - Sum_J Sum_i Db_J(t,x)/Du_i dt^du_i^du_J + terms with no dt,
so Qd(omega) = Sum_I(Int_0^1 Da_I(t,x)/Dt dt) du_I - Sum_J Sum_i (Int_0^1 Db_J(t,x)/Du_i dt) du_i^du_J
= Sum_I (a_I(1,x) - a_I(0,x)) du_I - Sum_J Sum_i (Int_0^1 Db_J(t,x)/Du_i dt) du_i^du_J.
On the other hand dQ(omega) = Sum_J Sum_i D(Int_0^1 b_J(t,x)dt)/Du_i du_i^ du_J.
Interchanging differentiation with respect to u_i and integration with respect to t makes this term equal and opposite to the second term in Qd(omega). So (dQ +Qd)(omega) = Sum_I (a_I(1,x) - a_I(0,x)) du_I.
Now the inclusion i_0 clearly satisfies i*_0(du_i) = du_i and i*_0(dt) = 0; consequently i*_0(omega) = Sum_I a_I o i_0 du_I = Sum_I a_I(0,x)du_I; similarly i*_1(omega) = Sum_I a_I(1,x)du_I. This proves the formula.

Finally suppose omega is a closed form representing a class in H^p(R x M); since domega=0, the formula gives dQ(omega) = i*_1(omega) - i*_0(omega); the two pulled-back forms differ by a coboundary; they are in the same cohomology class.

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