**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) --> ... .

0 --> H^0(M) --> H^0(U) + H^0(V) --> H^0(U inter V) --> H^1(M) i.e. 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) i.e. H^1(M) --> 0must 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) i.e. 0 --> R --> H^2(M) --> 0exactness implies H^2(S^2) = Z.

*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 d*omega*=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.

(Back to Week-by-week page)