MAT 531 (Spring 1996) Topology/Geometry II

Week 8
1. An n-dimensional topological manifold-with-boundary is a space locally homeomorphic to R^n- (the set of points in R^n with x_1 < = 0). The boundary DM is the set of points corresponding to x1 = 0 under one (and therefore any) of those local homeomorphisms. A smooth atlas on such a manifold is a set of coordinate charts of this type which are differentiably related. The only difference from the no-boundary definition occurs with the definition of Dy_i/Dx_1 at a point with x_1 = 0, where the x's and y's are coordinates in two systems overlapping at a boundary point. We take this to mean the one-sided derivative, taken from the side where x_1 < = 0.
The boundary DM is an (n-1)-dimensional manifold: it inherits a smooth atlas from M, given by the restriction to DM of the special smooth coordinate systems defining M as a smooth manifold-with-boundary. Since y_1 and x_1 are constant along the boundary, the functions (y_2,...,y_n) and (x_2,...,x_n) give smoothly related (n-1)-dimensional coordinates on DM.
Suppose M is oriented: an atlas has been chosen where all overlapping coordinates are positively related (the Jacobian matrix of the change-of-coordinates map has positive determinant). This also makes sense for a manifold-with-boundary; furthermore, if we assume as above that the boundary is given by x_1 = 0 (and the rest by x_1 < 0) in those coordinate charts intersecting the boundary, then when two such charts intersect the Jacobian D(y_2,...,y_n)/D(x_2,...,x_n) will be positive. So DM inherits an orientation from M.

1. M = [a,b] with a < b, a closed interval. This is clearly a 1-dimensional manifold-with-boundary. A boundary-defining coordinate near b is h_1(x) = x - b ; a boundary-defining coordinate near a is k_1(x) = a - x. These two coordinates are not positively related at an interior x. To study the orientation induced on the boundary, we make the convention that an orientation on a 0-dimensional manifold is an assignment of a sign, + or -, to each of its points; and that a boundary point receives the + orientation if it is defined as above by a positive boundary-defining coordinate, and - otherwise. Giving M the orientation inherited from the x-axis makes the coordinate h_1 positively oriented, and the coordinate k_1 negatively. Consequently D([a,b]) = {b} - {a}.
2. M = D^2, the disc {x^2 + y^2 <= 1}. The boundary is the circle S^1. Give M the orientation inherited from the standard x,y orientation of the plane. Then near the point (1,0) a positive, boundary defining coordinate system is, for example, h_1(x,y) = (x/sqrt(1-y^2) - 1), h_2(x,y)=y; so that h_1 <= 0 means x <= sqrt(1-y^2) as desired. The orientation induced on S^1 near (1,0) is that defined by increasing h_2, i.e. increasing y. This gives S^1 the counter-clockwise orientation.

Stokes' Theorem. With M, DM as above, consider a smooth (n-1)-form omega on M. The integral of d omega over M is equal to the integral of omega over DM.

             /                   /
             | d omega     =     | omega.
             /M                  /DM

Proof: following Bredon.

2. Differential forms and topology. Write Omega^p(M) for the (R-)vector-space of differential forms of degree p on the m-dimensional manifold M. This space is 0 for p> m. Since dd = 0 we can construct in each dimension < = m the vector-space H^p(M) = H^p = Z^p/B^p = (ker d)/(im d), where Z^p = ker d is the kernel of d: Omega^p --> Omega^(p+1) and B^p = im d is the image of d: Omega^(p-1) --> Omega^p. The elements of these sub-vector-spaces are called p-cocycles and p-coboundaries, respectively; while H^p(M) is called the p-th de Rham cohomology vector-space of M.

Examples. At this point we can calculate only a few examples of de Rham cohomology: for example M = {*} (a point) and M = S^1 (the circle).
M = {*} is a 0-dimensional manifold. The only non-zero Omega(M) is consequently Omega^0(M), the vector-space of real-valued functions; here functions on a point. This vector-space is clearly R. Since there are no other non-zero Omega(M)'s, Z^0 = Omega^0 and B^0 = 0; consequently H^0 = R, and all others are 0.
The circle is 1-dimensional, so there are only non-zero Omega^0 and Omega^1. Clearly B^0 = 0 and Z^1 = Omega^1. An element of Omega^0, i.e. a real-valued function f(theta), is in Z^0 if df = 0, so f is a constant, and the vector-space H^0 is again R. To calculate H^1, consider the map I: Z^1 = Omega^1 --> R which assigns to each 1-form f(theta)d(theta) its integral. Class exercise ker I = B^1. Consequently H^1 = Z^1/B^1 = Z^1/(ker I) = R.

The rest of the class was spent on the definitions of cochain complex, exact sequence, and homomorphism of cochain complexes. Examples: for a smooth manifold the sequence

                        d           d       d                     
          0 --> Omega^0 --> Omega^1 --> ... --> Omega^n --> 0
is a cochain complex (the de Rham complex of M, denoted Omega^*(M) ), and if f: M --> N is a smooth map, then the various f^*:Omega^k(N) --> Omega^k(M) fit together to give f^*:Omega^*(N) --> Omega^*(M), a homomorphism of cochain complexes.

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