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.
Examples.
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
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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