- Show that the
chain on the right hand side of (33.4) is a cycle.
- Check that
as defined by equation (33.5) satisfies (33.1).
- Show that the product in theorem (33.1) defines a bilinear map
.
- Determine explicitly the two/three chain
satisfying (33.4) when
(i)
and
.
(ii)
and
.
Hint: In the proof of lemma (32.2),
we chopped the square into two triangles. When
we need to chop a prism into three
pieces and map
affinely onto each of them.
- Use the map
of the previous lecture to calculate the generators of
.
- Use equation (33.1) to determine the image of the pair of generating one cycles of the previous exercise
under the map
.
in
Lecture - XXXIV Small simplicies
in
Recall that the Goursat lemma in complex analysis is proved by subdividing a triangle into four smaller triangles
determined by the midpoints of the sides of the given triangle.
The integral over the given triangle is then the sum of the integrals over the four little
pieces. Likewise, in the proof of
the classical Green's theorem (of which Cauchy's theorem is really a special case) one employs a subdivision into
tiny squares. The contributions to the integral from an edge common to a pair of abutting triangles/squares cancel out.
A similar idea underlies the method of small simplicies where we perform a systematic subdivision
operation known as barycentric subdivision. The barycentric subdivision enables us to replace a singular chain by a
homotopic one in which the constituent singular simplicies are small. A small simplex is
one whose image lies in an open set belonging to a prescribed open cover of the space. One achieves this
through iterated barycentric subdivisions a process reminiscent of one used in the proof of the Goursat lemma.
The fundamental theorem on small simplicies quickly leads us to the two fundamental results on algebraic topology - the
excision theorem discussed in lecture 39 and the Mayer Vietoris sequence that we shall derive here and use in the
next lecture.
Subsections
nisha
2012-03-20