Week 5
1. Definition (Transversal intersection). Consider a smooth map
f: M --> N and suppose N
has a submanifold P of co-dimension k.
This means that P is covered by coordinate charts (for N)
h: U --> R^n such that P intersect
U is (p o h)^(-1)(0), where
p: R^n --> R^k is projection on the first
k coordinates. We will
say that f is transverse to P (or that f(M)
intersects P
transversely) if each such (p o h o f):
f^(-1)(U) --> R^k has 0
as a regular value.
Proposition. In that case, f^(-1)(P) is a smooth submanifold (in
M) of codimension k.
Our next goal is a proposition which states that any smooth map R^n --> R^p can be approximated arbitrarily closely by one which has 0 as a regular value. This is the core of the proof of a theorem we will skip, which states that any f: M --> N as above can be approximated arbitrarily closely by one which in transverse to P (see Milnor, Thm. 1.35).
Theorem (Milnor, Lemma 1.19)
In M(n,p), the
np-dimensional vectorspace of n x p
matrices with real entries, the
subspace M(n,p;k) of matrices of rank k
is a smooth submanifold of dimension k(n+p-k).
Examples: In M(2,2) the space of matrices of rank 1 has dimension 3.
The three dimensions can be described as follows. One for the null-space,
a point in RP^1; one for the image, a point in RP^1;
and one for the
scaling between the 1-dimensional spaces (null-space complement)-->
(image). Similarly M(3,3;1) has dimension 5.
Definition: A subset A of R^n has measure zero if it can
be covered by a countable collection of cubes of arbitrarily
small total volume.
Class exercise: The x-axis in R^2 has measure zero.
Proposition: (Milnor, Lemma 1.14)
If f:R^n --> R^n is smooth, and A in R^n
has measure
zero, then f(A) has measure zero.
2. Theorem (Milnor, Thm. 1.21)
Consider a smooth f: U --> R^p, where
U is open in R^n. Then for any epsilon > 0
there exist an n x p matrix A and a
p-vector B, both with
maximum entry < epsilon, such that the map
g: U --> R^p
defined by g(x) = f(x) + Ax +
B has 0 as a regular value.
(Milnor shows
that for each k < p the map F_k: M
(n,p;k) x U --> M(n,p)
x R^p
given by F_k(K,x) =
(K-Df(x),-f(x)-
(K-Df(x))x) has image of
measure zero; it follows that there exist A,B within epsilon
of (0,0) and not in the image of any F_k. Use these to define
g(x) as above, and suppose 0 was not a regular value
for g. Then there would exist an x in U with
g(x)=0 and
Dg(x) of rank k < p.
Since Dg(x) = Df(x) + A, this means
that A is of the form K - Df(x),
where K is in M(n,p;k); since
g(x) =
f(x) + Ax + B = f(x) +
(K-Df(x))x +B = 0, this means
that B = -f(x)-(K-Df(x))x,
so (A,B) is in the
image of F_k, contrary to our choice.)
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