Re: compact manifold-submersion

brown042@xxxxxxxxx wrote:
On 7 26 , 8 40 , "W. Dale Hall"
<wdunderscorehallatpacbelldotnet@last> wrote:
brown...@xxxxxxxxx wrote:
Let M and N be compact connected manifold and f:M-->N be a
submersion(that is for all x in M, the derivative mapping df_x:T_xM--
T_f(x)N is surjective). Show that for any p,q in N, the "fiber"
f^-1(p) is diffeomorphic to f^-1(q).
All I know is that by preimage theorem, f^-1(p) and f^-1(q) are same
dimensional manifolds. Could anybody help me on doing this problem?
Thanks.
Take a smooth path

w:[0,1] ---> N

joining p to q (i.e., w(0) = p, w(1) = q). Perturb it so that
f is transverse to w(I) (oops, since f is a submersion, it's
transverse to everything!) and note several things:

1) W = f^(-1) w(I) is a smooth manifold
2) (the boundary of W:) dW = f^(-1)(p) U f^(-1)(q)
3) It isn't hard to construct a projection

g: W ----> I

for which

g| f^(-1)(p) = 0
g| f^(-1)(q) = 1
g has no critical points.

If you know Morse theory, you're done. If not, then pull back the
1-form dx from I to the 1-form g*w on W, and impose a metric to get
an equivalent vector field V on W (so that the dot product with V
is the 1-form g*w).

The vector field V will then have flow lines (i.e., trajectories)
that run from f^(-1)(p) to f^(-1)(q) , and that if V_t denotes
a trajectory of the ODE

x' = V(x)

then

d/dt(g(V_t)) = 1.

Now, define

F: f^(-1)(p) ---> f^(-1)(q)

as follows:

For x in f^(-1)(p), let V_t(x) denote the solution to
the DE

x' = V(x)

for which V_0(x) = x. Since d/dt(g(V_t)) = 1, we know that
y = V_1(x) is in f^(-1)(q), and we set

F(x) = y.

What needs to be shown is that F is 1:1, surjective, and has a
continuous inverse. The fact that we can play the vector field
backwards (i.e., take the vector field -V, use it to map f^(-1)(q)
to f^(-1)(p), and show that the compositions are the appropriate
identity maps) tells us all we need to know.

I think that about does it.

Dale

Thank you very much for your reply. It was very helpful.
But I am new in differential topology and I would like to know more
about what you noted:
1) W = f^(-1) w(I) is a smooth manifold
2) (the boundary of W:) dW = f^(-1)(p) U f^(-1)(q)
3) It isn't hard to construct a projection

g: W ----> I

for which

g| f^(-1)(p) = 0
g| f^(-1)(q) = 1
g has no critical points.

Are 1) and 2) general fact about smooth maps between smooth manifolds
or a consequence of trasversality?

Rgarding 3), I guess g is w^(-1)f if w is a diffeomorphism from [0,1]
to w([0,1]). Can we always find a path w from p to q on smooth
manifold N which is diffeomorphic to [0,1]?
Thanks.


(1) is a consequence of f being transverse to w (the map f is transverse
to the submanifold K of M if the image of the tangent map of f and
the tangent space of K together span the tangent space of K [at each
point where the concept makes any sense]). There is a theorem,
called (when I grew up, which was when dinosaurs ruled the earth)
"the transversality theorem", which states (informally) that a map
can be made transverse to a submanifold quite easily.

(2) I suppose would be a consequence of (1).

As for your remaining question: is N is a smooth manifold, then
(the image of) a smooth path connecting two distinct points p and
q is by definition?) a homeomorphic image of the interval I = [0,1].
The only way it could fail to be a diffeomorphic image of I is for
its differential to fail to behave properly at some point or points.
Since you always have smooth local coordinate neighborhoods, it is
never a problem to smooth out paths, so the answer to your question
is "yes, we can always find smooth paths in smooth manifolds, whenever
continuous paths exist".

Dale
.

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