Re: Flow of a differential equation

Victor B. wrote:
Hello,

here is a proposition I'm not sure of.


"Let f be a diffeomorphism of R^n. Prove that, given a _continuous_
family of diffeomorphisms (f_t)_t from R^n, with f_0 = Id and f_1=f,
f_t is the flow of a certain differential system".

Is this true ? How can I prove this ?

I'm unsure of what you're assuming and what you wish to prove.

If you are saying

"give me f : R^n --> R^n a diffeomorphism
and I'll conclude that there is a family
of diffeomorphisms

f_t : R^n --> R^n

(so that the rest follows...)"

then I'd say you're wrong (see explanation
below my .sig).

However, if you're assuming that you are given
a parametrized family {f_t | 0 <= t <= 1} of
self-diffeomorphisms of R^n, for which

f_0 = id_R^n, and f_1 = f,

then here's how to construct a vector field
V on R^n x [0,1], where coordinates of R^n x I
are given as (x,t):

V(x,t) = d/dt (f_t(x))

that is, V(x,t) is the velocity vector of the path
followed by x under the isotopy {f_t | t in [0,1]}.

Note that f_t(x) is, by construction, the solution
of the (time-dependent) DE

X' = V(X,t)

under the initial conditions X(0) = f_0(x).

Why? Because *that's what it means* to have
a solution of the time-dependent DE: a curve
for which the velocity vector at each point
is given by the vector field defining the DE.

That's all I know for now.

Dale

Here's why the first interpretation won't do:

It won't be true if f doesn't preserve orientation.

The diffeomorphism

f: (x1, x2, ..., xn) |--> (-x1, x2, x3, ..., xn)

is not isotopic (i.e. homotopic through homeomorphisms) to
the identity map. To prove it, note that one could take any
isotopy, extend in the obvious way to an isotopy of self-maps
of S^n [and if it ain't obvious: map the point at oo to itself
throughout]. The isotopy in S^n begins with a degree 1 self-
map (namely, the identity) and ends with a degree -1 self-
map. Since degree is a homotopy invariant (it's already a
homology invariant), this can't happen.


Thank you very much

Victor
.

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