Re: cohomotopy result
- From: "athan" <metanosis@xxxxxxxxx>
- Date: 28 Feb 2007 11:57:53 -0800
Zbigniew Karno wrote:
On 28 Lut, 04:39, "athan" <metano...@xxxxxxxxx> wrote:
if f and g are two smooth functions from a smooth manifold M^n->S^p,
and we have the condition
that |f-g|<2 then f and g are smoothly homotopic. I think I
understand
that in this case f and g
are both members of a cohomotopy group, but the significance of <2
evades me. Clearly, if
|f-g|>2, then either one of both of the maps don't map onto the n-
sphere, if |f-g|=2, then f,g
map to antipodal points, but don't see right away why that scotches
their being homotopic.
Thanks in advance for any suggestions on the significance of the
inequality condition.
Let B^{p+1} be the (p+1)-dimensional unit ball in R^{p+1}.
The boundary of B^{p+1} is the p-dimensional unit sphere S^p.
Let f,g : M^n --- > R^{p+1) be two smooth maps with property that
f(M^n) and g(M^n) are subsets of S^p, and ||f(x) - g(x)|| < 2 for
all x in M^n.
Consider a homotopy G : M^n x [0,1] --- > R^{p+1} defined by the
formula G(x,t) = (1-t)*f(x) + t*g(x), for x in M^n and t in [0,1].
Clearly, G is smooth, the image G(M^n x [0,1]) of G is contained
in B^{p+1}, and G joining f = G(.,0) and g = G(.,1). Moreover, for
each x in M^n the line segment [f(x),g(x)] with end-points f(x)
and g(x) is a track of x under homotopy G with the property that 0
- the center of B^{p+1} - does not belong to [f(x),g(x)], because
according to assumption ||f(x) - g(x)|| < 2. Hence 0 does not
belong to G(M^n x [0,1]).
Now consider a map r : R^{p+1} \ {0} --- > S^p defined by
r(x) = x/||x|| for x =/= 0. Clearly, r is smooth and r(x) = x for
x in S^p. In consequence, the map H : M^n x [0,1] --- > S^p, defined
as H(x,t) = r(G(x,t)), is a required homotopy joining f and g.
Regards,
Z. Karno
Many thanks for your response, I can see I am missing some obvious
insight
about what is going on geometrically. I think it has to do with the
general nature
of the ambiguity of the doman M^n. I used the dimension one case,
mapping into S^1,
so used an interval and the following functions f(x)=(cosx,sinx) and
g(x)=(cos(x+pi),sin(x+pi)).
then ||f-g||=2 unless I am mistaken.
Then the homotopy H: (0,pi) X (0,pi)--->S^1 by H(x,t)=(cos(x+t),sin(x
+t)),
so H(x,0)=f(x) and H(x,pi)=g(x), and basically the one function f is
pulled around
the circle until it becomes g by a phase change. There is no passing
through the origin, so "0" is not a problem.
I would have stopped, in your outline above, at the homotopy G(x,t)=(1-
t)*f(x)+t*g(x) and thought this was enough. Vaguely recall something
about not passing through the origin.
I guess really don't see why, if f,g map to the sphere, why any
homotopy of antipodal points would need to go thru the origin, since
all the action takes place on the surface of the sphere(yes??),
and away form the origin. Am visualizing the function f moving along
the surface of the sphere,
as it changes into the function g. Am sorry if this makes no sense,
but clearly the <2 condition constrains the tracks to miss the origin,
so I do see that, and thanks for that insight,
unfortunately can't quite grasp the remainder of whats going on.
Thanks again for your detailed response, and in advance if you are
able to somehow show me whats going on!!!
Best...athan
.
- References:
- cohomotopy result
- From: athan
- Re: cohomotopy result
- From: Zbigniew Karno
- cohomotopy result
- Prev by Date: Re: Circle Laying On Curves
- Next by Date: Re: Bell-curve distribution wanted
- Previous by thread: Re: cohomotopy result
- Next by thread: cohomotopy result
- Index(es):
Relevant Pages
|
