Does a function like this exist..?
Jace/TBL
I've got the following problem:
I am looking for a continuous function f: R3\{0} -> R3\{0} for
which no vector p exists with f(p) = t*p for some scalar t.
In other words, the function should change any vectors' direction (or
not have eigenvectors), and be continuous. Note that the function does
_not_ have to be linear at all!
As for the syntaxis: with R3\{0} I mean the 3-dimensional
euclid space, without the (0,0,0) vector.
At first I thought this to be quite simple, but after some tries I
found it surprisingly hard. Currently I tend to believe that such a
function is only possible for vectorspaces of _even_ dimension, but I
can't find a proof for this.
Can anyone construct such a function? Or better yet: prove that it
doesn't exist?
Noet that it would suffice to make a fuction only on the unit sphere
S3 (i.e. all 3-dimensional vectors p with |p| = 1), cause then we can
use f(p/|p|) in the general case.
Hope anyone can enlighten me!
Jace.
ptk
cont. function) has a fixed point. Furthermore, if n is odd, then there
is a fixed point free map of S^n --> S^n (here S^k is the set of unit
vectors in (k+1) Euclidean space)
W. Dale Hall
>
> Hi,
>
> I've got the following problem:
> I am looking for a continuous function f: R3\{0} -> R3\{0} for
> which no vector p exists with f(p) = t*p for some scalar t.
So, you want a map f: R^3 \ {0} --> R^3 \ {0}, for which no vector x is
mapped to a vector y with y parallel to x.
Depending on how much you know, or are prepared to assume, it's not
difficult to show that no such map can be continuous.
Note that the map can be thought of as a nonzero vector field on R^3 \
{0}, where you simply place the value f(x) as a vector based at the
point x. In that case, your condition on f translates into the
condition that at each point x, the vector f(x) is not pointed in the
radial direction.
Now, let's take your suggestion, placed somewhat farther down in the
article:
> Note that it would suffice to make a function only on the unit sphere
> S3 (i.e. all 3-dimensional vectors p with |p| = 1), cause then we can
> use f(p/|p|) in the general case.
>
> Hope anyone can enlighten me!
> Jace.
The restriction of our function to S^2 ( the name of the unit sphere in
R^n is S^(n-1) ) is not quite a vector field on S^2 (which is where I
mean to be heading). We can produce such a vector field by eliminating
the radial component of f(x) at each x in S^2, yielding what I'll call
g: that is,
g(x) = f(x) - (f(x).x) x
Note that as long as |x| = 1, the inner product g(x).x is zero, and thus
the vector g(x) is perpendicular to the radial direction. In other
words, g forms a vector field on the unit sphere S^2. Further, g is
continuous, and nowhere zero on the sphere.
That can't happen: every continuous vector field on the 2-sphere S^2 has
at least one point at which it takes the value zero (i.e., the zero
vector). This is a special case of the Poincare' - Hopf Theorem, which
states that the sum of indices of the zeros of a non- degenerate vector
field on a closed manifold M is equal to the Euler characteristic of M.
For the 2-sphere, that number is 2. [I'll leave the discussion of
Poincare' - Hopf, and of the index of a zero of a vector field for
others]
Note that there is nothing magical about the unit sphere. Any sphere
centered at the origin can be treated similarly, by projecting the
vector field f(x) orthogonally away from the radial direction.
Therefore, at each radial distance R from the origin, there must be at
least one x, with |x| = R, and with f(x) parallel to x.
Dale.
W. Dale Hall
No, Brouwer's Fixed Point Theorem doesn't say that at all. The antipodal
map (x |--> -x) has no fixed points on the n-sphere for any n.
Dale.
Ronald Bruck
the "To," "Cc," and "Newsgroups" headers for details. ]]
In article <8ad7b77f.02010...@posting.google.com>, Jace/TBL
<Jace_t...@yahoo.com> wrote:
Hmmm. You posted this to sci.math.num-analysis, where I replied, but
you apparently didn't CROSS-post it, since my answer didn't appear
here.
As I said there, look up the "hairy ball theorem," also known as the
"every red-headed boy has a cowlick" theorem. (This was the way a
popularizing math book described it--Time-Life?)
The idea is that there cannot be a non-vanishing (continuous) tangent
vector field on S^{2n}. But, as you note, there IS one in S^{2n+1},
(x1,x2,...) --> (-x2, x1,...) where the ... follow the same pattern.
The nonexistence on S^{2n} follows from e.g. the Brouwer fixed-point
theorem or the Borsuk-Ulam theorem (they're equivalent).
--Ron Bruck
Simeon Stefanov
> Note that there is nothing magical about the unit sphere. Any sphere
> centered at the origin can be treated similarly, by projecting the
> vector field f(x) orthogonally away from the radial direction.
> Therefore, at each radial distance R from the origin, there must be at
> least one x, with |x| = R, and with f(x) parallel to x.
>
> Dale.
We can even state that there is a connected set C of points with
f(x) parallel to x which is "connecting" 0 and infinity. There is a
theorem due to K.Borsuk stating that for any continuous map
f : R^(2n+1) -> R^(2n+1)
such that f(0)=0 there exists a connected unbounded set C of points
with f(x) parallel to x containing the origin 0. His arguments work
without changes for this situation as well. (I remember the proof, but
I have lost the references for Borsuk's theorem.)
Simeon