Re: I'm so confused...

On 2006-01-15, David C Ullrich <ullrich@xxxxxxxxxxxxxxxx> wrote:

> Say R(t) is the element of SO(3,R) defined by
> a rotation through an angle t about, say,
> the axis (0,0,1). Define c : [0,2pi] -> SO(3,R)
> by c(t) = R(t), and define d : [0,4pi] -> SO(3,R)
> by d(t) = R(t). Then
>
> (i) c is not null-homotopic (as a closed curve in
> SO(3,R)
>
> (ii) d _is_ null-homotopic in SO(3,R).
>
> At least I _think_ that that's what Penrose
> is claiming.

That's right. There is a neat way to visualize this. Identify SO(3) with
the closed unit ball in R^3. In spherical coordinates, the point
(r,n), where n is a unit vector, corresponds to a counter-clockwise
rotation through angle pi*r about the axis parallel to n. However a
counter-clockwise and a clockwise rotation through angle pi about the
same axis gives the same rotation. Hence we must identify the points
(pi,n) and (pi,-n) on the surface of the unit ball.

The curve c(t) that you defined in (i) is just the segment of the z-axis
intersected with the unit ball. As t varies, the curve passes through
the surface of the ball exactly once (keeping in mind the identification
of diametrically opposite points). On the other hand, it is easy to see
that any curve that is homotopic to a closed loop properly contained in
the unit sphere (or null-homotopic, as you say) will have to cross the
surface an even number of times. Hence c cannot be null-homotopic.

By now, you've probably already figured out how the curve d is
null-homotopic. But here's a picture to make it explicitly clear:

a b
********** a ********** b
*** || *** *** ***
** || ** ** \ / **
** || ** ** \ / **
* || * * | | *
** || ** ** / \ **
** || ** ** / \ **
*** || *** *** ***
********** b ********** a
b a

********** **********
*** *** *** ***
a ** _________** b ** ____b **
**\ / ** ** / \ **
* | | * * | | *
**/ \__________** ** \____/ **
b ** ** a ** a **
*** *** *** ***
********** **********

As has been pointed out in another message, this entire discussion works
for RP^2 as well.

> Related topic: Trying to think of another example
> of a curve which is not null-homotopic but which
> is homotopic to its inverse, in particular an
> example I could visualize, I came up with this:
>
> Let T be what you might call a twisted torus:
> Start with the unit square [0,1]^2. Identify
> the point (t,0) with (t,1), and identify the
> point (0,s) with (0,1-s). It's easy to see
> why the curve c has this curious property,
> if c(t) = (0,t) for t in [0,1].

You've just constructed the famed Klein bottle.

> But then this T thing has me puzzled. It seems
> to be a 2-manifold. But I thought I'd read that
> 2-manifolds are characterized by "number of handles";
> if so the T should be homeomorphic to an ordinary
> torus, which it seems it's not, since an ordinary
> torus does not have a non-trivial closed curve c
> which is homotopic to -c. (Huh? Maybe T is not
> really a 2-manifold? Or that characterization
> is only for orientable manifolds?)

The Klein bottle is a non-orientable compact 2-manifold. Hence it is not
covered by the classification of compact Riemann surfaces.

Hope this helps.

Igor
.

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