Re: I'm so confused...

José Carlos Santos wrote: 
David C. Ullrich 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.

At first this seemed clearly impossible since
d is just c + c, after all. But this morning
I think I see why (ii) holds: We only need to
show that c _is_ homotopic to -c (the same as
c except traversed in the opposite direction).
Say a(s) is a continuous path in the unit sphere in R^3 joining (0,0,1) to (0,0,-1), and
let c_s(t) be a rotation about a(s) through an
angle t (for t in [0,2pi].) Then (c_s) is a
continuous family of closed curves, c_0 = c
and c_1 = -c.

Right? If that is in fact a correct proof of
(ii) then how does one prove (i)? (I mean (i)
is obvious, but otoh (ii) is obviously false...)



Consider the multiplicative group G of the quaternions with norm
1 (which is isomorphic with SU(2)). Topologically, this is the
3-sphere and therefore it's simply-connected. Now consider the path

   g:[0,pi] ----> G
       t     |-> cos(t) + i sin(t),

which is clearly not a loop.

The group G acts naturally on the space H of purely imaginary
quaternions, because if g is in G and q is a purely imaginary
quaternion, then g*q*g^{-1} is also purely imaginary. It turns out
that the map from H into itself defined by q |-> g*q*g^{-1}
preserves the quaternionic norm; since H, as a real vector space, is
3-dimensional, this defines a homomorphism _f_ from G into SO(3,R)
and it turns out that it is a double covering.

Now, what's f o g? For each _t_, f(g(t)) is a rotation around
_i_ whose angle is 2*t. In particular, f o g is a loop. However,
this loop cannot be null-homotopic, since, if you lift it to G, you
get _g_, which is not even a loop.

I hope that this helps. 

This is a brilliant exposition.  Thanks.

All that is left is to show that gqg^{-1}=q if and only if g=-1 (i.e. the kernel of your map has two elements). By considering q=i,j, and k, this is easy.
.

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