Defeated by the Klein bottle! Its mapping class group is not Z_2, as I had erroneously claimed. As Allan and Andy pointed out, sources state that the mapping class group is Z_2 + Z_2. So, what are these four classes of self-homeomorphisms of K ??? Any homeomorphism h : X —> X of a space X to itself induces an automorphism on each homology group H_n(X). For the Klein bottle, the group H_1(K) = Z + Z_2. There are just 4 automorphisms of Z + Z_2. (The generator u of Z_2 must go to itself, and the generator 1 of Z can go to 1, -1, 1+u, or -1+u.) Can these 4 automorphisms all be induced by homeomorphisms K —> K ??? As it turns out, the answer is Yes. And by basic facts, homeomorphisms that induce unequal automorphisms on any H_n cannot be continuously deformable to each other, so (assuming that Z_2 + Z_2 is the correct group), these must be all the cases. * The 1 |—> 1 case is induced by the identity K —> K. * The 1 |—> -1 case is induced by reversing the cylinder before identification. But the other two cases? These come from homeomorphisms of K that in my naïveté I had not suspected: Follow either of the above maps on the cylinder with the operation of *twisting one end of the cylinder 360º while holding the other end fixed* . (Because of the orneriness of the Klein bottle, it doesn't matter which way you twist it.) This does not appear to be the same homeomorphism as just flipping the orientation of the cylinder. In fact, if flipping the orientation of the cylinder C means reflecting it about a plane P of symmetry that cuts C into two rectangles (and then identifying the ends of C to get K), this self-homeomorphism of K can be deformed to the identity by sliding the Klein bottle around itself once. —Dan PS For more fun with the Klein bottle, identify the exactly 5 classes of (unoriented) simple closed curves on it that cannot be deformed one to the other. Andy Latto wrote: ----- On Mon, Dec 24, 2018 at 10:35 PM Allan Wechsler <acwacw@gmail.com> wrote:
Wikipedia gives a different answer. See the article "Mapping Class Group", in the section "Examples", subsection "Nonorientable surfaces", where they say it's Z2 x Z2.
On Mon, Dec 24, 2018, 9:52 PM Dan Asimov <dasimov@earthlink.net wrote:
Think of the Klein bottle K as a cylinder S^1 x [0,1] after its two boundary circle S^1 x 0 and S^1 x 1 have been identified by a reflection of S^1 via (x, y) |—> (x, -y).
Then you cannot rotate the cylinder after identification (which has become K) around its axis, because its end circles must rotate in opposite directions. The two things you *can* do are combinations of
a) slide the cylinder along its length, and
b) reverse the cylinder's direction.
You can also flip the orientation of the circle. That gives the other Z/2Z Andy
All of the slides are continuously deformable to each other, so there are two equivalence classes, so the group is Z/2. (Not a proof, of course.)
—Dan
----- Here's a puzzle I think I know the answer to, but I don't have a proof:
Let K denote the Klein bottle, the ideal surface you get when you identify the top and bottom edges of the unit square [0,1] x [0,1] normally, by
(x,0) ~ (x,1),
but identify the left and right edges by a flip:
(0,y) ~ (1,1-y).
Puzzle: ----------------------------------------------------------------------- Consider the space Homeo(K) of self-homeomorphisms of the Klein bottle. I.e., Homeo(K) consists of all continuous bijections
h : K —> K
having a continuous inverse.
Two self-homeomorphisms h_0, h_1 of K are *in the same path component* of Homeo(K) if there is a continuous *family*
{h(t) | 0 <= t <= 1}
of homeomorphisms h(t) in Homeo(K) such that h(e) = h_e for e = 0, 1.
The continuity of this family just amounts to there being a continuous map
H : K x [0,1] —> K
such that the restriction of H to any time-slice K x {t}:
H | K x {t} —> K
is the homeomorphism h(t) : K —> K.
QUESTION: ————————— How many path components does Homeo(K) have? -----------------------------------------------------------------------