Re: [math-fun] Klein bottle puzzle (SPOILER)
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. 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? ----------------------------------------------------------------------- -----
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.
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? ----------------------------------------------------------------------- -----
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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? ----------------------------------------------------------------------- -----
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-- Andy.Latto@pobox.com
participants (3)
-
Allan Wechsler -
Andy Latto -
Dan Asimov