[math-fun] Klein bottle puzzles
Let K denote a Klein bottle. 1. The cartesian product S^1 x S^2 of a circle (S^1) and a sphere (S^2) is a certain 3-dimensional manifold. Puzzle: Does this manifold contain a Klein bottle K ⊂ S^1 x S^2 as a subset ??? 2. Call two simple closed curves C_0, C_1 on the Klein bottle "equivalent" if there is a continuous family {C_t ⊂ K | 0 ≤ t ≤ 1} of simple closed curves on K. How many inequivalent simple closed curves are there on K, and what is an example of each one ??? —Dan Appendix -------- A surface is called "nonorientable" if it has a subset that is a Möbius band. Otherwise it is called "orientable". The (compact) orientable and nonorientable surfaces (without boundary) were first classified up to topological equivalence by August Ferdinand Möbius in 1861, but this was not proved rigorously until Henry Roy Brahana did so in 1921. Any two surfaces can be combined topologically by the operation called "connected sum", which means to remove the interior of a disk from each surface and then connect the resulting two circular boundaries by a cylinder. Each orientable surface M is the connected sum of g tori (where the connected sum of 0 tori is defined to be the sphere). Each nonorientable surface M is the connected sum of g projective planes for some integer g ≥ 1, also called the "genus" of M. If S denotes the sphere, T denotes the torus and P denotes the projective plane, then every surface (compact and without boundary) can be expressed as the connected sum of S with finitely many copies of T and finitely many copies of P. This operation, denoted by #, has S as the identity element and is commutative with just one relation: P # P # P = P + T. It turns out that each compact surface M can be "triangulated" — expressed as the union of finitely many 2D triangles any two of which intersect in a common edge or vertex. Then if V, E, F denote the number of vertices, edges, and faces (triangles) of the surface, the computation 𝜒(M) = V - E + F is a topological invariant called the Euler characteristic of M. If M is orientable its genus satisfies g = 2 - 2 𝜒(M), and if M is nonorientable its genus satisfies g = 2 - 𝜒(M). Thus Euler characteristic and orientability determine the surface up to topological equivalence. The Klein bottle K shows up in this classification as K = P # P, the connected sum of two projective planes. It can also be viewed as the result of starting with a square and identifying the top and bottom sides in the same direction, but identifying the left and right sides in the reverse direction: |—————➞————| | | ↑ // ↓ | | |—————➞———— |
Regarding Dan’s second question, I kind of feel the answer should be 2, represented by one that joins the midpoints of the top and bottom edges and another that joins the midpoints of the left and right edges. But I’m not sure they’re genuinely different, and I don’t see why there couldn’t be other possibilities. Jim Propp On Wed, Nov 25, 2020 at 4:40 PM Dan Asimov <asimov@msri.org> wrote:
Let K denote a Klein bottle.
1. The cartesian product
S^1 x S^2
of a circle (S^1) and a sphere (S^2) is a certain 3-dimensional manifold.
Puzzle: Does this manifold contain a Klein bottle
K ⊂ S^1 x S^2
as a subset ???
2. Call two simple closed curves C_0, C_1 on the Klein bottle "equivalent" if there is a continuous family
{C_t ⊂ K | 0 ≤ t ≤ 1}
of simple closed curves on K.
How many inequivalent simple closed curves are there on K, and what is an example of each one ???
—Dan
Appendix -------- A surface is called "nonorientable" if it has a subset that is a Möbius band. Otherwise it is called "orientable".
The (compact) orientable and nonorientable surfaces (without boundary) were first classified up to topological equivalence by August Ferdinand Möbius in 1861, but this was not proved rigorously until Henry Roy Brahana did so in 1921.
Any two surfaces can be combined topologically by the operation called "connected sum", which means to remove the interior of a disk from each surface and then connect the resulting two circular boundaries by a cylinder.
Each orientable surface M is the connected sum of g tori (where the connected sum of 0 tori is defined to be the sphere).
Each nonorientable surface M is the connected sum of g projective planes for some integer g ≥ 1, also called the "genus" of M.
If S denotes the sphere, T denotes the torus and P denotes the projective plane, then every surface (compact and without boundary) can be expressed as the connected sum of S with finitely many copies of T and finitely many copies of P.
This operation, denoted by #, has S as the identity element and is commutative with just one relation:
P # P # P = P + T.
It turns out that each compact surface M can be "triangulated" — expressed as the union of finitely many 2D triangles any two of which intersect in a common edge or vertex. Then if V, E, F denote the number of vertices, edges, and faces (triangles) of the surface, the computation
𝜒(M) = V - E + F
is a topological invariant called the Euler characteristic of M. If M is orientable its genus satisfies
g = 2 - 2 𝜒(M),
and if M is nonorientable its genus satisfies
g = 2 - 𝜒(M).
Thus Euler characteristic and orientability determine the surface up to topological equivalence.
The Klein bottle K shows up in this classification as K = P # P, the connected sum of two projective planes. It can also be viewed as the result of starting with a square and identifying the top and bottom sides in the same direction, but identifying the left and right sides in the reverse direction:
|—————➞————| | | ↑ // ↓ | | |—————➞———— | _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On Wed, Nov 25, 2020 at 6:36 PM James Propp <jamespropp@gmail.com> wrote:
Regarding Dan’s second question, I kind of feel the answer should be 2, represented by one that joins the midpoints of the top and bottom edges and another that joins the midpoints of the left and right edges. But I’m not sure they’re genuinely different, and I don’t see why there couldn’t be other possibilities.
There are more than that. You can spiral around, connecting the top and bottom of the square as many times as you like. These represent different elements of the fundamental group, so they are not equivalent. That's enough to show that the literal answer to Dan's question is "Infinitely many", but this isn't all of them. Andy
On Wed, Nov 25, 2020 at 4:40 PM Dan Asimov <asimov@msri.org> wrote:
Let K denote a Klein bottle.
1. The cartesian product
S^1 x S^2
of a circle (S^1) and a sphere (S^2) is a certain 3-dimensional manifold.
Puzzle: Does this manifold contain a Klein bottle
K ⊂ S^1 x S^2
as a subset ???
My intuition is that an orientable 3-manifold can't contain a nonorientable compact surface, but I can't come up with a proof.
g = 2 - 2 𝜒(M),
This is backwards: it should be 𝜒(M) = 2 - 2 g
and if M is nonorientable its genus satisfies
g = 2 - 𝜒(M).
And I spent an embarrassingly long time trying to figure out whether this one was also backwards. Andy
--
Andy.Latto@pobox.com
Andy Latto wrote: 𝜒(M) = 2 - 2 g [for orientable surfaces M]. Yes, sorry about that. And as he almost wrote: 𝜒(M) = 2 - g for nonorientable compact surfaces. Examples: 𝜒(S^2) = 2 𝜒(Klein bottle) = 2 - 2 = 0. 𝜒(3-holed torus) = 2 - 2*3 = -4. 𝜒(P # P # P) = 2 - 3 = -1. —Dan
I'll just say that, so far, nothing correct has been mentioned about the answer to either puzzle. —Dan
As a corollary of Alexander duality, if M is a connected orientable, compact manifold of dimension n, and if H_1(M,Z) = (0) (the first homology group with coeffs in Z vanishes), then no nonorientable compact manifold N of dimension n -1 can be embebded in M. (for example, see Bredon, Topology and Geometry, Corollary 8.9, Page 353). This implies that K can’t be embedded in S^3. But this is not NOT Dan’s question! Unfortunately, by the Kunneth formula H_1(S^1xS^2, Z) = Z. (H_0(S^1,Z) = H_1(S^1,Z) = Z, H_0(S^2,Z) = Z, H_1(S^2,Z) = (0), H_2(S^2,Z) = Z). It appears that S^1 x S2 is a bit “bigger” than S^3, so I don’t know whether K can be embedded in S^1 x S^2. I would lean for no. Best, — Jean
On Nov 25, 2020, at 10:35 PM, Dan Asimov <asimov@msri.org> wrote:
I'll just say that, so far, nothing correct has been mentioned about the answer to either puzzle.
—Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I should at this point give some hints: 1. S^1 x S^2 in fact does have a Klein bottle as a subset. Can you find one? 2. The number of inequivalent simple closed curves on a Klein bottle is in fact finite (and less than 10). Can you find one example of each one? —Dan ————— PS The book "Topology and Geometry" mentioned by Jean is written at an advanced graduate-student level, but is excellent. William Goldman's review of it is around the middle of this page: <http://www.math.umd.edu/~wmg/reviews.html>.
On Friday/27November/2020, at 7:42 AM, Jean Gallier <jean@seas.upenn.edu> wrote:
As a corollary of Alexander duality, if M is a connected orientable, compact manifold of dimension n, and if H_1(M,Z) = (0) (the first homology group with coeffs in Z vanishes), then no nonorientable compact manifold N of dimension n -1 can be embebded in M. (for example, see Bredon, Topology and Geometry, Corollary 8.9, Page 353).
This implies that K can’t be embedded in S^3.
But this is not NOT Dan’s question!
Unfortunately, by the Kunneth formula H_1(S^1xS^2, Z) = Z. (H_0(S^1,Z) = H_1(S^1,Z) = Z, H_0(S^2,Z) = Z, H_1(S^2,Z) = (0), H_2(S^2,Z) = Z).
It appears that S^1 x S2 is a bit “bigger” than S^3, so I don’t know whether K can be embedded in S^1 x S^2. I would lean for no.
Best, — Jean
On Nov 25, 2020, at 10:35 PM, Dan Asimov <asimov@msri.org> wrote:
I'll just say that, so far, nothing correct has been mentioned about the answer to either puzzle.
—Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
For #1, take a family of great circles on S^2 that all intersect in the same two poles, like the meridional lines on a globe. Have the circle rotate 180 degrees, while its partner, a single point, goes all the way around S^1. For #2, I have a memory of having done this exercise as an undergraduate, and I think the answer was 4, but I don't remember the proof. It was something like showing that the curve that wraps twice around the "neck" could be deformed to a point by pulling one of the loops all the way around the long way. On Fri, Nov 27, 2020 at 10:56 AM Dan Asimov <asimov@msri.org> wrote:
I should at this point give some hints:
1. S^1 x S^2 in fact does have a Klein bottle as a subset. Can you find one?
2. The number of inequivalent simple closed curves on a Klein bottle is in fact finite (and less than 10). Can you find one example of each one?
—Dan ————— PS The book "Topology and Geometry" mentioned by Jean is written at an advanced graduate-student level, but is excellent. William Goldman's review of it is around the middle of this page: < http://www.math.umd.edu/~wmg/reviews.html>.
On Friday/27November/2020, at 7:42 AM, Jean Gallier <jean@seas.upenn.edu> wrote:
As a corollary of Alexander duality, if M is a connected orientable, compact manifold of dimension n, and if H_1(M,Z) = (0) (the first homology group with coeffs in Z vanishes), then no nonorientable compact manifold N of dimension n -1 can be embebded in M. (for example, see Bredon, Topology and Geometry, Corollary 8.9, Page 353).
This implies that K can’t be embedded in S^3.
But this is not NOT Dan’s question!
Unfortunately, by the Kunneth formula H_1(S^1xS^2, Z) = Z. (H_0(S^1,Z) = H_1(S^1,Z) = Z, H_0(S^2,Z) = Z, H_1(S^2,Z) = (0), H_2(S^2,Z) = Z).
It appears that S^1 x S2 is a bit “bigger” than S^3, so I don’t know whether K can be embedded in S^1 x S^2. I would lean for no.
Best, — Jean
On Nov 25, 2020, at 10:35 PM, Dan Asimov <asimov@msri.org> wrote:
I'll just say that, so far, nothing correct has been mentioned about the answer to either puzzle.
—Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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Yup, that's exactly it. A Klein bottle can be seen as a cylinder S^1 x [0,1] after the ends S^1 x {0} and S^1 x {1} have been identified by e.g. the map (cos(t), sin(t)) —> (cos(t), -sin(t)), and this is just what Allan's Klein bottle does. (More generally, the family of great circles on S^2 could rotate any odd multiple of 180º before returning to where it started, and their totality would be a Klein bottle.) —Dan
On Friday/27November/2020, at 10:15 AM, Allan Wechsler <acwacw@gmail.com> wrote:
For #1, take a family of great circles on S^2 that all intersect in the same two poles, like the meridional lines on a globe. Have the circle rotate 180 degrees, while its partner, a single point, goes all the way around S^1.
**********SPOILER********** for question 2. far below:
On Wednesday/25November/2020, at 1:39 PM, Dan Asimov <asimov@msri.org> wrote:
Let K denote a Klein bottle.
1. The cartesian product
S^1 x S^2
of a circle (S^1) and a sphere (S^2) is a certain 3-dimensional manifold.
Puzzle: Does this manifold contain a Klein bottle
K ⊂ S^1 x S^2
as a subset ???
2. Call two simple closed curves C_0, C_1 on the Klein bottle "equivalent" if there is a continuous family
{C_t ⊂ K | 0 ≤ t ≤ 1}
of simple closed curves on K.
How many inequivalent simple closed curves are there on K, and what is an example of each one ???
|—————➞————| | | ↑ // ↓ | | |—————➞———— | The Klein bottle has five different equivalence classes (of unoriented simple closed curves). Let the Klein bottle K be, as indicated by the picture, the result of identifying pairs of boundary points of the square [0,1]x[0,1] by * (x,0) ~ (x,1), 0 ≤ x ≤ 1 and * (0,y) ~ (1,1-y), 0 ≤ y ≤ 1. Then the five classes are represented by these sets: 1) {(1/2,t) | 0 ≤ t ≤ 1} 2) {(t,0) | 0 ≤ t ≤ 1} 3) {(t,1/2) | 0 ≤ t ≤ 1} 4) {(t,1/4) | 0 ≤ t ≤ 1} ∪ {(t,3/4) | 0 ≤ t ≤ 1} 5) {(1/2+(1/4)cos(t), 1/2+(1/4)sin(t) | 0 ≤ t ≤ 2π} —Dan
I am very puzzled by the purported answer to the puzzle about loop classes on the Klein bottle. I don't see the "identity" class, the one whose exemplar is a tiny local loop. If that one is included, the answer would make sense to me. The thing that is tripping me up is that the classes have to form a group, the first homotopy group of K. And the number of elements cited is prime, so the group would have to be cyclic. On Sat, Nov 28, 2020 at 3:27 PM Dan Asimov <asimov@msri.org> wrote:
**********SPOILER********** for question 2. far below:
On Wednesday/25November/2020, at 1:39 PM, Dan Asimov <asimov@msri.org> wrote:
Let K denote a Klein bottle.
1. The cartesian product
S^1 x S^2
of a circle (S^1) and a sphere (S^2) is a certain 3-dimensional manifold.
Puzzle: Does this manifold contain a Klein bottle
K ⊂ S^1 x S^2
as a subset ???
2. Call two simple closed curves C_0, C_1 on the Klein bottle "equivalent" if there is a continuous family
{C_t ⊂ K | 0 ≤ t ≤ 1}
of simple closed curves on K.
How many inequivalent simple closed curves are there on K, and what is an example of each one ???
|—————➞————| | | ↑ // ↓ | | |—————➞———— |
The Klein bottle has five different equivalence classes (of unoriented simple closed curves). Let the Klein bottle K be, as indicated by the picture, the result of identifying pairs of boundary points of the square [0,1]x[0,1] by
* (x,0) ~ (x,1), 0 ≤ x ≤ 1
and
* (0,y) ~ (1,1-y), 0 ≤ y ≤ 1.
Then the five classes are represented by these sets:
1) {(1/2,t) | 0 ≤ t ≤ 1}
2) {(t,0) | 0 ≤ t ≤ 1}
3) {(t,1/2) | 0 ≤ t ≤ 1}
4) {(t,1/4) | 0 ≤ t ≤ 1} ∪ {(t,3/4) | 0 ≤ t ≤ 1}
5) {(1/2+(1/4)cos(t), 1/2+(1/4)sin(t) | 0 ≤ t ≤ 2π}
—Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
The fundamental group considers only loops (not necessarily simple) starting and ending at a fixed choice of basepoint. This puzzle asks for only simple closed curves not necessarily starting & ending at any specific point. (The tiny local loop is the last one on the list of five classes I sent.) —Dan
On Saturday/28November/2020, at 1:49 PM, Allan Wechsler <acwacw@gmail.com> wrote:
I am very puzzled by the purported answer to the puzzle about loop classes on the Klein bottle. I don't see the "identity" class, the one whose exemplar is a tiny local loop. If that one is included, the answer would make sense to me.
The thing that is tripping me up is that the classes have to form a group, the first homotopy group of K. And the number of elements cited is prime, so the group would have to be cyclic.
On Sat, Nov 28, 2020 at 4:50 PM Allan Wechsler <acwacw@gmail.com> wrote:
The thing that is tripping me up is that the classes have to form a group, the first homotopy group of K. And the number of elements cited is prime, so the group would have to be cyclic.
They do not form a group, because of the requirement that these be *simple* closed curves, while the fundamental group looks at all maps from the circle to the space in question. For example, the fundamental group of the punctured plane is Z, but the number of inequivalent simple closed curves is 4; clockwise and counterclockwise circles that don't go around the removed point, and clockwise and counterclockwise circles that do go around the removed point. Any map of the circle to the punctured plane that goes twice around the missing point must self-intersect. Note that the requirement that the homotopy consist of simple closed curves means that the small clockwise and counterclockwise circle, both trivial as elements of the fundamental group, are not equivalent.
--
Andy.Latto@pobox.com
Just to be clear, the puzzle 2. that I proposed was only about subsets of K that are simple closed curves (without distinguishing anything about parametrizations). From that point of view the punctured plane has just two inequivalent families of simple closed curves. —Dan
On Saturday/28November/2020, at 4:35 PM, Andy Latto <andy.latto@pobox.com> wrote:
They do not form a group, because of the requirement that these be *simple* closed curves, while the fundamental group looks at all maps from the circle to the space in question. For example, the fundamental group of the punctured plane is Z, but the number of inequivalent simple closed curves is 4; clockwise and counterclockwise circles that don't go around the removed point, and clockwise and counterclockwise circles that do go around the removed point. Any map of the circle to the punctured plane that goes twice around the missing point must self-intersect.
Note that the requirement that the homotopy consist of simple closed curves means that the small clockwise and counterclockwise circle, both trivial as elements of the fundamental group, are not equivalent.
On Sat, Nov 28, 2020 at 3:27 PM Dan Asimov <asimov@msri.org> wrote:
**********SPOILER********** for question 2. far below:
How many inequivalent simple closed curves are there on K, and what is an example of each one ???
|—————➞————| | | ↑ // ↓ | | |—————➞———— |
The Klein bottle has five different equivalence classes (of unoriented simple closed curves)
How many different equivalence classes of oriented simple closed curves are there? The fifth of your curve (the one that represents the trivial element of the fundamental group) is equivalent to its reverse, because the Klein bottle is nonorientable. Are any of the others equivalent to their reverses? Showing that the five curves you exhibit are inequivalent is straightforward, since they all correspond to different elements of the fundamental group. Can you give a hint as to how you would go about showing that these 5 are all that there are? Andy
. Let the Klein bottle K be, as indicated by the picture, the result of identifying pairs of boundary points of the square [0,1]x[0,1] by
* (x,0) ~ (x,1), 0 ≤ x ≤ 1
and
* (0,y) ~ (1,1-y), 0 ≤ y ≤ 1.
Then the five classes are represented by these sets:
1) {(1/2,t) | 0 ≤ t ≤ 1}
2) {(t,0) | 0 ≤ t ≤ 1}
3) {(t,1/2) | 0 ≤ t ≤ 1}
4) {(t,1/4) | 0 ≤ t ≤ 1} ∪ {(t,3/4) | 0 ≤ t ≤ 1}
5) {(1/2+(1/4)cos(t), 1/2+(1/4)sin(t) | 0 ≤ t ≤ 2π}
—Dan _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
If you consider *oriented* simple closed curves on K, then the one that bounds a disk (5.) and the one that goes around once in the "y-direction" are each equivalent to their mirror images; the other three equivalence classes are not. They don't quite all correspond to different elements of the fundamental group of K, since they don't a priori share a basepoint. But they can be shown inequivalent by the use of Z_2 intersection number of homology classes.* The two classes that go once around horizontally each Z_2-intersect themselves once but not each other, so they're distinct. The horizontal class going twice around has 0 Z_2-intersection with each of these, so it's distinct from them. The other two classes are easy to distinguish from the above three classes and from each other using their homology class. To show there are no other classes is trickier and I'm not prepared to explain it here; it requires understanding covering spaces and the fundamental group of K, which has the presentation π_1(K,x_0) = 〈a,b | a^2 b^2〉. —Dan _____ * To get the Z_2 intersection number of two closed loops C and C', represent them by smooth curves that have a finite number of intersection points, at each of which they're not tangent to each other. Then just count them mod 2.
On Saturday/28November/2020, at 4:42 PM, Andy Latto <andy.latto@pobox.com> wrote:
How many different equivalence classes of oriented simple closed curves are there? The fifth of your curve (the one that represents the trivial element of the fundamental group) is equivalent to its reverse, because the Klein bottle is nonorientable. Are any of the others equivalent to their reverses?
Showing that the five curves you exhibit are inequivalent is straightforward, since they all correspond to different elements of the fundamental group. Can you give a hint as to how you would go about showing that these 5 are all that there are?
participants (6)
-
Allan Wechsler -
Andy Latto -
Dan Asimov -
Dan Asimov -
James Propp -
Jean Gallier