Hairy dog theorem
$begingroup$
I am interested in applications of the hairy ball theorem to non-closed surfaces, such as dogs and cats.
As has been pointed out to me in my previous question, it is possible to arrange the two cowlicks of a hairy ball so that they are on top of each other. One can then "cut out" these cowlicks, leaving a sphere with a hole that has been combed smooth with no cowlicks.
Let's consider the task of combing smooth an entire puppy. Topologically, animals are toruses, but if we seal shut the oral and anal orifices they become spheres. However, the fur (ie. fur-covered skin) of the puppy is not a closed sphere: There are holes in several places such as the paws, ears and eyes.
My conjecture therefore is that we are guaranteed to be able to comb the entire puppy smooth, because we can arrange for the [virtual] cowlicks to end up on these holes - for instance on the paws, where there is no fur. Real puppies, of course, have natural cowlicks, and presumably the tip of the tail would get complicated, but those are mere details anatomy. As I've mentioned above, with a single hole, it seems trivial to demonstrate that combing over is possible. I imagine that with multiple holes the situation is no different: They can be constructed from the one-hole case by cutting additional holes (which does not disrupt the combing).
Is it correct to conclude, therefore, that any compact surface can always be combed over with no cowlicks? Is it true that spheres are the only object that requires cowlicks?
differential-topology
asked Jul 25 '16 at 11:56
SuperbestSuperbest
2,50011120
$endgroup$
add a comment |
$begingroup$
I am interested in applications of the hairy ball theorem to non-closed surfaces, such as dogs and cats.
As has been pointed out to me in my previous question, it is possible to arrange the two cowlicks of a hairy ball so that they are on top of each other. One can then "cut out" these cowlicks, leaving a sphere with a hole that has been combed smooth with no cowlicks.
Let's consider the task of combing smooth an entire puppy. Topologically, animals are toruses, but if we seal shut the oral and anal orifices they become spheres. However, the fur (ie. fur-covered skin) of the puppy is not a closed sphere: There are holes in several places such as the paws, ears and eyes.
My conjecture therefore is that we are guaranteed to be able to comb the entire puppy smooth, because we can arrange for the [virtual] cowlicks to end up on these holes - for instance on the paws, where there is no fur. Real puppies, of course, have natural cowlicks, and presumably the tip of the tail would get complicated, but those are mere details anatomy. As I've mentioned above, with a single hole, it seems trivial to demonstrate that combing over is possible. I imagine that with multiple holes the situation is no different: They can be constructed from the one-hole case by cutting additional holes (which does not disrupt the combing).
Is it correct to conclude, therefore, that any compact surface can always be combed over with no cowlicks? Is it true that spheres are the only object that requires cowlicks?
differential-topology
asked Jul 25 '16 at 11:56
SuperbestSuperbest
2,50011120
$endgroup$
1
$begingroup$
If you want to be pedantic about it, mammals are not toruses, but have genus at least $3$, due to the nostrils.
$endgroup$
– Henning Makholm
Jul 25 '16 at 12:36
1
$begingroup$
Don't forget the ears.
$endgroup$
– Joel Reyes Noche
Aug 3 '16 at 0:35
add a comment |
$begingroup$
I am interested in applications of the hairy ball theorem to non-closed surfaces, such as dogs and cats.
As has been pointed out to me in my previous question, it is possible to arrange the two cowlicks of a hairy ball so that they are on top of each other. One can then "cut out" these cowlicks, leaving a sphere with a hole that has been combed smooth with no cowlicks.
Let's consider the task of combing smooth an entire puppy. Topologically, animals are toruses, but if we seal shut the oral and anal orifices they become spheres. However, the fur (ie. fur-covered skin) of the puppy is not a closed sphere: There are holes in several places such as the paws, ears and eyes.
My conjecture therefore is that we are guaranteed to be able to comb the entire puppy smooth, because we can arrange for the [virtual] cowlicks to end up on these holes - for instance on the paws, where there is no fur. Real puppies, of course, have natural cowlicks, and presumably the tip of the tail would get complicated, but those are mere details anatomy. As I've mentioned above, with a single hole, it seems trivial to demonstrate that combing over is possible. I imagine that with multiple holes the situation is no different: They can be constructed from the one-hole case by cutting additional holes (which does not disrupt the combing).
Is it correct to conclude, therefore, that any compact surface can always be combed over with no cowlicks? Is it true that spheres are the only object that requires cowlicks?
differential-topology
asked Jul 25 '16 at 11:56
SuperbestSuperbest
2,50011120
$endgroup$
I am interested in applications of the hairy ball theorem to non-closed surfaces, such as dogs and cats.
As has been pointed out to me in my previous question, it is possible to arrange the two cowlicks of a hairy ball so that they are on top of each other. One can then "cut out" these cowlicks, leaving a sphere with a hole that has been combed smooth with no cowlicks.
Let's consider the task of combing smooth an entire puppy. Topologically, animals are toruses, but if we seal shut the oral and anal orifices they become spheres. However, the fur (ie. fur-covered skin) of the puppy is not a closed sphere: There are holes in several places such as the paws, ears and eyes.
My conjecture therefore is that we are guaranteed to be able to comb the entire puppy smooth, because we can arrange for the [virtual] cowlicks to end up on these holes - for instance on the paws, where there is no fur. Real puppies, of course, have natural cowlicks, and presumably the tip of the tail would get complicated, but those are mere details anatomy. As I've mentioned above, with a single hole, it seems trivial to demonstrate that combing over is possible. I imagine that with multiple holes the situation is no different: They can be constructed from the one-hole case by cutting additional holes (which does not disrupt the combing).
Is it correct to conclude, therefore, that any compact surface can always be combed over with no cowlicks? Is it true that spheres are the only object that requires cowlicks?
differential-topology
differential-topology
asked Jul 25 '16 at 11:56
SuperbestSuperbest
2,50011120
asked Jul 25 '16 at 11:56
SuperbestSuperbest
2,50011120
edited Aug 2 '16 at 23:08
Superbest
asked Jul 25 '16 at 11:56
SuperbestSuperbest
2,50011120
asked Jul 25 '16 at 11:56
SuperbestSuperbest
2,50011120
asked Jul 25 '16 at 11:56
SuperbestSuperbest
2,50011120
2,50011120
1
$begingroup$
If you want to be pedantic about it, mammals are not toruses, but have genus at least $3$, due to the nostrils.
$endgroup$
– Henning Makholm
Jul 25 '16 at 12:36
1
$begingroup$
Don't forget the ears.
$endgroup$
– Joel Reyes Noche
Aug 3 '16 at 0:35
add a comment |
1
$begingroup$
If you want to be pedantic about it, mammals are not toruses, but have genus at least $3$, due to the nostrils.
$endgroup$
– Henning Makholm
Jul 25 '16 at 12:36
1
$begingroup$
Don't forget the ears.
$endgroup$
– Joel Reyes Noche
Aug 3 '16 at 0:35
1
1
$begingroup$
If you want to be pedantic about it, mammals are not toruses, but have genus at least $3$, due to the nostrils.
$endgroup$
– Henning Makholm
Jul 25 '16 at 12:36
$begingroup$
If you want to be pedantic about it, mammals are not toruses, but have genus at least $3$, due to the nostrils.
$endgroup$
– Henning Makholm
Jul 25 '16 at 12:36
1
1
$begingroup$
Don't forget the ears.
$endgroup$
– Joel Reyes Noche
Aug 3 '16 at 0:35
$begingroup$
Don't forget the ears.
$endgroup$
– Joel Reyes Noche
Aug 3 '16 at 0:35
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your conjecture is (mostly) correct, but perhaps stated badly:
every connected compact surfaces-with-boundary admits an everywhere nonzero vector fields.
On the other hand, there are compact surfaces-without-boundary, other than the sphere, that do not admit nonzero vector fields. A 2-holed torus, for example, does not.
As for combing cats and dogs, it may be that at the boundaries, the hair directions are predetermined, in much the same way that at the edges of your fingerprints (as you move from the front of the hand towards the back), the lines of the fingerprint are more or less perpendicular to the dividing-line between "fingerprinted" and "non-fingerprinted" areas. This constraint (along with those at the fingernails) turns out to put restrictions on the number of whorls, "triangles", and "folds" that can occur in the fingerprint pattern. If there were similar constraints on your cat or dog at every boundary piece (for instance, that at the "ankle", the hair is always growing pretty much downward toward the foot), then in asking whether there was a cowlick-free combing for the rest of the cat, the theorem in my first bullet is no longer relevant, and the question once again becomes interesting, much like the fingerprint question.
answered Jul 25 '16 at 12:22
John HughesJohn Hughes
64.4k24191
$endgroup$
add a comment |
$begingroup$
Connected compact manifolds are combable iff they have Euler characteristic zero. Cutting holes in a surface just makes it noncompact.
answered Jul 25 '16 at 12:06
AnonAnon
5,000516
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your conjecture is (mostly) correct, but perhaps stated badly:
every connected compact surfaces-with-boundary admits an everywhere nonzero vector fields.
On the other hand, there are compact surfaces-without-boundary, other than the sphere, that do not admit nonzero vector fields. A 2-holed torus, for example, does not.
As for combing cats and dogs, it may be that at the boundaries, the hair directions are predetermined, in much the same way that at the edges of your fingerprints (as you move from the front of the hand towards the back), the lines of the fingerprint are more or less perpendicular to the dividing-line between "fingerprinted" and "non-fingerprinted" areas. This constraint (along with those at the fingernails) turns out to put restrictions on the number of whorls, "triangles", and "folds" that can occur in the fingerprint pattern. If there were similar constraints on your cat or dog at every boundary piece (for instance, that at the "ankle", the hair is always growing pretty much downward toward the foot), then in asking whether there was a cowlick-free combing for the rest of the cat, the theorem in my first bullet is no longer relevant, and the question once again becomes interesting, much like the fingerprint question.
answered Jul 25 '16 at 12:22
John HughesJohn Hughes
64.4k24191
$endgroup$
add a comment |
$begingroup$
Your conjecture is (mostly) correct, but perhaps stated badly:
every connected compact surfaces-with-boundary admits an everywhere nonzero vector fields.
On the other hand, there are compact surfaces-without-boundary, other than the sphere, that do not admit nonzero vector fields. A 2-holed torus, for example, does not.
As for combing cats and dogs, it may be that at the boundaries, the hair directions are predetermined, in much the same way that at the edges of your fingerprints (as you move from the front of the hand towards the back), the lines of the fingerprint are more or less perpendicular to the dividing-line between "fingerprinted" and "non-fingerprinted" areas. This constraint (along with those at the fingernails) turns out to put restrictions on the number of whorls, "triangles", and "folds" that can occur in the fingerprint pattern. If there were similar constraints on your cat or dog at every boundary piece (for instance, that at the "ankle", the hair is always growing pretty much downward toward the foot), then in asking whether there was a cowlick-free combing for the rest of the cat, the theorem in my first bullet is no longer relevant, and the question once again becomes interesting, much like the fingerprint question.
answered Jul 25 '16 at 12:22
John HughesJohn Hughes
64.4k24191
$endgroup$
add a comment |
$begingroup$
Your conjecture is (mostly) correct, but perhaps stated badly:
every connected compact surfaces-with-boundary admits an everywhere nonzero vector fields.
On the other hand, there are compact surfaces-without-boundary, other than the sphere, that do not admit nonzero vector fields. A 2-holed torus, for example, does not.
As for combing cats and dogs, it may be that at the boundaries, the hair directions are predetermined, in much the same way that at the edges of your fingerprints (as you move from the front of the hand towards the back), the lines of the fingerprint are more or less perpendicular to the dividing-line between "fingerprinted" and "non-fingerprinted" areas. This constraint (along with those at the fingernails) turns out to put restrictions on the number of whorls, "triangles", and "folds" that can occur in the fingerprint pattern. If there were similar constraints on your cat or dog at every boundary piece (for instance, that at the "ankle", the hair is always growing pretty much downward toward the foot), then in asking whether there was a cowlick-free combing for the rest of the cat, the theorem in my first bullet is no longer relevant, and the question once again becomes interesting, much like the fingerprint question.
answered Jul 25 '16 at 12:22
John HughesJohn Hughes
64.4k24191
$endgroup$
Your conjecture is (mostly) correct, but perhaps stated badly:
every connected compact surfaces-with-boundary admits an everywhere nonzero vector fields.
On the other hand, there are compact surfaces-without-boundary, other than the sphere, that do not admit nonzero vector fields. A 2-holed torus, for example, does not.
As for combing cats and dogs, it may be that at the boundaries, the hair directions are predetermined, in much the same way that at the edges of your fingerprints (as you move from the front of the hand towards the back), the lines of the fingerprint are more or less perpendicular to the dividing-line between "fingerprinted" and "non-fingerprinted" areas. This constraint (along with those at the fingernails) turns out to put restrictions on the number of whorls, "triangles", and "folds" that can occur in the fingerprint pattern. If there were similar constraints on your cat or dog at every boundary piece (for instance, that at the "ankle", the hair is always growing pretty much downward toward the foot), then in asking whether there was a cowlick-free combing for the rest of the cat, the theorem in my first bullet is no longer relevant, and the question once again becomes interesting, much like the fingerprint question.
answered Jul 25 '16 at 12:22
John HughesJohn Hughes
64.4k24191
answered Jul 25 '16 at 12:22
John HughesJohn Hughes
64.4k24191
answered Jul 25 '16 at 12:22
John HughesJohn Hughes
64.4k24191
answered Jul 25 '16 at 12:22
John HughesJohn Hughes
64.4k24191
64.4k24191
add a comment |
add a comment |
$begingroup$
Connected compact manifolds are combable iff they have Euler characteristic zero. Cutting holes in a surface just makes it noncompact.
answered Jul 25 '16 at 12:06
AnonAnon
5,000516
$endgroup$
add a comment |
$begingroup$
Connected compact manifolds are combable iff they have Euler characteristic zero. Cutting holes in a surface just makes it noncompact.
answered Jul 25 '16 at 12:06
AnonAnon
5,000516
$endgroup$
add a comment |
$begingroup$
Connected compact manifolds are combable iff they have Euler characteristic zero. Cutting holes in a surface just makes it noncompact.
answered Jul 25 '16 at 12:06
AnonAnon
5,000516
$endgroup$
Connected compact manifolds are combable iff they have Euler characteristic zero. Cutting holes in a surface just makes it noncompact.
answered Jul 25 '16 at 12:06
AnonAnon
5,000516
answered Jul 25 '16 at 12:06
AnonAnon
5,000516
answered Jul 25 '16 at 12:06
AnonAnon
5,000516
answered Jul 25 '16 at 12:06
AnonAnon
5,000516
5,000516
add a comment |
add a comment |
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1
$begingroup$
If you want to be pedantic about it, mammals are not toruses, but have genus at least $3$, due to the nostrils.
$endgroup$
– Henning Makholm
Jul 25 '16 at 12:36
1
$begingroup$
Don't forget the ears.
$endgroup$
– Joel Reyes Noche
Aug 3 '16 at 0:35