Wikipedia:Reference desk/Archives/Mathematics/2009 January 11

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January 11

Group of points on the unit sphere

Let G be a finite non-trivial subgroup of SO(3). Let X be the set of points on the unit sphere in R^3 which are fixed by some non-trivial rotation in G. G acts on X - show that the number of orbits is either 2 or 3. What is G if there are only two orbits?

Could really just use a hand getting started on this one if nothing else. If the group is finite, I think all rotations have to be of the form 2pi/k for some integer k, and all axes of rotation have to be at an angle of 2pi/k to all other axes for a (different) integer k, right? That's about as much as I've accomplished so far. Cheers guys,

Spamalert101 (talk) 00:22, 11 January 2009 (UTC)Spamalert101

If the elements of G are not all rotations about the same axis then G must be the rotational symmetry group of one of the Platonic solids. There are three different "types" of points that are fixed by some non-trivial element of G and permuted by the other elements of G - these form the three orbits of G acting on X. One orbit is the vertices of the Platonic solid - I'll let you work out what the other two orbits are (hint: think about dual solids which have the same symmetry group).

On the other hand, if the elements of G are all rotations about the same axis then all non-trivial elements of G have the same two fixed points (what are they ?) and hence each of these two points is an orbit, and we have just two orbits. Gandalf61 (talk) 13:44, 11 January 2009 (UTC)

Is your first sentence obvious? Algebraist 14:13, 11 January 2009 (UTC)

Trivial to be precise (I was joking). :) PST —Preceding unsigned comment added by Point-set topologist (talk • contribs) 16:11, 11 January 2009 (UTC)

Having brushed up my geometry with our article Point groups in three dimensions, I can now say that it is not obvious, or even true. Gandalf has forgotten the rotation groups of the regular prisms. In this case, there are 3 orbits. Algebraist 17:22, 11 January 2009 (UTC)

To be more precise, if the elements of G are not all rotations about the same axis then G must be a subgroup of the rotational symmetry group of one of the Platonic solids. Moreover, considering only the icosahedron and octahedron suffice, as the dodecahedron has the same symmetry as the icosahedron, the cube the same symmetry as the octahedron, and tetrahedral symmetry is a subgroup of octahedral symmetry. Eric. 68.18.17.165 (talk) 16:27, 11 January 2009 (UTC)

(Reply to Algebraist) Not entirely obvious, I suppose - there is a proof in Chapter 15 of Neumann, Stoy and Thompson's Groups and Geometry. But maybe there is a simpler (although less geometric) solution to the original problem using Burnside's lemma. Each element of G fixes just 2 points of X, apart from the identity element which fixes all |X| points, so the sum over G of the points of X fixed by each element of G is |X| + 2|G| - 2. But by Burnside's lemma this is a multiple of |G| - in fact it is |G| times the number of orbits. I'll let the OP take it from there. Gandalf61 (talk) 17:25, 11 January 2009 (UTC)

Thanks for the reference. Google books reveals that the correct result is that the finite subgroups of SO(3) are exactly those you mentioned above plus the rotation groups of prisms (including the rectangle as a degenerate prism). The subgroups of these groups Eric alludes to in fact turn out to be on the list already. Algebraist 17:38, 11 January 2009 (UTC)

I agree with all of that - so which part of my original reply do you still think is incorrect ? Gandalf61 (talk) 17:54, 11 January 2009 (UTC)

The first sentence, since it excludes the prism/dihedron case, with dihedral symmetry groups. Algebraist 17:58, 11 January 2009 (UTC)

All of the non-trivial elements of the rotational symmetry group of a prism/dihedron have the same axis. So my first sentence "If the elements of G are not all rotations about the same axis then G must be the rotational symmetry group of one of the Platonic solids" is correct. The prism/dihedron case is covered in my second paragraph which starts "On the other hand, if the elements of G are all rotations about the same axis ...". In short, 3 orbits<=>Platonic sold, 2 orbits<=>prism/dihedron. Gandalf61 (talk) 18:07, 11 January 2009 (UTC)

No, they do not. For the prism associated with the regular n-gon, there are n-1 non-trivial rotations about an axis through the centres of the ends, plus 1 non-trivial order 2 rotation about the centre of each other face and about the midpoint of each edge running between the ends. Thus there are 3 orbits. Algebraist 18:13, 11 January 2009 (UTC)

If one took the prism with vertices (±1,±2,±3), then it has a symmetry group whose intersection with SO(3) has 8 elements, right? It is generated by rotations of 180° about the x, y, and z axes, right? It is not a platonic solid, right? I don't really get the geometric ideas though. I like your counting proof, since it only uses the fact that the elements fix at most 2 points. JackSchmidt (talk) 18:18, 11 January 2009 (UTC)

No, that group has 4 elements. It's the same as the rotation group of the rectangle-in-space, which I included in my list as a degenerate regular prism (the prism on a 'regular 2-gon'). It may gratify you to know that Gandalf's counting argument is used as the basis of the classification of finite rotation groups in the reference he supplied. Algebraist 18:24, 11 January 2009 (UTC)

I agree, 4 elements: the identity and the three 180° degree rotations about the x,y,z axes. I did get that it was the same as the flat rectangle, but I just misremembered you got the dihedral group of order 8. Having no mind for geometry, this seemed reasonable to me, though I was a little concerned that the group appeared to be abelian. I took out my very expensive model of a rectangular prism and applied the rotations carefully to check. Who says a geometry textbook is useless? Glad to know the classification uses ideas I could grasp. Even with 4 elements, this is one of your (A's) counterexamples to G's first sentence, right? Not that it matters that much, since the structure of G's proof is right, and checking the details should always be a part of reading someone's answer. JackSchmidt (talk) 18:37, 11 January 2009 (UTC)

It's not my counterexample, I'm hopeless at geometry. I lifted it from our article. Algebraist 19:58, 11 January 2009 (UTC)

(Reply to Algebraist) Yup, dumb mistake on my part. But it's always good to know that you are waiting to catch any little slips like that from us amateurs. Gandalf61 (talk) 18:44, 11 January 2009 (UTC)

Unidentified trig functions/identities?

Where $\Delta \theta =\theta _{f}-\theta _{s};\quad \Sigma \theta =\theta _{f}+\theta _{s}\,\!$, are

${\frac {\cos(\Sigma \theta )+\cos(\Delta \theta )}{\sin(\Delta \theta )}}={\frac {2\cos(\theta _{s})\cos(\theta _{f})}{\sin(\Delta \theta )}}={\frac {2}{\tan(\theta _{f})-\tan(\theta _{s})}}=?;\,\!$

and

${\frac {\cos(\Sigma \theta )-\cos(\Delta \theta )}{\sin(\Delta \theta )}}={\frac {2\sin(\theta _{s})\sin(\theta _{f})}{\sin(\Delta \theta )}}={\frac {-2}{\cot(\theta _{f})-\cot(\theta _{s})}}=?;\,\!$

(or their reciprocals) recognized as special functions, particularly in regards to spherical trigonometry? ~Kaimbridge~ (talk) 01:13, 11 January 2009 (UTC)

These are elementary consequences of the formulas for sin(A+B) and cos(A+B). What is your question exactly? McKay (talk) 01:26, 11 January 2009 (UTC)

Just if there is some elementary relationship/identity, like the "sine for sides", or in the same way that

${\frac {\lambda _{f}-\lambda _{s}}{{\mbox{arccosh}}(\sec(\phi _{f}))-{\mbox{arccosh}}(\sec(\phi _{s}))}}={\frac {\Delta \lambda }{\int _{\phi _{s}}^{\phi _{f}}\sec(\phi )d\phi }}=\tan(\alpha ){}_{\color {white}.}\,\!$

for loxodromic azimuth, which involves the inverse Gudermannian function. ~Kaimbridge~ (talk) 15:25, 11 January 2009 (UTC)

Conic section slope of directrix

Given a conic section $Ax^{2}+By^{2}+Cxy+Dx+Ey+F$ what is the slope of its directrix? Borisblue (talk) 01:21, 11 January 2009 (UTC)

Sadly, we don't do homework for people. Furthermore, this problem is trivial. Please think about it a bit more and tell us what you have done on the problem so far. PST —Preceding unsigned comment added by Point-set topologist (talk • contribs) 16:10, 11 January 2009 (UTC)

Cubic functions

I'm trying to use cubic (and similar) functions for a little model, but I really don't have much maths. Could a mathematician please explain whether one changes a, b or c to make the curves flatter/steeper, change the y intercept, etc.? The article on quadratic equations does this graphically through a clever image, but a few lines of text would be great. I'd like to add this info to the relevant article as well. --Matt's talk 14:09, 11 January 2009 (UTC) Edited to clarify which article is relevant --Matt's talk 14:17, 11 January 2009 (UTC)

I presume you mean that the cubic is of the form:

$f(x)=ax^{3}+bx^{2}+cx+d\,$

In that case, "flatness" of the curve is measured by its derivative which is (at x):

$f'(x)=3ax^{2}+2bx+c\,$

So only the coefficients a, b and c have an impact on the steepness of the curve (the greater these values are, the greater the steepness; the smaller these values are, the greater the flatness). The y-intercept is given by the image of 0 under f so the value of d equals the y-intercept. If d is 0, the curve passes through the origin. PST

The article elliptic curve might also be of interest to you. PST

And by the way, mathematicians usually use one branch of mathematics in another branch of mathematics. There are numerous examples of this (I might as well let someone else list these examples; there are so many that I can't be bothered!). One interesting example is applying graph theory and the theory of covering maps to prove the well known Nielson-Schreier theorem; i.e every subgroup of a free group is free.

On the same note, there are mathematicians who would prefer not applying mathematics to another field (theoretical mathematicians) and those who would prefer applying mathematics to another field (applied mathematician). From experience, applied mathematicians are generally not so interested in the theoretical parts of mathematics and thus do not choose to learn much theoretical mathematics. But there are special cases. PST —Preceding unsigned comment added by Point-set topologist (talk • contribs) 16:05, 11 January 2009 (UTC)

Elliptic curves aren't really relevant to what the OP is doing, and what does that last paragraph have to do with anything? --Tango (talk) 17:00, 11 January 2009 (UTC)

Have a look at his links to see the relevance of that part. PST

The steepness of the graph for large (either positive or negative) values of x is determined primarily by a. For smaller values of x, the graph will change direction a lot so it's rather more complicated. You may find it helpful to write the cubic as y=a(x-u)(x-v)(x-w), then the steepness for large values of x is, again, given by a, and u, v and w are the x-intercepts. The y-intercept would be -auvw. --Tango (talk) 17:00, 11 January 2009 (UTC)

Re-write equation as:

$y=a\left(x+{\frac {b}{3a}}\right)^{3}+\left(c-{\frac {b^{2}}{3a}}\right)\left(x+{\frac {b}{3a}}\right)+e$

where e is a function of a,b,c and d that I can't be bothered to write out. Then change co-ordinates:

$v=y-e\,;\,u=\left(x+{\frac {b}{3a}}\right)$

$v=au^{3}+\left(c-{\frac {b^{2}}{3a}}\right)u$

so we have placed the cubic's centre of symmetry at the origin. Now we can see that a determines the slope of the cubic far from its centre and $\left(c-{\frac {b^{2}}{3a}}\right)$ determines the slope at its centre, and the number of turning points. Gandalf61 (talk) 17:44, 11 January 2009 (UTC)

Sounds to me like you might be interested Bézier curve and splines in general.Dmcq (talk)

Thank you to all who answered, I've noted this down and plan to add it to the article later. Not as simple as I'd hoped though :-( --Matt's talk 19:47, 18 January 2009 (UTC)

Manifolds

I have a question for those people at the reference desk. This is not homework so you don't need to worry about that. I would like to know why manifolds (and more generally differentiable manifolds) are so important objects of study. I mean, they are in a sense so restrictive in nature (why exclude so many important topological spaces from study?). To me, it seems odd that the figure 8 is excluded from study (why can't you do calculus at the center of the figure 8?). I know that there are more general types of manifolds such as orbifolds and Banach manifolds but is every topological space some type of manifold? If not, what are the minimal conditions required for a space to be some type of manifold? I am pretty sure, for instance, that one can do calculus on the indiscrete space (any function between indiscrete spaces should have derivative 0). But on the other hand, I am also sure that the indiscrete space is not a type of manifold. I certainly understand that manifolds are important but unlike continuity and integration, I don't understand why differentiation cannot be generalized to arbitrary topological spaces (integration to measure spaces). Or it maybe just my lack of knowledge of the subject that I don't know that one can do calculus on arbitrary topological spaces. Thankyou for you help! —Preceding unsigned comment added by 129.143.15.142 (talk) 20:53, 11 January 2009 (UTC)

Note that the indiscrete space is indeed a (0-dimensional) manifold if the space in question has precisely one-point or vacuously a manifold if the space in question is empty. But those are just fine details for you. PST

It's not an area I've ever studied properly, but I can't see how you could define derivatives on a space unless the space was locally metrizable. That rules out some spaces (but not things like a figure of 8). Also relevant is that, depending on your definitions, the long line is not a manifold, but can have a differential structure put on it. --Tango (talk) 22:00, 11 January 2009 (UTC)

Thanks. The definition I follow is the "locally Euclidean" one so I allow the long line to be a manifold. I think one other important thing to consider is that in manifold theory there are important facts about the derivative of a smooth function between smooth manifolds (namely it is a linear isomorphism between tangent spaces). But on one of your notes, the indiscrete space is not locally metrizable (when it has more than one point as someone above mentioned) but I would expect that the derivative of a function between such indiscrete spaes to be 0. You could argue similarly that continuity needs a metric although that is not the case. Isn't there an 'open set formulation' of differentiability? —Preceding unsigned comment added by 129.143.15.142 (talk) 22:16, 11 January 2009 (UTC)

Just to clarify something I said, I think that people don't study calculus on locally metrizable spaces because often in manifold theory, one considers vector spaces also. These are basically my questions (by the way, a real-valued function defined on the figure 8 is differentiable if and only if it extends to a differentiable function on a open set containing the figure 8 but this cannot be generalized to topological spaces that cannot be embedded in Euclidean space; this brings me to another question):

1) Since all Hausdorff, second countable manifolds can be embedded in Euclidean space, why not define differentiability on them as I have just done (i.e differentiable iff extends to a differentiable function on an open superset?)? I know this would exclude all those interesting theories of differentiable structures (like one interesting theorem that there are only 28 non-equivalent differentiable structures on S^7), but for the most part, this definition would suffice.

2) Is the figure 8 some type of manifold?

3) What are minimal conditions for a space to be some sort of manifold?

4) Is the indiscrete space (with more than one point) some sort of manifold (I would expect this naturally although I don't think that it is)?

I have not been getting sleep over this since I first learnt about manifolds (why are they so restrictive in nature?) so I would appreciate any help from the knowledgeable people at WP! —Preceding unsigned comment added by 129.143.15.142 (talk) 22:26, 11 January 2009 (UTC)

1) While such manifolds can be embedded in Euclidean space, they can be embedded in lots of different ways which wouldn't yield the same definition of differentiability (for example, a 2-sphere can be embedded in R³ as a cube, in which case the image of a great circle wouldn't be differentiable since it would have corners in it, but with the standard embedding it clearly is). While you may be able to improve things by requiring the embedding to be smooth, you would end up with a circular definition. 2) A figure of 8 isn't a manifold simply because there is no open neighbourhood of the central point that is homeomorphic to the real line. Mathematicians define things the way they do because those definitions are useful. For example, if a space is locally Euclidean at a point you can define its tangent space at that point (which is, itself, a useful thing to do for all kinds of purposes), you can't define the tangent space to a figure of 8 at that central point (it has two tangents there, so you would end up with the union of two lines, which isn't a vector space). 3) A manifold is a space that is locally Euclidean everywhere (possibly with some extra conditions, depending on who you ask), that is the minimal condition (I'm not really sure what you mean by "some sort" of manifold, there are generalisations of manifolds, but they are really manifolds any more even if the word may appear in their name, you could argue that "topological space" is a generalisation of "manifold", but that doesn't mean much). 4) No, the only neighbourhood of any point in an indiscrete space is the whole space, which can't be homeomorphic to any Euclidean space because no Euclidean space (beyond R⁰, I guess) is indiscrete. --Tango (talk) 00:04, 12 January 2009 (UTC)

Thanks for the response! I guess you didn't understand what my question was. I wanted to know whether the figure 8, despite not being a manifold, is an orbifold, a Banach manifold etc... I understand your answer to question 1 (thanks!). But as I mentioned, I still don't know question 2 but I am pretty sure that the figure 8 is not an orbifold because it is not homogeneous whereas it could be a non-pure Banach manifold (as Banach spaces are always homogeneous)? Basically, I know that the indiscrete space is a Banach manifold (it is itself a Banach space with any vector space structure).

I understand that mathematicians define things that are useful but my question is: why are people still doing calculus?! I mean isn't it too late for that? There are so many important branches of mathematics (that are general such as measure theory and topology) but differential topology just seems to specialized for me. I just somehow think that broader definitions of manifolds (Banach and orbi-) allow a wider class of spaces but it would be great to know what this class is by purely topological methods. I have another question (thanks for answering all these questions for me; I appreciate it):

5) I would expect one to be able to do calculus on any locally path connected, first countable space but why is this not the case?

Thanks again!

This is the first interesting (compared to other questions) question that I have seen on this reference desk. And yet no-one except for Tango has answered it (disappointing!). But anyhow, differenciation involves the notion of 'direction' and arbitrary topological space don't have this notion unless you have a local vector space structure defined on them (for instance locally Euclidean). However, this is just the most likely reason as to why people don't study non-locally Euclidean topological spaces.

P.S What Tango said was correct but one could speak of 'calculus on the circle with respect to particular embeddings'. So this could make matters more interesting.

PST

You know, making differential calculus on a topological space it is more or less like skating on a football pitch. Not smooth enough, no fun. You ask, why the abstract definition of differentiable manifolds with atlases if they are essentially submanifold of R^m? Sometimes Whitney embedding is of help indeed, but in other cases it is not, and would only add an unnecessary additional structure, that makes things more complicated. For the same reason (finite dimensional, real) vector spaces are defined in abstract and nobody would define them as Rⁿ. As to the Banach manifolds, they are just manifolds modelled on Banach spaces, so the difference wrto the classical ones is just the possibly infinite dimension.... what should they have to do with a figure "8"? (a figure 8, by the way, is not a submanifold of R ² but can be seen as an injective immersion of R). Anyway, in general your query seems to address to the general issue: why certain structures are still studied if they are less general of others? The obvious reason is that they have nicer properties, so you can do more with them, even if you can apply them less. What makes successful a mathematical structure is not the maximal generality, but the right generality. That's why (for instance) Banach spaces are a winning specimen and you meet them everywhere, while the more general Topological Linear Spaces are still interesting, but not at all comparable, and usually you meet them only at the Zoo. And Differentiable Manifolds also; a great category, among other things it's an important bridge between Topology and Analysis. --78.13.140.37 (talk) 15:48, 12 January 2009 (UTC)

Thankyou for the response! I now have a somewhat better understanding of why manifolds are so restrictive. But I guess I still have a few questions...

1) The figure 8 is certainly not a submanifold of the plane, I agree. But I am still confused as to why you can't do calculus on it. I mean at the center, you can approach from four directions. If the change is the same in each direction, the function is differentiable. How about that? Similarly, the comb space is not a manifold (in any sense) but you can still do calculus on it. My other question is basically why so important spaces are excluded from study.

2) I like differentiation but continuity and integration are generalized as much as possible. I guess that perhaps this is the best you can generalize differentiation naturally but I feel (as in 1)) that you can generalize a bit further. So, why shouldn't it be possible to do calculus on a locally contractible first countable space?

3) My third question is why people are still doing calculus! I mean mathematics is about theory; not calculations and yet differential geometry seems to be precisely that.

4) What are the most general manifolds? Is every topological space some kind of manifold (can one, given a locally homogeneous space, find a Banach space that looks locally like it and sort of 'patch up this local pieces' to form a Banach space and conclude that every such space is a Banach manifold?)?

Thanks (I would greatly appreciate it if someone could answer each question at a time like Tango)!

I'm not entirely sure about 1) and 2). 3) The main reason for doing pure maths is because it is fun. You can often have more fun by restricting yourself to a specific subclass of objects since more theorems hold about them. For example, if you study right angled triangles, you can find Pythagoras' Theorem, if you study general triangles you can't. Often, once you've found theorems for a specific subclass you can try and generalise them, but often the generalised form isn't as elegant (compare Pythag to the cosine rule). It also helps if it has useful applications, since people are more likely to give you funding to do it. Calculus has enormous numbers of applications throughout the physical sciences and elsewhere, point-set topology and general measure theory don't (they have applications, certainly, but not to the same extent). 4) A manifold is a topological space that is locally Euclidean, nothing else is a manifold. Some generalisations include the word "manifold" in their name, but that doesn't mean much. If there was a type of object that all topological spaces fell into that that type of object would just be called "topological space". --Tango (talk) 18:59, 12 January 2009 (UTC)

As to point 1, I would add that even though the figure 8 is not a submanifold, it is an object that can be efficiently studied by means of differential geometry, and you can apply all differential calculus you want. As you are suggesting, locally at the crossing point it's just union of two nice transverse arcs. Notice that similar sets appear quite naturally in differential geometry: e.g as a level set of a function at a critical non degenerate value. Or it's a local embedding of S¹ etc. Notice also that many more or less classical curves are figure eight (e.g., the Lemniscate of Bernoulli, some Lissajous figures, the three body problem periodic solution by Chenciner & Montgomery [1], etc). Anyway, maybe an answer to your first question is just that you do not need to give a name to an object and to include it in a formal class in order to study it, and that, in particular, a space is not excluded from the methods of differential calculus just because it is not a smooth manifold. As to point 2, notice also that a generalized derivative of functions (always on Rⁿ or on smooth manifolds) is given by the very wide concept of weak differentiation of a distribution, that allows you to differentiate not only L¹_loc functions but even objects that are not even functions, like measures. You may reflect on the way this extension is done by means of functional analysis. You just restrict your attention on a very special and nice setting, that is the space $\scriptstyle {\mathcal {D}}:=C_{c}^{\infty }(\mathbb {R} ^{n})$ of all infinitely differentiable functions with compact support on Rⁿ; here you have the very nice and classical notion of partial derivatives $\scriptstyle \partial _{i}$. For the very reason that $\scriptstyle {\mathcal {D}}$ is so cute and "small", it's dual $\scriptstyle {\mathcal {D}}^{\prime }$ is a very very wide pot, that contains in a natural ways not only $\scriptstyle {\mathcal {D}}$ itself, but also functions of Sobolev classes, locally integrable functions, Radon measures, and all sort of weird and wild objects. Then, each partial derivative operator $\scriptstyle \partial _{i}$ on $\scriptstyle {\mathcal {D}}$ naturally gives rise to a transpose operator on $\scriptstyle {\mathcal {D}}^{\prime }$, in much the same way transposes of matrices are defined for the finite dimensional R^m. These transpose operators (with a change of sign) turn out to be an extension of the partial derivatives to $\scriptstyle {\mathcal {D}}^{\prime }$. What I mean is that there is a kind of moral in this (you are free to choose what moral). Now, this is not exactly the extension of calculus you are asking in 2, for you are thinking also to a more general kind of domain, rather than more general object to differentiate. Still I can imagine various situation and ways in which smooth differential calculus may be applied. This depend on what is your precise problem indeed... I do not agreee completely with your point 3, for theory and calculus are not two opposite issues after all. Abstract theories indeed reduce difficult problems to situations where you can do elementary computations (you can find yourself a lot of examples). I do not completely understand your 4... well, the only relevance of Banach spaces in this context that I can think is that every metric space can be isometrically embedded in a Banach space (this is usually referred to as the Frechét-Kuratowski embedding; a both easy and useful thing: once you put your metric space into a Banach, you can consider the operations or concepts there, like closure , that gives you a completion, convex hull, epsilon-neighbourhoods, etc). As to the question: what's the more general concept of manifolds, it seems a bit too generic. Not Banach manifolds however; you may introduce classes of topological spaces with atlases in order to describe global objects made patching together (in a given way) local objects (picked in a given category). In conclusion, I agree with everything Tango said, except his statement that people do pure math mainly for fun... although I do not have counter-examples to it. ;) --PMajer (talk) 11:17, 13 January 2009 (UTC)

PS: Just to speak, in fact if you put a metric space M into a Banach space X via the FK embedding, then you may define "differentiable functions" on M those functions that are restriction to M of functions on some open nbd of M in the Banach space. The Whitney extension theorem, that can be stated also in the context of Banach spaces, will then give you a characterization of these functions (but do not ask me what to do with this definition) --PMajer (talk) 15:39, 13 January 2009 (UTC)

Thanks! I appreciate your help. I know that calculus has lots of applications everywhere but maths is for the sake of it. First priority should be maths itself. Although non-Hausdorff spaces, for instance, pretty much have no applications, I consider them interesting in their own right: or even orthocompact spaces and such. Anyway, thanks again!