Talk:Hyperbolic geometry/Archive 2

Lorentzian ≠ hyperbolic
This edit introduces the sentence "The geometry turns out to be hyperbolic because of Lorentz invariance." What is the intent? As it stands, it is false: the geometry of the four-dimensional Minkowski spacetime is not hyperbolic. Even a space-like flat section of it is a Euclidean 3-space. The hyperboloid model shows that a complete hyperbolic 3-space can be embedded isometrically in a Minkowski space, but that is clearly not the intent. The space of subluminal velocities (or equivalently, the space of time-like points at infinity) is a hyperbolic space, but this is as abstruse as saying that the space of directions in a Euclidean space is an elliptic geometry. So what is it that the source is saying? —Quondum 15:06, 28 June 2015 (UTC)


 * Undone now M&and;Ŝc2ħεИτlk 16:46, 28 June 2015 (UTC)


 * From the edit summary: the sources talk about the line element and its hyperbolic nature. I do not have the source, but I guess this might be referring to the geometry of the space of lines.  It would correspond to my point about the space of points at infinity.  Since the article is about hyperbolic geometry, it may make sense to include a corrected form of this statement, along the lines of the space of velocities that I mentioned above – if that is what the source was saying.  —Quondum 17:43, 28 June 2015 (UTC)


 * Okay, I've found a mention in the reference. In referring to a "line element", it is speaking of the (infinitesimal) difference of velocities in the space of velocities.  I've re-added the statement as a stand-alone paragraph (it is quite distinct from the other geometries being referred to), reworded so that hopefully it will make sense to the lay reader., any comment would be welcome so that I can massage this statement for clarity and ease of understanding.  —Quondum 14:36, 4 July 2015 (UTC)


 * Sorry for the very late reply, but my edit summary already summarized the response at least vaguely. The sources just talk about the invariance of the line element, that Lorentz boosts are hyperbolic rotations in spacetime, and mention that SR involves a non-Euclidean spacetime, without mentioning the geometry is hyperbolic. I have nothing more to say or intentions to clear up, sorry. M&and;Ŝc2ħεИτlk 17:45, 4 July 2015 (UTC)


 * L&L is a bit more specific than that, now that I've managed to find the reference. To quote: "The required line element dl, is the relative velocity of two points with velocities v and v + dv" and "From the geometrical point of view this is the line element in three-dimensional Lobachevskii space—the space of constant negative curvature".  There is also a formula for calculating dl2 in terms of v and dv only.  To me, my edited re-insertion captures this clearly enough – they are saying that the distance measure makes the space of velocities into a three-dimensional Lobachevskian (i.e. hyperbolic) space, without any ambiguity.  This is not to be confused with four-dimensional spacetime.  It also fits exactly with the hyperboloid model, so it is a correct interpretation.  The group of Lorentz transformations (including rotations) does act on this velocity space, just not in the same way as we are familiar with for normal spacetime. I was not asking for further help, but was merely asking whether someone feels that it needs further clarification, which I feel I could do if so.  If not, it can stay as it is now.  —Quondum 21:09, 4 July 2015 (UTC)


 * From your honored lay reader :) WillemienH (talk) 21:26, 4 July 2015 (UTC)
 * I do think that by now I knew something about hyperbolic geometry, but maybe you are talking about general relativity or space time, in which case you are correct, I try to stay away of it (I interest is in Geometry, not in Physics :), I think it all boils down to a confusion between Minkowski geometry and hyperbolic geometry, Minkowski geometry is an affine geometry, which hyperbolic geometry is not, And I am wondering is space time isotropic? (I guess not)
 * I stay away from editing the sections on "Geometry of the universe (including relativity)" and "The hyperboloid model" just because I am not familiar with them.
 * (I only started the section "Geometry of the universe (including relativity)" to save the "Geometry of the universe (pre dating relativity)" from edit wars, confusing the two.
 * I do think the hyperboloid model is quite useless for hyperbolic plane geometry as it is not embedded in euclidean space. (it is embedded in Minkowski space, but what does that mean, is Euclidean space not an Minskowski space ? , see here my knowledge is really in deficient) Lorentsian or not? I don't know, i don't even know what it means , so back to you , the not-so-lay person. :) WillemienH (talk) 21:26, 4 July 2015 (UTC)


 * I'm only a lay person myself, just with an interest in physics and mathematical structure, though I only discovered abstract geometry very recently, in particular Klein's approach. Physics gives only one application for geometry; one can tackle this entirely from a geometric perspective.
 * A four-dimensional Euclidean space can be projectively completed; the additional points (the points at infinity) form a three-dimensional elliptic space; I mention this as a framework that I assume you are familiar with. In a geometric sense, four-dimensional Minkowski space is not isotropic (in the linked sense; there is another sense of the word which means the opposite), since there is a light cone which separates timelike directions from spacelike directions.  Similarly to the Euclidean case, the points at infinity form a geometry of is own, in this case two geometries: the one (timelike) set of points at infinity form a three-dimensional hyperbolic space, and the other (spacelike) set forms a three-dimensional de Sitter space.  The crucial point as that the points at infinity are a different set (or space) from the points of the affine geometry in both the Euclidean and Minkowski cases.  A Euclidean space is not a Minkowski space, even though both are affine spaces of the same dimension.  A spacelike section of Minkowsi space is a three-dimensional Euclidean space, so in that sense Minkowski space contains Euclidean subspaces.  It would be nice if somehow we can turn this discussion into text in the article to make it simple and understandable.  Perhaps we should refer to the points at infinity rather than the space of velocities (these being equivalent)?  —Quondum 04:51, 5 July 2015 (UTC)

A so we are both non cognanti in this. I am wondering about the whole relation between Minkowski geometry and Minkowski space-time and I don't think any of them gets any good of mixing the two up.

As you explain it now doesn't make any sense either (sorry I am being very critical here) If the hyperboloid model is embedded in an geometry that is not Euclidean, then the relations between the models loses its meaning. (because the other models are embedded in Euclidean geometry) Maybe I just don't understand minkowski geometry I am now reading a bit on it. but it looks like it has nothing to do with what is written at Minkowski space (if only lets start with it as only a geometry start with the 2 dimensional case, its relations with affine geometry and euclidean geometry. forgetting all about special relativity. If you like I can start with a stub idea. i found some (not a lot) links for a start of an articcle WillemienH (talk) 15:41, 5 July 2015 (UTC)


 * Minkowski spacetime and Minkowski geometry are the same thing. The hyperboloid model is isometrically embedded in Minkowski space.  The Poincaré and Klein models are non-isometrically embedded in Euclidean space, which is to say the "euclidean" part of it is irrelevant.  So in a sense, the hyperboloid model is more faithful than the Poincaré and Klein models are.  My description should not be at odds with Minkowski space.  —Quondum 17:50, 5 July 2015 (UTC)

I disagree Minkowski spacetime and Minkowski geometry are definitly not the same thing. (if anything Minkowski spacetime  has 4 dimensions while Minkowski geometry also exists in 2 dimensions, found a nice link :  see nothing to do with spacetime at all. I am intrigued what do you mean by "The hyperboloid model is isometrically embedded in Minkowski space" ? (Do you mean the length of the curve is a measure of the distance between the points? or something else) WillemienH (talk) 22:12, 5 July 2015 (UTC)


 * Reviewing Cayley-Klein metric may assist in resolving geometric differences.Rgdboer (talk) 22:34, 5 July 2015 (UTC)


 * Rgdboer, sorry, that does not help me much, even though the content relates.
 * WillemienH, are you saying that you want to reserve "Minkowski space" to mean specifically the four-dimensional case, and "Minkowski geometry" to mean any number of dimensions)? (Or only two dimensions?)  This is a rather trivial point of terminology and is not of concern to me.  Yes, I mean the length of any curve on the hyperboloid is the same as measured in the hyperbolic space and as measured in the embedding space, using the Minkowski metric.  The link you gave is quite cute, but I haven't read through all of it.  But the gryphon is clearly talking about the same thing as I am.  —Quondum 23:37, 5 July 2015 (UTC)

Hi Quondum, I moved (or restarted ) the discussion about Minkowski geometry at Talk:Minkowski space I think that is the better place to discuss it. (hope you can agree that there is the better place) I am not sure about your "Yes, I mean the length of any curve on the hyperboloid is the same as measured in the hyperbolic space and as measured in the embedding space, using the Minkowski metric." That's all to advanced for me. I noticed that something similar is written at the hyperboloid model but that did not mean anything to me either. (while I think I do understand distances in the other models, a bit) WillemienH (talk) 09:43, 7 July 2015 (UTC)


 * Actually, I need to check that statement on isometric embedding, though I have no reason to doubt it. I'll have to spend a little time doing the math from first principles; not complicated, but I'm rusty.  It would help if my statement is checked and made explicit at Hyperboloid model.  I'll leave some time for replies at the Minkowski space talk page; yes, that is a better space for such a discussion, though obviously you already know my feeling on the subject . —Quondum 13:01, 7 July 2015 (UTC)

Justifying the classification of isometries
In a hidden comment in the text of section Hyperbolic geometry, questions whether the classification includes all isometries which can be achieved with three reflections. I will explain the derivation of our classification.

First, I take it as obvious that the isometries of S1 (a one-dimension "sphere", i.e. a circle) are: A third reflection just cancels one of the reflections already being used, changing the rotation back to a reflection. Therefor, the isometries of the hyperbolic plane (or of the Euclidean plane or of the 2-sphere) which hold a given point fixed will have the same classification.
 * identity (zero reflections, zero degrees of freedom),
 * reflection (one reflection, one degree of freedom), and
 * rotation (any two reflections separated by half the angle of the turn, one degree of freedom).

Now consider an arbitrary isometry U of the hyperbolic plane. Choose any point p. If U(p)=p, then we are done because the above classification is a sub-classification of the one given in this article. If U(p)≠p, then draw a line segment from p to U(p). Find its midpoint and construct the perpendicular bisector. Let R be the reflection through that perpendicular bisector. R(U(p))=p by the construction of R. Since R is a reflection, the composition of R with itself R∘R=I is the identity. Thus U=U∘R∘R. Let us now consider U∘R, see that U∘R(U(p))=U(p). That is, U∘R holds U(p) fixed. So we can apply the above classification to U∘R.

If U∘R is the identity, then U=U∘R∘R=I∘R=R which is just a reflection through a line.

If U∘R is a reflection Q, then U=Q∘R. Consider the two lines through which the reflections reflect. They cannot be the same because the line for R does not pass through U(p), but the line for Q does pass through U(p).

If the lines for Q and R intersect at the midpoint (between p and U(p)), then they are perpendicular and U is the inversion through that midpoint.

If the lines intersect elsewhere, then U is a rotation around that intersection.

If the lines do not intersect but are asymptotically parallel, then U is a "rotation" around the ideal point to which the lines tend.

If the lines are ultraparallel, then construct a third line perpendicular to both of them. Then U will be a translation along that third line.

If U∘R is a rotation around U(p), then we can choose two lines through U(p) separated by half the turn angle. Let the first of them be the line (of reflection S) which passes through p (and the midpoint). Then U∘R=T∘S, so U=T∘S∘R. Then S∘R will be the inversion through the midpoint. Drop a perpendicular from the midpoint onto the line of T. Let the lines of V and W be perpendicular intersecting at the midpoint (so that S∘R=W∘V) with the line of V being the perpendicular dropped onto the line of T. Then U=T∘W∘V and T∘W  is a translation along the line of V since the lines of T and W are both perpendicular to the line of V.

Is that clear? JRSpriggs (talk) 05:09, 18 July 2016 (UTC)


 * Sorry your answer is not very clear but also that was not my question (sorry I was not very clear either).


 * My question is more specific "Is every combination of 3 reflections a single reflection or a glide reflection, a glide reflection described as a combined reflection through a line and translation along the same line?".
 * Especially how can you prove that 3 reflections that don't combine to a single reflection (because the reflection lines are part of the same pencil) always combine to a glide reflection, a combined reflection through a line and translation along the same line? WillemienH (talk) 22:08, 18 July 2016 (UTC)


 * Yes, every composition of three reflections in the hyperbolic plane is equivalent to either a single reflection or a glide reflection.
 * I already proved it above, especially in the last paragraph. Just take U to be your composite of three reflections.
 * I thought about trying to give another kind of argument for you, but there are just too many cases to consider. You really need to use p and R to make the simpler argument work. JRSpriggs (talk) 05:21, 19 July 2016 (UTC)


 * Maybe it would help if I give an overview or executive summary of the relevant part of my proof. To your three reflections, I add a fourth carefully selected reflection. This reduces them to just a rotation, i.e. two reflections. To cancel out the change that I made when I added the fourth, I add it again (since it is its own inverse). Now we have a reflection and a rotation. I break the rotation into a reflection perpendicular to the reflection I added and another reflection. The two perpendicular reflections form an inversion. So we now have that your three reflections are equivalent to an inversion and a reflection. The inversion can be broken into a reflection perpendicular to the reflection and another reflection. So we now have two reflections, both perpendicular to a third reflection. The two form a translation along the line of the third, i.e. it adds up to a glide reflection. OK? JRSpriggs (talk) 08:46, 19 July 2016 (UTC)

Lemma: If A and B are reflections and p is any point, then there are reflections C, D and E such that p lies on the line of D, and A∘B=C∘D=D∘E.

Any two reflections form a rotation, horolation, or translation. If a rotation, let the line of D be the line connecting p to the center of the rotation. If a horolation, let the line of D connect p to the ideal point shared by A and B. If a translation, then let the line of D be the line one gets by dropping a perpendicular from p to the line along which A and B are translating. JRSpriggs (talk) 17:23, 20 July 2016 (UTC)


 * Suppose we have three reflections which I will call 1, 2, 3 (applied in that order). Pick any point q on the line of 3, but not on the line of 2. Using the lemma, shift 1 and 2 to 1' and 2' so that 2' goes thru q. Then 2' and 3 form a rotation about q (or the identity which reduces this to just a single reflection 1'). Drop a perpendicular from q onto the line of 1'. Call the foot of that perpendicular m. Now shift 2' and 3 to 2' ' and 3' so that 2' ' goes thru m. Then 1' and 2' ' meet perpendicularly at m, so they form an inversion thru m. Drop a perpendicular from m onto the line of 3', and call its foot n. Shift 1' and 2' ' to 1' ' and 2' ' ' so that 1' ' goes thru n. Then 1' ' and 2' ' ' still meet perpendicularly at m, but now 1' ' and 3' also meet perpendicularly at n. Thus 3∘2∘1=3∘2'∘1'=3'∘2' '∘1'=3'∘2' ' '∘1' ' is the composition of a reflection 1' ' and a translation 3'∘2' ' ' along the line of 1' ', i.e. they form a glide reflection. Do you understand now? JRSpriggs (talk) 13:33, 28 July 2016 (UTC)
 * Since the objective is to prove that the three reflections are equivalent to either one reflection or a glide reflection, I will assume that they are not equivalent to a single reflection. Thus 1≠2, because otherwise 3∘2∘1=3. And 2≠3, because otherwise 3∘2∘1=1. There is a point q on 3 which is not on 2, because otherwise 3=2. q is not the common element of 1 and 2, because otherwise it would be on 2. 2' is not 3, because otherwise 3∘2∘1=3∘2'∘1'=1'. m is not q, because otherwise 1' and 2' would share both the common element of 1 and 2 and the distinct point m=q so that 1'=2' and thus 1=2. n is not m, because otherwise 2' '=3' so 2'=3. I hope that fills in the gaps in the proof by eliminating any special cases which might cause problems. JRSpriggs (talk) 05:09, 29 July 2016 (UTC)

I copied the section Hyperbolic geometry to Hyperbolic motion that page needed a more general (model independent) section and this section was just that. I guess in the long run the two copies will differ but thart is no worry to me. WillemienH (talk) 23:24, 8 September 2016 (UTC)

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