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2004

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Don't you miss (the capacity of) orientation - a basic condition of a sane person?

Apogr 10:53, 2 Sep 2004 (UTC)

No. Elroch 18:12, 15 February 2006 (UTC)[reply]

Explanation of correction in triangulation section

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The previous version of this page stated that, to orient a triangulated surface, it is equivalent to orient each triangle so that identified edges point in the same direction. In fact, they should point in the opposite direction. The picture to have in your head is that all the triangles are oriented clockwise (for example). If you have two clockwise oriented triangles in the plane, and you glue them along an edge, the edge will be oriented in opposite directions in first triangle and the second.

18.87.1.112 (talk) 16:04, 6 May 2010 (UTC)[reply]

Orientable characters

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The article earlier used an intuitive description of orientablility of surfaces based on moving a letter around the surface. To be rigourous, the letter needs to be considered as a graph made up of line segments. In most fonts, no letters (even ones like Courier R) are strictly suitable for this purpose, as they may be deformed as graphs in the plane into their mirror image (Check this!). I have now updated the article to refer to moving a small image (which has a clear handedness) around a surface. I hope this makes it clear to the reader, as well as more precise. Elroch 18:12, 15 February 2006 (UTC)[reply]

Wow, nice job! This was one of the first articles I edited anonymously years ago, and it's come a long way since. The article looks great! --C S (Talk) 04:23, 12 April 2006 (UTC)[reply]

I think using images in the text is a bad idea. I would like to change it back to text. Can you explain to me why you think that the letter R is not handed? It seems to me that you can profitably view the upper left corner of the "R" as a vertex, which I think will keep you from being able to deform it into its mirror. If that doesn't satisfy, why not consider rigid tranformations only? I really don't like at all filling up text with little pie charts. They don't scale with the font sizes, they can't be seen with text browsers, and they just look unprofessional. -lethe talk + 05:58, 13 April 2006 (UTC)[reply]

"R" (in the font I see on the uploaded version of this page) has three vertices (including the one you want) right? So it's actually topologically the same as "A". The top half is a triangle with two legs on the bottom half. Anyway, as for the rigid deformations, I believe that's probably what was intended with the very initial version of the page, and it's fine for intuition's sake, mostly. I personally didn't think it was that big a deal, but now that's it's been mentioned, since in most of that content, everything's being done topologically, I think it's a little confusing to talk about rigid moves. Especially since it's only "rigid" in a small neighborhood of the "R" anyway, and there's also the problem that this assumes the surface is geometric in some fashion.
As for the problem with accessibility, can't that be fixed somehow without removing the image? I don't know what "unprofessional" means here. I might use an illustration in the text like that in a paper, and certainly I've seen plenty of papers that use something like that. It's a matter of style. I don't think we're being less professional than the norm. --C S (Talk) 13:06, 13 April 2006 (UTC)[reply]
I'm inclined to agree that the pie doesn't look very "nice"; however, I agree that using a simple graphic is more intuitive than using a letter from the alphabet (and in particular, stands out nicely in the text). My proposal would be to use a simple black and white image instead of the pie (Maybe an umbrella? I don't know) so that it looks more professional. It would be a large contribution if somebody would take that image and produce a small animation showing how handedness can be changed on some non-orientable surface. Meekohi (talk) 17:16, 10 June 2008 (UTC)[reply]

Where the text says: "Equivalently, a surface is orientable if a two-dimensional figure such as Small pie.svg in the space cannot be moved (continuously) around the space and back to where it started so that it looks like its own mirror image Pie 2.svg." While these images are technically mirror images, the mirror axis is from upper-right to lower-left, rather than being the simple vertical or horizontal mirror most readers will expect in an introductory paragraph. I'm hoping this can be improved. 24.57.239.43 (talk) 12:27, 30 November 2010 (UTC)[reply]

I think that the svg looks perfect (that is also an usual example btw) but i prefer subtly looking and subtly formatted texts. How about "d" can be rotated to/from "p" but never match "b" and "q". --14.198.221.19 (talk) 15:43, 30 August 2013 (UTC)[reply]

I agree that it would be better to use typeface (or possibly typeface in addition to the svg). I think we should be careful to choose a font so that in fact d, p, b, q are all isomorphic under the dihedral group. This seems to be the case for the sans font that I personally see as default, but may differ depending on browser and user preferences. I'm pretty sure these are no longer isomorphic in serif fonts (and in italic sans fonts). Sławomir Biały (talk) 20:35, 30 August 2013 (UTC)[reply]

Merge?

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There has been a merge tag on Orientable manifold for a long time. I don't know the subject well enough to comment, but could someone who does start a discussion to resolve the proposition? Kcordina 12:29, 17 March 2006 (UTC)[reply]

Comment for readers: This has already been done; orientable manifold redirects here. --C S (Talk) 04:25, 12 April 2006 (UTC)[reply]

universe is closed

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As far as I know, the difference from the critical density of the universe has not yet been measured, thus whether the universe is closed or not is completely up in the air. Anyway, it's not really relevant, so I'm removing it. -lethe talk + 06:03, 13 April 2006 (UTC)[reply]

It appears to me also that this is a controversial claim to make, so it's good to remove it. --C S (Talk) 13:09, 13 April 2006 (UTC)[reply]

tangent bundle always oriented?

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I was also looking suspiciously at the recent anonymous edit that Oleg reverted, which changed "manifold orientable iff its tangent bundle is" to "tangent bundle always orientable". Oleg says that Mathworld's article agrees with the original. I myself am a little bit confused. I know for sure that the cotangent bundle of any manifold is always orientable, indeed even carries a choice of orientation. I think it's believable that the tangent bundle could also be orientable, Does not the orientation on the cotangent bundle also induce an orientation on the tangent bundle? I'm gonna poke in some books (mathworld isn't definitive). -lethe talk + 04:17, 22 May 2006 (UTC)[reply]

Yes, the cotangent bundle is always orientable, per planetmath, see at the bottom. I don't know details though. :) Oleg Alexandrov (talk) 04:22, 22 May 2006 (UTC)[reply]
The proof that the cotangent bundle is easy: there exists a canonical symplectic form on any cotangent bundle. Any symplectic manifold is oriented. How's this for a proof for tangent bundles: every second-countable Hausdorff manifold admits a Riemannian metric. Every Riemannian metric induces an isomorphism between the cotangent bundle and the tangent bundle. The cotangent bundle is always oriented, so the tangent bundle is too. Mathworld is wrong (haven't found it in my books yet though). -lethe talk + 04:26, 22 May 2006 (UTC)[reply]
My theory is that since the orientation of a manifold is defined as an orientation of tangent vector spaces, someone over at mathworld got confused and thought that the orientation of a manifold and its tangent bundle (which is after all a disjoint union of tangent vector spaces) were the same thing. This isn't right: the orientation of a manifold is defined in terms of its tangent bundle. The orientation of the tangent bundle is defined in terms of the tangent bundle of the tangent bundle, which is a different object. Someone at mathworld made that mistake and it propagated to us, that's my theory. Anyway, I like my easy proof above well enough to be convinced of its rightness. Comments welcome. -lethe talk + 04:35, 22 May 2006 (UTC)[reply]

A new anon changed it back. Without a real reference in hand, I'm unsure what to do, though I'm pretty confident in my argument above. -lethe talk + 19:46, 23 May 2006 (UTC)[reply]

If I may inject: I think there is some confusion here between orientation of a vector bundle and orientation of a manifold. A vector bundle is orientable if it admits a smooth choice of orientation (or if its structure group can be reduced to GL+(k)). A manifold is orientable iff its tangent bundle is orientable as a vector bundle. As Lethe points out, the tangent bundle is always orientable as a manifold, meaning the tangent bundle of the tangent bundle is orientable (as a vector bundle). Am I making any sense? The article should clarify this. -- Fropuff 21:04, 23 May 2006 (UTC)[reply]

Ah yes. That makes perfect sense, I probably should have known that right away. OK, yes, that needs to be made clear in the article. So anyway, the statement "a manifold is true iff its tangent bundle is orientable" is trivially true because it's the definition of orientability for manifolds. -lethe talk + 00:11, 24 May 2006 (UTC)[reply]
I've reworked that short section. Comments welcome. -lethe talk + 00:28, 24 May 2006 (UTC)[reply]

oriented vs orientable

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I don't understand the difference : "An orientable surface is an abstract surface that admits an orientation, while an oriented surface is a surface that is abstractly orientable, and has the additional datum of a choice of one of the 2 possible orientations." —Preceding unsigned comment added by 82.252.12.212 (talk) 11:42, 11 July 2008 (UTC)[reply]

Consider a sphere in R3. An orientation on the sphere is given by a unit vector normal to the surface. The sphere is orientable since it is possible to do this. However, it is not oriented until you say which unit vector you are going to use: the outward normal or the inward normal. siℓℓy rabbit (talk) 13:09, 11 July 2008 (UTC)[reply]

Merging of article on orientation algorithm

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I noticed that someone has effectively deleted my article (Orientation (topology)) on how to determine the orientation of a simple closed polygon represented as an ordered set of points, and pointed the article to redirect to Orientability instead, with the justification of "no original research". This article is definitely NOT original research, it simply needs to be properly cited. The algorithms described appear in a number of computer graphics textbooks. As an engineer, I resent that the article I worked so hard on was hijacked by a bunch of theoreticians. This algorithm deserves to appear somewhere on Wikipedia, even if not in this particular article. James Monroe (talk) 00:10, 29 August 2008 (UTC)James Monroe[reply]

Hum, there does seem to be a need for material on algorithms for calculating Curve orientation. I would say that article is the most appropriate place for that material. Orientation (topology) is not the appropriate place for that material and the redirect here is appropriate.
Now the question is whether the algorithm is the standard way of doing things. There is a problem with it, consider a figure eight shape with four points.
    |\    /|
    | \ /  |
    | / \  |
    |/    \|
Each point is on the convex hull so as the algorithm states they would be valid points for calculating the cross-products, however we have different signs, giving different orientations. Of course this curve is non orientable.
Ideally it would be best to find a published algorithm for the algorithm. --Salix alba (talk) 08:18, 29 August 2008 (UTC)[reply]
Alright, I've moved the algorithm to the Curve orientation page, and updated the title to "Determining the orientation of a simple closed planar polygon. Your figure-8, or similar non-orientable polygons, are not simple and planar. I can try to provide more sources for the algorithm once I find my textbooks :/ James Monroe (talk) 20:52, 29 August 2008 (UTC)James Monroe[reply]

Another approach for co-dimension 1 surfaces

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We can use normal vectors to define orientablity in an intuitive way. It's better to mention at this article.--刻意 09:57, 25 December 2008 (UTC)[reply]

very confusing.

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I found this article's beginning to be quite confusing. It does not make sense to those not experienced in the field of mathematics. Would it be possible to create a "common language" explanation as well?

What do I mean by that? I am referring to some of the sections in other mathematical articles, such as the division by zero article. They have a complex explanation of the problem as well as a basic one comprehendable to those who have at least some basic understanding of math. Could this article have a similar section too? Fusion7 (talk) 19:20, 23 August 2009 (UTC)[reply]

Unclear statement

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In the paragraph "topological definitions" it is stated: an orientable manifold is one whose structure group [...] can be reduced to the subgroup GL+(n) [...]. The only way I can understand this is by thinking of the tangent bundle, but in a paragraph about topological, not differentiable, manifolds, this would not make sense. Should the sentence be removed, perhaps? --83.228.202.2 (talk) 13:13, 11 March 2012 (UTC)[reply]

What's Euclid Got To Do With It

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In mathematics, orientability is a property of surfaces in Euclidean space ...

I've just read The Shape of Space by Jeffrey Weeks, in which orientability is presented as a purely topological property, independent of geometry (Euclidean or otherwise); and indeed our article says later on, An abstract surface (i.e., a two-dimensional manifold) is orientable if a consistent concept of clockwise rotation can be defined on the surface in a continuous manner. So why make Euclidean space part of the definition?

... More generally, orientability of an abstract surface, or manifold, measures whether one can consistently choose a "clockwise" orientation for all loops in the manifold.

Including loops that cannot be shrunk to a point? What's clockwise for a non-shrinkable loop on a torus? —Tamfang (talk) 05:32, 30 October 2014 (UTC)[reply]

I think the article starts off in Euclidean space because it's easier for most people to visualize, though as you say there's no mathematical reason for doing so.
Focusing on particular loops doesn't really capture the essence of orientability; orientability requires that you simultaneously define "clockwise" for all loops on a surface at once. Imagine that we have a donut covered in sticky, liquid sugar glaze. Imagine that the glaze flows down through the hole in the donut. It sticks to the donut, but instead of settling at the bottom, we blow on it in an outwards direction (away from the hole) to make it flow up around the outside of the donut. Finally it loops back into the hole again. This defines a vector field on the torus. There's another vector field defined by rotation: Looking down at the torus from the top, it's mostly circular, so we imagine a vector field which rotates this circle clockwise. Together, these two vector fields define an orientation of the torus: If we write down a closed loop on the torus (contractible or not), then we can imagine traveling along the loop in the direction given by these two vector fields. There's never any ambiguity. Using the metric inherited from the ambient Euclidean space, one can turn the vector fields into differential forms, and then their wedge product defines a non-zero top degree differential form, which is one of the equivalent conditions listed under "Orientation of differential manifolds" in the article. Ozob (talk) 13:54, 30 October 2014 (UTC)[reply]

Is it correct to talk about tangent spaces in the section "Topological definitions"?

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Does every manifold have tangent spaces at it every point? Not only the differentiable ones? There is a separate section for differentiable manifolds. Wouldn't be better to move the things belong to only differentiable manifolds there? 89.135.8.194 (talk) 10:55, 14 February 2016 (UTC)[reply]

No, not really. The tangent bundle is really a construction for differentiable manifolds, and it just doesn't work in the topological or PL cases. Milnor's microbundles, however, are usually an adequate substitute. While I'm not very familiar with microbundles I would presume that they can be used to define local orientations as well.
I agree that this paragraph of the article is questionable as it stands. I'll have to think about what I'd like done to it. Ozob (talk) 18:41, 14 February 2016 (UTC)[reply]
After some thought, I think the "orientability of manifolds" section has problems running all through it. Firstly, the section begins by saying that non-orientability of an n-manifold is equivalent to being able to reflect an n-ball by moving it around the manifold. I think that "moving" should mean isotopy, but without a reference I'm not sure. I'm convinced that this definition works in the PL case (though I haven't tried to work on the details), but on a general manifold I can't tell. I'd like a reference, and unfortunately the article doesn't have one. Second, the section doesn't distinguish well between the different types of manifolds (topological, PL, and differentiable) and the definitions that work for each of them. For instance, "reducing the structure group to GL+(n)" requires that the manifold was differentiable to begin with, an unstated hypothesis. And yet the article isn't entirely wrong; one has, for instance, that every topological n-manifold has a tangent microbundle, and by a theorem of Kister and Mazur, such a microbundle is equivalent to a fiber bundle with fiber Rn and structure group the group of homeomorphisms of Rn that fix the origin. While I haven't checked the details (again), I'm confident that the manifold is orientable if and only if this structure group reduces to the group of orientation preserving homeomorphisms (where "orientation preserving" is defined in terms of homology). (I would also guess that the tangent microbundle has a "structure group" which is the local topological group of germs of origin-fixing homeomorphisms of Rn and that orientability is equivalent to reducing this structure group to germs of origin-fixing orientation-preserving homeomorphisms of Rn.)
I think the right definition of orientation, philosophically, is that the top homology group is isomorphic to Z. This works for all (connected) manifolds (and ought to work even more generally). The article shouldn't indulge my speculations above about reducing structure groups unless references can be found, and it shouldn't conflate the topological, PL, and differentiable cases. I'll see if I can rewrite this section appropriately. Ozob (talk) 01:51, 15 February 2016 (UTC)[reply]
I've rewritten this section entirely. I believe the new version should address the concerns raised above. However, there's always the possibility that I've made mistakes, so additional eyes are welcome. Ozob (talk) 02:37, 18 February 2016 (UTC)[reply]
Good job! Thanks! Could you provide also some references for this? 89.135.8.194 (talk) 04:53, 19 February 2016 (UTC)[reply]
I added a reference. There are plenty; any basic book on manifolds or algebraic topology should discuss this subject. Ozob (talk) 15:34, 20 February 2016 (UTC)[reply]
Yet one question. Your write "M is orientable if and only if the nth homology group is isomorphic to the integers Z.". I think that it's valid only for closed and connected manifolds. Am I right? 89.135.8.194 (talk) 07:26, 22 February 2016 (UTC)[reply]
Good catch. I had meant for the connectedness of M to be a standing hypothesis throughout the section, but I had forgotten the compactness requirement. I believe compactness can be dropped by substituting Borel–Moore homology, but I don't remember for sure right now. Ozob (talk) 13:42, 22 February 2016 (UTC)[reply]
I continue to believe that I'm correct about Borel–Moore homology, but I haven't managed to find a reference, and I'm finding the details elusive. It is a theorem that in some circumstances, if M can be embedded in a compact space N, then the Borel–Moore homology is isomorphic to the relative homology . If we can choose N to be a manifold with boundary whose interior is M, then an orientation of N is the same as an orientation of M; orientability of M therefore implies orientability of N and hence . Since M is assumed not compact, the long exact sequence for relative homology implies that contains a copy of Z; and I'd guess that one can show that's the whole homology group. But this argument relies on the existence of N. If we really need the conditions listed at the Borel–Moore homology page, then such an N doesn't always exist: M might be R2 with infinitely many punctures, and then its complement is not a finite CW-complex. Presumably one can work around this with better technique, but I haven't managed so far. Ozob (talk) 04:16, 23 February 2016 (UTC)[reply]
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Non orientablitlity

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Why are smooth surfaces which contain an open set diffeomorphic to mobius band , non-orientable ? Abhinna Sundar (talk) 14:33, 10 November 2018 (UTC)[reply]