I'm trying to piece the equator of the hemisphere representation together but so far unsuccessful.
Keiji wrote:http://teamikaria.com/hddb/dl/FHEVA987RR8ZM8SBEXEYC18QJY.png[/url]
Secret wrote:does it required to be non orientable no matter which way you walk?
Keiji wrote: As you can see, it appears to be "locally orientable" - as do all surfaces.
Keiji wrote:You'd have to collapse that into a line, giving you something that looked like a mutilated torus, and then into a point, to get an accurate immersion, which would probably give you one of the immersions we already know of.
Secret wrote:Keiji wrote:You'd have to collapse that into a line, giving you something that looked like a mutilated torus, and then into a point, to get an accurate immersion, which would probably give you one of the immersions we already know of.
here's the resulted object after doing the above instructions
https://docs.google.com/leaf?id=0B11-z8yC5qelNGMyOWFlMjAtN2RhNy00YjkyLTljMjUtZWU0ZTZlMTk1MTU0&hl=zh_TW&authkey=CIvK-uEI (For the 3D model)
Secret wrote:Myabe I'm a really stubbon visual person, but the reason why I'm not satisfied with the hemisphere representation is that the circular boundary remains unglued in the image. Because it is unglued, it looks incomplete. To me, even though I understand how they switch sides and connect as illustrated, it looks as if it 'magically' jumped from one end to another. The RP2 doesn't look like a 'closed surface' (layman) in the hemisphere representation (as there's clearly an opening visible in the representation)
Secret wrote:Seeing that nearly all immersions contains point intersection, is the RP2 actually pointed even in 4D?
Or is it mathematically impossible to have RP2 immersions without point intersections?
(In case you wonder why I still ask this question: The ans to this can tell me whether i should continue to pusurade that elusive immersion. Because if that is true but I manage to make that immersion, that means maths (at least in this context) is false. But I think this is very unlikely, thus I'll stop looking for that immersion if the ans to the question is true(and leave it until I got to uni))
Keiji wrote:[...]so this tells us that Boy's surface must be RP2.
Keiji wrote:Do note, there's nothing "pointy" about RP2 - only that it happens to contain a point where three disconnected regions of the plane intersect (as in Boy's surface) or a point where a circular sweeping section has been inverted (as in the second variation of your latest rendering that I verified above).
Similarly to how you can't find a way to have three perpendicular planes which don't intersect at a point, you also can't find an immersion of RP2 without such a point.
Keiji wrote:If you take the Klein bottle and pull it "taut" to get rid of the circle of self-intersection, what you end up with is a torus with a pinch point where the circular sweeping section inverts.[...]
Keiji wrote:you can see it doesn't work for genus 0, so the non-orientability must come from a pinch point.
Secret wrote:In respond to the boy's surface:
For what I know about ant walks, you only walk striaght forward (locally) on the surface. Therefore on the pic where you approches the lobes and suddenly make a 90^{o} turn into the lobe looks weird to me (as you supposed to go straight forward into another lobe nearby). And that's why I haven't considered this path.
Secret wrote:Btw if I understand your "3 perpendicular plane theory" correctly, does that mean RP2 self intersect/the planes touches at a point even in 4D?
wendy wrote:...[in complete (f)]...Since spheric space exists, it is orientable, but lines cross twice, ie it's oS. Elliptic space corrects the single crossing for axiomic reasons, but causes it to be non-orientable: it's mS...One supposes the existence of similar complete geometries for E and for H. The geometry derived from the artist's projections follow the general rule of mE, which means that it is non-orientable.
wendy wrote:...The out-vector can reverse over an 'askew marginoid', usually such occurs where surfaces cross. The density of the surface itself does not change, but the across-vector reverses.
) at the "askew marginoid", and then lengthen but with the direction now pointing 'inward' RP2 after passing through the "askew marginoid"?...there should be no net transverse outvector...
wendy wrote:...Thah...
wendy wrote:...One could, i suppose, build it into a polytope, but the thing would have holes of various kinds, in much the same way that an a reef knot exists on the surface of a torus without intersection...
wendy wrote:I usually hold the projective geometries to be an unproductive dead end...The nature of the projective plane, then is set by the presentation of the projective point 'u' (point defining straight lines).
wendy wrote:...and it is only when the infinite is brought to bear on problem, that the geometry divides.
wendy wrote:Some theroms, like the Desarges alignments, work in project geometry, and hence works in all three spaces.
wendy wrote:But when one tries to realise a projective plane, of any kind, one is left with the realisation of 'u', variously as a "pencil of lines" through a centre of the circle (the projective S geometry), or a line of infinity (in the euclidean geometry), or the bounding conic (in the hyperbolic), which seeks to cause problems.
Mrrl wrote:[...]But you also need that the last ring is a neighbour of itself, like "halves" of Moebius stripe, and its outside boundary should be a non-orientable curve ("projective line" of the plane) - with the same color on both sides. If your painting has these properties, than yes, the surface is a projective plane. [...]
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