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How do axioms for Euclidean geometry exclude non-trivial topology?

  1. Dec 26, 2013 #1
    Think for example of the torus as a square with the proper edges identified. Viewed like this (i.e. using the flat metric), it clearly has zero curvature everywhere. More specifically, it seems Euclid's axioms are satisfied. But however we have non-trivial topology. So what's up?

    Or is Euclid's circle axiom not satisfied? But anyway my question in general is: is it clear Euclid's axioms forbid non-trivial topology? (In other words: I see that they imply Euclidean geometry locally, but why globally?) Or is it not necessary?
     
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  3. Dec 26, 2013 #2

    Simon Bridge

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    Strictly, the answer to your question is "no": Euclid's axioms do not forbid non-trivial topology.

    Anyway: the term "non-trivial" as applies to topology, in general, may be a synonym for "non-eucliedean".
    Otherwise what counts as "non-trivial" depends on the context.
    i.e. a torus may be non-trivial to a sphere in the sense that it would be to a coffee cup.
    Where have you seen the claim being made?

    Also see:
    https://www.physicsforums.com/showthread.php?t=676534
    http://math.stackexchange.com/questions/438055/non-trivial-topology
     
  4. Dec 26, 2013 #3
    Sorry, it was just a thought I was having, so "non-trivial" in this context was my own use of words. Indeed by "non-trivial topology" in this context I just mean "not being the Euclidean plane". I have a physics background so that probably explains the abuse in terminology :)

    But regarding the main topic: thanks for answering the original question!
     
  5. Dec 26, 2013 #4

    mathwonk

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    this is a very interesting question. look at euclid's postulate 2, that every segment can be extended arbitrarily into a line. do you think a "line" should be allowed to double back on itself or not? if not, then a torus is disqualified. i.e. depending on your answer to this, you either get simply connected spaces, i.e. trivial topology examples, or else you may get more complex ones with the usual plane as universal covering space.
     
  6. Dec 26, 2013 #5

    Simon Bridge

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    I agree it's interesting - it's the sort of thing that can get mathematicians killed on pedestrian crossings.
    I had to restrain myself from trying to come up with violations... if we accept that lines double back on themselves smoothly, what about lines that intersect with themselves?

    I was wondering if the parallel lines postulate was obeyed on a torus but decided to have lunch instead... it's been too long.

    A simple confusion in terminology can lead to questions that we may not think up by ourselves :)

    Unfortunately I could only answer the original question in the trivial sense of "it depends" ... i.e. on what you think "trivial" means.
     
  7. Dec 26, 2013 #6
    I think a line "should" be allowed to double back on itself, otherwise the sphere doesn't satisfy Euclid's second axiom which I would just find weird.

    What I also find weird is that both of you seem to count the sphere as topologically trivial? For me topologically trivial means trivial in an algebraic topology sense, so I suppose I mean "contractible". Basically my question was about: does satisfying Euclid's axioms mean you're on a (Euclidean) plane (or something homeomorphic (or homeotopic?) to it)?

    But whilst we're trying to vet the different axioms: what about the circle axiom? So take the torus with the flat metric. Is axiom 3 satisfied? "It is possible to construct a circle with any point as its center and with a radius of any length"
     
  8. Dec 26, 2013 #7

    Simon Bridge

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    Both of who?
    I believe I am on the record (above) twice stating that "trivial" is a matter of context.
    Nobody else has mentioned a sphere.

    "Weird" is even more subjecting than "trivial".

    Why should a sphere satisfy the second axiom? What's weird about not doing so?
    You'll notice that it also does not satisfy the parallel lines axiom.

    In what way do you find a sphere non-trivial in an algebraically topological sense?

    Did you have a look at the links I gave you?
     
  9. Dec 26, 2013 #8
    I did take a look yes. Okay my apologies for including you as well, it seems you only said a sphere is topologically trivial in one of two contexts you mentioned. But spheres were mentioned: "you either get simply connected spaces, i.e. trivial topology examples" (MathWonk) which implied that a sphere was regarded as topologically trivial.

    I'm not sure why you are so harsh on my use of "weird" though. I think that's perfectly acceptable. I merely alluded to the idea that usually an example of something obeying all but the fifth of Euclid's axioms is geometry on the surface of a sphere. Of course "weird" is not an exact word in this context, because of course there is no proper exact word in this case, since I'm not trying to prove that double backing on itself is fine for lines, I'm merely argueing it's natural to allow it given how we look at spheres. (No need to italicize "should", I had already put it in quotation marks in my post; it feels like you are looking for things to be picky about.)

    And spheres are non-trivial in an algebraic topological sense since they have non-trivial/non-zero homology and homotopy groups (other than the zeroth ones).
     
  10. Dec 28, 2013 #9

    mathwonk

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    this is why euclidean geometry is so interesting.
     
  11. Dec 29, 2013 #10

    AlephZero

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    I think the basic issue here is that Euclid's axioms, as stated in the Elements, are so vaguely stated and incomplete that it doesn't really make and sense to ask the OP's question.

    Take the first proposition of Euclid book 1, for example: to construct an equilateral triangle given a straight line AB as one side.

    Euclid does the obvious thing: draw two circles, radius AB, with centers A and B. His axioms say that is always possible. The third vertex C of the triangle is then the intersection of the two circles.

    But hey, wait a minute - what if the circles don't intersect? That is not one of the axioms, and worse, there is nothing in the axioms that you can prove it from.

    Compare Euclid's axioms with "modern" axiom systems for geometry, for example http://en.wikipedia.org/wiki/Hilbert's_axioms

    A translation of Hilbert's "Foundations of Geometry" here: https://archive.org/details/abr1237.0180.001.umich.edu
     
  12. Dec 29, 2013 #11

    Simon Bridge

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    Do mathematicians usually insist that Euclidean Geometry exactly follow what is written in Elements - so needing a different name for anything that explicitly includes the unspoken assumptions?

    I thought the circle axiom requires a center and a length but the length is left arbitrary - defined on a line segment. In the equilateral triangle proposition - the "length" is the line segment AB (the only defined length in the statement) ... which becomes the default radius for any circles unless otherwise specified ... i.e. the radius of the circle is included in the definition of how to construct a circle. Thus - following the instructions, the two circles will always intersect at points which allow the construction of an equilateral triangle.

    And, of course, the construction is the proof from the axioms.
    It was a different way of thinking.
     
  13. Dec 30, 2013 #12

    lavinia

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    When I learned Eudlidean geometry, it was postulated that two lines can intersect in at most one point. If one takes topologically trivial to mean contractible the this requirement may guarantee topological triviality. Proof?

    However if one only requires that the Pythagorean theorem be true then many topologically non-trivial spaces - such as the flat torus - are possible. These are the flat Riemannian manifolds and non-trivial ones exist in all dimensions.

    Every flat Riemannian manifold is obtained from flat Eulidean space by a group of covering transformations that are isometries. This means that all of the length and angle relations are preserved so locally the geometry is Euclidean. Such a manifold can be cut with a knife a finite number of times and unrolled into a chunk of Euclidean space. For instance, the flat torus requires two cuts.
     
    Last edited: Dec 30, 2013
  14. Dec 30, 2013 #13

    lavinia

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    A little on the idea of a trivial topology.

    Technically the trivial topology on a set is the topology where the only open sets are the null set and the whole space. The only manifold that has this topology is the single point.

    However, one might generalize this idea to say that a space is in some sense topologically trivial if it can be continuously deformed to a single point. By this is meant a continuous function

    F:space x [0,1] -> space such that F is the identity map at time zero and maps the space to a single point at time one.

    Such a space is called contractible.

    Euclidean space is contractible.

    Compact manifolds without boundary ,such as the sphere, are never contractible.

    The sphere is simply connected which means that every closed loop can be shrunk to a point but it is not contractible.
     
    Last edited: Dec 31, 2013
  15. Dec 30, 2013 #14
    Ah interesting side-note there Lavinia.

    As a physicist, when I used "trivial topology" I meant trivial as an adjective in the sense of trivial objects. ( http://en.wikipedia.org/wiki/Triviality_(mathematics) ) This is vague on my end, and of course the more mathematically minded interpret "trivial topology" in the strict/exact sense as: topology where every point is open.

    Interestingly it seems you managed to connect the two and make them one an the same, if we define the former (i.e. "no interesting topology") as being homeotopic to a space with trivial topology (in the strict sense). It's a bit of a weaker concept than what I had in mind, since I suppose I intuitively meant "something homeomorphic to affine space", but your notion has the benefit of connecting the two notions of "trivial topology" :)
     
  16. Dec 31, 2013 #15

    lavinia

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    I can see why one might call the topology where every point is open a trivial topology but I think in mathematics the trivial topology would be the one where there are no open sets at all except the mandatory whole space and null set.

    The topology of Euclidean space is highly non-trivial and many deep theorems are needed to understand it. However, its homotopy type is the same as that of a single point. One might say that it has trivial homotopy type.

    You are correct that homotopy type is weaker the homeomorphic but it is not a weak concept and plays an important role in topology.

    BTW: The Wikipedia link to the mathematical idea of trivial that you provided seems correct.
    According to the link, the sphere would have trivial fundamental group since every loop can be deformed to a point, but it does not have trivial homotopy type. Euclidean space has trivial homotopy type since it can be deformed to a point.

    Here is a somewhat different example. A vector bundle over a topological space is called "trivial" if it is homeomorphic to the Cartesian product of the space with some vector space. Such a bundle has no structure that depends upon the topology of the base space. So for instance the tangent space of Euclidean space is a trivial bundle as is the tangent space of the torus. However, the tangent space of the sphere is not trivial.


    Aside: I do not know whether in the Theory of Relativity, Space-Time is diffeomorphic to Euclidean space but that seems to be what the texts assume. A mathematician told me once that in order for there to exist a magnetic monopole that space would have to have "non -trivial topology" which I guess means handles that give it a non-trivial homotopy type. Can you elaborate on this?
     
    Last edited: Dec 31, 2013
  17. Dec 31, 2013 #16

    lavinia

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    Plane Geometries

    To me, plane geometry does not allow lines to self-intersect. One thinks of a line as dividing the plane into two disjoint half planes. Self intersection seems to prevent this. If one adds that two lines can intersect in at most one point, one gets the Euclidean and Gaussian planes as the only two possibilities. Both of these are contractible. The sphere is not a model of plane geometry under these axioms because great circles intersect in two points. The projective plane is not a model because a line does not separate the projective plane into two disjoint half planes.
     
    Last edited: Dec 31, 2013
  18. Dec 31, 2013 #17

    lavinia

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    Euclid's Second Axiom

    In Differential Geometry, one might restate Euclid's second axiom to say that any unit speed geodesic can be extended to an arbitrary parameter value. The Hopf-Rinow Theorem guarantees this when the manifold (no boundary) is a complete metric space. Conversely, if every unit speed geodesic can be extended indefinitely, the manifold is a complete metric space. So in this context one can think of Euclid's second axiom as stating that the manifold is complete.

    So from the Differential Geometric point of view, you are correct to think that the sphere should satisfy Euclid's second axiom.

    I wonder whether Euclid's intuition of the plane was that it is complete even though he was not aware of the mathematical idea.
     
    Last edited: Dec 31, 2013
  19. Jan 6, 2014 #18

    lavinia

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    Can you give an example?
     
  20. Jan 6, 2014 #19
    Why do you all have the interesting conversations when I'm in school? :tongue:

    This is a fun question. I'll see if I can come up with a worthwhile answer when I wake up tomorrow.
     
  21. Jan 6, 2014 #20

    lavinia

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    I read Euclid's actual axioms and was confused by what the third axiom actually meant. What does he mean by a circle? If it is a closed loop of constant distance to a central point then the third axiom probably eliminates non-trivial homotopy type. For instance on the sphere, draw the segment from the north to south pole. The south pole is the only point of distance pi to the north pole so one can not draw the circle. Similarly if the surface contains a segment that is a closed loop, then no circle can be drawn. It seems then that one needs a theorem that says that any 2 dimensional Riemannian manifold with non-trivial homotopy type has a closed geodesic. Not sure if that is true. This also excludes the possibility that a line can intersect itself since that would create a closed segment.

    I think the second axiom lets you assume that the surface is complete as a metric space and has no boundary which simplifies the problem.
     
    Last edited: Jan 6, 2014
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