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Intersection of simply connected

  1. Feb 2, 2010 #1
    Hi everybody!

    I have a question, if I have A and B simply connected subspaces of a geodesic space X, what can be said about their intersection?
    When is it simply connected? Are there rules for this?

    I need to prove it in a special case, but I am not able to do it and I was wondering if there are general rules for this statement.

    Thanks!
     
  2. jcsd
  3. Feb 2, 2010 #2
    what is a geodesic space?
     
  4. Feb 2, 2010 #3
    A geodesic space is a space where for all two points there exists at least one geodesic joining them.

    A geodesic is the shortest path between points, but I am considering a general case where there might be infinite shortest paths between two points.
     
  5. Feb 2, 2010 #4
    so you mean a closed Riemannian manifold?

    Two spheres intersect in a circle. If these two spheres are subspaces of R^3 then they are simply connected subspaces with non-simply connected intersection. Do you mean that the subspaces are also geodesic in the sense that any two points can be connected by a geodesic that is also a geodesic in the big manifold?
     
    Last edited: Feb 2, 2010
  6. Feb 2, 2010 #5
    "Do you mean that the subspaces are also geodesic in the sense that any two points can be connected by a geodesic that is also a geodesic in the big manifold?"

    Exactly!

    I know that in general the intersection of two simply connected spaces is not simply connected. I was wondering if, considering some more hypothesis, I could prove that it is.
     
  7. Feb 2, 2010 #6
    Are you sure that's how you want to define geodesics and geodesic spaces? Because it seems in that definition, any closed path-connected space is a geodesic space (just optimize over all paths between two points, and label the winner "geodesic").

    For example, I don't see how your definition would prevent me from labeling the union of two (intersecting) spheres a geodesic space, and then considering the intersection of the individual spheres.

    If you are forcing yourself to consider all spaces as subspaces of Euclidean space, "geodesic spaces" turn into "convex sets." And it is true that the intersection of two convex sets is convex, and convex sets are trivially simply connected.

    Aren't geodesics typically defined in terms of covariant derivatives or some other local (rather than global) criterion? For example, the great circle between two points on a sphere splits up into a long and a short arc segment (unless the two points are antipodal). The long arc is usually considered a geodesic, although it is not the shortest path between the two points.
     
  8. Feb 2, 2010 #7
    In the 3 sphere take two -2 spheres that intersect in a great circle. These are simply connected totally geodesic sub manifolds that intersect in a geodesic that is closed and thus not simply connected.
     
  9. Feb 3, 2010 #8
    I know that in general the intersection of two simply connected spaces is not simply connected.

    But for example, if I consider R^2 equipped with the metric associated to the l^1 norm, then any path whose image is the graph of a monotonic function, is a geodesic segment. Therefore there are infinitely many geodesic segments joining any two points in this space.

    Now if I consider three points a,b,c in this space, and X_{ab} the subspace containing all the geodesics between a and b, and X_{ab} the subspace containing all the geodesics between b and c, they are simply connected.

    In this case they are convex and therefore also their intersection is simply connected.

    But in general I cannot suppose that if something is simply connected then it is also convex. I would like to find examples as the previous one in which X_{ab} is not convex and understand if then the intersection could be also simply connected.
     
  10. Feb 3, 2010 #9
    In the example of the spheres they are convex
     
  11. Feb 4, 2010 #10
    sorry... what do you mean by "In the example of the spheres they are convex"?

    Do you have suggestions, how to find other examples where between points there are infinitely many geodesics?
     
  12. Feb 4, 2010 #11
    The 3 sphere is a manifold whose geodesics are great circles. The 2 sphere subspaces are totally geodesics i.e. any two points on them can be connected by a great circle - they are therefore convex.
     
  13. Feb 6, 2010 #12
    At the risk of interrupting an interesting discussion on convexity, there's an interesting example to consider (especially for people involved in geodesic spaces!):

    Fact: Any topological space X is the intersection of two simply connected (even contractible) spaces.
    Hint: Take a double cone over X.
     
  14. Feb 6, 2010 #13
    At the risk of interrupting an interesting discussion on convexity, there's an interesting example to consider (especially for people involved in geodesic spaces!):

    Fact: Any topological space X is the intersection of two simply connected (even contractible) spaces.
    Hint: Take a double cone over X.
     
  15. Feb 8, 2010 #14
    So my example of a sphere would be a double cone over a 1 less dimensional sphere.

    In general the double cone would have singularities at the cone points and geodesics would not pass through them?
     
  16. Feb 9, 2010 #15
    In rereading your post I realized that I may have misunderstood your question. All I said was that convex simply connected subspaces can have non-simply connected convex intersection.

    This snippet from your last post seems to say that X_{ab}must be simply connected so I am confused. In general it does not have to be simply connected.
     
  17. Feb 9, 2010 #16
    Yes, you could think of it that way.

    In general, geodesic spaces need not be smooth manifolds. A double cone is a perfectly allowable geodesic space.
     
  18. Feb 9, 2010 #17
    What is the definition of a geodesic space?
     
  19. Feb 9, 2010 #18
    It's a metric space with some added structure, i.e. there's a notion of 'length of path' and a variational principle related to arc length. It need not be a smooth manifold, or finite dimensional, etc. It's quite natural. If you abstract away from geodesics on riemannian manifolds, we're left with a variational principle on an infinite-dimensional path space.
     
  20. Feb 17, 2010 #19
    Hi guys....

    I was thinking about my questions and another question arises...

    if I have two simply connected spaces with path-connected intersection, could I conclude that it is also simply connected?

    Intuitively I think so, but I am not able to prove it or to find a counterexample!
     
  21. Feb 17, 2010 #20
    find a counter example.
     
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