$4D$ analog of Alexander's Horned Sphere

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Is it a trivial exercise to find a $4D$ analog to Alexander's Horned Sphere? In other words, is there a manifold homeomorphic to $S^3$, embedded in $mathbb R^4$ that has a wild complement?



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  • I don't think anything concerning Alexander's horned sphere is trivial...
    – Lord Shark the Unknown
    Sep 2 at 9:39














up vote
0
down vote

favorite












Is it a trivial exercise to find a $4D$ analog to Alexander's Horned Sphere? In other words, is there a manifold homeomorphic to $S^3$, embedded in $mathbb R^4$ that has a wild complement?



Thank you.






general-topology







share|cite|improve this question























  • I don't think anything concerning Alexander's horned sphere is trivial...
    – Lord Shark the Unknown
    Sep 2 at 9:39












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Is it a trivial exercise to find a $4D$ analog to Alexander's Horned Sphere? In other words, is there a manifold homeomorphic to $S^3$, embedded in $mathbb R^4$ that has a wild complement?



Thank you.






general-topology







share|cite|improve this question















Is it a trivial exercise to find a $4D$ analog to Alexander's Horned Sphere? In other words, is there a manifold homeomorphic to $S^3$, embedded in $mathbb R^4$ that has a wild complement?



Thank you.






general-topology




general-topology






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edited Sep 2 at 7:56









Jneven

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asked Sep 2 at 7:43









Barnaby Finch

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  • I don't think anything concerning Alexander's horned sphere is trivial...
    – Lord Shark the Unknown
    Sep 2 at 9:39
















  • I don't think anything concerning Alexander's horned sphere is trivial...
    – Lord Shark the Unknown
    Sep 2 at 9:39















I don't think anything concerning Alexander's horned sphere is trivial...
– Lord Shark the Unknown
Sep 2 at 9:39




I don't think anything concerning Alexander's horned sphere is trivial...
– Lord Shark the Unknown
Sep 2 at 9:39










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I recommend to have a look into



Rushing, T. Benny. Topological embeddings. Vol. 52. Academic Press, 1973.



Theorem 2.6.1 gives a positive answer to your question. But this is not a "trivial exercise".



However, if you have wild sphere $S subset mathbbR^n$, then you get an obvious embedding of the suspension $Sigma S$ into $mathbbR^n+1$.



If $mathbbR^n backslash S$ is not uniformly locally 1-connected (1-ULC, see Rushing), then one can show that $mathbbR^n+1 backslash Sigma S$ also fails to be 1-ULC. Thus $Sigma S subset mathbbR^n+1$ is again a wild sphere. This applies to the horned sphere.






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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

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    active

    oldest

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    active

    oldest

    votes








    up vote
    0
    down vote













    I recommend to have a look into



    Rushing, T. Benny. Topological embeddings. Vol. 52. Academic Press, 1973.



    Theorem 2.6.1 gives a positive answer to your question. But this is not a "trivial exercise".



    However, if you have wild sphere $S subset mathbbR^n$, then you get an obvious embedding of the suspension $Sigma S$ into $mathbbR^n+1$.



    If $mathbbR^n backslash S$ is not uniformly locally 1-connected (1-ULC, see Rushing), then one can show that $mathbbR^n+1 backslash Sigma S$ also fails to be 1-ULC. Thus $Sigma S subset mathbbR^n+1$ is again a wild sphere. This applies to the horned sphere.






    share|cite|improve this answer
























      up vote
      0
      down vote













      I recommend to have a look into



      Rushing, T. Benny. Topological embeddings. Vol. 52. Academic Press, 1973.



      Theorem 2.6.1 gives a positive answer to your question. But this is not a "trivial exercise".



      However, if you have wild sphere $S subset mathbbR^n$, then you get an obvious embedding of the suspension $Sigma S$ into $mathbbR^n+1$.



      If $mathbbR^n backslash S$ is not uniformly locally 1-connected (1-ULC, see Rushing), then one can show that $mathbbR^n+1 backslash Sigma S$ also fails to be 1-ULC. Thus $Sigma S subset mathbbR^n+1$ is again a wild sphere. This applies to the horned sphere.






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        I recommend to have a look into



        Rushing, T. Benny. Topological embeddings. Vol. 52. Academic Press, 1973.



        Theorem 2.6.1 gives a positive answer to your question. But this is not a "trivial exercise".



        However, if you have wild sphere $S subset mathbbR^n$, then you get an obvious embedding of the suspension $Sigma S$ into $mathbbR^n+1$.



        If $mathbbR^n backslash S$ is not uniformly locally 1-connected (1-ULC, see Rushing), then one can show that $mathbbR^n+1 backslash Sigma S$ also fails to be 1-ULC. Thus $Sigma S subset mathbbR^n+1$ is again a wild sphere. This applies to the horned sphere.






        share|cite|improve this answer












        I recommend to have a look into



        Rushing, T. Benny. Topological embeddings. Vol. 52. Academic Press, 1973.



        Theorem 2.6.1 gives a positive answer to your question. But this is not a "trivial exercise".



        However, if you have wild sphere $S subset mathbbR^n$, then you get an obvious embedding of the suspension $Sigma S$ into $mathbbR^n+1$.



        If $mathbbR^n backslash S$ is not uniformly locally 1-connected (1-ULC, see Rushing), then one can show that $mathbbR^n+1 backslash Sigma S$ also fails to be 1-ULC. Thus $Sigma S subset mathbbR^n+1$ is again a wild sphere. This applies to the horned sphere.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Sep 2 at 13:05









        Paul Frost

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