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  • one year ago

Is it possible to move the band on the surface so that it's inside out? http://prntscr.com/7pa4qf Also, the red edges are connected and the blue edges are connected, so you can stretch like this: http://prntscr.com/7pa5i8

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  1. geerky42
    • one year ago
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    *

  2. ganeshie8
    • one year ago
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    isn't it same as wearing a t shirt the wrong way which is definitely possible ?

  3. Empty
    • one year ago
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    :) Show me a series of steps that can take you from one to the other, if I need to clarify anything please ask, I came up with this question on my own and found it interesting!

  4. ganeshie8
    • one year ago
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    lol i have a shirt which i wear it the wrong way some times

  5. Empty
    • one year ago
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    Haha well I say this because this was similar to my intuition, but it turns out that this is false. It's not possible!

  6. Empty
    • one year ago
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    There is a clear geometric reason why it's impossible as well. Remember, you can't lift the band off the surface, you can only slide it around on it.

  7. dan815
    • one year ago
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    it would be the same as turning a sphere inside out, which is not possible in 3D

  8. ganeshie8
    • one year ago
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    yeah i saw flatland movie the other day you will need 1 extra dimension i think

  9. anonymous
    • one year ago
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    Can the band intersect with itself? i.e. is the situation below possible?|dw:1436156822443:dw|

  10. anonymous
    • one year ago
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    First move to the corner |dw:1436171267339:dw| |dw:1436171302595:dw| Connect 1 and 1' |dw:1436171394472:dw| Now you can connect 2 and 2' |dw:1436171503732:dw|

  11. anonymous
    • one year ago
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    |dw:1436171561702:dw|

  12. anonymous
    • one year ago
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    It highly depends on what happens a the corners, for example: |dw:1436171729088:dw| There are some topologies where this would or wouldn't happen.

  13. anonymous
    • one year ago
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    |dw:1436171915304:dw| If this were to happen, then this method won't work.

  14. anonymous
    • one year ago
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    |dw:1436171983615:dw||dw:1436172064741:dw|

  15. anonymous
    • one year ago
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    |dw:1436172227836:dw|

  16. anonymous
    • one year ago
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    We're basically on a torus, if you connect up the red sides: |dw:1436173616476:dw|

  17. Empty
    • one year ago
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    These last two pictures are the only correct cases.

  18. anonymous
    • one year ago
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    |dw:1436174290291:dw| Anyway, going to the outer square will put you in the inner square and vise versa, so previous method won't work.

  19. Empty
    • one year ago
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    |dw:1436174478954:dw| My reasoning for why it's impossible is because this single hole right here acts as a barrier. If we filled in the hole to make it a sphere, then we could do it however. My reasoning for why we can't find another way is as long as we slide the circle around the outer most rim, the case is the same due to symmetry: |dw:1436174755725:dw| Now suppose you put the circle in the center part and suspect there is some way we can "pull it out" inside out from here: |dw:1436174840614:dw| We acknowledge that these are all symmetric as well, and although this particular location appears different, it's actually the exact same location except rotating the circle by 90 degrees as indicated by how the lines on the torus turn by 90 degrees when we turn the manifold itself inside-out: |dw:1436174956104:dw| I think this is a fairly rigorous argument for why it's impossible, but if anyone has a more solid explanation I'd love to hear it. :D

  20. anonymous
    • one year ago
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    Your explanation doesn't seem very rigorous.

  21. anonymous
    • one year ago
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    |dw:1436178050601:dw||dw:1436178081156:dw||dw:1436178111428:dw|

  22. anonymous
    • one year ago
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    |dw:1436178155510:dw|

  23. anonymous
    • one year ago
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    |dw:1436178198991:dw||dw:1436178225271:dw||dw:1436178252909:dw|

  24. anonymous
    • one year ago
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    |dw:1436178583739:dw|

  25. anonymous
    • one year ago
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    |dw:1436179146170:dw|

  26. anonymous
    • one year ago
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    |dw:1436180310900:dw| The sum of the interior angles of the green square is \(2\pi\). There is no possible operation that changes this property, because all transformations in this system will preserve it. Even with circles, we can randomly select 4 points on the circle to represent the \(\pi/2\) angles. The only way to remove corners is to go through a corner, and this configuration prohibits such an operation.

  27. Empty
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @wio Your explanation doesn't seem very rigorous. \(\color{blue}{\text{End of Quote}}\) Why?

  28. anonymous
    • one year ago
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    It doesn't make much sense to begin with. It seems like a proof by contradiction where even when someone pulls it inside out, you tell them it's not due to symmetry. You don't really know what someone might try to do to pull it inside out.

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