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Morse theory can be used to prove that certain solutions exist/don’t exist. A simple example is: you have an angular robot arm that can turn 360 degrees. Can y

by lixtra 7y ago

Morse theory can be used to prove that certain solutions exist/don’t exist. A simple example is: you have an angular robot arm that can turn 360 degrees. Can you come up with a smooth algorithm that moves it from any current position to any other position? Morse theory answers this because your configuration space is a torus. Edit: Maybe one could say that Morse theory is the intermediate value [1] theorem for manifolds on steroids. [1] https://en.m.wikipedia.org/wiki/Intermediate_value_theorem https://en.m.wikipedia.org/wiki/Intermediate_value_theorem

breadandcrumbel 7y ago
So basically morse theory can help you to make faster validation conclusions? Does developers/engineers use it?
krick 7y ago
Not sure if I understand you correctly. Do I need to prove that out of mathematical curiosity or for some profit? I mean, if it's the latter, isn't that kind of obvious (and the solution seems to be an exercise in geometry, not differential topology)? Maybe there are examples, where it would be much harder to see without some kind of sophisticated mathematical tools, but I don't see how it helps me in your example.
- lixtra 7y ago
  > Do I need to prove that out of mathematical curiosity or for some profit? Knowing that a solution (or how many) exists or not is valuable because it tells you wether it’s worthwhile to search for one. The robot arm example above helped an engineer friend of mine because his boss was not happy with his non-smooth solution. I could provide him with the papers that you cannot do better. This saved him time. Is it worth the overhead to learn Morse theory as an engineer? Probably not. Is it worthwhile to know it exists and ask a mathematician? I would argue yes.
  - krick 7y ago
    Ok, that makes sense: apparently, I don't understand what is meant by "smooth algorithm". I assumed that it simply means you don't have to stop the manipulator on the path from A to B, but if it's something different, that might be helpful indeed. I'm failing to find a clear definition of what it actually meant when used like that, though.
    - lixtra 7y ago
      Say you want to move from A to B on a circle. A simple algorithm would be to just go the shortest way from A to B which is well defined if A and B are not antipodal. If they are, you could define to always go against the clock. Now this algorithm is fine, but it is not continuous (smooth), because a small variation in an antipodal point gives two vastly different paths.

by lixtra 7y ago

breadandcrumbel 7y ago

So basically morse theory can help you to make faster validation conclusions? Does developers/engineers use it?

krick 7y ago

Not sure if I understand you correctly. Do I need to prove that out of mathematical curiosity or for some profit? I mean, if it's the latter, isn't that kind of obvious (and the solution seems to be an exercise in geometry, not differential topology)? Maybe there are examples, where it would be much harder to see without some kind of sophisticated mathematical tools, but I don't see how it helps me in your example.

lixtra 7y ago
> Do I need to prove that out of mathematical curiosity or for some profit? Knowing that a solution (or how many) exists or not is valuable because it tells you wether it’s worthwhile to search for one. The robot arm example above helped an engineer friend of mine because his boss was not happy with his non-smooth solution. I could provide him with the papers that you cannot do better. This saved him time. Is it worth the overhead to learn Morse theory as an engineer? Probably not. Is it worthwhile to know it exists and ask a mathematician? I would argue yes.
- krick 7y ago
  Ok, that makes sense: apparently, I don't understand what is meant by "smooth algorithm". I assumed that it simply means you don't have to stop the manipulator on the path from A to B, but if it's something different, that might be helpful indeed. I'm failing to find a clear definition of what it actually meant when used like that, though.
  - lixtra 7y ago
    Say you want to move from A to B on a circle. A simple algorithm would be to just go the shortest way from A to B which is well defined if A and B are not antipodal. If they are, you could define to always go against the clock. Now this algorithm is fine, but it is not continuous (smooth), because a small variation in an antipodal point gives two vastly different paths.