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Hairy billiard balls and eigenvectors

  1. Mar 9, 2005 #1

    mathwonk

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    this is no doubt older than dirt, but it just popped into my head when trying to think of how to convince my beginning linear algebra class that every linear isometry of R^3 has an eigenvector:

    i.e. it follows from the theorem that one cannot comb the hair on a billiard ball. do you see how?

    it might not be totally well known, at least to students, because we so seldom mingle topology and linear algebra in the same course, unlike in "real life" i.e. research.

    anyway they seemed to like it, perhaps for its freshness, after all the tedious matrices. of course when i tried to ilustrate the theorem with my own head, my bald spot made it seem I was giving myself an unfair advantage.
     
    Last edited: Mar 9, 2005
  2. jcsd
  3. Mar 10, 2005 #2
    Isn't this an example of Brouwer's fixed point theorem (often famously called the "hairy ball" theorem)?

    I agree that it's an interesting analogy.
     
  4. Mar 10, 2005 #3

    mathwonk

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    brouwers fixed point theorem says that a continuous self map of a closed ball has a fixed point. how is that equivalent to the theorem that a self map of the sphere has either a fixed point or a point mapped to its antipode?

    i believe you, since all these results seem to be equivalent, but it is not immediately obvious to me.

    and i have never heard brouwers theorem called the hairy ball theorem.
     
  5. Mar 11, 2005 #4
    Think of it like this. Imagine a ball that has hair sprouting out of it. Each of these hairs will be "combed" down, i.e., not mapped to themselves, but mapped to some other point on the sphere. But if each of these hairs is combed down, there must exist one hair that stands completeley straight up: that is mapped onto itself. The other "combed" hairs start at their origin but end up at a different point on the sphere.

    Basically, if you look at your own head of hair, it is equivalent to the (at least) one hair on your head that is sticking up: because it it mapped to no other "point" on your head but the point that it originates from.

    I'm not totally sure that I explained that correctly, but I'm pretty sure. Brouwer's fixed point theorem is referred to as the "hairy ball" theorem in L.C. Thomas' Games, Theory and Applications. It's a very good introductory game theory text (a very good balance of quantitative and qualitative aspects) that uses this theorem as a leading up to Nash's theorem about equilibria in n-person games.

    Like I saie, I'm not totally sure that Brouwer's hairy ball theorem fits in, but that's what jumped at me when I first saw your thread.
     
  6. Mar 11, 2005 #5

    mathwonk

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    actually i remember now they are different. i will explain later when i have more time, but the brouwer theorem is easier because it follows from the elementary fact that 1 ≠ 0. the hairy billiard ball result requires the much subtler fact that 1 ≠ -1! [for experts, the first result follows already from mod 2 cohomology while the latter requires something like integral cohomology.]
     
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