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When do SO(2) actions on the circle in the plane determine a metric?

  1. Mar 5, 2013 #1

    lavinia

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    a metric on the plane determines an action of SO(2) on is unit circle by rotation.

    Suppose one starts with a free transitive action of SO(2) on a circle. When does this come from a metric? Always?
     
  2. jcsd
  3. Mar 5, 2013 #2
    Here is a rough idea without calculations. I have not proved anything, so I might be misunderstanding something somewhere, or I might have just made a mistake in my thinking. Also, since you did not state the nature of the metric, I assumed the easiest thing which is that the metric is an inner product metric. Please clarify if you meant something more general.

    Lets first consider a free transitive action of SO(2) on the circle. Both can be denoted R/2piZ.

    Then we can parametrize these actions by strictly increasing functions F(s) satisfying F(0)=0, F(2pi)=2pi and whose derivative F' is periodic. Then the action of s on t is given by

    [itex] s*t = F(s+F^{-1}(t)) [/itex]

    Now let us consider the action induced by a metric. This is vague because I dont' know what you mean exactly by metric. So I will assume you mean an inner product metric. This is given by a quadratic form: Ax^2+2Bxy+Cy^2. I think it is clear that the resulting action of SO(2) is unchanged by scaling the coefficients of this inner product, so such actions are determined by 2 numbers. In more geometric terms, these two numbers are:
    1. The angular inclination of the eigenvector that lies in the first quadrant,
    2. The ratio of the eigenvalues.

    If your metric is Finsler, then I don't understand it well enough to describe the resulting action of SO(2) on the circle. But I would be interested to know more about that.
     
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