Re: circle homeomorphism


"David C. Ullrich" <ullrich@xxxxxxxxxxxxxxxx> wrote in message
news:8ul7l1p6ljvhud9v55hubqsl8t7o3qq47f@xxxxxxxxxx
> On Sun, 16 Oct 2005 20:53:50 +0100, "Thomas Novascott"
> <Thomas.novascott@xxxxxxxxxxxxxxxxxx> wrote:
>
>>
>>"David C. Ullrich" <ullrich@xxxxxxxxxxxxxxxx> wrote in message
>>news:ft65l1tc7gd90vvjt159t7a12ebo5fi7ea@xxxxxxxxxx
>>> On Sun, 16 Oct 2005 11:46:21 +0100, "Thomas Novascott"
>>> <Thomas.novascott@xxxxxxxxxxxxxxxxxx> wrote:
>>>
>>>>> It seems we have an increasing homeomorphism F of R,
>>>>> with period 2\pi, such that
>>>>>
>>>>> f(eit(t)) = eit(F(t))
>>>>>
>>>>> (writing eit(t) = exp(2 Pi i t)); F is a "lift"
>>>>> of f (ie a lifting of f to R wrt the covering
>>>>> map eit:R -> S^1.)
>>>>>
>>>>> and the rotation number is the limit of
>>>>>
>>>>> (*) (F^n(t) - t)/n,
>>>>>
>>>>> (where F^n denotes iterated composition).
>>>>>
>>>>> It follows that
>>>>>
>>>>> eit(F(F(t)) = f(eit(F(t)) = f(f(eit(t));
>>>>>
>>>>> that is, F^2 is a lift of f^2. If we're willing
>>>>> to believe that the limit (*) exists the result
>>>>> you ask about follows easily...
>>Thank you for your indepth explanation,
>>i think i should have made it clearer, but i dont see how
>>eit(F(F(t)) = f(eit(F(t)) = f(f(eit(t));
>>
>>proves p(f^n)=n.p(f)
>>(sorry for not making this clearer)
>
> It proves it for n = 2. Since F^2 is a lift
> of f^2 it follows that
>
> p(f^2) = lim ((F^2)^n(t) - t)/n
>
> = lim (F^(2n)(t) - t)/n
>
> = 2 lim (F^(2n)(t) - t)/(2n)
>
> = 2 lim (F^n(t) - t)/n,
>
> where the last equality is because if a sequence
> convergese then any subsequence converges to the
> same limit.
>
seems i forgot the last statement! i see this proof would work for n > 0 and
n=0 is obvious, out of curiousity could i use this approach to try and solve
n <0 ?


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