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Dynamics and algebra

  1. Jan 26, 2004 #1
    it's an interaction between dynamics and algebra.

    let f and g be elements of C[x] where C is complex numbers (or any alg. closed field).

    denote the nth iterate of f by f^n, though that notation is often also used for the nth power of f. (this is where the dynamics of f is involved.)

    consider the ideal (and affine variety) generated by f-g, f^2-g, f^3-g, ... . denote this by I.

    the hilbert basis theorem states that this can be generated by a finite number of elements in I. denote this finite generating set by [f,g], not to imply that the generators are f and g. then the variety associated with I is the same as the one associated with <[f,g]>, the "span" of [f,g]. denote the variety by V[f,g].

    i want to investigate properties of V[f,g] depending on f and g. the easiest case is if f is a constant function. then its hilbert dimension is easy to calculate. if f is linear, things get a lot more interesting already. i imaging that for most or all g, V[f,g] is empty, but i'm not sure.

    any thoughts?

    well, i was hoping this approach might shed some light on the dynamics of f and my ultimate goal would be to say something like f "converges" to g if V[f,g] is of maximal dimension and "does not converge" if for all g, V[f,g] is empty. something like that.
  2. jcsd
  3. Jan 26, 2004 #2
    if we let g=x then could something interesting come out of that? perhaps the result that if f has any period r orbits where r is finite then all points have a bounded period (i guess that follows because f is in C[x])? this is all in the "making mistakes in different directions" so i hope you'll excuse the conjecture-like, random nature of my questions.
  4. Jan 26, 2004 #3

    matt grime

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    Deleted that post cos most of it was unimportant and misleading, but the idea that varieties do not distinguish between the ideals I and I^2 is important. In fact the reason I deleted it was because I realized in fact that my assertion that something was going to be non-empty was probably wrong. There was no reason at all for the rad ideal generated to be prime.

    A prime ideal is maximal in C[x] and the only such is <(x-a)> for some a (well, non-trivial I mean).

    In general, considering the difference of two elements, f^r - f^n is in the ideal. so a zero of this is a point with period n-r. this is true for all r,n, and in particular for n-r=1. So the zeroes of the ideal are some subset of the points with f(x)=f(f(x)). But for the reverse to be true, it must be that f(x)=g(x)

    So it must be that f-g has a unique zero that satisfies f(x)=f(f(x)).

    and i think that is enough to characterize when varieties occur, and what they are.
  5. Jan 26, 2004 #4
    in that case, f is like a projection operator in that f(x)=f(f(x)). hence f must map some subset of C to itself. for example, f could map the unit disk to itself or C to itself. i mean that these are conditions on f that imply f(x)=f(f(x)) and that the ideal has zeros, i think.

    i spotted your misprint as well, or i shall say i doubted it, but i think i'd like to just speak my mind regardless of whether it's correct. when one of us makes an error, we can correct each other. that way, the imagination can be free to roam though checked by the balance of the other person. it's up to you.
  6. Jan 26, 2004 #5

    matt grime

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    this f though is just a polynomial, and every polynomial has some set of x with f(x)=x (with the obvious exception), approx deg f of them give or take. And
    checking f=g has at most max(deg f,deg g) solutions.

    So it is nothing special to require f(x)=f(f(x)) cos every f will have some (finite number of points) where this is true. (at most [deg f] squared). So I wouldn't say 'in that case' because that is almost always the case. interestingly the exception is f(x)=x+b for b non-zero. but if b's non-zero then the previous argument applies (from the deleted post) that the ideal contains the functions

    x+b -g

    x+2b-g, and thus all the constant functions, and the ideal is C[x]
  7. Jan 26, 2004 #6
    right because for that function, when b!=0, there are no fixed points.
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