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[cross posted from math.se due to lack of answers]

I'm attempting to compute the group of continuous (or more generally holomorphic) $\phi: \mathbb{C} \rightarrow \mathbb{C}$ such that $f(\phi(x)) = f(x)$.

In the linear setting, $f = ax + b$ we don't see anything interesting, the group $\phi$ is trivial.

In the quadratic setting, we note that $f = x^2$ has symmetry group $\lbrace x, -x \rbrace$, has $x^2 = (-x)^2$ and to go a step further, we can complete the square on any $ax^2 + bx + c$, to derive:

$$ ax^2 + bx + c = a \left(x^2 + \frac{b}{a} x + \frac{c}{a}\right) = a\left( x + \frac{b}{2a} \right)^2 + c - \frac{b^2}{4a} $$

And thus generally it has symmetry group $\lbrace x, -x-\frac{b}{2a} \rbrace$,

Now the cubic case becomes much trickier. we can extrapolate from $f = x^3$ to the general $e_0(x+e_1)^3 + e_2$ but this doesn't cover all possible cubic polynomials.

Any idea how to dig deeper here and compute the symmetries of all cubic polynomials?


Some additional observations, in the quadratic case: the solutions of $ax^2 + bx + c = t$ must necessarily be permuted by such a symmetry, i.e. i was looking for a function that sends $-\frac{b}{2a} + \frac{\sqrt{b^2 - 4a(c-t)}}{2a} \rightarrow -\frac{b}{2a} - \frac{\sqrt{b^2 - 4a(c-t)}}{2a}$

In the cubic case, I can do a similar thing, writing down the form of the roots, then deducing ANY function that sends one form to the other, but such functions seem like they will be very unwieldy to write. And i'm not sure how to guess a form of such a function.

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For any polynomial $f$ of degree $n$, $$f(\phi(x)) - f(x) = (\phi(x) - x) g(\phi(x),x)$$ where $g$ is a bivariate polynomial of degree $n-1$, so the symmetries will be $\phi(x) = x$ and the roots of $g(\cdot, x)$. In general, the latter will not be entire functions. Thus if $f(x) = x^3 + a x$, we get $$ \phi(x) = \frac{-x\pm\sqrt {-3\,{x}^{2}-4\,a}}{2} $$

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  • $\begingroup$ I have used this identity in a recent note - do you happen to know its origin? $\endgroup$ – Franz Lemmermeyer Apr 4 '17 at 19:28
  • $\begingroup$ Which identity? $\endgroup$ – Robert Israel Apr 4 '17 at 20:13
  • $\begingroup$ The one in the second line. $\endgroup$ – Franz Lemmermeyer Apr 4 '17 at 20:23
  • $\begingroup$ Surely it was known to Newton, but I don't know details of the history. $\endgroup$ – Robert Israel Apr 5 '17 at 1:29
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Here is a general remark about the equation $f(\phi(x)) = f(x)$ for a function $f: {\bf R}^d \rightarrow {\bf R}^d$. This equation is equivalent to the fact that the level sets $f^{-1}(y)$ of $f$, for all $y \in {\bf C}$, are left invariant by $\phi$. If $f$ is a degree 3 polynomial on ${\bf C}$, then most level sets have 3 elements and $\phi$ acts as a permutation on these sets. Any family of bijections on the levels is valid if there are no regularity assumptions on $\phi$.

If the roots are simple and distincts and $\varphi$ is smooth, then it is possible to start with a permutation say at the zero level and prolong continuously (even analytically applying some inversion theorem) the permutation to the adjacent levels in a unique way until we meet a level where two or more roots collide in which case there may be several continuous choices for the roots but analyticity is lost.

So for example assuming $f$ smooth proper, there should be a unique extension of the original permutation on the connected components of the reciprocal of the regular set. For a polynomial on ${\bf C}$, there are only a finite set of points whose image is not regular, and that's where the extension may fail. Still there should exist a Riemann surface over the complement of the critical points on which a global analytic extension exists (e.g. in the example of Robert Israel, obtained from the Riemann surface of the square root).

This can be illustrated in the real case with $f(x) = x^2$. The two maps $x\mapsto x$ and $x\mapsto -x$ are continuous solutions for $\phi$, but you also have $x\mapsto |x|$, which is obtained by gluing the two previous solutions at the singular zero level.

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