9
$\begingroup$

Problem

Suppose $f^{(t)}(z)=a_0^{(t)}+\dotsb+a_{n-1}^{(t)}z^{n-1}+z^n\in\mathbb C[z]$ for all $t\in\mathbb R$, where $a_0^{(t)},\dotsc,a_{n-1}^{(t)}\colon\mathbb R\to\mathbb C$ are continuous on $t$. Is it true that there is always a complex-valued continuous function $\phi^{(t)}\colon\mathbb R\to\mathbb C$ such that $f^{(t)}(\phi^{(t)})=0\;(\forall t\in\mathbb R)$?

Topologically,

Let $S=\big\{\,(a_0,\dotsc,a_{n-1},z)\in\mathbb C^{n+1}\,\big\vert\,a_0+a_1z+\dotsb+a_{n-1}z^{n-1}+z^n=0\,\big\}$ and $\pi\colon S\to\mathbb C^n,(a_0,\dotsc,a_{n-1},z)\mapsto(a_0,\dotsc,a_{n-1})$. Is it true that $\pi$ has path lifting property (i.e. for each continuous map $p\colon[0,1]\to\mathbb C^n$, there exists a continuous map $\tilde p\colon[0,1]\to S$ such that $p=\pi\circ\tilde p$? (Sorry, I cannot find a good reference for that term. In my definition, there's no assumption of uniqueness.)

Or algebraically,

Let $R=\mathcal C(\mathbb R,\mathbb C)$ denote the ring of complexed-valued real continuous functions. Is it true that any monic polynomial over $R$ has a root in $R$?

Discussion

It's certainly true that there is a function $\phi_{t_0}^{(t)}$ continuous at $t=t_0$ such that $f(\phi_{t_0}^{(t)})=0\;(\forall t\in\mathbb R)$, no matter whether $t$ is a real or complex parameter, or a parameter from some Hausdorff space. It follows directly from, say, Rouché's theorem. For an elementary proof, see Michael Artin's Algebra, proposition 5.2.1(b). A sharper proposition of the original problem, i.e. replacing the real parameter $t\in\mathbb R$ with a complex parameter $w\in\mathbb C$, is generally demonstrably false, i.e. there could be no continuous function to be a root of the polynomial. Here's a simple counterexample: $f^{(w)}(z)=z^2-w$. Note that there's a branch point at $w=0$, and if we draw an arbitrary circle around the origin in the $w$-plane, we'll see that a root shouldn't be continuous on the whole $w$-plane $\mathbb C$, for otherwise letting $w$ travel the circle would lead to a contradiction.

It seems true when $t$ is a real parameter, since the dimension is lower. I have no idea how to attack this. I expect your grateful ideas or hints. Thanks!

Postscript

There's an old post related, inequivalent but informative and interesting.

$\endgroup$
  • $\begingroup$ I don't understand your counterexample. Surely for small $|w|$, any roots of $z^2-w$ will be close to the origin as well? $\endgroup$ – Jack M Jan 30 '14 at 6:40
  • $\begingroup$ @jack I've edited that paragraph to reduce some ambiguity. On your question directly: the discontinuity doesn't appear at the origin. I used the adjective small because I wanted to emphasize that it works even if we only consider the local existence of the continuous function. The difference between continuity at a point and continuity in a neighborhood here is substantial. $\endgroup$ – Yai0Phah Jan 30 '14 at 11:54
3
+50
$\begingroup$

Let $Pol_n$ denote the space of degree $n$ monic polynomials. You have the natural continuous map $R: Pol_n \to Q= {\mathbb C}^n/S_n$ sending each polynomial to its set of (unordered) roots; here $S_n$ is the permutation group on $n$ letters. Let $q: {\mathbb C}^n\to Q$ denote the quotient map. The key observation is that the map $q$ has the path-lifting property (no uniqueness of the lift is assumed, only existence). This is a special case of the path-lifting property for orbifold-coverings, Lemma 4.1.3 here, or Lemma 2 in

M. Armstrong, The fundamental group of the orbit space of a discontinuous group. Proc. Cambridge Philos. Soc. 64 (1968) 299–301.

Note that Armstrong works even in greater degree of generality, namely with proper (but not, in general, free) discrete group actions on locally compact metrizable topological spaces.

Now, we can prove the path-lifting property you are asking for. Take a map $f: {\mathbb R}\to Pol_n$, compose it with $R$. The result is a map $$ g: {\mathbb R}\to Q. $$ Applying the above path-lifting property to $g$ we obtain a lift $$ \tilde g: {\mathbb R}\to {\mathbb C}^n. $$ Let $\tilde g_1$ denote the first component of this lift (the "1st root" of the polynomial $f(t)$). Then the map $$ \tilde f: t\mapsto (f(t), \tilde g_1(t))\in Pol_n \times {\mathbb C} $$ is the lift you want.

Edit: Armstrong's proof depends on Theorem 2 (path lifting property for "open light maps") in

E.E. Floyd, "Some characterizations of interior maps". Ann. of Math. 51 (1950), 571-575.

Armstrong simply observes that quotient maps like $q$ in your case, are "light and open": Open is clear (since it is a quotient map), "light" follows from the fact that point preimages are discrete (finite in your case).

Hence, Floyd's theorem applies. (One needs a tiny compactness argument since Floyd assumes that the domain and the range are compact, but the path-lifting property is a purely local issue, so it works for locally compact spaces.)

Floyd's paper is available (for free) via Jstor, once you open a free account with them.

$\endgroup$
  • $\begingroup$ Thanks for your profound answer. Could you please explicate a more elementary proof of the existence of path lifting for $\mathbb C^n\to Q$ or some elementary notes on the underlying idea of the profound proof? The paper is far beyond my ability, but as I skimmed, the lemma somewhat proclaims the uniqueness. $\endgroup$ – Yai0Phah Feb 6 '14 at 10:14
  • $\begingroup$ @FrankScience: You will have easier time reading Armstrong's paper since the proof there is less technical. If you understand the homotopy-lifting property for ordinary covering maps, you will understand his proof too, since there is not much difference. $\endgroup$ – Moishe Kohan Feb 6 '14 at 10:17
  • $\begingroup$ Well, thanks. I googled but it seems inaccessible to me. I found one in jstor, but when I use the proxy of university e-library, it claims 'not found' by jstor. I'll ask for somebody to obtain one. $\endgroup$ – Yai0Phah Feb 6 '14 at 10:23
  • $\begingroup$ @FrankScience: See the edit: What you really need to read is Floyd's paper, if you also cannot get it via jstor, let me know. $\endgroup$ – Moishe Kohan Feb 6 '14 at 10:31
  • $\begingroup$ Thanks. I went to school and downloaded two papers directly. It seems a bug of the proxy system. I'll read them as soon as possible. $\endgroup$ – Yai0Phah Feb 6 '14 at 15:11
0
$\begingroup$

There are two reasons that $\pi$ doesn't have the path-lifting property.

First, if $a_n$ is allowed to vanish, one of the $n$ complex roots (or more, if more coefficients vanish simultaneously) will go off to infinity. For example, consider the polynomial $P_t(z) = tz + 1$ : when $t \neq 0$, it has a unique root $z = -1/t$, but when $t = 0$, $z$ diverges to infinity. This problem can be solved by either restricting $a_n$ to nonzero values, or extending the range of $z$ to the Riemann sphere $\Bbb C \cup \{ \infty \}$

Secondly, the path may not be unique. This happens when you pass through a polynomial that has a double (or triple or more ...) root. For example, if $P_t(z) = z^2 + t$, if you start at $t= -1$ and $z(-1)= 1$, you have a unique path only up to $t=0$. Then you have two choices : you can have (for $t > 0$) either $z(t) = +\sqrt t i $, either $z(t) = - \sqrt t i$. You can solve this problem by restricting your polynomials to polynomials whose discriminant doesn't vanish.

If you stick with polynomials with nonzero discriminant, essentially nothing bad can happen. You should be able to apply some version of the inverse function theorem to a simply connected neighbourhood of $P$ to show that the roots of $P$ vary continuously in the coefficients and unambiguously on that neighbourhood.

$\endgroup$
  • $\begingroup$ I have reread the wiki page and unfortunately I found that it's not exactly the one I wanted. I only need the existence, not the uniqueness. Now I'll edit my post. $\endgroup$ – Yai0Phah Feb 2 '14 at 16:28
  • 1
    $\begingroup$ And in all my expression of the problem, all polynomials are monic and therefore $a_n=1$. $\endgroup$ – Yai0Phah Feb 2 '14 at 16:30
0
$\begingroup$

Isn't a consequence of the continuity of the roots respect of the coefficients?

http://www.ams.org/journals/proc/1987-100-02/S0002-9939-1987-0884486-8/S0002-9939-1987-0884486-8.pdf

Possible problem: the "space of the roots" isn't ${\Bbb C}$, but a quotient.

$\endgroup$
  • $\begingroup$ That's different. Scrutiny my counterexample in the original post: $z^2-w=0$. Intuitively, when $w$ runs through the $w$-plane, the two roots $z$ run quite continuously, but unfortunately, we cannot extract a continuous root. $\endgroup$ – Yai0Phah Feb 2 '14 at 16:39
  • $\begingroup$ OK, I overlooked the discussion. $\endgroup$ – Martín-Blas Pérez Pinilla Feb 2 '14 at 16:40
  • $\begingroup$ Well, but what you've posted is related to: math.stackexchange.com/a/63206/23875 $\endgroup$ – Yai0Phah Feb 2 '14 at 16:44
  • $\begingroup$ I went into the details of the proof by that paper, and found that it's incomplete (the part that the map $\hat\sigma$ is open). See ams.org/journals/proc/1988-102-04/S0002-9939-1988-0934880-2 for details. $\endgroup$ – Yai0Phah Feb 6 '14 at 6:01
0
$\begingroup$

I think I have an operator-theoretic formulation of your problem:

Given an $n \times n$ matrix-valued function $B(t), \ t\in \mathbb{R}$ whose entries are continuous, complex-valued functions on $\mathbb{R}$, is there a continuous function $\lambda(t), \ t \in \mathbb{R}$ such that $\lambda(t)$ is an eigenvalue of $B(t)$ for all $t \in \mathbb{R}$?

This is equivalent to your analytic question as every characteristic polynomial of such $B(t)$ is of the form of your function $f^{(t)}$, and every polynomial of the RHS of your equation can be interpreted as a characteristic polynomial of a companion matrix. This formulation of the question seems to be answered in an older question: Eigenvalues of matrix with entries that are continuous functions.

Further results in that direction can be found in T. Kato, Perturbation theory of linear operators, Chapter Two, §5.

$\endgroup$
  • 1
    $\begingroup$ The proof of the old question seems wrong. It's abusing the theorem mentioned in that paper. I think I've explained enough in the discussion part of my post. $\endgroup$ – Yai0Phah Feb 6 '14 at 15:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.