Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

What can be said about rational self-maps of $\mathbb P^1$ for which all critical points are also fixed points ?

If all but one of the fixed points are critical, there is a characterization in http://arxiv.org/abs/math/0411604v1 ( see Corollary 1 and the discussion just after the statement ).

Still assuming that all critical points are fixed: Is it possible to bound the degree of the rational map if all but two of the fixed points are critical ?

I think that the answer is probably no, but I would really love to hear the contrary.


Motivation. The question is motivated by a rather specific problem I like to think about from time to time. It concerns the classification of some special arrangements of lines on the projective plane. More specifically, I would like to classify arrangements of $3d$ lines(or rather hyperplanes through the origin of $\mathbb C^3$) invariant by degree $d$ homogeneous polynomial vector fields on $\mathbb C^3$. Given one arrangement like that one can produce a degree $d$ rational map having all its critical points fixed.


Update. (08/28/2013) The paper On the classification of critically fixed rational maps by Cordwell, Gilbertson, Nuechterlein, Pilgrim, and Pinella classifies rational maps for which all critical points are fixed. In particular, for every degree $d\ge 3$ there exists a rational such that all but two of the fixed points are critical.

share|improve this question
1  
Do you want almost all the critical points to be fixed, or almost all the fixed points to be critical? In your title and first question you ask for the first, but in the arxiv link and the third paragraph you ask for the latter. –  David Speyer Nov 9 '09 at 2:11
    
I am interested in maps having all its critical points fixed. In the Arxiv link the paragraph following the statement of the Corollary discuss exactly this. In the third paragraph I am implicitly assuming that all the critical points are fixed. Thanks. –  jvp Nov 9 '09 at 2:37

5 Answers 5

up vote 8 down vote accepted

Consider $$z \mapsto \frac{(n-2) z^n + n z}{n z^{n-1} + (n-2)}.$$ This has $n+1$ fixed points, at $0$, $\infty$, and the $(n-1)$-st roots of $-1$. The only critical points are the roots of $-1$, each of which is ramified of index $3$. So this is a map with all critical points fixed, and all fixed points but two critical.

I am tempted to leave it at that. But, being a nice person, I will explain how I found this. Moreover, I will show that this is (up to conjugacy), the only degree $n$ map with $n+1$ distinct fixed points, two of which are not critical and the rest of which are critical with multiplicity $2$ (ramification index $3$.)

We can take the two noncritical fixed points to be $0$ and $\infty$. So our map is of the form $$z \mapsto z - z p/q,$$ where $p$ and $q$ are of degree $n-1$ and relatively prime. The $n-1$ fixed points other than $0$ and $\infty$ are the roots of $p$.

The derivative of this map is $$\frac{q^2 - zqp' + zpq' - pq}{q^2}.$$ The condition that the fixed points other than $0$ and $\infty$ be critical means $p$ divides the numerator. So $p$ divides $q (q-zp')$ and, as $p$ and $q$ are relatively prime, we conclude that $q - z p' = kp$. Checking degrees, $k$ has degree $0$ and is thus a constant, to be determined later.

Now, we want to impose the stronger condition that every zero of $p$ be doubly a critical point, so the numerator is $\ell p^2$ for some constant $\ell$. Plugging in $q = kp + z p'$, and simplifying $$\ell p^2 = p \left( k(k-1) p + 2 k z p' + z^2 p'' \right).$$ Cancelling $p$ from both sides, $$\ell p = k(k-1) p + 2 k z p' + z^2 p''.$$ Plugging in $z=0$, and noting that $p(0) \neq 0$, we get $\ell = k(k-1)$. So $$2 kz p' + z^2 p'' =0.$$ The solution to this differential equation is $p = C z^{1-2k} + D$. But we know that $z$ has degree $n-1$, so $1-2k=n-1$ and $k = -(n-2)/2$.

Taking the simplest choices $C=D=1$ and plugging back in gives the above solution. All other solutions are related to this one by rescaling the variable $z$.

share|improve this answer

Theorem: for each n, there are up to Mobius conjugacy a finite, nonzero, number of rational maps of degree n such that all of the critical points are fixed.

Proof: Nonempty is clear: z^n. For finiteness: the degree bounds the number of critical points. Given a bound on the degree and the size of the postcritical set, with the usual set of flexible Lattes exceptions (which have non-fixed critical points), the set of such rational maps up to Mobius conjugacy is finite; this is a consequence of Thurston's Rigidity Theorem [Douady, Adrien; Hubbard, John H. A proof of Thurston's topological characterization of rational functions. Acta Math. 171 (1993), no. 2, 263–297.]; cf. also section 3 of: [E. Brezin, R. Byrne, J. Levy, K. Pilgrim, and K. Plummer). Conf. Geom. & Dynam. Sys. (electronic) 4(2000), 35-74. ]

share|improve this answer

I realize that the question is about rational functions whose critical points are all fixed points, but it might be useful to note that polynomials with this property are called conservative polynomials. They have been studied by Tischler and Pakovich. In particular, Tischler proved that there are $\binom{2d-2}{d-1}$ normalized conservative polynomials of degree $d$, where a polynomial $C$ is normalized if it is monic and satisfies $C(0)=0$. An example of a normalized conservative polynomial is $x^d + \frac{d}{d-1}x$. Here are the references for Tischler's and Pakovich's articles.

  • David Tischler, Critical points and values of complex polynomials, Journal of Complexity 5 (1989), 438-456, MR1028906.

  • Fedor Pakovich, Conservative polynomials and yet another action of ${\rm Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ on plane trees, Journal de Theorie des Nombres de Bordeaux 20 (2008), 205-218, MR2434164.

share|improve this answer
    
Thanks for the references. I will take a look. –  jvp Jul 9 '11 at 2:01

I've been thinking about this a little. I would guess that, for any sufficiently large $n$, there is a finite, nonzero, number of rational maps of degree $n$ such that all of the critical points are fixed. Here is my heuristic argument.

Fix a partition of $2n-2$ into $n+1$ parts: $2n-2 = \lambda\_1 + \lambda\_2 + \cdots \lambda_{n+1}$. For any $n+1$ points $z\_1$, $z\_2$, ..., $z\_{n+1}$ on $\mathbb{CP}^1$, there are finitely many degree $n$ covers of $\mathbb{CP}^1$ which are ramified over the $z\_i$, with the ramified point over $z\_i$ being ramified of index $\lambda\_i+1$, and no other ramification. (You just need to choose which $\lambda\_i$ sheets will be permuted by the monodromy around $z\_i$.) Some of these covers will be disconnected, but all the connected ones will have genus $0$ by the Riemmann-Hurwitz formula. FIRST NONRIGOROUS STEP: I expect that, for most choices of the $\lambda\_i$, there will be a nonzero number of connected covers. Let D be the number of these connected covers.

Now, in each of these connected covers, the covering curve has genus $0$, and is thus isomorphic to $\mathbb{CP}^1$. Let $w\_i$ be the ramified preimage of $z\_i$. The $n+1$ points $w\_i$ give us a point in $M\_{0,n+1}$, and the points $z\_i$ give another point of $M_{0,n+1}$. Plotting the pairs $((w\_1, w\_2, \ldots, w\_n), (z\_1, z\_2, \ldots, z\_n))$ gives us a subvariety of $M\_{0, n+1} \times M\_{0,n+1}$ of dimension equal to that of $M\_{0,n+1}$; the projection onto the second factor is generically $D$ to $1$. Let's call this subvariety $X$. You goal is to understand the intersection of $X$ with the diagonal.

Now, here is the VERY NONRIGOROUS STEP. $X$ has dimension $(n+1)-3=n-2$. So does the diagonal. Our ambient space, $M\_{0, n+1} \times M\_{0,n+1}$, has dimension $2n-4$. In the absence of any other information, the intersection is probably finite and nonempty. :-)

I expect we may be able to extend all of these ideas to work with subvarieties of the compactification $\overline{M}\_{0,n+1}$. That would be good because then we could hope to compute the cohomology class of $X$, and show that it cannot miss the diagonal.

Filling in the gaps here sounds like a really nice problem. Unfortunately, I have too many nice problems, but I wish you luck.

share|improve this answer
    
Hmmm. There is something wrong with my heuristic above. The same argument suggests that there should be a degree $n$ map from $P^1$ to itself with simple branching only, and all $2n-2$ branched points fixed. But, of course, a degree $n$ map fixes at most $n+1$ points. Presumably, understanding the solution would mean understanding the obstructions in the paper of Douady and Hubbard referenced by Agol. –  David Speyer Nov 9 '09 at 16:15

David Speyer's answer is right on the money. Douady and Hubbard proved that a map of the sphere to itself whose postcritical set is finite has a unique "uniformization" as a rational map (up to conjugation by Mobius transformations). For a given degree, then, there are finitely many rational maps with all critical points fixed points, since these are postcritically finite.

I haven't thought about the second part of the question about whether one can bound the degree if there are only two non-critical fixed points. It might be possible to determine this from the branching data and the Lefschetz fixed-point formula.

share|improve this answer
    
Thanksfor the reference. –  jvp Nov 9 '09 at 14:58

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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