Studying a problem in conformal geometry, I am facing to the following interpolation problem.

Let $P$ and $Q$ two coprime polynomials. Then let $A$ and $B$ two coprime polynomials such that $$\frac{A}{B} = \left( \frac{P}{Q}\right)'.$$

We denote by $(z_i)_{0\leq i\leq n}$ some distinct complex numbers with $n=deg(A)$ and which are not some roots of $Q$. Then we consider some perturbations of the $\left( \frac{P}{Q}\right)'(z_i)$ denoted $\tilde{a}_i$ and we try find a perturbation of $P$ and $Q$ denoted $\tilde{P}$ and $\tilde{Q}$ such that

$$\left( \frac{\tilde{P}}{\tilde{Q}}\right)'(z_i)=\tilde{a}_i$$

The problem is easy to solve when $Q\equiv 1$, it is classical interpolation on $P'$ and then we integrate. But When $Q$ is not trivial we can't just perturb $A$, finding $\tilde{A}$ such that

$$\tilde{A}(z_i)= \tilde{a_i} B(z_i)$$

because when integrating $\frac{\tilde{A}}{B}$ there is no reason to get a rational fraction.

I have tried to perturb successively $P$ and $Q$(or $A$ and $B$) and to get some estimate to make a fixed point but without success? did any one get some references or ideas on that topic?

The Geometric point of view is the following, you consider a (finite) multiple conformal covering of $S^2$, then identifying $S^2$ with $\hat{\mathbb{C}}$ it is parametrized by a rational fraction. Then you can define the vector field $ \left( \frac{P}{Q}\right)'$. And the question is, if you perturb this vector field at $n+1$ points which are not poles, can you reach this perturbation by perturbing $P$ and $Q$?

Edit: The first question is of course the existence of $\tilde{P}$ and $\tilde{Q}$? But I would like that $\tilde{P}$ and $\tilde{Q}$ will be closed to $P$ and $Q$ with respect to $\sum \vert a_i -\tilde{a}_i\vert$. Closed means for me, with the same degree and closed in $\mathbb{C}_d[X]$ with $d$ the degree considered.

  • 1
    $\begingroup$ The argument I posted was incorrect. The interpolation problem does not always have a solution. For exaple if more than $d$ values are equal. $\endgroup$ Jul 16, 2013 at 22:58
  • $\begingroup$ Of course my first idea was to linearized: $(P,Q)-> (P'Q-Q'P-a_iQ^2)(z_i)$, but i didn't succeed in proving that the linearized operator is invertible. $\endgroup$
    – Paul
    Jul 17, 2013 at 2:11
  • 2
    $\begingroup$ The problem is that it is sometimes invertible sometimes not, and it is hard to tell. An special case of this problem is to interpolate all zero values. That is to find a rational function with prescribed zeros has been solved (existence proved) but almost nothing is known about the Jacobian and small perturbation, except special cases. $\endgroup$ Jul 17, 2013 at 7:14

1 Answer 1


This is an answer to a slightly different question.

Let $\deg P = p$ and $\deg Q = q$. The polynomials $P$ and $Q$ have $(p+1)+(q+1) = p+q+2$ coefficients between them. However, rescaling $P$ and $Q$ by the same number does not effect the rational function, so there are actually $p+q+1$ parameters available. In addition, if $\deg P \geq \deg Q$ and $a$ is a constant then $(P+a Q)/Q$ is a rational function with numerator and denominator of the same degree, and $(P/Q)' = ((P+aQ)/Q)'$. So there are actually $p+q$ parameters in this case. In short:

We should expect to be able to interpolate $p+q+1$ points if $p<q$, or $p+q$ if $p \geq q$.

The original question asked us to interpolate at $\deg A$ points. We have $\deg A = p+q-1$ if $p \neq q$, or $p+q-2$ if $p=q$ (because the leading terms of $P'Q$ and $P Q'$ cancel.) So the original question is asking for less interpolation than we would expect to be able to achieve. I will show that we cannot always achieve as much interpolation as we would naively expect; I do not know if we can always achieve the smaller amount you ask for.

It is not always possible to perturb an interpolation. The simplest example is to start with $p(x)/q(x) = 1/(x^2+1)$ and $(w_1, w_2, w_3) = (-1, 0, 1)$. So $(a_1, a_2, a_3) = (1/2, 0, -1/2)$. I claim that we cannot perturb to $(1/2, 0, -1/2+\epsilon)$ for $\epsilon \neq 0$, with $\deg p=0$ and $\deg q = 2$. Proof: Such a perturbation is of the form $\phi(x)=1/q(x)$ for some quadratic polynomial $q$. Then $\phi'(x) = -q'(x)/q(x)^2$ and we have imposed that the derivative vanish at $0$. So $q'(x)=0$ and $q(x) = a x^2 + b$ for some $a$ and $b$. This shows that $\phi(x)$ is an even function, so $\phi'(x)$ is odd, and we have $\phi'(1) = - \phi'(-1)$, making it impossible to interpolate $(-1, 0, 1+\epsilon)$.

If you want a counterexample with $p=q$, take $p=q=2$ with $(w_1, w_2, w_3, w_4) = (1, -1, 1/2, 2)$. Try to perturb $(x^2+1)/(x^2+x+1)$ while keeping $a_1=a_2=0$. The condition $\phi'(\pm 1) =0$ implies that $\phi(x) = \phi(1/x)$ so $\phi'(x) = (-1/x^2) \phi'(1/x)$ and we get $\phi'(2) = (-1/4) \phi'(1/2)$, and we can use this to show the impossibility of perturbing to $(0,0,a_3, a_4)$ for generic $(a_3, a_4)$.

I don't have any counterexamples where the $a_i$ are all nonzero. I also don't have any counter-examples which involve interpolation at only $\deg A$ points.

  • $\begingroup$ I understand your counterexample. But here $(p/q)' =a/b$ with deg(a)=1 and you try to interpol with 3 points while i only ask with 2. $\endgroup$
    – Paul
    Jul 17, 2013 at 4:54
  • $\begingroup$ Right. Added a sentence (in bold) to clarify. $\endgroup$ Jul 17, 2013 at 10:44

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.