4 added 8 characters in body

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 $n-3$. (n+1)-3=n-2$. So does the diagonal. Our ambient space,$M_{0, n+1} \times M_{0,n+1}$, has dimension$2n-6$. 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.

3 added 4 characters in body

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 $z_1$, $z_2$, ..., $z_n$ 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$n-3$. So does the diagonal. Our ambient space,$M_{0, n+1} \times M_{0,n+1}$, has dimension$2n-6$. 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. 2 added 1 characters in body 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$z_1$,$z_2$, ...,$z_n$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$n-3$. So does the diagonal. Our ambient space,$M_{0, n+1} \times M_{0,n+1}$, has dimension$2n-6$. 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 man many nice problems, but I wish you luck.

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