Return to Answer

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 $n-3$. 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.

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.

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 many nice problems, but I wish you luck.

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 $n-3$. 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.

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 nice problems, but I wish you luck.

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 many nice problems, but I wish you luck.