Birational maps

Dynamical degree

The most naive definition of a rational map $f$ from ${\Bbb C}^n$ to itself is $f=(f_1,\dots,f_n)$, where each coordinate $f_j=p_j/q_j$ is a quotient of two polynomials, i.e., a rational function. A rational map defines a holomorphic function outside the zeros of the denominators $q_j$. Thus the iterates $f^n:=f\circ\cdots\circ f$ are well defined on a dense, open set. We say that $f$ is birational if it has a rational inverse, which is to say that there is a rational map $g$ of ${\Bbb C}^n$ such that $f\circ g$ and $g\circ f$ are the identity on a dense, open set.

Let us consider the imbedding of ${\Bbb C}^n\ni z=(z_1,\dots,z_n)\mapsto [1:z_1:\cdots:z_n]\in{\Bbb P}^n$. We may then write $f=[f_0:\cdots:f_n]:{\Bbb P}^n\to{\Bbb P}^n$, where the coordinates $f_j$ are homogeneous polynomials, all of the same degree $d$, and without common factor. We say, then, that $d$ is the degree of $f$. If we wish to consider $f$ as a dynamical system, we wish to study the behavior of the iterates $f^n$ as $n\to\infty$. If $\varphi$ is birational, then the iterates $\{f^n\}$ are conjugate to the iterates of $\{g^n\}$, where $g:=\varphi^{-1}\circ f\circ\varphi$.

The most naive question is whether $f$ is nontrivial or not. A birational map $L$ is linear if and only if it has degree 1. However, if $\varphi$ is birational, $\varphi^{-1}\circ L\circ\varphi$ is likely not to be linear, which reflects the fact that degree is not a birational invariant. However, the dynamical degree, defined by $$\text{ddeg}(f) := \lim_{n\to\infty}(\text{deg}(f^n))^{1/n}$$ is birationally invariant. The dynamical degree is a measure of "dynamical nontriviality" of $f$. The dynamical degree has been discussed in many places and we refer to this expository article for more information and references.

Mappings related to matrix inversion

We consider the space ${\cal M}_q$ of complex $q\times q$ matrices. For $x=(x_{i,j})\in{\cal M}_q$, we let $I(x) = (x_{i,j})$ denote matrix inversion. Thus $I$ is a birational map of ${\cal M}_q\cong {\Bbb C}^{q^2}$ and in fact is an involution, $I^2=\text{identity}$. Another involution is given by Hadamard involution $J(x)=(X_{i,j}^{-1})$, which takes the inverse of the individual entries in the matrix. This is also known as the Cremona involution. The involutions $I$ and $J$ arise as natural symmetries in Lattice Statistical Mechanics, and a question that had arisen was the behavior of the composition $K:=I\circ J$ on the projectivized space ${\Bbb P}({\cal M}_q)$. In particular, the projectivized subspaces ${\Bbb P}({\cal S}_q)$ (and resp. ${\Bbb P}({\cal C}_q)$) of symmetric (resp. cyclic) matrices are also of interest. Based largely on numerical work, it was conjectured by Angles d'Auriac, Maillard and Viallet that $$\text{ddeg}(K|_{{\Bbb P}({\cal M}_q)})= \text{ddeg}(K|_{{\Bbb P}({\cal S}_q)}) \ \ \text{ and }\ \ \text{ddeg}(K|_{{\Bbb P}({\cal M}_q)})= \text{ddeg}(K|_{{\Bbb P}({\cal C}_q)})$$ It is interesting for the mathematician to try to imagine how one could do numerical experiments to estimate any of these quantities. And it is doubly surprising because their numerical results were so accurate. For instance, if $q=10$ (not yet a large number), then the projectivized space of $q\times q$ matrices has dimension $q^2-1=99$; the degree of $I$ is $q-1=9$, and the degree of $J$ is the same as the dimension of the space, again $=99$. So a direct calculation of the degree of $K^n$ is already difficult if $n>1$.

The approach that these authors took was to start with a matrix $m=(m_{i,j})$ with coprime integer entries. They then computed the growth rate of the denominators in $K^n(m)$. In other words, they are using the fact that $K$ is a rational map with integer coefficients, and they are computing the exponential rate of growth of height. Subsequently, it has been shown by Kawaguchi and Silverman that the height is in fact bounded above by the dynamical degree. The reverse inequality (for a Zariski open set of points) is still open.

The left hand equality conjectured above was proved by Tuyen Trung Truong and the right hand equality was proved by Bedford and Truong

There are other subspaces of the space of matrices that are of interest. One of them is the space ${\cal CS}_q:={\Bbb P}({\cal S}_q\cap {\cal C}_q)$. It was observed by Angl\`es d'Auriac, Maillard and Viallet that $\text{ddeg}(K|_{\cal CS})$ must be smaller than the general number. Then Bedford and Kim found a formula for the dynamical degree $\text{ddeg}(K|_{\cal CS})$ in terms of $q$. It turns out that the dependence on $q$ is rather complicated and splits into 3 quite different cases: $(i)$ $q$ is odd, $(ii)$ $q$ is 2 times odd, and $(iii)$ $q$ is divisible by 4.

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