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Let $V\subset \Bbb R^n$ be an irreducible affine variety of degree $\ge 2$ and $U_V\subseteq V$ a (Euclidean) open subset. Suppose that $U_V$ is the graph of a rational function, that is, there is an orthogonal decomposition $\Bbb R^n=X\oplus Y$, an open subset $U_X\subseteq X$ and a rational function $f:U_X\to Y$, so that

$$\hat f: U_X\to \Bbb R^n, \;x\mapsto x+f(x)$$

defines a bijection between $U_X\subseteq X$ and $U_V\subseteq V$. In particular, $\operatorname{im}\hat f= U_V$.

Question: Given $V$ and $U_V\subseteq V$ as above, are the decomposition $X\oplus Y$ and map $f$ unique?

Clearly, if $V$ is linear then this is not the case. What would be the simplest example where this fails in higher degree? Also, for clarity, by $f:U_X\to Y$ being a rational function I mean that I identify $X\simeq \Bbb R^m$ and $Y\simeq \Bbb R^k$ and then have $f=(f_1,...,f_k)$ with $f_i=P_i/Q_i$ for polynomials $P_i,Q_i\in\Bbb R[X_1,...,X_m]$. Where I am coming from everything is real valued, but I understand that it might be more convenient to give an answer in $\Bbb C$. If you do so, please indicate whether the real case is equivalent.

Updates

  • I added the condition that the decomposition $X\oplus Y$ should be orthogonal after Jason Starr noted in the comments that otherwise one could always "linearly augment" the decomposition and function $f$ to yield another solution.

  • As noted by Sam Hopkins in the comments, if $f$ has a rational inverse (e.g. if it is a fractional linear functions) then swapping $X$ and $Y$ yields another solution. So the updated question shoudl be whether this is essentially the only thing that can happen.

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    $\begingroup$ The graph of $y=x^2$ is also the graph of $z=x^2 + ax$ for the coordinate $z=y+ax$. $\endgroup$
    – Jason Starr
    Commented Oct 2 at 12:27
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    $\begingroup$ What if $f$ is invertible, like a fractional linear transformation? Then can't we swap the role of $X$ and $Y$? $\endgroup$
    – Sam Hopkins
    Commented Oct 2 at 14:00
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    $\begingroup$ @M.Winter: There are many invertible rational maps. In fact, there are even many invertible polynomial maps (see en.wikipedia.org/wiki/Jacobian_conjecture). $\endgroup$
    – Andy Putman
    Commented Oct 2 at 14:45
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    $\begingroup$ Also the set of $x,y,z$ such that $\frac{1}{x}+ \frac{1}{y} + \frac{1}{z}=0$ can be expressed in three ways as the graph of a function since each of $x,y,z$ may be expressed as a rational function in the other two. $\endgroup$
    – Will Sawin
    Commented Oct 2 at 14:51
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    $\begingroup$ @WillSawin Even though I can think of ways to modify my question to avoid such examples, I also think your example was sufficiently unexpected for me that I would accept it as an answer. $\endgroup$
    – M. Winter
    Commented Oct 2 at 18:29

2 Answers 2

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Let $W$ be a vector space and let $f_1,f_2,f_3\colon W\to W$ be birational maps. Then $\{ x,y,z \in W^3 \mid f_1(x)+f_2(y)+f_3(z)=0\}$ can be expressed in three ways as a graph of a function since each of the $x,y,z$ may be expressed as a rational function of the other two.

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I found that already a quadratic surface can be a graph in uncountably many different ways. Consider the graph of the function $z=xy$. By the substitution

\begin{align} x &= a\bar x - b\bar z, \\ y&=\bar y, \\ z&=b\bar x + a\bar z \end{align}

for $a,b\in\Bbb R$ (not both zero), we obtain a new graph representation of the same variety

$$ b\bar x+a\bar z = (a\bar x-b\bar z)y \quad\implies\quad \bar z = \frac{a \bar y-b}{b\bar y+a}\bar x. $$

The first representation corresponds to the decomposition

$$\Bbb R^3=X\oplus Y:=\langle (1,0,0),(0,1,0)\rangle\oplus\langle(0,0,1)\rangle,$$

and if I am not mistaken, the second one corresponds to

$$\Bbb R^3=\bar X\oplus \bar Y:=\langle(a,0,-b),(0,1,0)\rangle \oplus \langle(b,0,a)\rangle.$$

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  • $\begingroup$ Good point. The way I would summarize this is that if we fix the $y$ variable it's linear, and as already mentioned changing the coordinates in something linear gives something linear which is therefore a graph, so in any system of coordinates in which the $y$ variable is included this is still a graph. $\endgroup$
    – Will Sawin
    Commented Oct 15 at 14:54

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