Return to Question

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.

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.

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

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

satisfies $\operatorname{im}(\hat f)= U_V$ and defines a bijection between $U_X\subseteq X$ and $U_V\subseteq V$.

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.

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.

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 a decomposition $\Bbb R^n=A\oplus B$, an open subset $U_A\subseteq A$ and a rational function $f:U_A\to B$, so that

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

satisfies $\operatorname{im}(\hat f)= U_V$ and defines a bijection between $U_A\subseteq A$ and $U_V\subseteq V$.

Question: Given $V$ and $U_V\subseteq V$ as above, is the decomposition $A\oplus B$ 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 $f:U_A\to B$ being a rational function I mean that I identify $A\simeq \Bbb R^m$ and $B\simeq \Bbb R^k$ 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.

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

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

satisfies $\operatorname{im}(\hat f)= U_V$ and defines a bijection between $U_X\subseteq X$ and $U_V\subseteq V$.

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.