Skip to main content
2 of 4
added 16 characters in body
M. Winter
  • 13.6k
  • 3
  • 28
  • 70

Can a variety be the graph of a function in more than one way?

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.

M. Winter
  • 13.6k
  • 3
  • 28
  • 70