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Given a vectorspace $V$ (over a field $F$) with a specified basis $b_1, \ldots b_n$ and a set $S \subset V$ with two properties:

1) $S$ is a union of lines through the origin (so for all $s \in S$ we have that $Fs \subset S$)

2) For every affine line line parallel to one of the basis vectors intersects $S$ in exactly one point. (So for every $i \in \{1, \ldots n\}$ and every $v \in V$ we have $|(v + Fb_i) \cap S| = 1$.)

Is $S$ necessarily a codimension 1 linear subspace or are there other examples?

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  • $\begingroup$ Well, we can assume that $V\simeq F^n$, and so the affine lines parallel to the basis vectors are given by the system of $n$ equations $x_1 = c_1,\ldots, x_i = x_i, \ldots, x_n = c_n$, for some $i\in \{1,\ldots,n\}$. What about some sort of ruled surface? $\endgroup$
    – David Roberts ♦
    Commented Jan 4, 2012 at 23:48
  • $\begingroup$ Also, you can take $S \subset \mathbb{P}^{n-1}$, and translate the problem to projective space. $\endgroup$
    – David Roberts ♦
    Commented Jan 4, 2012 at 23:49
  • $\begingroup$ In 2), should "For every affine" be replaced with "Every affine"? $\;$ $\endgroup$
    – user5810
    Commented Jan 5, 2012 at 1:47

2 Answers 2

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In ${\mathbb R}^3$ with the standard basis try the surface $ x^3 + y^3 + z^3 = 0$.

alt text

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  • $\begingroup$ Right. Or more generally $x^n+y^n+z^n=0$ for any odd $n>1$. $\endgroup$
    – Noam D. Elkies
    Commented Jan 5, 2012 at 5:42
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I think this is a counterexample in $\mathbb{R}^3$:

Start with the plane with normal direction $(1,1,1)$. Tilt each radial ray of that plane by an angle $\alpha = \epsilon \sin 3\phi$ towards $(1,1,1)$ (or away if negative), where $\phi$ is the angle the ray makes with the ray through, say, $(1,-1,0)$. For small $\epsilon$, the resulting surface is still a graph of a single valued function $x=f_1(y,z)$ or $y=f_2(z,x)$ or $z=f_3(x,y)$.

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