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1

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?

flag	asked Jan 4 2012 at 23:38 Vincent 101●2

	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? – David Roberts Jan 4 2012 at 23:48
	Also, you can take $S \subset \mathbb{P}^{n-1}$, and translate the problem to projective space. – David Roberts Jan 4 2012 at 23:49
	In 2), should "For every affine" be replaced with "Every affine"? $\;$ – Ricky Demer Jan 5 2012 at 1:47

Robert Israel · Answer 1 · 2012-01-05 00:59:01Z UTC

5

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

$alt text$

edited Jan 5 2012 at 0:59

answered Jan 5 2012 at 0:08

Robert Israel
10.8k●8●22

Right. Or more generally $x^n+y^n+z^n=0$ for any odd $n>1$. – Noam D. Elkies Jan 5 2012 at 5:42

Yoav Kallus · Answer 2 · 2012-01-05 00:10:25Z UTC

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 $z=f_3(x,y)$.

Naive tomography question

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