Finite field Kakeya and Nikodym sets in three dimensions

NEW

Problem: Reconstructed statement: for a family of line sets $L_q$ in affine space $\mathbb F_q^3$, if $|L_q|=\Omega(q^3)$ and every affine plane contains only $O(q)$ lines of $L_q$ (the standard reading of “no plane contains $\omega(q)$ lines”), then the conjecture claims

$$|P(L_q)|=\left|\bigcup_{\ell\in L_q}\ell\right|\ge (1-o(1))q^3 .$$

This is the literal quoted conjecture, with $P(L)$ the union of the lines.

Result: The conjecture is false, even with at most $q+1$ lines in every plane.

Let $q$ be odd, choose a nonsquare $\nu\in\mathbb F_q^\times$, and put

$$Q(x,y)=x^2-\nu y^2 .$$

For each $(a,b)\in\mathbb F_q^2$ and each projective direction $[u:v]\in\mathbb P^1(\mathbb F_q)$ with $Q(u,v)$ a nonzero square, define the affine line

$$\ell_{a,b,[u:v]} =\{(a+tu,\ b+tv,\ Q(a,b)+2t(au-\nu bv)):t\in\mathbb F_q\}.$$

Let $L_q$ be the set of all these lines.

Since $Q$ is the norm form $N(u+v\sqrt\nu)$ from $\mathbb F_{q^2}$ to $\mathbb F_q$, each nonzero value has $q+1$ representations. Hence exactly $(q+1)/2$ projective directions have square $Q(u,v)$. Also each such line meets the paraboloid $z=Q(x,y)$ only at its base point, because

$$Q(a+tu,b+tv)-\bigl(Q(a,b)+2t(au-\nu bv)\bigr)=t^2Q(u,v),$$

and $Q(u,v)\ne0$. Thus

$$|L_q|=q^2\frac{q+1}{2}=\Omega(q^3).$$

Every affine plane contains at most $q+1$ lines of $L_q$. Indeed, a line of $L_q$ contained in a plane $\Pi$ has its unique tangency point in $\Pi\cap\{z=Q(x,y)\}$. If $\Pi$ is not the tangent plane at that point, at most one such tangent line arises from that point. If $\Pi$ is the tangent plane, anisotropy of $Q$ gives intersection with the paraboloid only at that point. Finally, any plane intersects the paraboloid in at most $q+1$ points: vertical planes give $q$ points, and nonvertical planes reduce after completing the square to $Q(X,Y)=c$, which has $1$ or $q+1$ solutions. Hence the plane condition holds.

Now compute the union. For $X=(x,y,z)$, define

$$D(X)=Q(x,y)-z.$$

On every selected line,

$$D(X)=t^2Q(u,v),$$

so $D(X)$ is either $0$ or a square. Conversely, if $D(X)=r^2\ne0$, then $X$ lies on the selected line with base $(x-r,y)$ and direction $[1:0]$; if $D(X)=0$, then $X$ lies on the paraboloid and hence on selected tangent lines. Therefore

$$P(L_q)=\{(x,y,z):Q(x,y)-z\in \{0\}\cup(\mathbb F_q^\times)^2\}.$$

For each $(x,y)$, exactly $(q+1)/2$ values of $z$ occur, so

$$|P(L_q)|=q^2\frac{q+1}{2} =\left(\frac12+o(1)\right)q^3.$$

This contradicts the claimed lower bound $(1-o(1))q^3$.

Verification audit: the construction works for infinitely many $q$ tending to infinity, satisfies $|L_q|=\Omega(q^3)$, satisfies the plane hypothesis with the explicit bound $q+1=O(q)$, and has union density tending to $1/2$, not $1$.

Citation: No external resolution is invoked; the counterexample above is elementary, using only the standard finite-field norm count.