Blaschke’s characterization of ellipsoids found in Kreis und Kugel (1916) page 158 states:
Theorem: If every shadow boundary of a closed, bounded convex body $K \subset \mathbb{R}^3$ is planar, then $K$ is an ellipsoid.
A proof of the compact case (detailed below) relies on the compactness of the body to force the symmetry group to be conjugate to $SO(2)$.
My Question: Does this characterization extend to non-compact convex body? Specifically, if a non-compact strictly convex body $S \subset \mathbb{R}^3$ has a planar shadow boundary for every direction in which a shadow is defined, must $\partial S$ be an ellipsoid, an elliptic paraboloid or one sheet of a two-sheeted hyperboloid? The proof below breaks down without compactness because the symmetry group is no longer forced to be compact (it could admit parabolic shears or hyperbolic stretches).
Motivation: The Polar Plane Property of Quadrics
The forward implication holds for all non-degenerate quadrics, regardless of compactness. In projective geometry, the set of points on a quadric surface $Q$ where the tangent plane passes through a given point $L$ (the light source) lies on the polar plane of $L$ with respect to $Q$.
When the light source moves to infinity (parallel illumination), the "shadow boundary" is simply the intersection of the quadric with the polar plane of that point at infinity. Since the polar of a point with respect to a quadric is always a plane, every shadow boundary of a quadric (whether an ellipsoid, paraboloid, or hyperboloid) is planar.
Since this algebraic property is independent of compactness, it is natural to expect the converse characterization to hold for non-compact convex surfaces as well.
Here is a proof for the compact case:
1. Central Symmetry and Homothety of Cross-Sections
Let $G$ be an arbitrary plane. Let $p$ and $q$ be the two points on the boundary of $K$ where the tangent planes are parallel to $G$.
Affine Normalization: We apply an affine transformation to the space that maps the chord connecting $p$ and $q$ to a line perpendicular to $G$, while mapping the tangent planes at $p$ and $q$ to planes parallel to $G$.
Consider the cross-sections of the transformed body parallel to $G$, denoted as $K_t$ (where $t$ parametrizes the position along the $pq$ axis). Let $o_t$ be the unique intersection point of the convex slice $K_t$ with the chord $pq$.
The Shadow Geometry:
Assume the light rays are parallel to $G$. Since the tangent planes at $p$ and $q$ are parallel to $G$, the light rays graze $p$ and $q$, placing them on the shadow boundary.
By hypothesis, the shadow boundary is a planar curve. Since this plane contains the chord $pq$ (which is perpendicular to $G$), the shadow plane must be perpendicular to $G$.
Parallel Tangents Condition:
The intersection of this shadow plane (perpendicular to $G$) with any slice $K_t$ (parallel to $G$) is a line segment passing through $o_t$. The endpoints of this segment are the "grazing points" for that specific light direction.
Since the light rays are parallel to the slice plane, the tangents to the boundary curve $\partial K_t$ at these two endpoints must be parallel to the light rays, and thus parallel to each other.
Since the light direction parallel to $G$ can be arbitrary, we deduce:
For every line passing through $o_t$, the tangents to $\partial K_t$ at the two intersection points are parallel.
Proof of Central Symmetry:
Let the boundary curve $\partial K_t$ be parameterized in polar coordinates centered at $o_t$ by $r(\theta)$. Let $\tau(\theta)$ be the angle of the tangent line.
The relationship between the radial slope and the tangent angle is:
$$\frac{r'(\theta)}{r(\theta)} = \cot(\tau(\theta) - \theta)$$
Consider a chord through $o_t$ at angle $\theta$, intersecting the boundary at $P_1(r(\theta), \theta)$ and $P_2(r(\theta+\pi), \theta+\pi)$.
The condition of parallel tangents implies $\tau(\theta) = \tau(\theta+\pi)$ (modulo $\pi$).
Substituting this into the differential equation:
$$\frac{r'(\theta)}{r(\theta)} = \cot(\tau(\theta) - \theta) \quad \text{and} \quad \frac{r'(\theta+\pi)}{r(\theta+\pi)} = \cot(\tau(\theta+\pi) - (\theta+\pi))$$
Using $\tau(\theta) = \tau(\theta+\pi)$ and the periodicity of cotangent ($\cot(\alpha - \pi) = \cot(\alpha)$):
$$\frac{r'(\theta)}{r(\theta)} = \frac{r'(\theta+\pi)}{r(\theta+\pi)}$$
Integrating both sides yields $\ln r(\theta) = \ln r(\theta+\pi) + C \implies r(\theta+\pi) = k \cdot r(\theta)$.
Applying this twice ($r(\theta+2\pi) = k^2 r(\theta)$) implies $k^2=1$. Since $r>0$, $k=1$.
Therefore, $r(\theta+\pi) = r(\theta)$, proving that every cross-section $K_t$ is centrally symmetric about the chord $pq$.
Proof of Homothety:
Since $r(\theta+\pi) = r(\theta)$, the shape of $\partial K_t$ is strictly determined by the mapping $\theta \mapsto \tau(\theta)$. Since the shadow condition enforces that the "grazing points" align on lines perpendicular to $G$ for every direction $\theta$, this mapping is identical for all slices.
Two centrally symmetric convex curves with the same tangent-normal mapping are identical up to homothety. Since they share the center axis $pq$, the cross-sections $K_t$ are homothetic.
2. The shadow curve has many affine symmetries
After we undo the affine normalization in Step 1, the body (and thus the shadow curve $\mathfrak{C}$) has affine-reflectional symmetry with respect to the axis $pq$.
Since the choice of the plane $G$ was arbitrary, we could have oriented the body relative to any pair of points with parallel tangent planes. Thus
The shadow curve $\mathfrak{C}$ is affine-reflectional symmetric with respect to every diameter (the chord connecting any two points with parallel tangent lines).
3. Constructing the Symmetry Group
We define $\Phi$ as the affine transformation mapping $\mathfrak{C}$ to itself. We now construct a dense set of rotational symmetries by taking square roots of these transformations via continuity.
Setup: Fix a diameter $L_0$ and let $P_0$ be an intersection point of the curve $\mathfrak{C}$ with $L_0$. Let $\Phi_n$ be a known affine-reflectional symmetry from Step 2. Let $P_{target} = \Phi_n(P_0)$.
Variable Construction: Consider a variable diameter $L$ rotating continuously through the center. We define a test transformation dependent on $L$:
$$T(L) = (R_L \circ R_{L_0})^2$$
Let $P(L) = T(L)(P_0)$ be the image of our fixed point under this transformation.
Applying IVT to the Curve:
Since $L_0$ and $L$ are both valid symmetry axes (by Step 2), the transformation $T(L)$ maps $\mathfrak{C}$ to itself. Therefore, the point $P(L)$ is constrained to move along the perimeter of the curve.
When $L = L_0$, then $P(L_0) = P_0$.
As $L$ rotates, $P(L)$ moves continuously along the boundary.
By the Intermediate Value Theorem applied to the position on the curve, there exists a specific diameter $L_{mid}$ such that $P(L_{mid}) = P_{target}$.
Since an orientation-preserving linear symmetry of a closed curve $\mathfrak{C}$ is uniquely determined by the image of a single boundary point, the condition $T(L_{mid})(P_0) = \Phi_n(P_0)$ implies that the operators are identical:
$$(R_{L_{mid}} \circ R_{L_0})^2 = \Phi_n$$
We define the new symmetry $\Phi_{n+1} = R_{L_{mid}} \circ R_{L_0}$, which is a square root of $\Phi_n$.
Closure to a Continuous Group: Recursive application of this procedure generates a sequence of symmetries $\{\Phi_n\}_{n=0}^\infty$ where $\Phi_n=\Phi_{n+1}^2$. Since the dyadic rationals are dense in $\mathbb{R}$, the closure of this generated group is a continuous 1-parameter subgroup of linear transformations.
4. Compactness and the Ellipse
- Compactness of the Group: Because $K$ is bounded, the group of its linear symmetries must be compact. This forces the eigenvalues of $\Phi_n$ to lie on the unit circle (Elliptic Affinities), thereby excluding parabolic shears or hyperbolic stretches which are possible for unbounded curves (e. g., parabolas).
- The Shadow is an Ellipse: The closure of the group generated by $\{\Phi_n\}_{n=0}^\infty$ forms a continuous 1-parameter group, which is conjugate to $SO(2)$ by Every compact Lie subgroup of $GL(n, \mathbb{R})$ is conjugate into $O(n)$. There exists a change of basis $S$ such that every affine symmetry $\Phi_n$ becomes a standard orthogonal matrix ($S \Phi_n S^{-1} \in O(2)$), which forces the curve to be a circle in the new coordinates. Transforming back, the original shadow boundary $\mathfrak{C}$ must be an ellipse.
5. The Body is an Ellipsoid
Since the shadow boundary $\mathfrak{C}$ is an ellipse (for the specific illumination in Step 1), and the horizontal cross-sections $K_t$ are homothetic to $\mathfrak{C}$ (by Step 1), it follows that all cross-sections parallel to the chord $pq$ are ellipses.
We apply an affine transformation to the body that maps the elliptical horizontal shadow $\mathfrak{C}$ into a perfect circle and maps the chord $pq$ to be perpendicular to the plane $G$. In this transformed state, the horizontal cross-sections are circles, making the body a surface of revolution.
Similarly the vertical cross-sections (meridians) are ellipses. A surface of revolution whose meridians are ellipses is an ellipsoid of revolution. Since the original body is an affine image of this ellipsoid of revolution, the original body $K$ is an ellipsoid.
The Issue with Non-Compactness
Step 3 established that the shadow curve $\mathfrak{C}$ (and by homothety, every cross-section) admits a continuous 1-parameter group $\overline{\langle\{\Phi_n:n\ge0\}\rangle}$ of affine symmetries.
In the compact $K$ case, since the compact curve $\mathfrak{C}$ is invariant under $\overline{\langle\{\Phi_n:n\ge0\}\rangle}$, this group $\overline{\langle\{\Phi_n:n\ge0\}\rangle}$ must be compact, implying it is conjugate to $SO(2)$. The orbits of $\overline{\langle\{\Phi_n:n\ge0\}\rangle}$ are ellipses. So $\mathfrak{C}$ is an ellipse. When strict convexity is maintained but compactness is dropped, the symmetry group can be non-compact, allowing for parabolic and hyperbolic orbits.
In the non-compact $K$ case, we look to the classification of planar curves invariant under continuous 1-parameter subgroups of the affine group $\mathrm{Aff}(2, \mathbb{R})$,
Theorem: If a strictly convex planar curve $\gamma$ is invariant under a continuous 1-parameter group of affine transformations that acts transitively on the curve, then $\gamma$ is a conic section.
If a continuous 1-parameter group acts transitively on the curve $\gamma$, then every point on the curve is geometrically indistinguishable from every other point. Therefore, any local affine-invariant scalar must be constant along the curve. The fundamental local invariant in this geometry is the affine curvature $\kappa$. Due to the transitive symmetry, $\kappa$ must be constant. By the last part of Chapter 1, Volume 2, of Spivak's A Comprehensive Introduction to Differential Geometry (roughly corresponding to pp. 36-48). It is known that, for planar curves, the only curves of constant affine curvature, up to a transformation in $SL(2,\mathbb{R})$, are the conic sections.
Is there a known proof that handles this case?