Baker's Dozen — A totally skew disc

Definitions and mathematical statement

Let $D^3=\{u\in\mathbb R^3:\|u\|\le1\}$. A smooth embedding $f:D^3\to\mathbb R^7$ is understood to be smooth on a neighborhood of the disk, injective on it, and of derivative rank three. At $u$ its affine tangent plane is $A_u=f(u)+df_u(\mathbb R^3)$. The image is called totally skew when, for every pair $u\ne v$, \[ \dim\operatorname{aff}(A_u\cup A_v)=7. \] Equivalently, the $7\times7$ matrix formed by concatenating the two $7\times3$ derivative matrices and the displacement column, $[\,df_u\mid df_v\mid f(v)-f(u)\,]$, is nonsingular. This condition includes both the relative directions of the tangent spaces and the displacement between their base points; disjointness of the affine planes by itself is insufficient.

The question is whether such an embedding exists. Dimension seven is the smallest ambient dimension in which two affine three-planes can have an affine span of dimension $2\cdot3+1$. A construction must satisfy the condition for every distinct pair, including arbitrarily close points. A negative answer must obstruct all smooth embeddings, not merely polynomial examples of a fixed degree.

Short English statement

Can a three-dimensional disk sit smoothly in seven-dimensional space so that every two tangent planes, together with their relative position, span all seven dimensions?

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, supplied problems.json, record 600014 (AMR-005-0014), AMR Open Problem Lists. Catalogue identity and source transcription; CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Tabachnikov - A Baker's Dozen of Problems (2015), Problem 6. Original source indexed by AMR-005-0014. https://amj.math.stonybrook.edu/html-articles/Files-2015-2024/14-01/.