Projective self-duals

OpenYear of origin: 2023

Posted online: 2024-03-15 15:59:01Z by SciLag Admin3

Cite as: P-240315.1

  • Algebraic Geometry
  • Geometric Topology
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Problem's Description

Consider a vector space $V$. The projective space $\mathbb{P}$ consists of all 1-dimensional subspaces of $V$. The projective dual space $\mathbb{P}^*$ is obtained by projectivizing $V^*$, where $V^*$ denotes the vector space containing all linear functions over $V$. For every subspace $A$ in $V$, we can associate its annihilator $A^*$ in $V^*$.

V. I. Arnold's problem book [1] problem 1994-17:

Find all projective curves that are projectively equivalent to their duals. Even in the real projective plane $\mathbb{RP}^2$, the answer to this question remains unknown.

A discrete version of Arnold’s question has been treated by Chavez-Caliz, Ana in [2]: Find all $m$-self-dual projective polygons. A closed $n$-gon $P$ in a $k$-dimensional projective space is a sequence of ordered vertices $A_1, A_3, A_5, \ldots \in \mathbb{P}^k$ such that $A_i = A_{i+2n}$ for all $i$. Its dual polygon $P^*$ is a polygon in the dual space with vertices $B^*_k, B^*_{k+2}, B^*_{k+4}, \ldots \in (\mathbb{P}^k)^*$, where $B^*_i$ is the annihilator of the subspace $B_i := \text{span} \{ A_i - (k-1), A_i - (k-3), \ldots, A_i + (k-3), A_i + (k-1) \}$.

The authors also conjecture the following.

Conjecture 1

Let $P$ be a $(k + 3)$-gon in $\mathbb{R}\mathbb{P}^k$ and $J = (1, \ldots, 1)$.

A) If $k$ is an even number and $I = (1, \ldots, 1, 2, 1, \ldots, 1)$ with the same number of ones at the beginning and at the end of $I$, or

B) if $k$ is odd and $I = (1, 1, \ldots, 1, 2, 1, \ldots, 1)$ with one more $1$ at the beginning of $I$ than at the end,

then $P$ and $T_{I,J}(P)$ are projectively equivalent.

While this conjecture lacks a formal proof, computational findings validate it for dimensions up to $k = 200$. Additionally, there's always the possibility of identifying a pair of indices $I$ and $J$ such that a $(k + 3)$-gon $P$ and $T_{I,J}(P)$ are not projectively equivalent.

Conjecture 2

If $I = (1, \ldots, 1)$, $J = (2, 1, \ldots, 1)$ and $k \geq 4$, then there is a $(k + 3)$-gon $P$ such that $P$ and $T_{I,J}(P)$ are not projectively equivalent.

Conjecture 2 has also been verified up to dimension $k = 200$.

  1. BookIs an originArnold’s problems

    Springer 2004

  2. ArticleIs an originProjective self-dual polygons in higher dimensions

    Advances in Geometry 23, 567-582, 2023fulltext


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  • Created at: 2024-03-15 15:59:01Z