Timeline for Maximal Number of Pairs of Orthogonal vectors in a set of $n$ vectors in $\mathbb{R}^3$
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 14, 2015 at 10:38 | vote | accept | batconjurer | ||
| Aug 13, 2015 at 17:58 | comment | added | Sean Eberhard | @Seva I think most of it carries over, assuming "line" means $(r-1)$-dimensional hyperplane. Realize our copy of $\mathbb{F}_q^r$ as an affine hyperplane $P$ of $\mathbb{F}_q^{r+1}$. For each point in our collection we get a point in $\text{PG}(r,q)$ by projection, and for each "line" in our collection we get an $r$-dimensional hyperplane in $\mathbb{F}_q^{r+1}$ and hence a point in $\text{PG}(r,q)$ by duality. As before incidence corresponds to orthogonality. The number of points we get is between $\max(n,l)$ and $n+l$ and the number of orthogonal pairs is at least the number of incidences. | |
| Aug 13, 2015 at 17:39 | comment | added | Seva | I wonder whether this correspondence works for the finite vector space ${\mathbb F}_q^r$. Assuming we have a system of $n$ points and $l$ lines in ${\mathbb F}_q^r$ that determines $I$ incidences, how many points we get in ${\rm PG}(r,q)$ and how many pairs of them will be orthogonal? | |
| Aug 13, 2015 at 15:22 | comment | added | Sean Eberhard | @YoavKallus Construct different sets of points $x$ and $y$. As you say, it only affects the constant. | |
| Aug 13, 2015 at 14:20 | comment | added | Yoav Kallus | When we read the paragraph backwards, are we to construct different sets of points $x$ and $y$ on $S$ for the points $p_x$ and lines $l_y$ on $P$? Or is there some way of ensuring that every point $p_x$ agrees with some line $l_x$ on the resulting point $x$? Either way, you would still get $\sim n^{4/3}$, but I couldn't understand what "read the paragraph backwards" meant. | |
| Aug 13, 2015 at 12:37 | history | answered | Sean Eberhard | CC BY-SA 3.0 |