Timeline for Is finite field version kakeya conjecture still true when changing the line of every direction with only 2(or several but not the full line)element?
Current License: CC BY-SA 4.0
9 events
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| Nov 17, 2020 at 6:15 | comment | added | katago | The counterexample I construct is wrong, I still can not figure out weather there exist a $(1,q^{−(1−c)})$-kakeya set $K$ such that $|K|=O(q^{(1-\epsilon)n})$($\epsilon>0$ relay on $c$) and K lie in a suffice large $(Z/qZ)^n, i.e. q\to \infty$ or not. | |
| Nov 16, 2020 at 17:37 | vote | accept | katago | ||
| Nov 16, 2020 at 17:37 | comment | added | katago | Sorry, I am fool, there exist easy example to explain the result of Divr is optimal. | |
| Nov 16, 2020 at 16:27 | comment | added | katago | You are totally right, and in general in the article using polynomial method we can only get a bound of $(1,q^{-(1-c)})$-kakeya set $K$ has lower bound $|K|=O(q^{cn})$, which coincide with the obersevation in the last of my question. And in general wheather we have $(1,q^{-(1-c)})$-kakeya set also has full dimension or there is a counterexample? This is not discuss in the original article of Divr. | |
| Nov 16, 2020 at 15:34 | comment | added | Seva | @xyh: Dvir's result seems to answer this; I inserted an explanation into my answer. | |
| Nov 16, 2020 at 15:33 | history | edited | Seva | CC BY-SA 4.0 |
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| Nov 16, 2020 at 14:39 | comment | added | katago | thank you for the answer, the point is if change $\gamma q$ to $\gamma q^c, 0<c<1$, is the kakeya conjecture still true, becasue if a line in $(Z/pZ)^n$ is too long, it is unreasonable the analoge in $R^n$ is a segement with length 1 | |
| Nov 16, 2020 at 13:59 | history | edited | Seva | CC BY-SA 4.0 |
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| Nov 16, 2020 at 13:47 | history | answered | Seva | CC BY-SA 4.0 |