Timeline for A mixing property for finite fields of characteristic $2$
Current License: CC BY-SA 3.0
16 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Aug 7, 2012 at 19:38 | history | edited | Seva | CC BY-SA 3.0 |
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| Aug 7, 2012 at 19:36 | vote | accept | Seva | ||
| Jul 28, 2012 at 8:23 | history | edited | Seva | CC BY-SA 3.0 |
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| Jul 25, 2012 at 12:29 | answer | added | Peter Mueller | timeline score: 10 | |
| Jul 24, 2012 at 16:08 | answer | added | Boris Bukh | timeline score: 6 | |
| Jul 24, 2012 at 15:06 | answer | added | David E Speyer | timeline score: 10 | |
| Jul 24, 2012 at 3:51 | comment | added | David E Speyer |
Do we have to look at the graphs of functions? Let $\beta \in \mathbb{F}_q$ be such that $x^2+x+\beta$ doesn't split in $\mathbb{F}_q$. Then $X:=\{ (x,y): x^2+xy+\beta y^2=1 \}$ has $q+1$ elements and, for every nonzero linear map $\phi: (x,y) \mapsto ax+by$, I believe that $|\phi(X)| = q/2+1$. (I posted an incorrect version of this comment before.)
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| Jul 23, 2012 at 20:04 | comment | added | David E Speyer | Observation that might help someone else: If $\phi(x) = x^{-1}$ and $\phi(0)=0$, then $|\phi_a(\mathbb{F})| =(q-1)/2$ for all $a \neq 0$. So it's easy to get down to the tight bound for almost all $a$. | |
| Jul 23, 2012 at 13:45 | answer | added | Peter Mueller | timeline score: 15 | |
| Jul 22, 2012 at 18:28 | comment | added | Seva | @Xander: no restrictions are imposed on $\varphi_0$. Taking $\varphi_0(x)=x$ is a bad choice, as in this case $\max_a|\varphi_a({\mathbb F})|$ is large. The question is, how small can one make this maximum choosing $\varphi_0$ appropriately. | |
| Jul 22, 2012 at 18:00 | comment | added | Xander Faber | What hypotheses are you requiring for $\varphi_0$? As you point out, there are functions $\varphi_0$ such that the image of $\varphi_a$ has cardinality $\ll 2q/3$, independent of the choice of $a$. But you could also take $\varphi_0(x) = x$ (or any permutation polynomial) and get the image to be all of $\mathbb{F}$. | |
| Jul 21, 2012 at 16:44 | history | edited | Seva | CC BY-SA 3.0 |
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| Jul 21, 2012 at 6:04 | comment | added | Seva | @Victor: for $q$ odd, taking $\varphi_0(x)=x^2$ leads to all $\varphi_a({\mathbb F})$ being of size $(q+1)/2$. On the other hand, it is not difficult to see that for any choice of $\varphi_0$, there exists $a$ with $|\varphi_a({\mathbb F})|>q/2$. | |
| Jul 20, 2012 at 23:11 | comment | added | Victor Protsak | Can you, please, comment on the restriction "$q$ is even"? Is the answer known or expected to be different when $q$ is odd? | |
| Jul 20, 2012 at 16:54 | history | asked | Seva | CC BY-SA 3.0 |