Timeline for Non-standard Gauss sums
Current License: CC BY-SA 3.0
24 events
| when toggle format | what | by | license | comment | |
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| May 28, 2015 at 16:45 | comment | added | Will Sawin | @Liss Yes, the same bound from the same proof in both cases. In fact it's the same sum- to see this, multiply the variable by $l$, and you get an $l^2$ in the exponent. | |
| May 28, 2015 at 14:49 | comment | added | Liss | @WillSawin, maybe this is a trivial question, but the same bound $2\sqrt{p}$ also holds if it was $R(t)=t(t+\ell),$ too? I made a mistake in my question, so I need to have $k(k+l)$, or and $k(k+1)$ is just the case of $l=1,$ in which in the exponent there should be also no $l...$ But I guess the bounds , both lower and upper do not change from this fact? | |
| May 19, 2015 at 21:01 | comment | added | Liss | thank you, @GHfromMO, too :) I will comment also below on your response | |
| May 19, 2015 at 20:59 | comment | added | Liss | thank you, @KConrad, my interest is coming from signal processing (discrete time-frequency analysis), and it is just very beautiful "coincidence" that I am dealing with quadratic characters and uncommon questions for those sums.. I can prove that this particular type of sums is non-zero, but it would have been nice to show that it is away from zero in absolute value.. | |
| May 19, 2015 at 4:02 | comment | added | GH from MO | Your sum in absolute value is at least $(4p)^{(2-p)/2}$. See my response below. | |
| May 19, 2015 at 2:49 | answer | added | GH from MO | timeline score: 3 | |
| May 18, 2015 at 18:41 | comment | added | KConrad | Some Kloosterman sums involving a quadratic character are zero. I'm not saying yours might be zero, but if you are looking for some off-the-shelf result that will prove character sums $S$ in some family do not equal $0$ then you will be disappointed. The standard task for exponential sums is to get sharp upper bounds, not lower bounds away from $0$. Saying there is a lower bound that is a negative number is not really the point of these bounds which is to say $|S|$ is less than or equal to some sharp value. | |
| May 18, 2015 at 12:58 | comment | added | Liss | thank you a lot, @KConrad and Will Sawin. It looked like these bounds can be sufficient for what I need, but when all expression taken together (with $\chi(X)$ as in the question posed), I get only $\geq -2\sqrt{p}-2$ which is useless, because I need to show that their absolute values are away from zero. | |
| May 18, 2015 at 12:53 | history | edited | Liss | CC BY-SA 3.0 |
rephrase the question (i need a lower bound of the whole expression)
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| May 18, 2015 at 2:42 | answer | added | Alexey Ustinov | timeline score: 6 | |
| May 18, 2015 at 2:19 | history | edited | Alexey Ustinov |
edited tags
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| May 17, 2015 at 18:06 | comment | added | Will Sawin | @Liss It's an upper bound on the absolute value. For a reference you can go back to Weil: "On Some Exponential Sums". Example 1 after equation (5) on the bottom of page 206 is the bound, where $\mathfrak d=\{0,-1\}$ so $R_{\mathfrak d}(t) = t(t+1)$, $\chi$ is the quadratic character, and $\psi(x) = \omega_p^{lx}$. | |
| May 17, 2015 at 17:11 | comment | added | Liss | @WillSawin, thank you. Is that an upper or lower bound? Could you give me some reference? KConrad, what I try to do is to find a lower bound for a sum of roots of unity taken over a set $k \in K, k-q \in K,$ for fixed $q,$ where $K$ is a quadratic difference set, i.e. elements of the form $t^2, t \in Z_p^*.$ I need it for an estimate of the coherence of a Gabor system generated by difference sets... | |
| May 17, 2015 at 0:06 | comment | added | KConrad | Whoops, my previous comment on Jacobi sums was incorrect since the sum is not just $\sum_{k \in \mathbf Z/(p)} (\frac{k(k+1)}{p})$ but has the factor $\omega_p^{kl}$ in there too. Replacing $k$ with $k/l$ makes the sum $\sum_{k \in \mathbf Z/(p)} (\frac{k(k+l)}{p})\omega_p^k$. I agree with Will that it is hopeless to expect an exact formula for this but it gets a bound like $2\sqrt{p}$ from the Weil conjectures. | |
| May 16, 2015 at 23:50 | comment | added | Will Sawin | This is a finite field hypergeometric sum. There is no simple formula, but it's bounded by $2\sqrt{p}$. | |
| May 16, 2015 at 22:30 | comment | added | KConrad | Are you trying to count how often consecutive numbers $k$ and $k+1$ in $\mathbf Z/(p)$ are perfect squares? | |
| May 16, 2015 at 21:21 | comment | added | Liss | Ok, thank you! I understand. I it true I didn't estimate the hardness of the question correctly, I will start with math.stackexchange next time. Thanks a lot for your help and explanations! | |
| May 16, 2015 at 21:09 | comment | added | KConrad | @Liss, first of all you are right that $(\frac{l}{p})^{-1}$ comes out, but the character is quadratic so the exponent doesn't matter: $a^{-1} = a$ if $a = \pm 1$. Concerning a sum with $(\frac{k^2+k}{p}) = (\frac{k(k+1)}{p})$ in it, replacing $k$ with $-k$ makes that $(\frac{-1}{p})(\frac{k(1-k)}{p})$ and then you basically have a Jacobi sum, which you can look up elsewhere. This is not really a research-level question. I suggest if you have similar questions that you ask them on math.stackexchange. | |
| May 16, 2015 at 21:02 | history | edited | Liss | CC BY-SA 3.0 |
put the accent on a slightly different question that the main posed before
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| May 16, 2015 at 20:52 | comment | added | Liss | Could we keep the question open for the case $(\frac{k^2+k}{p})$? I could edit the question.. Similar trick I think can not be done in that case.. | |
| May 16, 2015 at 20:50 | comment | added | Liss | Great, thanks a lot! this was super-easy :) also, instead of $(\frac{l}{p})$ I believe what comes out is $(\frac{l^{-1}}{p})$ | |
| May 16, 2015 at 20:47 | comment | added | GH from MO | @KConrad: I suggest you give your comment as a response below, and then this question can be closed. | |
| May 16, 2015 at 20:27 | comment | added | KConrad | The missing term at $k = 0$ is $1$, so your sum is $\sum_{k \in \mathbf Z/(p)} (\frac{k+1}{p})\omega_p^{kl} - 1$. Now you're summing over an additive group, so replacing $k$ with $k-1$ makes it $\sum_{k \in \mathbf Z/(p)} (\frac{k}{p})\omega_p^{(k-1)l} - 1 = \omega_p^{-l}\sum_{k \in \mathbf Z/(p)} (\frac{k}{p})\omega_p^{kl} - 1$, and omitting the $k=0$ term (which is $0$) and then doing a multiplicative change of variable you get $\omega_p^{-l}(\frac{l}{p})\sum_{k \not\equiv 0 \bmod p} (\frac{k}{p})\omega_p^k - 1$. This last sum is a standard Gauss sum of the Legendre symbol. | |
| May 16, 2015 at 20:22 | history | asked | Liss | CC BY-SA 3.0 |