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I am faced with the variant of the quadratic Gauss sum which has just half the usual exponent (a $\pi$ in place of $2\pi$). Explicit computer checks suggest that $$ k \in 2 \mathbb{N}_{>0} \;\;\;\;\;\;\;\; \overset{?}{\Rightarrow} \;\;\;\;\;\;\;\; \sum_{n=0}^{k-1} e^{ \tfrac{\pi \mathrm{i}}{k} n^2 } \;=\; \sqrt{k} \, (-1)^{1/4} \,, $$ but what would be a proof?

More generally, is there an evaluation formula for the generalized quadratic Gauss sum with half the usual exponent: $$ \sum_{n=0}^{k-1} e^{ q \tfrac{\pi \mathrm{i}}{k} n^2 } \;\; \in \;\; \mathbb{C} \,, \;\;\;\;\;\;\;\;\;\;\text{where}\;\; (k,q) \in \mathbb{N}_{>0} \times \mathbb{Z} \;\;\;\text{with}\;\;\; k q \in 2\mathbb{Z} \,, $$ or at least a criterion (on $k,q$) for this to be non-vanishing? (The latter is maybe easy by the usual route of computing instead the square?)

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    $\begingroup$ This seems relevant: math.stackexchange.com/questions/4232267/… $\endgroup$ Commented Mar 21 at 19:50
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    $\begingroup$ @MarkLewko For a self-contained treatment, see my answer below (including the short note quoted there). $\endgroup$ Commented Mar 21 at 19:53
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    $\begingroup$ Serge Lang, Algebraic Number Theory, chapter on Gauss sums, defines $$G(a,b)=\sum_{x\bmod b}e^{{2\pi i\over b}ax^2}$$ for $a,b$ non-zero integers, $b>0$, $\gcd(a,b)=1$, and shows $G(1,b)$ is zero if $b\equiv2\bmod4$, $(1+i)\sqrt b$ if $b\equiv0\bmod4$. $\endgroup$ Commented Mar 22 at 1:09
  • $\begingroup$ @MarkLewko thanks, that's a good proof. Have recorded it here: ncatlab.org/nlab/show/… $\endgroup$ Commented Mar 22 at 9:03
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    $\begingroup$ Yes, but if $b\equiv0\bmod4$, then the $2$ cancels, and you're left with an exponent of the form ${\pi i\over k}x^2$ with $k$ even, as in your question. $\endgroup$ Commented Mar 22 at 12:02

2 Answers 2

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This follows from a well-known reciprocity relation of Gauss sums. Let us denote $$G(a,b,c):=\frac{1}{\sqrt{c}}\sum_{n=0}^{c-1}e^{\frac{\pi i}{c}(an^2+2bn)}.$$ If $a$ and $c$ are positive integers, and one of them is even, then for any integer $b$ we have that $$G(a,b,c)=e^{\frac{\pi i}{4}-\frac{\pi i b^2}{ac}}\overline{G(c,b,a)}.$$ For a proof, see here. In particular, $$\sum_{n=0}^{k-1}e^{\frac{\pi i}{k}n^2}=G(1,0,k)\sqrt{k}=e^{\frac{\pi i}{4}}\overline{G(k,0,1)}\sqrt{k}=e^{\frac{\pi i}{4}}\sqrt{k}.$$

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  • $\begingroup$ Thanks, that's useful! Have recorded it here: ncatlab.org/nlab/show/… $\endgroup$ Commented Mar 22 at 9:00
  • $\begingroup$ @UrsSchreiber Thank you. I am glad you liked it! $\endgroup$ Commented Mar 22 at 10:12
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    $\begingroup$ In my decade and a half floating around SE I somehow never knew what the intitials GH stood for til I read those PDFs at ncatlab. It feels like I have acquired forbidden knowledge. $\endgroup$ Commented Mar 22 at 12:18
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The reciprocity relation of Gauss sums mentioned by GH from MO is also known as Landsberg – Schaar identity $$\frac{1}{\sqrt{p}} \sum_{x=0}^{p-1} e\left(\frac{x^{2} q}{p}\right)=\frac{e\left(1/8\right)}{\sqrt{2 q}} \sum_{x=0}^{2 q-1} e \left(-\frac{x^{2} p}{4 q}\right), $$ where $p$, $q$ are positive integers and $e(t)=e^{2\pi it}$. The required formula is a special case ($p=1$) of this identity.

The identity has many different proofs. The original Schaar’s approach employs the Poisson summation formula, see Mémoire sur la théorie des résidus biquadratiques. It can easily be proved if we know the explicit values for the Gauss sums of general form $$S(a,p)=\sum\limits_{x=1}^{p}e(ax^2/p).$$ In such an approach, all difficulties are reduced to the calculation of the sums $S(a, p)$. Another version of the proof is based on the the functional equation $$\theta(t^{-1})=\sqrt{t}\,\theta(t)$$ for the theta function $$\theta(t)=\sum\limits_{k=-\infty}^{\infty}e^{-\pi tk^2} \qquad(\mathrm{Re}\, t>0),$$ see Dym H., McKean H. P. Fourier series and integrals, 1972, sec. 4.6. In Gauss Sums and Quantum Mechanics, the Landsberg–Schaar formula is proved by methods of quantum mechanics. The article A Short Proof of the Landsberg–Schaar Identity gives the proof based on finite Fourier transform.

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    $\begingroup$ Thanks! I have recorded that here: ncatlab.org/nlab/show/Gauss+sum#LandsbergSchaarIdentity And this also answers the second part of my question, at least for odd $p$, because it reduces it to the standard case. $\endgroup$ Commented Mar 24 at 11:28
  • $\begingroup$ I suppose this is a special case of the relation in GH's answer mathoverflow.net/a/489863/381 $\endgroup$ Commented Mar 24 at 11:52
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    $\begingroup$ @UrsSchreiber Yes, but it is better to know that this formula has some history. GH from MO cited the paper that ignored the whole story. The contour integration method is also a well-known approach to computing Gauss sums. $\endgroup$ Commented Mar 24 at 12:37
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    $\begingroup$ sure, and I appreciate that. The references you gave are most valuable. Was just trying to see how things hang together. $\endgroup$ Commented Mar 24 at 12:43

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