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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTimeMar 15th 2025
   • (edited Mar 15th 2025)
   • PermaLink
   Author: Urs
   Format: MarkdownItexThis may just show my ignorance, but I am wondering how to see the following: $\,$ For $k \in \mathbb{N}_{\gt 0}$ and $a \in \mathbb{N}$, the expression $$ S_k(a) \,\coloneqq\, \sum_{n = 0}^{k-1} e^{ \tfrac{\pi \mathrm{i}}{k} (n + a)^2 } $$ is independent of $a$ if $k$ is even: $$ k\;\text{is even} \;\;\;\Rightarrow\;\;\; \underset{a}{\forall} \big( \; S_k(a) = S_k(0) \; \big) \,. $$ In a roundabout way this follows from the modular transformation law of conformal characters of the $\mathrm{U}(1)$ WZW model. But what's an elementary arithmetic way to see this? (If the exponent had another factor of "2", then it would follow easily, also for odd $k$, by modular arithmetic, now in the other sense of "modular". But the question is for the exponent as given.)

   This may just show my ignorance, but I am wondering how to see the following:

   $\,$

   For $k \in \mathbb{N}_{\gt 0}$ and $a \in \mathbb{N}$, the expression

   $$S_k(a) \,\coloneqq\, \sum_{n = 0}^{k-1} e^{ \tfrac{\pi \mathrm{i}}{k} (n + a)^2 }$$

   is independent of $a$ if $k$ is even:

   $$k\;\text{is even} \;\;\;\Rightarrow\;\;\; \underset{a}{\forall} \big( \; S_k(a) = S_k(0) \; \big) \,.$$

   In a roundabout way this follows from the modular transformation law of conformal characters of the $\mathrm{U}(1)$ WZW model.

   But what’s an elementary arithmetic way to see this?

   (If the exponent had another factor of “2”, then it would follow easily, also for odd $k$, by modular arithmetic, now in the other sense of “modular”. But the question is for the exponent as given.)

2. • CommentRowNumber2.
   • CommentAuthorDavid_Corfield
   • CommentTimeMar 15th 2025
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexI guess the squares repeat with period $k$. If $k$ is even then $(k + a)^2 - a^2 = 2a k + k^2$ is $0$ mod $2 k$.

   I guess the squares repeat with period $k$. If $k$ is even then $(k + a)^2 - a^2 = 2a k + k^2$ is $0$ mod $2 k$.

3. • CommentRowNumber3.
   • CommentAuthorUrs
   • CommentTimeMar 15th 2025
   • PermaLink
   Author: Urs
   Format: MarkdownItexI am prepared to find that I am missing the obvious -- but could you expand on how this answers the question?

   I am prepared to find that I am missing the obvious – but could you expand on how this answers the question?

4. • CommentRowNumber4.
   • CommentAuthorUrs
   • CommentTimeMar 15th 2025
   • PermaLink
   Author: Urs
   Format: MarkdownItexOh, I think I see what you mean. Right.

   Oh, I think I see what you mean. Right.

5. • CommentRowNumber5.
   • CommentAuthorUrs
   • CommentTimeMar 15th 2025
   • PermaLink
   Author: Urs
   Format: MarkdownItexRight, so the summands are $k$-periodic after all, because $$ e^{ \tfrac{\pi \mathrm{i}}{k} (n + k)^2 } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k} (n^2 + 2 n k + k^2) } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k} n^2 } \underset{ 1\;if\;k\;even }{ \underbrace{ e^{ \pi \mathrm{i}(2n + k) } } } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k} n^2 } $$ and so if we shift $n \mapsto n + a$ in the exponent, then we may just relabel to get rid of the $a$. Thanks! :-)

   Right, so the summands are $k$-periodic after all, because

   $$e^{ \tfrac{\pi \mathrm{i}}{k} (n + k)^2 } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k} (n^2 + 2 n k + k^2) } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k} n^2 } \underset{ 1\;if\;k\;even }{ \underbrace{ e^{ \pi \mathrm{i}(2n + k) } } } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k} n^2 }$$

   and so if we shift $n \mapsto n + a$ in the exponent, then we may just relabel to get rid of the $a$.

   Thanks! :-)

6. • CommentRowNumber6.
   • CommentAuthorDavid_Corfield
   • CommentTimeMar 15th 2025
   • PermaLink
   Author: David_Corfield
   Format: MarkdownItexSorry for being cryptic. A snatched moment when supposed to be supporting my wife in her clothes shopping for our daughter's wedding!

   Sorry for being cryptic. A snatched moment when supposed to be supporting my wife in her clothes shopping for our daughter’s wedding!

7. • CommentRowNumber7.
   • CommentAuthorUrs
   • CommentTimeMar 18th 2025
   • (edited Mar 18th 2025)
   • PermaLink
   Author: Urs
   Format: MarkdownItexBest wishes to your daughter! Meanwhile, I have another puzzlement: If in my construction of irreps of $\widehat{\mathbb{Z}^2} \rtimes SL_2(\mathbb{Z})$ ([here](https://ncatlab.org/nlab/show/integer+Heisenberg+group#LinearRepresentations)) I replace the factor $$ \zeta \,\coloneqq\, e^{ \tfrac{\pi \mathrm{i}}{k} } $$ throughout by $e^{ \tfrac{\pi \mathrm{i}}{k}q }$, for $q \in \{1,2, \cdots, k-1\}$, then all conclusions seem to remain valid. Indeed, it seems to me that if $q$ is chosen to be even, then we no longer need to assume that $k$ is even, because the analogue of the above periodicity of summands (used [here](https://ncatlab.org/nlab/show/integer+Heisenberg+group#eq:PeriodicityOfSummands)) follows already as $$ e^{ \tfrac{\pi \mathrm{i}}{k}q (n + k)^2 } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k}q (n^2 + 2 n k + k^2) } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k}q n^2 } \underset{ 1\;since\;q\;even }{ \underbrace{ e^{ \pi \mathrm{i}q(2n + k) } } } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k} q n^2 }$$ This is not a contradiction in itself, but it does seem to be in contradiction to Will Sawin's reply to me on MO [here](https://mathoverflow.net/a/489582/381), where he seems to exclude such irreps of odd $k$. I must again be missing something.

   Best wishes to your daughter!

   Meanwhile, I have another puzzlement:

   If in my construction of irreps of $\widehat{\mathbb{Z}^2} \rtimes SL_2(\mathbb{Z})$ (here) I replace the factor

   $$\zeta \,\coloneqq\, e^{ \tfrac{\pi \mathrm{i}}{k} }$$

   throughout by $e^{ \tfrac{\pi \mathrm{i}}{k}q }$, for $q \in \{1,2, \cdots, k-1\}$, then all conclusions seem to remain valid.

   Indeed, it seems to me that if $q$ is chosen to be even, then we no longer need to assume that $k$ is even, because the analogue of the above periodicity of summands (used here) follows already as

   $$e^{ \tfrac{\pi \mathrm{i}}{k}q (n + k)^2 } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k}q (n^2 + 2 n k + k^2) } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k}q n^2 } \underset{ 1\;since\;q\;even }{ \underbrace{ e^{ \pi \mathrm{i}q(2n + k) } } } \;=\; e^{ \tfrac{\pi \mathrm{i}}{k} q n^2 }$$

   This is not a contradiction in itself, but it does seem to be in contradiction to Will Sawin’s reply to me on MO here, where he seems to exclude such irreps of odd $k$. I must again be missing something.

8. • CommentRowNumber8.
   • CommentAuthorDavid_Corfield
   • CommentTimeMar 20th 2025
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   Author: David_Corfield
   Format: MarkdownItexThanks, wedding's this Saturday! Sorry, I can't see where the possible mismatch lies from Sawin's answer.

   Thanks, wedding’s this Saturday!

   Sorry, I can’t see where the possible mismatch lies from Sawin’s answer.

9. • CommentRowNumber9.
   • CommentAuthorUrs
   • CommentTimeMar 21st 2025
   • (edited Mar 21st 2025)
   • PermaLink
   Author: Urs
   Format: MarkdownItexYet another small issue I have is the further evaluation of the normalization factor $c_k$ ([here](https://ncatlab.org/nlab/show/integer+Heisenberg+group#eq:NormalizationFactorForTOperator)) $$ c_k \;=\; \Big( \tfrac{1}{\sqrt{k}}\textstyle{\sum_{n=0}^{k-1}} e^{ \tfrac{\pi \mathrm{i}}{k} n^2 }\Big)^{1/3} \,. $$ [edit: the first version of this comment had a misprint in this formula, with $\sqrt{k}$ appearing instead of $1/\sqrt{k}$. Now that this is fixed, the following actually comes out as expected] This wouldn't be a big deal (except if it came out zero, which it doesn't) were it not for the fact that the literature seems to think it equals 1 (again [Manoliu 98, p. 67](https://ncatlab.org/nlab/show/abelian+Chern-Simons+theory#Manoliu98a)). I don't know yet how to evaluate this analytically (probably needs some simple trick) [edit: I see now that this and the following expressions are known as "[[generalized Gauss sums]]"] but computer experiment suggests that for *odd* $k$ we have $$ k\;\;odd \;\;\;\; \Rightarrow \;\;\;\; \sum_{n=0}^{k-1} e^{ \tfrac{\pi \mathrm{i}}{k} n^2 } \;=\; 1 $$ while for the pertinent case of even $k$ we seem to have $$ k\;\;even \;\;\;\; \Rightarrow \;\;\;\; \sum_{n=0}^{k-1} e^{ \tfrac{\pi \mathrm{i}}{k} n^2 } \;=\; \sqrt{k \mathrm{i}} \,. $$ If we now generalize to $\zeta^q$ as in #7 above and look at $q = 2$, then computer experiment suggests that $$ k = 2 r + 1 \;\;\;\; \Rightarrow \;\;\;\; \sum_{n=0}^{k-1} e^{ \tfrac{2 \pi \mathrm{i}}{k} n^2 } \;=\; \left\{ \begin{array}{ll} \sqrt{k} & if\; r \; even \\ \sqrt{k} \, \mathrm{i} & if\; r \; odd \end{array} \right. $$ and $$ k = 2 r \;\;\;\; \Rightarrow \;\;\;\; \sum_{n=0}^{k-1} e^{ \tfrac{2 \pi \mathrm{i}}{k} n^2 } \;=\; \left\{ \begin{array}{ll} \propto \sqrt{i} & if\; r \; even \\ 0 & if\; r \; odd \end{array} \right. $$ If my computations [here](https://ncatlab.org/nlab/show/integer%20Heisenberg%20group#ModularAutomorphisms) are correct, then we get an irrep of $\widehat{\mathbb{Z}^2} \rtimes SL_2(\mathbb{Z})$ whenever this sum is not zero (as it is in the very last case above). It's not clear to me how this squares with the mentioned discussion on MO. Maybe I have a mistake, somewhere.

   Yet another small issue I have is the further evaluation of the normalization factor $c_k$ (here)

   $$c_k \;=\; \Big( \tfrac{1}{\sqrt{k}}\textstyle{\sum_{n=0}^{k-1}} e^{ \tfrac{\pi \mathrm{i}}{k} n^2 }\Big)^{1/3} \,.$$

   [edit: the first version of this comment had a misprint in this formula, with $\sqrt{k}$ appearing instead of $1/\sqrt{k}$. Now that this is fixed, the following actually comes out as expected]

   This wouldn’t be a big deal (except if it came out zero, which it doesn’t) were it not for the fact that the literature seems to think it equals 1 (again Manoliu 98, p. 67).

   I don’t know yet how to evaluate this analytically (probably needs some simple trick)

   [edit: I see now that this and the following expressions are known as “generalized Gauss sums”]

   but computer experiment suggests that for odd $k$ we have

   $$k\;\;odd \;\;\;\; \Rightarrow \;\;\;\; \sum_{n=0}^{k-1} e^{ \tfrac{\pi \mathrm{i}}{k} n^2 } \;=\; 1$$

   while for the pertinent case of even $k$ we seem to have

   $$k\;\;even \;\;\;\; \Rightarrow \;\;\;\; \sum_{n=0}^{k-1} e^{ \tfrac{\pi \mathrm{i}}{k} n^2 } \;=\; \sqrt{k \mathrm{i}} \,.$$

   If we now generalize to $\zeta^q$ as in #7 above and look at $q = 2$, then computer experiment suggests that

   $$k = 2 r + 1 \;\;\;\; \Rightarrow \;\;\;\; \sum_{n=0}^{k-1} e^{ \tfrac{2 \pi \mathrm{i}}{k} n^2 } \;=\; \left\{ \begin{array}{ll} \sqrt{k} & if\; r \; even \\ \sqrt{k} \, \mathrm{i} & if\; r \; odd \end{array} \right.$$

   and

   $$k = 2 r \;\;\;\; \Rightarrow \;\;\;\; \sum_{n=0}^{k-1} e^{ \tfrac{2 \pi \mathrm{i}}{k} n^2 } \;=\; \left\{ \begin{array}{ll} \propto \sqrt{i} & if\; r \; even \\ 0 & if\; r \; odd \end{array} \right.$$

   If my computations here are correct, then we get an irrep of $\widehat{\mathbb{Z}^2} \rtimes SL_2(\mathbb{Z})$ whenever this sum is not zero (as it is in the very last case above).

   It’s not clear to me how this squares with the mentioned discussion on MO. Maybe I have a mistake, somewhere.

10. • CommentRowNumber10.
    • CommentAuthorUrs
    • CommentTimeMar 21st 2025
    • (edited Mar 21st 2025)
    • PermaLink
    Author: Urs
    Format: MarkdownItexFound my mistake: The previous post had started with a misprint: Instead of $\sqrt{k}$ the first formular has a $1/\sqrt{k}$. With that, the following factors of $\sqrt{k}$ cancel out and the end result for the basic case is $c_k = (-1)^{1/12}$, which is actually the expected central charge! (cf. Gannon02 (3.1b) ([p. 9](https://arxiv.org/pdf/math/0103044#page=9))

    Found my mistake: The previous post had started with a misprint: Instead of $\sqrt{k}$ the first formular has a $1/\sqrt{k}$. With that, the following factors of $\sqrt{k}$ cancel out and the end result for the basic case is $c_k = (-1)^{1/12}$, which is actually the expected central charge! (cf. Gannon02 (3.1b) (p. 9)