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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 and , the expression

   is independent of if is even:

   In a roundabout way this follows from the modular transformation law of conformal characters of the 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 , 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 . If is even then is mod .

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 -periodic after all, because

   and so if we shift in the exponent, then we may just relabel to get rid of the .

   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 (here) I replace the factor

   throughout by , for , then all conclusions seem to remain valid.

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

   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 . 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 (here)

   [edit: the first version of this comment had a misprint in this formula, with appearing instead of . 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 we have

   while for the pertinent case of even we seem to have

   If we now generalize to as in #7 above and look at , then computer experiment suggests that

   and

   If my computations here are correct, then we get an irrep of 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 the first formular has a . With that, the following factors of cancel out and the end result for the basic case is , which is actually the expected central charge! (cf. Gannon02 (3.1b) (p. 9)