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1. • CommentRowNumber1.
   • CommentAuthorUrs
   • CommentTime4 days ago
   • (edited 4 days ago)
   • 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
   • CommentTime4 days ago
   • 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
   • CommentTime4 days ago
   • 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
   • CommentTime4 days ago
   • 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
   • CommentTime4 days ago
   • 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
   • CommentTime4 days ago
   • 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
   • CommentTime1 day ago
   • (edited 1 day ago)
   • 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.