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I am self-reading P. Coleman's Introduction to Many-Body Physics. In Exercise 2.2, we are asked to prove the following orthogonality relation \begin{align} \frac{1}{N} \sum_{j=0}^{N-1} e^{i(q_{n}-q_{m})ja} = \frac{(-1)^{(n-m)(N-1)}}{N} \frac{\sin{\pi(n-m)}}{\sin{\frac{\pi}{N}(n-m)}} = \delta_{n,m}, \end{align} where $q_{n} = n \frac{2\pi}{Na}$ is the discrete wave vector. I proceed as in the instruction by treating this summation as a geometric series and I have \begin{align} \frac{1}{N} \sum_{j=0}^{N-1} e^{i\frac{2\pi(n-m)}{N}j} &= \frac{1}{N} \frac{1-e^{i{2\pi(n-m)}}}{1-e^{i\frac{2\pi(n-m)}{N}}} = \frac{1}{N} \frac{e^{i\pi(n-m)}}{e^{i\pi(n-m)/N}} \frac{e^{-i\pi(n-m)}-e^{i\pi(n-m)}}{e^{-i\pi(n-m)/N}-e^{i\pi(n-m)/N}}=\frac{1}{N} \frac{e^{i\pi(n-m)}}{e^{i\pi(n-m)/N}} \frac{\sin\pi(n-m)}{\sin{\frac{\pi}{N}(n-m)}} \\ & = \frac{1}{N} e^{i\pi(n-m)(1-\frac{1}{N})}\frac{\sin\pi(n-m)}{\sin{\frac{\pi}{N}(n-m)}}=\frac{1}{N} e^{i\pi(n-m)(\frac{N-1}{N})}\frac{\sin\pi(n-m)}{\sin{\frac{\pi}{N}(n-m)}}. \end{align} But I am not sure whether this is equal to \begin{align} \frac{(-1)^{(n-m)(N-1)}}{N} \frac{\sin{\pi(n-m)}}{\sin{\frac{\pi}{N}(n-m)}} \end{align} since I think \begin{align} e^{i\pi(n-m)(\frac{N-1}{N})} \neq (-1)^{(n-m)(N-1)}. \end{align} Is there any suggestion? Thanks!

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    $\begingroup$ It's possible this is a typo in the book, since you would end up with the orthogonality relationship regardless of whether the exponent had a factor of $N-1$ or $(N-1)/N$ $\endgroup$ – Andrew Feb 26 at 21:16
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    $\begingroup$ For what it's worth, I didn't see a problem with any of your steps, so I think you are right. At the very least, your approach is sound and should work. $\endgroup$ – Andrew Feb 26 at 21:27
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    $\begingroup$ Since you're self-studying Coleman's book, it's worth mentioning that it has very many typos. It might be helpful to have one or two alternative resources at your side, in case you run into a trickier typo. $\endgroup$ – Zack Feb 26 at 23:10
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    $\begingroup$ It's not uncommon for books or lectures to have mistakes precisely in the intermediate steps. It's easy to be blind to them as an author/lecturer, as long as the end result is what you know it should be. $\endgroup$ – Anyon Feb 27 at 3:28
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    $\begingroup$ The intermediate result is correct, but probably a typo, and completely unnecessary. $\endgroup$ – user200143 Feb 27 at 21:33

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