Tahara refers to the "Gauss symbol" in the article, On the second cohomology groups of semidirect products, Math. Z. 129 (1972) 365--379. For a fixed $n$, let $S_{ij}$ be the expression \begin{equation} \left[ \frac{i+j}{n}\right] -\left[ \frac{i}{n}\right] - \left[ \frac{j}{n}\right] \end{equation} which comes up often in the paper. What does $\left[ \text{ }\right] $ mean here? Also, what then are $S_{k,-k}$ and $S_{k,k}$, assuming $k$ and $n$ are coprime?
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$\begingroup$ I don't know. I'm familiar with the use of $[x]$ to mean the integer part of $x$, that is, the integer such that $0\le x-[x]<1$. Does that work in the paper? $\endgroup$– Gerry MyersonCommented Aug 31, 2019 at 1:59
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1$\begingroup$ Is it the quadratic reciprocity symbol?? I don't know if people call it Gauss symbol.. it is usually called as Legendre symbol... $\endgroup$– Praphulla KoushikCommented Aug 31, 2019 at 2:11
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2$\begingroup$ Seeing the context, this might in fact be just the integer part function. Moreover, Wikipedia tells me that the notation $[x]$ for integer part was actually introduced by Gauss, hence it is not unreasonable to call it the Gauss symbol (but as witnessed here, calling it as such results in nothing but confusion). $\endgroup$– Emil JeřábekCommented Aug 31, 2019 at 12:08
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1$\begingroup$ Which would imply that $S_{k,-k}=0$ if $m\mid k$ (which, in view of coprimality of $m$ and $k$, means $m=1$), and $S_{k,-k}=1$ otherwise. $\endgroup$– Emil JeřábekCommented Aug 31, 2019 at 12:16
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1$\begingroup$ Why I think this question should not be closed: In my own research, I have asked other mathematicians for help with unclear notation in published papers---to me, that makes the question topical for MO. While the answer here is not profound, note that there were various initial guesses in the comments, so the answer was not necessarily obvious. $\endgroup$– Brian HopkinsCommented Sep 3, 2019 at 13:54
1 Answer
Gerry and Emil in the comments are correct: it's the integer floor function.
Mastnak (citation below) considers Tahara's work in an appendix, "An excursion into group cohomology." On p433 in Example A.6, working in the order $n$ cyclic group generated by $x$, Mastnak uses
...the cocycle defined by $c_a(x^i, x^j) = a^{\lfloor \frac{i+j}{n}\rfloor}$ for $i,j \in \{0,1,\ldots,n-1\}$.
This is followed by
Remark. We can drop the condition $i,j \in \{0,1,\ldots,n-1\}$ in definition of $c_a$ if we replace $\lfloor \frac{i+j}{n}\rfloor$ by $\lfloor \frac{i+j}{n}\rfloor - \lfloor \frac{i}{n}\rfloor - \lfloor \frac{j}{n}\rfloor$.
For $i=k, j=-k$, suppose $k = an+b > 0$ with $a \in \Bbb{Z}$ and $b \in \{1,\ldots,n-1\}$ (if $b=0$ then $n|k$ contrary to your coprime condition, and the case $n=1$ doesn't make much sense in this context). Then \begin{equation} \left\lfloor \frac{0}{n}\right\rfloor - \left\lfloor \frac{k}{n}\right\rfloor - \left\lfloor \frac{-k}{n}\right\rfloor = 0 - a - (-a-1) = 1.\end{equation}
For the $i=k, j=k$ question added later, use the same set up. Then \begin{equation} \left\lfloor \frac{2k}{n}\right\rfloor - 2\left\lfloor \frac{k}{n}\right\rfloor = \left\lfloor \frac{2an+2b}{n}\right\rfloor - 2a = \begin{cases} 2a-2a = 0 & \text{if } 2b < a, \\ 2a+1 - 2a = 1 & \text{if } 2b \ge a. \end{cases}\end{equation}
As to Tahara calling it the Gauss symbol, here's a passage from Wolfram MathWorld on the floor function and its evolving notation.
Unfortunately, in many older and current works (e.g., Honsberger 1976, p. 30; Steinhaus 1999, p. 300; Shanks 1993; Ribenboim 1996; Hilbert and Cohn-Vossen 1999, p. 38; Hardy 1999, p. 18), the symbol $[x]$ is used instead of $\lfloor x \rfloor$ (Graham et al. 1994, p. 67). In fact, this notation harks back to Gauss in his third proof of quadratic reciprocity in 1808. However, because of the elegant symmetry of the floor function and ceiling function symbols $\left\lfloor x \right\rfloor$ and $\lceil x \rceil$, and because $[x]$ is such a useful symbol when interpreted as an Iverson bracket, the use of $[x]$ to denote the floor function should be deprecated.
Mitja Mastnak (2002), Hopf algebra extensions arising from semi-direct products of groups, J. Alg. 251: 413--434 (MSN).
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1$\begingroup$ Gerry in the comments was also correct. mathoverflow.net/questions/339529/… $\endgroup$ Commented Aug 31, 2019 at 23:40
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$\begingroup$ Thank you very much "Brian". what do you think about the value of the number $S_{k; k}$. $\endgroup$ Commented Sep 1, 2019 at 2:45
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2$\begingroup$ @GerryMyerson Yes, sorry, you too---edited. I got caught up in the subtleties between least integer and integer part for negative numbers $\endgroup$ Commented Sep 1, 2019 at 11:51