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$\newcommand{\GL}{\operatorname{GL}} \newcommand{\R}{{\Bbb R}} \newcommand{\C}{{\Bbb C}} $For a natural number $n$, let $z\in \GL(n,\C)$ be a 1-cocycle of $G=\GL_{n,\R}\,$, that is, an invertible complex matrix such that $$z\cdot\bar z=1$$ where the bar denotes complex conjugation. By a generalization of Hilbert's Theorem 90, we have $H^1(k,\GL_n)=1$ for any field $k$, and in particular, $H^1(\R, \GL_n)=1$. It follows that there exists an element $g\in\GL(n,\C)$ such that \begin{equation}\label{e:1} g^{-1}\cdot \bar g =z,\quad\text{that is,}\quad \bar g =g\cdot z.\tag{$*$} \end{equation}

For example, let $n=2$, $z=\begin{pmatrix} 0&1\\1&0\end{pmatrix}$. Write $g=\begin{pmatrix} a&b\\c&d\end{pmatrix}$. Then \eqref{e:1} gives $$ \begin{pmatrix} \bar a&\bar b\\ \bar c&\bar d\end{pmatrix}= \begin{pmatrix} b&a\\d&c\end{pmatrix},$$ that is, $b=\bar a$ and $c=\bar d$. Thus $$ g=\begin{pmatrix} a&\bar a\\ \bar d&d\end{pmatrix}.$$ We choose $a=1$, $d=i$, and we obtain \begin{equation}\label{e:2} g= \begin{pmatrix} 1& 1\\ -i&i\end{pmatrix}.\tag{$**$} \end{equation}

Question. I need an algorithm that, for given $n$ and a 1-cocycle $z\in \GL(n,\C)$, will give $g\in\GL(n,\C)$ satisfying \eqref{e:1}. Moreover, I want the algorithm to give a nice element $g$. For example, for $z=\begin{pmatrix} 0&1\\1&0\end{pmatrix}$, I want to get something like \eqref{e:2}.

Remark. We (my coauthor and I) already have an algorithm and a computer program for this problem. However, for $z$ as above, our program gives \begin{equation}\label{e:3} \tag{$*{*}*$} \begin{pmatrix} \frac12+i&\tfrac12-i\\ \tfrac12-i&\tfrac12+i\end{pmatrix}, \end{equation} which I like less than \eqref{e:2}.

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  • $\begingroup$ This seems hard to answer without knowing what a "nice" element is (or, perhaps equivalently, how much you like which elements). $\endgroup$
    – LSpice
    Commented Dec 3, 2022 at 18:14
  • $\begingroup$ @LSpice: My cocycle $z$ lives in the stabilizer $H_t$ of a real tensor $t$. I use $g$ in order to obtain a twisted form $t'=g\cdot t$ of $t$. An easy calculation shows that $t'$ is again real: $\overline{t'}=t'$. My original tensor $t$ has all coefficients equal to 0 or 1 only. I want to obtain $t'$ with small coefficients, if possible with coefficients 0, $1$, and $-1$ only (up to a scalar). This works with \eqref{e:2}, but not with \eqref{e:3}. $\endgroup$
    – Mikhail Borovoi
    Commented Dec 3, 2022 at 18:46
  • $\begingroup$ In the context of the question, what are the $t$ and $t'$ in your comment? $\endgroup$
    – LSpice
    Commented Dec 3, 2022 at 19:19
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    $\begingroup$ Actually I consider the group ${\rm GL}_{5,\Bbb R}\times {\rm GL}_{4,\Bbb R}$ acting on the vector space $\bigwedge^2 {\Bbb R}^5\otimes {\Bbb R}^4$. My cocycle $z$ is a real block-diagonal matrix consisting of 2-dimensional blocks \eqref{e:1}, one-dimensional blocks 1, and a one-dimensional block $-1$. $\endgroup$
    – Mikhail Borovoi
    Commented Dec 3, 2022 at 19:53

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