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I would like to lift an arbitrary one-parameter subgroup $e^{t K}$ with $K\in\mathfrak{sp}(2N,\mathbb{R})$ to the universal cover $\widetilde{\mathrm{Sp}}(2N,\mathbb{R})$ (or at least its two-fold cover, i.e., the metaplectic group).

I follow the paper of John Rawnsley On the universal covering group of the real symplectic group, where an element of the universal covering group $\widetilde{\mathrm{Sp}}(2N,\mathbb{R})$ is represented as pair \begin{align} \widetilde{\mathrm{Sp}}(2N,\mathbb{R})=\left\{(g,c)\in\mathrm{Sp}(2N,\mathbb{R})\times\mathbb{R}\,\big|\,e^{ic}=\varphi(g)\right\}\,, \end{align} where $\varphi: \mathrm{Sp}(2N,\mathbb{R})\to S^1\subset\mathbb{C}$ is a normalized circle function defined as follows. We start with a complex structure $J: \mathbb{R}^{2N}\to \mathbb{R}^{2N}$ that is compatible with the symplectic form $\Omega$ on $\mathbb{R}^{2N}$. For every group element $g\in\mathrm{Sp}(2N,\mathbb{R})$, we then define $C_g=\frac{1}{2}(g-JgJ)$, which commutes with $J$. We can therefore identify $C_g$ with a $N$-by-$N$ complex matrix, which we can use to compute a determinant. We then define the circle function as \begin{align} \varphi(g)=\frac{\det{C_g}}{|\det{C_g}|}\,, \end{align} where the determinant is meant in the above sense (of a complex matrix, rather than of real $2N$-by-$2N$ matrix). The universal covering group is then defined with the group multiplication \begin{align} (g_1,c_1)\cdot(g_2,c_2)=(g_1\cdot g_2,c_1+c_2+\eta(g_1,g_2))\,, \end{align} where $\eta:\mathrm{Sp}(2N,\mathbb{R})\times \mathrm{Sp}(2N,\mathbb{R})\to\mathbb{R}$ is the unique smooth function, such that $\varphi(g_1g_2)=\varphi(g_1)\varphi(g_2)e^{i\eta(g_1,g_2)}$ everywhere.

My question: How can I find the unique continuous function $c_K: \mathbb{R}\to\mathbb{R}$ that satisfies \begin{align} \varphi(e^{tK})=e^{i c_K(t)}\,. \end{align} Essentially, I would like to lift the curve $e^{tK}$ to its double cover. Of course, I could just numerically evaluate $\varphi(e^{tK})$ and correct by an offset of $2\pi$, whenever I go around the circle, but I am hoping that there is a smarter and MORE EXPLICIT way!

More thoughts: I believe $c_K$ should satisfy the differential equation $\dot{c}_K(t)=-i\frac{d}{dt}\log\varphi(e^{tK})$. Maybe this can be solved somehow or used to write a formal solution!?

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I made some progress in the sense that I believe that I could reduce it to a more standard problem: Morally speaking, I have \begin{align} c_K(t)=\mathrm{Im}\log\det\left(\frac{e^{tK}-Je^{tK}J}{2}\right)=\mathrm{Im}\mathrm{Tr}\log\left(\frac{e^{tK}-Je^{tK}J}{2}\right)\,. \end{align} The tricky thing is that $\log{(e^x)}$ is only equal to $x$ in certain patches. Consider the special case, where $J$ commutes with $K$, such that $[J,K]=0$. In this case, we can simplify to find \begin{align} c_K(t)=t\,\mathrm{Im}\,\mathrm{Tr}(K) \end{align} and everything is good. However, I'm not aware if there is similar simplification for more general expressions.

Question: Is there any way to describe $c_K(t)$ more explicitly with analytic functions, rather than just defining it to incorporate the Winding number by hand?

Ok, I solved the problem. We need to use the cocycle function $\eta(M_1,M_2)$, which is defined to satisfy $\varphi(M_1M_2)=\varphi(M_1)\varphi(M_2)e^{i\eta(M_1,M_2)}$. The idea is that we write $K=u\tilde{K}u^{-1}$, such that $c_{\tilde{K}}(t)=t\mathrm{Im}\mathrm{Tr}(\tilde{K})$. This can always be found by using a transformation $u$ that brings $K$ into a standard block diagonal form with respect to $J$, i.e., both of them are block diagonal (they may not quite commute, but almost). We can then use the cocycle relation to see that $c_K(t)=c_{\tilde{K}}(t)+\eta(u,e^{K})+\eta(ue^{K},u^{-1})$. This can possibly be simplified further, but the idea should be clear.

I hope this helps somebody with a similar problem in the future...

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    $\begingroup$ I think that this is probably part of the question, rather than an answer. $\endgroup$ – LSpice Apr 17 at 21:07
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    $\begingroup$ Well, this was just my approach - adding this to the question would make it rather lengthy and originally I thought that people would probably know what to do. Anyway, now that I solved the problem, my post actually expanded into a brief summary of the full solution... $\endgroup$ – LFH Apr 28 at 16:15

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