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](https://www.maths.ed.ac.uk/~v1ranick/papers/rawnsley.pdf), 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!?



Lifting one parameter subgroup $e^{t K}$ to the universal cover of $\mathrm{Sp}(2N,\mathbb{R})$
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...


