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

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!?

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.