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.

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

| cite | improve this answer | |
  • 1
    $\begingroup$ I think that this is probably part of the question, rather than an answer. $\endgroup$ – LSpice Apr 17 at 21:07
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.