$\newcommand{Tr}{\operatorname{Tr}}$ Is there a continuous map $(p,t) \mapsto \lambda(p,t)$ which, given a path $p: [0,1] \to M(2,\mathbb R)$ and a $t \in \mathbb [0,1]$, gives back an eigenvalue of $p(t)$? Additionally, I'd like it that if $p(a) = -p(b)$ for some $a, b \in [0,1]$, then $\lambda(p,a) = -\lambda(p,b)$.

A naive possibility would be $\lambda(p, t) = \frac{\Tr(p(t)) \pm \sqrt{\Tr(p(t))^2 - 4\det(p(t))} }2$, using the quadratic formula and the characteristic polynomial. But this has got $\pm$ in it, which is not a function. Changing $\pm$ into $+$, the image of the path $$p(t) = \begin{pmatrix}0 & 1-2t \\ 2t - 1 & 0 \end{pmatrix}$$ is seen to violate the second condition: namely, $\lambda(p,0)=\lambda(p,1)=i$ when what's needed is $\lambda(p,0)=-\lambda(p,1)$.

This problem is unsolvable when $M(2, \mathbb R)$ is changed to $M(2, \mathbb C)$, as the existence of $\lambda(-,-)$ would violate the simple-connectivity of $SL(2,\mathbb C)$.

  • $\begingroup$ You switch between $\lambda$ and $L$. It's also not clear what $\sqrt{\cdot}$ should mean in your proposed definition of $L$, as a well defined function $\mathbb R \to \mathbb C$ (although I guess any naïve candidate is fine?). What is the topology in which $\lambda$ is continuous? $\endgroup$
    – LSpice
    May 8, 2019 at 8:45
  • $\begingroup$ Please clarify the domain of your function $\lambda$. Does it take in the whole path, or is it defined as $\lambda:M(2,\Bbb R)\times\Bbb R\to\Bbb C$? If it takes in the whole path, then we really need the topology you use on the space of paths. $\endgroup$
    – M. Winter
    May 8, 2019 at 9:52
  • $\begingroup$ @M.Winter I take your point. I only understand these concepts naively at the moment. It might be premature to be asking these questions if I don't know which topology to use for $M(2, \mathbb R)^{[0,1]}$, for instance. The motivation was to see if I could complete a proof of the lack of simple-connectivity of $SL(2, \mathbb R)$. And indeed $\lambda$ should take in the whole path, and the second condition should be in hindsight that $t \mapsto \lambda(p, t)$ is smooth whenever $p$ is smooth. But I'm even less sure how to solve that $\endgroup$
    – wlad
    May 8, 2019 at 15:04
  • $\begingroup$ By the way, the discriminant should be $\operatorname{Tr}(p(t))^2 - 4\det(p(t))$, not $\operatorname{Tr}(p(t))^2 + 4\det(p(t))$. $\endgroup$
    – LSpice
    May 10, 2019 at 1:44
  • $\begingroup$ @LSpice Fixed.. $\endgroup$
    – wlad
    May 10, 2019 at 9:13

1 Answer 1


I originally gave a wrong candidate for an answer, but in fact no such path can exist (at least, assuming that your topologies are such that $t \mapsto \lambda(p, t)$ is continuous for all paths $p$). Indeed, define $p : [0, 1] \to \operatorname M(2, \mathbb R)$ by $$ p(t) = \begin{pmatrix} \cos(\pi t) & \sin(\pi t) \\ \sin(\pi t) & -\cos(\pi t) \end{pmatrix} $$ for all $t \in [0, 1]$. Then $\lambda(p, t) \in \{\pm1\}$ for all $t \in [0, 1]$, but $\lambda(p, 1) = -\lambda(p, 0)$, which is impossible if $t \mapsto \lambda(p, t)$ is continuous.

EDIT: I suspected from your mention of the simple connectedness of $\operatorname{SL}(2, \mathbb C)$, and see definitely from your comment, that you are interested in the non-simple connectedness of $\operatorname{SL}(2, \mathbb R)$. It may be that a particularly simple way to see this is to consider the metaplectic group, which is a non-split double cover of $\operatorname{SL}(2, \mathbb R)$. Notice that your original example can be adapted to give a polynomial path in $\operatorname{SL}(2, \mathbb R)$ (not just in $\operatorname M(2, \mathbb R)$, or even just in $\operatorname{GL}(2, \mathbb R)$ as my original path was): $$ t \mapsto p(t) = \begin{pmatrix} t(1 - t) & 1 - t^2(6 - 4t) \\ -1 + t^2(6 - 4t) & 4t(1 - t)(3 - 2t)(1 + 2t) \end{pmatrix}. $$

Notice that the cover of $\operatorname{SL}(2, \mathbb R)$ implicit in your question, namely $$ E \mathrel{:=} \{(g, \lambda) : \text{$g \in \operatorname{SL}(2, \mathbb R)$ and $\lambda$ is an eigenvalue of $g$}\}, $$ is a branched cover; it is a double cover only on the regular semisimple set (of elements with distinct eigenvalues). Its restriction to the regular semisimple set is isomorphic (as a cover of the rss set) to the restriction there of the branched cover $$ \{(g, B) : \text{$g \in \operatorname{SL}(2, \mathbb R)$ and $B$ is a Borel subgroup of $\operatorname{SL}(2, \mathbb C)$ containing $g$}\} $$ (namely, associate $(g, \lambda)$ and $(g, B)$ if and only if the non-trivial eigenvalue of $\operatorname{Ad}(g)$ on $\operatorname{Lie}(B)$ is $\lambda^2$). However, I think that the restriction is not isomorphic to the restriction of the metaplectic group. Indeed, let $\lambda$ and $\mu$ be distinct elements of $\mathbb R$ satisfying $\lambda\mu = 1$. Then any lift to $E$ of the path $$ t \mapsto \operatorname{Int}\begin{pmatrix} \cos(\pi t) & \sin(\pi t) \\ -\sin(\pi t) & \sin(\pi t) \end{pmatrix}\begin{pmatrix} \lambda & 0 \\ 0 & \mu \end{pmatrix} $$ projects, via $E \to \mathbb C$, to $\{\lambda, \mu\}$, hence is constantly $\lambda$ or constantly $\mu$; but, if I've done my computation properly, then $$ \operatorname{Int}(w(t))\Bigl(\begin{pmatrix} \lambda & 0 \\ 0 & \mu \end{pmatrix}, \sqrt\mu\Bigr), $$ where $w(t) = \Bigl(\begin{pmatrix} \cos(\pi t) & \sin(\pi t) \\ -\sin(\pi t) & \cos(\pi t) \end{pmatrix}, \epsilon_t\Bigr)$, with $\epsilon_t(i) = e^{\pi i t/2}$, is a path in the restriction to the rss set of the metaplectic cover that connects the two pre-images of $\begin{pmatrix} \lambda & 0 \\ 0 & \mu \end{pmatrix}$. (I'm using the description of points in the metaplectic group from Wikipedia.)

  • $\begingroup$ The topologies are indeed the usual Euclidean topologies. The $\sqrt : \mathbb R \to \mathbb C$ is really any continuous function such that $(\sqrt x)^2 = x$ $\endgroup$
    – wlad
    May 8, 2019 at 9:22
  • $\begingroup$ It makes sense to speak of the continuity of a map $e$ like mine in the Euclidean topologies, but your map is defined on paths, which don't carry a Euclidean topology. Probably something like the compact-open topology there is what you want. Anyway, for any reasonable choice of topology on paths, at least the sort of composition I've suggested should be continuous. $\endgroup$
    – LSpice
    May 8, 2019 at 9:25
  • $\begingroup$ @manonlaptop, sorry, this isn't continuous; consider its composition with $t \mapsto \begin{pmatrix} t & -1 \\ 1 & 0 \end{pmatrix}$ at $t = 0$. I'll think further. $\endgroup$
    – LSpice
    May 8, 2019 at 9:29
  • $\begingroup$ @manonlaptop, in fact I now think that there is no such path, as evidenced by a small variant of your proposed example. Please see the new edit. Sorry for my mistake! $\endgroup$
    – LSpice
    May 8, 2019 at 9:56
  • $\begingroup$ I think I asked the wrong question. But it's a nice example. My second condition should've been that if $p$ is smooth then $t \mapsto \lambda(p, t)$ should also be smooth. This is violated for $p(t) = \begin{pmatrix} \cos(2\pi t) & -\sin(2\pi t) \\ \sin(2\pi t) & \cos(2\pi t) \end{pmatrix}$ and my naive definition of $\lambda$. After $t=0.5$, the trajectory of $t\mapsto\lambda(p,t)$ suddenly reverses. This is also why I used $L$ originally instead of $\lambda$, so I could talk about my example without confusing it with the imagined $\lambda$.I may reverse the edit that proposed the name change $\endgroup$
    – wlad
    May 8, 2019 at 14:43

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