# Eigenvalue-taking operator?

$$\newcommand{Tr}{\operatorname{Tr}}$$ 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 $$SL(2,\mathbb C)$$.

• 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? May 8, 2019 at 8:45
• 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. May 8, 2019 at 9:52
• @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
May 8, 2019 at 15:04
• By the way, the discriminant should be $\operatorname{Tr}(p(t))^2 - 4\det(p(t))$, not $\operatorname{Tr}(p(t))^2 + 4\det(p(t))$. May 10, 2019 at 1:44
• @LSpice Fixed..
May 10, 2019 at 9:13

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.)

• The topologies are indeed the usual Euclidean topologies. The $\sqrt : \mathbb R \to \mathbb C$ is really any continuous function such that $(\sqrt x)^2 = x$
• 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. May 8, 2019 at 9:25
• @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. May 8, 2019 at 9:29
• 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