0
$\begingroup$

Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on an infinite-dimensional complex Hilbert $F$.

Let $T,S\in\mathcal{B}(F)$. The pair $(T,S)$ is said to $\lambda$-commute if there exists $\lambda\in \mathbb{C}^*$ such that $TS=\lambda ST$. I am looking for necessary and sufficient conditions on the operators $S$ and $T$ such that $(T,S)$ $\lambda$-commute, then they already commute.

I know already that $S,T \geq 0$ or $S,T \leq 0$ is a sufficient condition for $S$ and $T$ to commute if they $\lambda$-commute, but this is far from necessary.

Improve this question
$\endgroup$
10
  • 7
    $\begingroup$ In finite dimensions, tr(TS)=tr(ST), so if tr(TS) is non-zero, then $\lambda=1$. $\endgroup$
    – Michael Renardy
    Commented Feb 26, 2018 at 14:52
  • $\begingroup$ I think you just want "if" in your guess, not "only if". The "only if" direction fails trivially, for example for any $S$ and $T$ which satisfy $TS \neq \lambda ST$ and $TS \not\geq 0$. $\endgroup$
    – Nik Weaver
    Commented Feb 26, 2018 at 17:21
  • $\begingroup$ I guess the question is meant to be: Let $TS = \lambda ST$. Then $[T,S] = 0$ if and only if $TS \geq 0$ and $ST \geq 0$. By the way, $TS \leq 0$ and $ST \leq 0$ is also sufficient for $[T,S] = 0$ then, no? $\endgroup$
    – Hannes
    Commented Feb 26, 2018 at 17:40
  • $\begingroup$ Well, that version is also obviously false: you can certainly find $S$ and $T$ which commute but for which $TS \not\geq 0$. $\endgroup$
    – Nik Weaver
    Commented Feb 26, 2018 at 17:47
  • 1
    $\begingroup$ If $ST$ is not quasi-nilpotent and either $|\lambda| \neq 1$ or the spectrum of $ST$ (with or without zero) is not rotation invariant for every angle (including rational multiples of $\pi$), then $ST =\lambda TS$ is either impossible or entails $\lambda = 1$. I don't see what else can be said ... $\endgroup$
    – David Handelman
    Commented Feb 27, 2018 at 0:08

1 Answer 1

Reset to default
11
$\begingroup$

I don't see which kind of condition you are looking for, as there are a lot of pairs $T,S$ such that $TS=\lambda ST$ and $\lambda\ne1$, even in finite dimension. Such pairs are said to $\lambda$-commute.

An interesting case happens when $\lambda$ is root of unity, say of order $r$. Then (Potter's Theorem) $(T+S)^r=T^r+S^r$.

Still when $\lambda^r=1$, here is a construction. Choose $T={\rm diag}(I_{n_1},\lambda I_{n_2},\ldots,\lambda^{r-1}I_{n_r})$ and $S$ a blockwise cyclic matrix $$S=\begin{pmatrix} 0_{n_1} & M_1 & 0 & \cdots & 0 \\ \vdots & \ddots & \ddots & \ddots & \vdots \\ \vdots & & \ddots & \ddots & 0 \\ 0 & & & \ddots & M_{r-1} \\ M_r & 0 & \cdots & \cdots & 0_{n_r} \end{pmatrix}.$$ This pair $\lambda$-commutes.

Improve this answer
$\endgroup$
2
  • 2
    $\begingroup$ It is amazing how many lovely facts are in the world but of which I am completely ignorant. For anyone else so ignorant, Potter's theorem apparently dates back to Potter - On the latent roots of quasi-commutative matrices (MSN). $\endgroup$
    – LSpice
    Commented Feb 27, 2018 at 19:04
  • 2
    $\begingroup$ Incidentally, this finite dimensional case is a lovely first instance of the general fact that semisimple centralisers in simply connected algebraic groups like $\mathrm{SL}_n$ are connected, but those in non-simply connected groups like $\mathrm{PGL}_n$ need not be. $\endgroup$
    – LSpice
    Commented Feb 27, 2018 at 19:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .