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\documentclass{article} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{amsthm} \newtheorem{proposition}{Proposition} \begin{document} \begin{proposition} Any one-parameter semigroup $S_t$ of partial isometries is centered. \end{proposition} \begin{proof} Write $P_t= S_tS_t^*$, $Q_t=S_t^*S_t$, $t\ge0$. Since $S_t$ is a partial isometry, $P_t$ and $Q_t$ are projections. We need to prove that this family of projections is commuting. a) First show that $P_t$ and $S_t$, $t\ge0$ are commuting families. In fact, we show that $P_tP_s = P_s$, $Q_tQ_s = Q_s$ for $t<s$. Then passing to adjoints we get the needed commutation. Let $t<s$. Then \[ P_tP_s = P_tS_{s}S_{s}^* = P_t S_tS_{s-t}S_{s}^*. \] But since $P_t$ is a projection onto the image of $S_t$, we have $P_tS_t= S_t$, and therefore, $P_tP_s= P_s$. For $Q_t$, the proof is quite similar. Now we show that $P_tQ_s = Q_tP_s$ for $t$, $s\ge0$. We have \[ Q_{t+s}=S_{t+s}^*S_{t+s} = S_t^*Q_sS_t. \] Since $Q_{t+s}$ is a projection, $Q_{t+s} = Q_{t+s}^2$, or \[ S_t^*Q_sS_t = S_t^*Q_sS_tS_t^* Q_s S_t = S_t^* Q_s P_tQ_sS_t. \] Multiplying this equality by $S_t$ from the left, and by $S_t^*$ from the right, we get \[ P_tQ_sP_t = P_tQ_sP_tQ_sP_t = (P_tQ_sP_t)^2, \] which is possible if and only if $P_t$ and $Q_s$ commute. \end{proof} \end{document} %%% Local Variables: %%% mode: latex %%% TeX-master: t %%% End: