Let $\pi:\mathscr M\to\mathscr M$ be a normal $*$-homomorphism between a von Neumann algebra $\mathscr M.$ Assume $\mathscr M$ has a normal semifinite faithful trace. Does $\pi$ extend as a bounded map between noncommutative $L_p$-spaces?
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asked Jul 29, 2019 at 10:58
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1 Answer
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No, this already fails in the abelian case. Take $M = L^\infty[0,1]$ with the trace coming from integration against Lebesgue measure. Let $\pi$ take $f(x)$ to $f(x^2)$, i.e., composition with the squaring map. Then $f(x) = x^{-3/4}$ belongs to $L^2[0,1]$ but $\pi(f)(x) = x^{-3/2}$ does not.
answered Jul 29, 2019 at 11:27
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$\begingroup$ Can $\pi$ be extended to the algebra of all measurable operators? $\endgroup$ Commented Jul 29, 2019 at 12:48
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1$\begingroup$ Nope. This is a good exercise for you. Take $M = L^{\infty}(\mathbb{R})$ with integration against Lebeague measure. $\endgroup$ Commented Jul 29, 2019 at 13:02