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Let $G$ be a locally compact group with modular function $\delta_G$ and consider $\mathcal{K}(G)$, the set of compactly supported continuous functions $G\to \mathbb{C}$, with the $*$-algebra structure given by $$(\xi\eta)(s) = \int_G \xi(t)\eta(t^{-1})dt, \quad \xi^\sharp(s) = \delta_G(s^{-1})\overline{\xi(s^{-1})}.$$ This $*$-algebra becomes a left Hilbert algebra for the inner product on $L^2(G)$ (w.r.t. left Haar measure) by viewing $\mathcal{K}(G)\subseteq L^2(G)$.

Consider now the following fragment from Takesaki's book "Theory of operator algebras II":

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Two questions:

(1) Is $\Delta_G$ the modular operator associated with the left Hilbert algebra $\mathcal{K}(G)$? I.e. is $\Delta_G = S^*S$ where $S$ is the closure of the involution $\sharp$?

(2) To be a Tomita algebra, we must have that $\Delta_G^\alpha$ maps $\mathcal{K}(G)$ into itself. Why is this the case?

Thanks in advance for your help!

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First note that there is a typo: the complex one-parameter group should be $(\Delta_G^{i\alpha})$ instead of $(\Delta_G^\alpha)$.

(1) Yes. This follows from abstract theory, because $\Delta$ is the closure of $U(-i)$ for a Tomita algebra (Theorem VI.2.2) and obviously $\mathcal K(G)$ is a core for $\Delta$. But in this case it's also not hard to show directly: If $\xi,\eta\in\mathcal K(G)$, then $$ \langle\eta^\sharp,\xi^\sharp\rangle=\int_G \delta_G(g^{-1})^2\overline{\xi(g^{-1})}\eta(g^{-1})\,dg=\int_G\delta_G(g)\overline{\xi(g)}\eta(g)\,dg=\langle\Delta_G\xi,\eta\rangle. $$ Thus $S_0^\ast S_0\subset \Delta_G$. Then you just have to show that $\mathcal K(G)$ is a core for $\Delta_G$ to get equality.

(2) The modular function is continuous and takes values in $(0,\infty)$, hence $g\mapsto \delta_G(g)^{i\alpha}=e^{i\alpha\log\delta_G(g)}$ is also continuous. And $\Delta_G^{i\alpha}$ simply acts by $\Delta_G^{i\alpha}\xi(g)=\delta_G(g)^{i\alpha}\xi(g)$. Thus $\Delta_G^{i\alpha}(\mathcal K(G))\subset\mathcal K(G)$.

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  • $\begingroup$ Isn't this a bit circular? To answer (1) you use the Tomita theory machinery but this is established in (2)? $\endgroup$
    – Andromeda
    Commented Apr 19, 2023 at 10:11
  • $\begingroup$ Well I meant with the circularity to invoke abstract theory. $\endgroup$
    – Andromeda
    Commented Apr 19, 2023 at 10:50
  • $\begingroup$ I don't think it is. I don't use that $\Delta_G=S^\ast S$ for (2). But you are right that one should answer (2) first and then use it for (1). $\endgroup$
    – MaoWao
    Commented Apr 19, 2023 at 11:23
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    $\begingroup$ Yes, but for multiplication operators, this all simplifies. If $M_f$ is the operator of multiplication by the measurable function $f$ (with maximal domain), then $M_f$ is self-adjoint and the spectral measure is given by $E(A)=M_{1_A\circ f}$. By approximation with simple functions one can then see that $\phi(f)=M_{\phi\circ f}$ for measurable $\phi$. This should be among the first examples one sees for the spectral theorem. $\endgroup$
    – MaoWao
    Commented Apr 20, 2023 at 21:10
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    $\begingroup$ You can also formulate this without any reference to quadratic forms: $C_c(G)$ is a core for $S$ by definition, hence also for $(S^\ast S)^{1/2}$ (remember that $D(S)=D(|S|)$ and $\|S\xi\|=\||S|\xi\|$). Since its also a core for $\Delta_G^{1/2}$ and both operators coincide on $C_c(G)$, we have $(S^\ast S)^{1/2}=\Delta_G^{1/2}$ and hence also $S^\ast S=\Delta_G$. $\endgroup$
    – MaoWao
    Commented Apr 24, 2023 at 9:51

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