Original text: CC0 1.0. Prerequisite proofs and component terms.

Changing reference for a weight cocycle

The ordered chain rule

Theorem. Let $\alpha,\beta,\gamma$ be faithful normal semifinite weights on a von Neumann algebra $M$. For every real $t$, $$[D\gamma:D\alpha]_t =[D\gamma:D\beta]_t[D\beta:D\alpha]_t. \tag{CH.1}$$ The factors need not commute.

Proof. Represent $M$ faithfully and nondegenerately on a Hilbert space $H$, and put $N=M'$. Choose a faithful normal semifinite weight $\omega$ on $N$. All three spatial derivatives can then be taken on the same $H$: $$A=\frac{d\alpha}{d\omega},\qquad B=\frac{d\beta}{d\omega},\qquad C=\frac{d\gamma}{d\omega}.$$ They are positive, injective, self-adjoint operators. Their imaginary powers are bounded unitaries on all of $H$, even if the derivatives and their inverses are unbounded. The spatial formula for the balanced-weight cocycle gives $$\begin{aligned} [D\gamma:D\beta]_t[D\beta:D\alpha]_t &=(C^{it}B^{-it})(B^{it}A^{-it})\\ &=C^{it}A^{-it}\\ &=[D\gamma:D\alpha]_t. \end{aligned}$$ Only the adjacent factors involving $B$ cancel. No unbounded operators are moved across one another, and no common domain for their products is needed. Each cocycle in this computation lies in $M$ and is independent of the commutant weight and of the faithful representation. Faithfulness of the representation proves the identity in $M$. The zero algebra has the same identity in its zero-dimensional representation. $\square$

Reversal and paths of reference changes

Taking all three weights equal in the theorem gives the identity cocycle. Taking $\gamma=\alpha$ then gives the reverse comparison: $$[D\alpha:D\alpha]_t=1,\qquad [D\alpha:D\beta]_t=[D\beta:D\alpha]_t^*. \tag{CH.2}$$ Indeed, the spatial formula yields $A^{it}A^{-it}=1$, and each cocycle is unitary.

For faithful normal semifinite weights $\varphi_0,\ldots,\varphi_n$, apply the chain rule repeatedly: $$[D\varphi_n:D\varphi_0]_t =[D\varphi_n:D\varphi_{n-1}]_t \cdots[D\varphi_1:D\varphi_0]_t. \tag{CH.3}$$ Induction gives this product in precisely the displayed order. If $n=0$, it is the empty product $1$. Thus an intermediate reference can be inserted or removed without changing the total comparison.

The same order appears in modular covariance. Set $u_t=[D\beta:D\alpha]_t$ and $v_t=[D\gamma:D\beta]_t$. The spatial construction gives $\sigma_t^\beta=\operatorname{Ad}(u_t)\circ\sigma_t^\alpha$. Using the cocycle laws for $u$ and $v$, $$\begin{aligned} v_{s+t}u_{s+t} &=v_s\sigma_s^\beta(v_t)\,u_s\sigma_s^\alpha(u_t)\\ &=v_su_s\sigma_s^\alpha(v_t)u_s^*u_s\sigma_s^\alpha(u_t)\\ &=(v_su_s)\sigma_s^\alpha(v_tu_t). \end{aligned}$$ Hence $vu$ is a $\sigma^\alpha$-cocycle. Both factors are strongly* continuous unitary families, so their product is strongly* continuous as well.

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